3.14 \(\int \text{csch}^2(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=46 \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{2 a b \tanh (c+d x)}{d}+\frac{b^2 \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0567826, antiderivative size = 46, 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} \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{2 a b \tanh (c+d x)}{d}+\frac{b^2 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*Coth[c + d*x])/d) + (2*a*b*Tanh[c + d*x])/d + (b^2*Tanh[c + d*x]^3)/(3*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 )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^2}{x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a b+\frac{a^2}{x^2}+b^2 x^2\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{a^2 \coth (c+d x)}{d}+\frac{2 a b \tanh (c+d x)}{d}+\frac{b^2 \tanh ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.448521, size = 43, normalized size = 0.93 \[ \frac{b \tanh (c+d x) \left (6 a-b \text{sech}^2(c+d x)+b\right )-3 a^2 \coth (c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*Coth[c + d*x] + b*(6*a + b - b*Sech[c + d*x]^2)*Tanh[c + d*x])/(3*d)

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Maple [A]  time = 0.048, size = 68, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -{a}^{2}{\rm coth} \left (dx+c\right )+2\,ab\tanh \left ( dx+c \right ) +{b}^{2} \left ( -{\frac{\sinh \left ( dx+c \right ) }{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{\tanh \left ( dx+c \right ) }{2} \left ({\frac{2}{3}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{3}} \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)^2,x)

[Out]

1/d*(-a^2*coth(d*x+c)+2*a*b*tanh(d*x+c)+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)))

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Maxima [B]  time = 1.05402, size = 184, normalized size = 4. \begin{align*} \frac{2}{3} \, 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{4 \, a b}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} + \frac{2 \, a^{2}}{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.85856, size = 683, normalized size = 14.85 \begin{align*} -\frac{4 \,{\left ({\left (3 \, a^{2} + b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (3 \, a^{2} + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \,{\left (3 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{3} +{\left (9 \, a^{2} - b^{2}\right )} \cosh \left (d x + c\right ) + 2 \,{\left (3 \,{\left (3 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 3 \, a b - b^{2}\right )} \sinh \left (d x + c\right )\right )}}{3 \,{\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + d \sinh \left (d x + c\right )^{5} + d \cosh \left (d x + c\right )^{3} +{\left (10 \, d \cosh \left (d x + c\right )^{2} + 3 \, d\right )} \sinh \left (d x + c\right )^{3} +{\left (10 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 2 \, d \cosh \left (d x + c\right ) +{\left (5 \, d \cosh \left (d x + c\right )^{4} + 9 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\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)^2,x, algorithm="fricas")

[Out]

-4/3*((3*a^2 + b^2)*cosh(d*x + c)^3 + 3*(3*a^2 + b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + 2*(3*a*b + b^2)*sinh(d*x
+ c)^3 + (9*a^2 - b^2)*cosh(d*x + c) + 2*(3*(3*a*b + b^2)*cosh(d*x + c)^2 + 3*a*b - b^2)*sinh(d*x + c))/(d*co
sh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5 + d*cosh(d*x + c)^3 + (10*d*cosh(d*x + c
)^2 + 3*d)*sinh(d*x + c)^3 + (10*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 - 2*d*cosh(d*x + c) +
(5*d*cosh(d*x + c)^4 + 9*d*cosh(d*x + c)^2 + 2*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 )^{2} \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)**2,x)

[Out]

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

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Giac [A]  time = 1.34978, size = 116, normalized size = 2.52 \begin{align*} -\frac{2 \,{\left (\frac{3 \, a^{2}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + \frac{6 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b + b^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}\right )}}{3 \, 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)^2,x, algorithm="giac")

[Out]

-2/3*(3*a^2/(e^(2*d*x + 2*c) - 1) + (6*a*b*e^(4*d*x + 4*c) + 3*b^2*e^(4*d*x + 4*c) + 12*a*b*e^(2*d*x + 2*c) +
6*a*b + b^2)/(e^(2*d*x + 2*c) + 1)^3)/d