### 3.191 $$\int \frac{\tanh ^5(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx$$

Optimal. Leaf size=109 $-\frac{a^2}{4 b^2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{a (a+2 b)}{2 b^2 d (a+b)^2 \left (a+b \tanh ^2(c+d x)\right )}+\frac{\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^3}+\frac{\log (\cosh (c+d x))}{d (a+b)^3}$

[Out]

Log[Cosh[c + d*x]]/((a + b)^3*d) + Log[a + b*Tanh[c + d*x]^2]/(2*(a + b)^3*d) - a^2/(4*b^2*(a + b)*d*(a + b*Ta
nh[c + d*x]^2)^2) + (a*(a + 2*b))/(2*b^2*(a + b)^2*d*(a + b*Tanh[c + d*x]^2))

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Rubi [A]  time = 0.17042, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {3670, 446, 88} $-\frac{a^2}{4 b^2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{a (a+2 b)}{2 b^2 d (a+b)^2 \left (a+b \tanh ^2(c+d x)\right )}+\frac{\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^3}+\frac{\log (\cosh (c+d x))}{d (a+b)^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[Tanh[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

Log[Cosh[c + d*x]]/((a + b)^3*d) + Log[a + b*Tanh[c + d*x]^2]/(2*(a + b)^3*d) - a^2/(4*b^2*(a + b)*d*(a + b*Ta
nh[c + d*x]^2)^2) + (a*(a + 2*b))/(2*b^2*(a + b)^2*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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{\left (1-x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1-x) (a+b x)^3} \, dx,x,\tanh ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{(a+b)^3 (-1+x)}+\frac{a^2}{b (a+b) (a+b x)^3}-\frac{a (a+2 b)}{b (a+b)^2 (a+b x)^2}+\frac{b}{(a+b)^3 (a+b x)}\right ) \, dx,x,\tanh ^2(c+d x)\right )}{2 d}\\ &=\frac{\log (\cosh (c+d x))}{(a+b)^3 d}+\frac{\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^3 d}-\frac{a^2}{4 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{a (a+2 b)}{2 b^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.9841, size = 91, normalized size = 0.83 $-\frac{\frac{a^2 (a+b)^2}{b^2 \left (a+b \tanh ^2(c+d x)\right )^2}-\frac{2 a (a+2 b) (a+b)}{b^2 \left (a+b \tanh ^2(c+d x)\right )}-2 \log \left (a+b \tanh ^2(c+d x)\right )-4 \log (\cosh (c+d x))}{4 d (a+b)^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Tanh[c + d*x]^5/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

-(-4*Log[Cosh[c + d*x]] - 2*Log[a + b*Tanh[c + d*x]^2] + (a^2*(a + b)^2)/(b^2*(a + b*Tanh[c + d*x]^2)^2) - (2*
a*(a + b)*(a + 2*b))/(b^2*(a + b*Tanh[c + d*x]^2)))/(4*(a + b)^3*d)

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Maple [B]  time = 0.027, size = 234, normalized size = 2.2 \begin{align*} -{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{2\,d \left ( a+b \right ) ^{3}}}+{\frac{{a}^{3}}{2\,d \left ( a+b \right ) ^{3}{b}^{2} \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{3\,{a}^{2}}{2\,d \left ( a+b \right ) ^{3}b \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{a}{d \left ( a+b \right ) ^{3} \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{{a}^{4}}{4\,d \left ( a+b \right ) ^{3}{b}^{2} \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{a}^{3}}{2\,d \left ( a+b \right ) ^{3}b \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{a}^{2}}{4\,d \left ( a+b \right ) ^{3} \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{\ln \left ( a+b \left ( \tanh \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ( a+b \right ) ^{3}}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x)

[Out]

-1/2/d/(a+b)^3*ln(tanh(d*x+c)+1)+1/2/d/(a+b)^3*a^3/b^2/(a+b*tanh(d*x+c)^2)+3/2/d/(a+b)^3*a^2/b/(a+b*tanh(d*x+c
)^2)+1/d/(a+b)^3*a/(a+b*tanh(d*x+c)^2)-1/4/d/(a+b)^3*a^4/b^2/(a+b*tanh(d*x+c)^2)^2-1/2/d/(a+b)^3*a^3/b/(a+b*ta
nh(d*x+c)^2)^2-1/4/d/(a+b)^3*a^2/(a+b*tanh(d*x+c)^2)^2+1/2*ln(a+b*tanh(d*x+c)^2)/(a+b)^3/d-1/2/d/(a+b)^3*ln(ta
nh(d*x+c)-1)

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Maxima [B]  time = 1.24097, size = 508, normalized size = 4.66 \begin{align*} \frac{d x + c}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} + \frac{4 \,{\left ({\left (a^{2} + a b\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a^{2} - 2 \, a b\right )} e^{\left (-4 \, d x - 4 \, c\right )} +{\left (a^{2} + a b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5} + 4 \,{\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \,{\left (3 \, a^{5} + 7 \, a^{4} b + 6 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 7 \, a b^{4} + 3 \, b^{5}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \,{\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, d x - 6 \, c\right )} +{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac{\log \left (2 \,{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

(d*x + c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d) + 4*((a^2 + a*b)*e^(-2*d*x - 2*c) + (a^2 - 2*a*b)*e^(-4*d*x - 4*
c) + (a^2 + a*b)*e^(-6*d*x - 6*c))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5 + 4*(a^5 + 3*a^4*
b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*e^(-2*d*x - 2*c) + 2*(3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a
*b^4 + 3*b^5)*e^(-4*d*x - 4*c) + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*e^(-6*d*x - 6*c) +
(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*e^(-8*d*x - 8*c))*d) + 1/2*log(2*(a - b)*e^(-2*d*x -
2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d)

