### 3.181 $$\int \frac{\tanh ^4(c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx$$

Optimal. Leaf size=89 $-\frac{\sqrt{a} (a+3 b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{2 b^{3/2} d (a+b)^2}+\frac{a \tanh (c+d x)}{2 b d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac{x}{(a+b)^2}$

[Out]

x/(a + b)^2 - (Sqrt[a]*(a + 3*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(2*b^(3/2)*(a + b)^2*d) + (a*Tanh[c
+ d*x])/(2*b*(a + b)*d*(a + b*Tanh[c + d*x]^2))

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Rubi [A]  time = 0.116831, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.217, Rules used = {3670, 470, 522, 206, 205} $-\frac{\sqrt{a} (a+3 b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{2 b^{3/2} d (a+b)^2}+\frac{a \tanh (c+d x)}{2 b d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac{x}{(a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Tanh[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

x/(a + b)^2 - (Sqrt[a]*(a + 3*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(2*b^(3/2)*(a + b)^2*d) + (a*Tanh[c
+ d*x])/(2*b*(a + b)*d*(a + b*Tanh[c + d*x]^2))

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\tanh ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{a \tanh (c+d x)}{2 b (a+b) d \left (a+b \tanh ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{a+(-a-2 b) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 b (a+b) d}\\ &=\frac{a \tanh (c+d x)}{2 b (a+b) d \left (a+b \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b)^2 d}-\frac{(a (a+3 b)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{2 b (a+b)^2 d}\\ &=\frac{x}{(a+b)^2}-\frac{\sqrt{a} (a+3 b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{2 b^{3/2} (a+b)^2 d}+\frac{a \tanh (c+d x)}{2 b (a+b) d \left (a+b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.484395, size = 90, normalized size = 1.01 $\frac{-\frac{\sqrt{a} (a+3 b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{b^{3/2}}+\frac{a (a+b) \sinh (2 (c+d x))}{b ((a+b) \cosh (2 (c+d x))+a-b)}+2 (c+d x)}{2 d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Tanh[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

(2*(c + d*x) - (Sqrt[a]*(a + 3*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/b^(3/2) + (a*(a + b)*Sinh[2*(c + d*
x)])/(b*(a - b + (a + b)*Cosh[2*(c + d*x)])))/(2*(a + b)^2*d)

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Maple [B]  time = 0.026, size = 172, normalized size = 1.9 \begin{align*}{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{2\,d \left ( a+b \right ) ^{2}}}+{\frac{{a}^{2}\tanh \left ( dx+c \right ) }{2\,d \left ( a+b \right ) ^{2}b \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{a\tanh \left ( dx+c \right ) }{2\,d \left ( a+b \right ) ^{2} \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{{a}^{2}}{2\,d \left ( a+b \right ) ^{2}b}\arctan \left ({b\tanh \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,a}{2\,d \left ( a+b \right ) ^{2}}\arctan \left ({b\tanh \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x)

[Out]

