### 3.153 $$\int \coth ^5(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx$$

Optimal. Leaf size=72 $-\frac{a^2 \coth ^4(c+d x)}{4 d}-\frac{a (a+2 b) \coth ^2(c+d x)}{2 d}+\frac{(a+b)^2 \log (\tanh (c+d x))}{d}+\frac{(a+b)^2 \log (\cosh (c+d x))}{d}$

[Out]

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

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Rubi [A]  time = 0.0988802, antiderivative size = 72, 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} $-\frac{a^2 \coth ^4(c+d x)}{4 d}-\frac{a (a+2 b) \coth ^2(c+d x)}{2 d}+\frac{(a+b)^2 \log (\tanh (c+d x))}{d}+\frac{(a+b)^2 \log (\cosh (c+d x))}{d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.390955, size = 58, normalized size = 0.81 $-\frac{a^2 \coth ^4(c+d x)+2 a (a+2 b) \coth ^2(c+d x)-4 (a+b)^2 (\log (\tanh (c+d x))+\log (\cosh (c+d x)))}{4 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.054, size = 91, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{2} \left ({\rm coth} \left (dx+c\right ) \right ) ^{4}}{4\,d}}-{\frac{ab \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}}{d}}+2\,{\frac{ab\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^2*ln(sinh(d*x+c))-1/2*a^2*coth(d*x+c)^2/d-1/4*a^2*coth(d*x+c)^4/d-1/d*a*b*coth(d*x+c)^2+2/d*a*b*ln(sinh(
d*x+c))+1/d*b^2*ln(sinh(d*x+c))

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Maxima [B]  time = 1.03643, size = 319, normalized size = 4.43 \begin{align*} a^{2}{\left (x + \frac{c}{d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac{4 \,{\left (e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d{\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 2 \, a b{\left (x + \frac{c}{d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac{2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} + \frac{b^{2} \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}{d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.18359, size = 4165, normalized size = 57.85 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^5*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + (a^2 + 2
*a*b + b^2)*d*x*sinh(d*x + c)^8 - 4*((a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^6 + 4*(7*(a^2 + 2*a*b
+ b^2)*d*x*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d*x + a^2 + a*b)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*d
*x*cosh(d*x + c)^3 - 3*((a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(3*(a^2 + 2*a*
b + b^2)*d*x - 2*a^2 - 4*a*b)*cosh(d*x + c)^4 + 2*(35*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 3*(a^2 + 2*a*b
+ b^2)*d*x - 30*((a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^2 - 2*a^2 - 4*a*b)*sinh(d*x + c)^4 + 8*(7
*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^5 - 10*((a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^3 + (3*(a^2
+ 2*a*b + b^2)*d*x - 2*a^2 - 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*d*x - 4*((a^2 + 2*a*b
+ b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^6 - 15*((a^2 + 2*a*b + b
^2)*d*x - a^2 - a*b)*cosh(d*x + c)^4 - (a^2 + 2*a*b + b^2)*d*x + 3*(3*(a^2 + 2*a*b + 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)*c
osh(d*x + c)*sinh(d*x + c)^7 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^8 - 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 4
*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - a^2 - 2*a*b - b^2)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d
*x + c)^3 - 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 2*(
35*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 - 30*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 3*a^2 + 6*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 + 2*a*b + b^2)*cosh(d*x + c)^3 + 3*(a^2 + 2*a*
b + b^2)*cosh(d*x + c))*sinh(d*x + c)^3 - 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*cos
h(d*x + c)^6 - 15*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 9*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - a^2 - 2*a*b -
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 + 2*a*b + b^2)*cosh
(d*x + c)^5 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*
sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^7 - 3*((a^2 + 2*a*b
+ b^2)*d*x - a^2 - a*b)*cosh(d*x + c)^5 + (3*(a^2 + 2*a*b + b^2)*d*x - 2*a^2 - 4*a*b)*cosh(d*x + c)^3 - ((a^2
+ 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^8 + 8*d*cosh(d*x + c)*sinh(d*x
+ c)^7 + d*sinh(d*x + c)^8 - 4*d*cosh(d*x + c)^6 + 4*(7*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^6 + 8*(7*d*cosh(d
*x + c)^3 - 3*d*cosh(d*x + c))*sinh(d*x + c)^5 + 6*d*cosh(d*x + c)^4 + 2*(35*d*cosh(d*x + c)^4 - 30*d*cosh(d*x
+ c)^2 + 3*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c)^5 - 10*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x +
c)^3 - 4*d*cosh(d*x + c)^2 + 4*(7*d*cosh(d*x + c)^6 - 15*d*cosh(d*x + c)^4 + 9*d*cosh(d*x + c)^2 - d)*sinh(d*
x + c)^2 + 8*(d*cosh(d*x + c)^7 - 3*d*cosh(d*x + c)^5 + 3*d*cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x + c) +
d)

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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(coth(d*x+c)**5*(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.31524, size = 165, normalized size = 2.29 \begin{align*} -\frac{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (d x + c\right )}}{d} + \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{d} - \frac{4 \,{\left ({\left (a^{2} + a b\right )} e^{\left (6 \, d x + 6 \, 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 (2 \, d x + 2 \, c\right )}\right )}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2 + 2*a*b + b^2)*(d*x + c)/d + (a^2 + 2*a*b + b^2)*log(abs(e^(2*d*x + 2*c) - 1))/d - 4*((a^2 + a*b)*e^(6*d
*x + 6*c) - (a^2 + 2*a*b)*e^(4*d*x + 4*c) + (a^2 + a*b)*e^(2*d*x + 2*c))/(d*(e^(2*d*x + 2*c) - 1)^4)