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Fricas [B]  time = 2.45023, size = 6044, normalized size = 55.45 \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)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/2*(2*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^8 + 16*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + 2
*(a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c)^8 + 8*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b
+ b^2)*d*x*cosh(d*x + c)^2 + (a^2 - b^2)*d*x - a^2 - a*b)*sinh(d*x + c)^6 + 16*(7*(a^2 + 2*a*b + b^2)*d*x*cos
h(d*x + c)^3 + 3*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 4*((3*a^2 - 2*a*b + 3*b^2)*d*x
- 2*a^2 + 4*a*b)*cosh(d*x + c)^4 + 4*(35*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + (3*a^2 - 2*a*b + 3*b^2)*d*
x + 30*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^2 - 2*a^2 + 4*a*b)*sinh(d*x + c)^4 + 16*(7*(a^2 + 2*a*b + b
^2)*d*x*cosh(d*x + c)^5 + 10*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^3 + ((3*a^2 - 2*a*b + 3*b^2)*d*x - 2*
a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(a^2 + 2*a*b + b^2)*d*x + 8*((a^2 - b^2)*d*x - a^2 - a*b)*cosh
(d*x + c)^2 + 8*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^6 + 15*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^4
+ (a^2 - b^2)*d*x + 3*((3*a^2 - 2*a*b + 3*b^2)*d*x - 2*a^2 + 4*a*b)*cosh(d*x + c)^2 - a^2 - a*b)*sinh(d*x + c)
^2 - ((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^7 + (a^2 + 2*a*b
+ b^2)*sinh(d*x + c)^8 + 4*(a^2 - b^2)*cosh(d*x + c)^6 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - b^2
)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + 3*(a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c)^5 +
2*(3*a^2 - 2*a*b + 3*b^2)*cosh(d*x + c)^4 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 30*(a^2 - b^2)*cosh(d*
x + c)^2 + 3*a^2 - 2*a*b + 3*b^2)*sinh(d*x + c)^4 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 10*(a^2 - b^2)*
cosh(d*x + c)^3 + (3*a^2 - 2*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(a^2 - b^2)*cosh(d*x + c)^2 + 4*(
7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 15*(a^2 - b^2)*cosh(d*x + c)^4 + 3*(3*a^2 - 2*a*b + 3*b^2)*cosh(d*x +
c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^7 + 3*(a^2 - b^2)
*cosh(d*x + c)^5 + (3*a^2 - 2*a*b + 3*b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*(
(a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) +
sinh(d*x + c)^2)) + 16*((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^7 + 3*((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x + c
)^5 + ((3*a^2 - 2*a*b + 3*b^2)*d*x - 2*a^2 + 4*a*b)*cosh(d*x + c)^3 + ((a^2 - b^2)*d*x - a^2 - a*b)*cosh(d*x +
c))*sinh(d*x + c))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^8 + 8*(a^5 + 5*
a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)*sinh(d*x + c)^7 + (a^5 + 5*a^4*b + 10*a^3*b^2
+ 10*a^2*b^3 + 5*a*b^4 + b^5)*d*sinh(d*x + c)^8 + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d
*cosh(d*x + c)^6 + 4*(7*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^2 + (a^5 + 3
*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d)*sinh(d*x + c)^6 + 2*(3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^
3 + 7*a*b^4 + 3*b^5)*d*cosh(d*x + c)^4 + 8*(7*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh
(d*x + c)^3 + 3*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(
35*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^4 + 30*(a^5 + 3*a^4*b + 2*a^3*b^2
- 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c)^2 + (3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 + 3*b^5)*
d)*sinh(d*x + c)^4 + 4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c)^2 + 8*(7*(a^5 +
5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^5 + 10*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*
b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c)^3 + (3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 + 3*b^5)*d*cosh(d*
x + c))*sinh(d*x + c)^3 + 4*(7*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^6 + 1
5*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c)^4 + 3*(3*a^5 + 7*a^4*b + 6*a^3*b^2 +
6*a^2*b^3 + 7*a*b^4 + 3*b^5)*d*cosh(d*x + c)^2 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d)*s
inh(d*x + c)^2 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*d + 8*((a^5 + 5*a^4*b + 10*a^3*b^2
+ 10*a^2*b^3 + 5*a*b^4 + b^5)*d*cosh(d*x + c)^7 + 3*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*
cosh(d*x + c)^5 + (3*a^5 + 7*a^4*b + 6*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 + 3*b^5)*d*cosh(d*x + c)^3 + (a^5 + 3*a^4
*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*d*cosh(d*x + c))*sinh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)**5/(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.26539, size = 339, normalized size = 3.11 \begin{align*} \frac{\log \left ({\left | a{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b \right |}\right )}{2 \,{\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} - \frac{3 \, a{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 3 \, b{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} - 4 \, a{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 12 \, b{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 4 \, a + 12 \, b}{4 \,{\left (a^{2} d + 2 \, a b d + b^{2} d\right )}{\left (a{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(d*x+c)^5/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/2*log(abs(a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a - 2*b))/(a^3
*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) - 1/4*(3*a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c))^2 + 3*b*(e^(2*d*x + 2*c) +
e^(-2*d*x - 2*c))^2 - 4*a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) - 12*b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) -
4*a + 12*b)/((a^2*d + 2*a*b*d + b^2*d)*(a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + b*(e^(2*d*x + 2*c) + e^(-2*d*
x - 2*c)) + 2*a - 2*b)^2)