1/2/d/(a+b)^2*ln(tanh(d*x+c)+1)+1/2/d/(a+b)^2*a^2/b*tanh(d*x+c)/(a+b*tanh(d*x+c)^2)+1/2/d/(a+b)^2*a*tanh(d*x+c
)/(a+b*tanh(d*x+c)^2)-1/2/d/(a+b)^2*a^2/b/(a*b)^(1/2)*arctan(tanh(d*x+c)*b/(a*b)^(1/2))-3/2/d/(a+b)^2*a/(a*b)^
(1/2)*arctan(tanh(d*x+c)*b/(a*b)^(1/2))-1/2/d/(a+b)^2*ln(tanh(d*x+c)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.10094, size = 4703, normalized size = 52.84 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[1/4*(4*(a*b + b^2)*d*x*cosh(d*x + c)^4 + 16*(a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(a*b + b^2)*d*x
*sinh(d*x + c)^4 + 4*(a*b + b^2)*d*x + 4*(2*(a*b - b^2)*d*x - a^2 + a*b)*cosh(d*x + c)^2 + 4*(6*(a*b + b^2)*d*
x*cosh(d*x + c)^2 + 2*(a*b - b^2)*d*x - a^2 + a*b)*sinh(d*x + c)^2 + ((a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)^4 +
4*(a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 4*a*b + 3*b^2)*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b
- 3*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b - 3*b^2)*sinh(d*x + c)^2
+ a^2 + 4*a*b + 3*b^2 + 4*((a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)^3 + (a^2 + 2*a*b - 3*b^2)*cosh(d*x + c))*sinh(d
*x + c))*sqrt(-a/b)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x +
c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x
+ c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2
)*cosh(d*x + c))*sinh(d*x + c) - 4*((a*b + b^2)*cosh(d*x + c)^2 + 2*(a*b + b^2)*cosh(d*x + c)*sinh(d*x + c) +
(a*b + b^2)*sinh(d*x + c)^2 + a*b - b^2)*sqrt(-a/b))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d
*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d
*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) - 4*a^2 - 4*a*b + 8*(2
*(a*b + b^2)*d*x*cosh(d*x + c)^3 + (2*(a*b - b^2)*d*x - a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c))/((a^3*b + 3*a
^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^4 + 4*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)*sinh(d*x + c
)^3 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*sinh(d*x + c)^4 + 2*(a^3*b + a^2*b^2 - a*b^3 - b^4)*d*cosh(d*x + c
)^2 + 2*(3*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^2 + (a^3*b + a^2*b^2 - a*b^3 - b^4)*d)*sinh(d*x
+ c)^2 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d + 4*((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^3 + (
a^3*b + a^2*b^2 - a*b^3 - b^4)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(2*(a*b + b^2)*d*x*cosh(d*x + c)^4 + 8*(a*
b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a*b + b^2)*d*x*sinh(d*x + c)^4 + 2*(a*b + b^2)*d*x + 2*(2*(a*b
- b^2)*d*x - a^2 + a*b)*cosh(d*x + c)^2 + 2*(6*(a*b + b^2)*d*x*cosh(d*x + c)^2 + 2*(a*b - b^2)*d*x - a^2 + a*
b)*sinh(d*x + c)^2 - ((a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)^4 + 4*(a^2 + 4*a*b + 3*b^2)*cosh(d*x + c)*sinh(d*x +
c)^3 + (a^2 + 4*a*b + 3*b^2)*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b - 3*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 4*a*b +
3*b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b - 3*b^2)*sinh(d*x + c)^2 + a^2 + 4*a*b + 3*b^2 + 4*((a^2 + 4*a*b + 3*b^2)
*cosh(d*x + c)^3 + (a^2 + 2*a*b - 3*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a/b)*arctan(1/2*((a + b)*cosh(d*x
+ c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(a/b)/a) - 2*a^2 - 2*a*b
+ 4*(2*(a*b + b^2)*d*x*cosh(d*x + c)^3 + (2*(a*b - b^2)*d*x - a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c))/((a^3*
b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^4 + 4*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)*sinh(
d*x + c)^3 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*sinh(d*x + c)^4 + 2*(a^3*b + a^2*b^2 - a*b^3 - b^4)*d*cosh(
d*x + c)^2 + 2*(3*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c)^2 + (a^3*b + a^2*b^2 - a*b^3 - b^4)*d)*s
inh(d*x + c)^2 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d + 4*((a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*cosh(d*x + c
)^3 + (a^3*b + a^2*b^2 - a*b^3 - b^4)*d*cosh(d*x + c))*sinh(d*x + c))]

