3.8 \(\int \text{csch}^4(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=44 \[ \frac{(a-b) \coth (c+d x)}{d}-\frac{a \coth ^3(c+d x)}{3 d}-\frac{b \tanh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0431187, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3663, 448} \[ \frac{(a-b) \coth (c+d x)}{d}-\frac{a \coth ^3(c+d x)}{3 d}-\frac{b \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)*Coth[c + d*x])/d - (a*Coth[c + d*x]^3)/(3*d) - (b*Tanh[c + d*x])/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 448

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

Rubi steps

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

Mathematica [A]  time = 0.0678965, size = 61, normalized size = 1.39 \[ \frac{2 a \coth (c+d x)}{3 d}-\frac{a \coth (c+d x) \text{csch}^2(c+d x)}{3 d}-\frac{b \tanh (c+d x)}{d}-\frac{b \coth (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 55, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{3}} \right ){\rm coth} \left (dx+c\right )+b \left ( -{\frac{1}{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }}-2\,\tanh \left ( dx+c \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+b*(-1/sinh(d*x+c)/cosh(d*x+c)-2*tanh(d*x+c)))

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Maxima [B]  time = 1.16375, size = 153, normalized size = 3.48 \begin{align*} \frac{4}{3} \, a{\left (\frac{3 \, e^{\left (-2 \, d x - 2 \, 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 \, b}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.04028, size = 657, normalized size = 14.93 \begin{align*} -\frac{8 \,{\left ({\left (a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a + 3 \, b\right )} \sinh \left (d x + c\right )^{2} + a - 3 \, b\right )}}{3 \,{\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} - 2 \, d \cosh \left (d x + c\right )^{4} +{\left (15 \, d \cosh \left (d x + c\right )^{2} - 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \,{\left (5 \, d \cosh \left (d x + c\right )^{3} - 2 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )^{2} +{\left (15 \, d \cosh \left (d x + c\right )^{4} - 12 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{2} + 2 \,{\left (3 \, d \cosh \left (d x + c\right )^{5} - 4 \, d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 2 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/3*((a + 3*b)*cosh(d*x + c)^2 + 4*a*cosh(d*x + c)*sinh(d*x + c) + (a + 3*b)*sinh(d*x + c)^2 + a - 3*b)/(d*co
sh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 - 2*d*cosh(d*x + c)^4 + (15*d*cosh(d*x +
 c)^2 - 2*d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 - 2*d*cosh(d*x + c))*sinh(d*x + c)^3 - d*cosh(d*x + c)^2
 + (15*d*cosh(d*x + c)^4 - 12*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^5 - 4*d*cosh(d*x +
 c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + 2*d)

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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 ) \operatorname{csch}^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tanh(c + d*x)**2)*csch(c + d*x)**4, x)

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Giac [A]  time = 1.2148, size = 108, normalized size = 2.45 \begin{align*} \frac{2 \,{\left (\frac{3 \, b}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - \frac{3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a + 3 \, b}{{\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)^4*(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

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