### 3.22 $$\int \text{csch}^2(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx$$

Optimal. Leaf size=64 $\frac{3 a^2 b \tanh (c+d x)}{d}-\frac{a^3 \coth (c+d x)}{d}+\frac{a b^2 \tanh ^3(c+d x)}{d}+\frac{b^3 \tanh ^5(c+d x)}{5 d}$

[Out]

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

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Rubi [A]  time = 0.0639957, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.087, Rules used = {3663, 270} $\frac{3 a^2 b \tanh (c+d x)}{d}-\frac{a^3 \coth (c+d x)}{d}+\frac{a b^2 \tanh ^3(c+d x)}{d}+\frac{b^3 \tanh ^5(c+d x)}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3663

Int[sin[(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^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.678704, size = 70, normalized size = 1.09 $\frac{b \tanh (c+d x) \left (15 a^2-b (5 a+2 b) \text{sech}^2(c+d x)+5 a b+b^2 \text{sech}^4(c+d x)+b^2\right )-5 a^3 \coth (c+d x)}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-5*a^3*Coth[c + d*x] + b*(15*a^2 + 5*a*b + b^2 - b*(5*a + 2*b)*Sech[c + d*x]^2 + b^2*Sech[c + d*x]^4)*Tanh[c
+ d*x])/(5*d)

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Maple [B]  time = 0.053, size = 141, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ( -{a}^{3}{\rm coth} \left (dx+c\right )+3\,{a}^{2}b\tanh \left ( dx+c \right ) +3\,a{b}^{2} \left ( -1/2\,{\frac{\sinh \left ( dx+c \right ) }{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+1/2\, \left ( 2/3+1/3\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2} \right ) \tanh \left ( dx+c \right ) \right ) +{b}^{3} \left ( -{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{3\,\tanh \left ( dx+c \right ) }{8} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) } \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-a^3*coth(d*x+c)+3*a^2*b*tanh(d*x+c)+3*a*b^2*(-1/2*sinh(d*x+c)/cosh(d*x+c)^3+1/2*(2/3+1/3*sech(d*x+c)^2)*
tanh(d*x+c))+b^3*(-1/2*sinh(d*x+c)^3/cosh(d*x+c)^5-3/8*sinh(d*x+c)/cosh(d*x+c)^5+3/8*(8/15+1/5*sech(d*x+c)^4+4
/15*sech(d*x+c)^2)*tanh(d*x+c)))

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Maxima [B]  time = 1.11298, size = 470, normalized size = 7.34 \begin{align*} \frac{2}{5} \, b^{3}{\left (\frac{10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{5 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{1}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + 2 \, a b^{2}{\left (\frac{3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac{1}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac{6 \, a^{2} b}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} + \frac{2 \, a^{3}}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.87471, size = 1449, normalized size = 22.64 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-4/5*((5*a^3 + 5*a*b^2 + 2*b^3)*cosh(d*x + c)^5 + 5*(5*a^3 + 5*a*b^2 + 2*b^3)*cosh(d*x + c)*sinh(d*x + c)^4 +
(15*a^2*b + 10*a*b^2 + 3*b^3)*sinh(d*x + c)^5 + (25*a^3 + 5*a*b^2 - 2*b^3)*cosh(d*x + c)^3 + (45*a^2*b + 10*a*
b^2 - 3*b^3 + 10*(15*a^2*b + 10*a*b^2 + 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + (10*(5*a^3 + 5*a*b^2 + 2*b^3
)*cosh(d*x + c)^3 + 3*(25*a^3 + 5*a*b^2 - 2*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^3 - a*b^2)*cosh(d*x
+ c) + (5*(15*a^2*b + 10*a*b^2 + 3*b^3)*cosh(d*x + c)^4 + 30*a^2*b + 10*b^3 + 3*(45*a^2*b + 10*a*b^2 - 3*b^3)*
cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*cosh(d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 3
*d*cosh(d*x + c)^5 + (21*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^5 + 5*(7*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)
)*sinh(d*x + c)^4 + d*cosh(d*x + c)^3 + (35*d*cosh(d*x + c)^4 + 50*d*cosh(d*x + c)^2 + 9*d)*sinh(d*x + c)^3 +
3*(7*d*cosh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^2 - 5*d*cosh(d*x + c) + (7*d*co
sh(d*x + c)^6 + 25*d*cosh(d*x + c)^4 + 27*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \operatorname{csch}^{2}{\left (c + d x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tanh(c + d*x)**2)**3*csch(c + d*x)**2, x)

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Giac [B]  time = 1.71288, size = 273, normalized size = 4.27 \begin{align*} -\frac{2 \,{\left (\frac{5 \, a^{3}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + \frac{15 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 15 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 5 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 30 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 20 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 10 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} b + 5 \, a b^{2} + b^{3}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}\right )}}{5 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(5*a^3/(e^(2*d*x + 2*c) - 1) + (15*a^2*b*e^(8*d*x + 8*c) + 15*a*b^2*e^(8*d*x + 8*c) + 5*b^3*e^(8*d*x + 8*
c) + 60*a^2*b*e^(6*d*x + 6*c) + 30*a*b^2*e^(6*d*x + 6*c) + 90*a^2*b*e^(4*d*x + 4*c) + 20*a*b^2*e^(4*d*x + 4*c)
+ 10*b^3*e^(4*d*x + 4*c) + 60*a^2*b*e^(2*d*x + 2*c) + 10*a*b^2*e^(2*d*x + 2*c) + 15*a^2*b + 5*a*b^2 + b^3)/(e
^(2*d*x + 2*c) + 1)^5)/d