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Sympy [A]  time = 101.773, size = 2179, normalized size = 24.48 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)**4/(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((x - tanh(c + d*x)**3/(3*d) - tanh(c + d*x)/d)/a**2, Eq(b,
0)), (x/b**2, Eq(a, 0)), (x*tanh(c)**4/(a + b*tanh(c)**2)**2, Eq(d, 0)), (2*I*a**(5/2)*b*sqrt(1/b)*tanh(c + d
*x)/(4*I*a**(7/2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt
(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sq
rt(1/b)*tanh(c + d*x)**2) + 4*I*a**(3/2)*b**2*d*x*sqrt(1/b)/(4*I*a**(7/2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3
*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2
+ 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sqrt(1/b)*tanh(c + d*x)**2) + 2*I*a**(3/2)*b**2*sqrt(1/b
)*tanh(c + d*x)/(4*I*a**(7/2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)
*b**3*d*sqrt(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(
a)*b**5*d*sqrt(1/b)*tanh(c + d*x)**2) + 4*I*sqrt(a)*b**3*d*x*sqrt(1/b)*tanh(c + d*x)**2/(4*I*a**(7/2)*b**2*d*s
qrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt(1/b) + 8*I*a**(3/2)*b**4*
d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sqrt(1/b)*tanh(c + d*x)**2)
- a**3*log(-I*sqrt(a)*sqrt(1/b) + tanh(c + d*x))/(4*I*a**(7/2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b
)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(
3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sqrt(1/b)*tanh(c + d*x)**2) + a**3*log(I*sqrt(a)*sqrt(1/b) + tanh(c
+ d*x))/(4*I*a**(7/2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d
*sqrt(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5
*d*sqrt(1/b)*tanh(c + d*x)**2) - a**2*b*log(-I*sqrt(a)*sqrt(1/b) + tanh(c + d*x))*tanh(c + d*x)**2/(4*I*a**(7/
2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt(1/b) + 8*I*a**
(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sqrt(1/b)*tanh(c
+ d*x)**2) - 3*a**2*b*log(-I*sqrt(a)*sqrt(1/b) + tanh(c + d*x))/(4*I*a**(7/2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*
b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x
)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sqrt(1/b)*tanh(c + d*x)**2) + a**2*b*log(I*sqrt(a)*s
qrt(1/b) + tanh(c + d*x))*tanh(c + d*x)**2/(4*I*a**(7/2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh
(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b
**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sqrt(1/b)*tanh(c + d*x)**2) + 3*a**2*b*log(I*sqrt(a)*sqrt(1/b) + tanh(c +
d*x))/(4*I*a**(7/2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*s
qrt(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d
*sqrt(1/b)*tanh(c + d*x)**2) - 3*a*b**2*log(-I*sqrt(a)*sqrt(1/b) + tanh(c + d*x))*tanh(c + d*x)**2/(4*I*a**(7/
2)*b**2*d*sqrt(1/b) + 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt(1/b) + 8*I*a**
(3/2)*b**4*d*sqrt(1/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sqrt(1/b)*tanh(c
+ d*x)**2) + 3*a*b**2*log(I*sqrt(a)*sqrt(1/b) + tanh(c + d*x))*tanh(c + d*x)**2/(4*I*a**(7/2)*b**2*d*sqrt(1/b)
+ 4*I*a**(5/2)*b**3*d*sqrt(1/b)*tanh(c + d*x)**2 + 8*I*a**(5/2)*b**3*d*sqrt(1/b) + 8*I*a**(3/2)*b**4*d*sqrt(1
/b)*tanh(c + d*x)**2 + 4*I*a**(3/2)*b**4*d*sqrt(1/b) + 4*I*sqrt(a)*b**5*d*sqrt(1/b)*tanh(c + d*x)**2), True))

________________________________________________________________________________________

Giac [B]  time = 1.21292, size = 274, normalized size = 3.08 \begin{align*} -\frac{{\left (a^{2} + 3 \, a b\right )} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt{a b}}\right )}{2 \,{\left (a^{2} b d + 2 \, a b^{2} d + b^{3} d\right )} \sqrt{a b}} + \frac{d x + c}{a^{2} d + 2 \, a b d + b^{2} d} - \frac{a^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + a b}{{\left (a^{2} b d + 2 \, a b^{2} d + b^{3} d\right )}{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^4/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

-1/2*(a^2 + 3*a*b)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^2*b*d + 2*a*b^2*d
+ b^3*d)*sqrt(a*b)) + (d*x + c)/(a^2*d + 2*a*b*d + b^2*d) - (a^2*e^(2*d*x + 2*c) - a*b*e^(2*d*x + 2*c) + a^2
+ a*b)/((a^2*b*d + 2*a*b^2*d + b^3*d)*(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*
d*x + 2*c) + a + b))
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