3.155 \(\int \text{csch}^2(c+d x) (a+b \sinh ^3(c+d x))^2 \, dx\)

Optimal. Leaf size=82 \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{2 a b \cosh (c+d x)}{d}+\frac{b^2 \sinh ^3(c+d x) \cosh (c+d x)}{4 d}-\frac{3 b^2 \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{3 b^2 x}{8} \]

[Out]

(3*b^2*x)/8 + (2*a*b*Cosh[c + d*x])/d - (a^2*Coth[c + d*x])/d - (3*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + (b
^2*Cosh[c + d*x]*Sinh[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.0979725, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3220, 3767, 8, 2638, 2635} \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{2 a b \cosh (c+d x)}{d}+\frac{b^2 \sinh ^3(c+d x) \cosh (c+d x)}{4 d}-\frac{3 b^2 \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{3 b^2 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*x)/8 + (2*a*b*Cosh[c + d*x])/d - (a^2*Coth[c + d*x])/d - (3*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) + (b
^2*Cosh[c + d*x]*Sinh[c + d*x]^3)/(4*d)

Rule 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rubi steps

\begin{align*} \int \text{csch}^2(c+d x) \left (a+b \sinh ^3(c+d x)\right )^2 \, dx &=-\int \left (-a^2 \text{csch}^2(c+d x)-2 a b \sinh (c+d x)-b^2 \sinh ^4(c+d x)\right ) \, dx\\ &=a^2 \int \text{csch}^2(c+d x) \, dx+(2 a b) \int \sinh (c+d x) \, dx+b^2 \int \sinh ^4(c+d x) \, dx\\ &=\frac{2 a b \cosh (c+d x)}{d}+\frac{b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}-\frac{1}{4} \left (3 b^2\right ) \int \sinh ^2(c+d x) \, dx-\frac{\left (i a^2\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \coth (c+d x))}{d}\\ &=\frac{2 a b \cosh (c+d x)}{d}-\frac{a^2 \coth (c+d x)}{d}-\frac{3 b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}+\frac{1}{8} \left (3 b^2\right ) \int 1 \, dx\\ &=\frac{3 b^2 x}{8}+\frac{2 a b \cosh (c+d x)}{d}-\frac{a^2 \coth (c+d x)}{d}-\frac{3 b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.284753, size = 92, normalized size = 1.12 \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{2 a b \sinh (c) \sinh (d x)}{d}+\frac{2 a b \cosh (c) \cosh (d x)}{d}+\frac{3 b^2 (c+d x)}{8 d}-\frac{b^2 \sinh (2 (c+d x))}{4 d}+\frac{b^2 \sinh (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*(c + d*x))/(8*d) + (2*a*b*Cosh[c]*Cosh[d*x])/d - (a^2*Coth[c + d*x])/d + (2*a*b*Sinh[c]*Sinh[d*x])/d -
(b^2*Sinh[2*(c + d*x)])/(4*d) + (b^2*Sinh[4*(c + d*x)])/(32*d)

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Maple [A]  time = 0.046, size = 65, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -{a}^{2}{\rm coth} \left (dx+c\right )+2\,ab\cosh \left ( dx+c \right ) +{b}^{2} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8}} \right ) \cosh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07606, size = 153, normalized size = 1.87 \begin{align*} \frac{1}{64} \, b^{2}{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + a b{\left (\frac{e^{\left (d x + c\right )}}{d} + \frac{e^{\left (-d x - c\right )}}{d}\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*sinh(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

1/64*b^2*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) + a*b*(e
^(d*x + c)/d + e^(-d*x - c)/d) + 2*a^2/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [A]  time = 1.91237, size = 360, normalized size = 4.39 \begin{align*} \frac{b^{2} \cosh \left (d x + c\right )^{5} + 5 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - 9 \, b^{2} \cosh \left (d x + c\right )^{3} +{\left (10 \, b^{2} \cosh \left (d x + c\right )^{3} - 27 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 8 \,{\left (8 \, a^{2} - b^{2}\right )} \cosh \left (d x + c\right ) + 8 \,{\left (3 \, b^{2} d x + 16 \, a b \cosh \left (d x + c\right ) + 8 \, a^{2}\right )} \sinh \left (d x + c\right )}{64 \, d \sinh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(b^2*cosh(d*x + c)^5 + 5*b^2*cosh(d*x + c)*sinh(d*x + c)^4 - 9*b^2*cosh(d*x + c)^3 + (10*b^2*cosh(d*x + c
)^3 - 27*b^2*cosh(d*x + c))*sinh(d*x + c)^2 - 8*(8*a^2 - b^2)*cosh(d*x + c) + 8*(3*b^2*d*x + 16*a*b*cosh(d*x +
c) + 8*a^2)*sinh(d*x + c))/(d*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(csch(d*x+c)**2*(a+b*sinh(d*x+c)**3)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.22934, size = 224, normalized size = 2.73 \begin{align*} \frac{3 \,{\left (d x + c\right )} b^{2}}{8 \, d} + \frac{{\left (64 \, a b e^{\left (5 \, d x + 5 \, c\right )} - 64 \, a b e^{\left (3 \, d x + 3 \, c\right )} - 9 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2} - 8 \,{\left (16 \, a^{2} - b^{2}\right )} e^{\left (4 \, d x + 4 \, c\right )}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d{\left (e^{\left (d x + c\right )} + 1\right )}{\left (e^{\left (d x + c\right )} - 1\right )}} + \frac{b^{2} d^{3} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, b^{2} d^{3} e^{\left (2 \, d x + 2 \, c\right )} + 64 \, a b d^{3} e^{\left (d x + c\right )}}{64 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(d*x + c)*b^2/d + 1/64*(64*a*b*e^(5*d*x + 5*c) - 64*a*b*e^(3*d*x + 3*c) - 9*b^2*e^(2*d*x + 2*c) + b^2 - 8*
(16*a^2 - b^2)*e^(4*d*x + 4*c))*e^(-4*d*x - 4*c)/(d*(e^(d*x + c) + 1)*(e^(d*x + c) - 1)) + 1/64*(b^2*d^3*e^(4*
d*x + 4*c) - 8*b^2*d^3*e^(2*d*x + 2*c) + 64*a*b*d^3*e^(d*x + c))/d^4