3.2.55 \(\int \text {csch}^2(c+d x) (a+b \sinh ^3(c+d x))^2 \, dx\) [155]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 = {3299, 3852, 8, 2718, 2715} \begin {gather*} -\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} \end {gather*}

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 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 3299

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 3852

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.20, size = 92, normalized size = 1.12 \begin {gather*} \frac {3 b^2 (c+d x)}{8 d}+\frac {2 a b \cosh (c) \cosh (d x)}{d}-\frac {a^2 \coth (c+d x)}{d}+\frac {2 a b \sinh (c) \sinh (d x)}{d}-\frac {b^2 \sinh (2 (c+d x))}{4 d}+\frac {b^2 \sinh (4 (c+d x))}{32 d} \end {gather*}

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 = 1.99, size = 124, normalized size = 1.51

method	result	size
risch	\(\frac {3 b^{2} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} b^{2}}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}+\frac {a b \,{\mathrm e}^{d x +c}}{d}+\frac {a b \,{\mathrm e}^{-d x -c}}{d}+\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} b^{2}}{64 d}-\frac {2 a^{2}}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )}\)	\(124\)

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,method=_RETURNVERBOSE)

[Out]

3/8*b^2*x+1/64/d*exp(4*d*x+4*c)*b^2-1/8/d*exp(2*d*x+2*c)*b^2+a*b/d*exp(d*x+c)+a*b/d*exp(-d*x-c)+1/8/d*exp(-2*d
*x-2*c)*b^2-1/64/d*exp(-4*d*x-4*c)*b^2-2*a^2/d/(exp(2*d*x+2*c)-1)

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Maxima [A]
time = 0.28, size = 113, normalized size = 1.38 \begin {gather*} \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 {gather*}

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 = 0.44, size = 142, normalized size = 1.73 \begin {gather*} \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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [A]
time = 0.46, size = 149, normalized size = 1.82 \begin {gather*} \frac {24 \, {\left (d x + c\right )} b^{2} + b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 64 \, a b e^{\left (d x + c\right )} + \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 )}}{{\left (e^{\left (d x + c\right )} + 1\right )} {\left (e^{\left (d x + c\right )} - 1\right )}}}{64 \, d} \end {gather*}

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]

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

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Mupad [B]
time = 0.74, size = 123, normalized size = 1.50 \begin {gather*} \frac {3\,b^2\,x}{8}-\frac {2\,a^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}+\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {b^2\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}+\frac {a\,b\,{\mathrm {e}}^{c+d\,x}}{d}+\frac {a\,b\,{\mathrm {e}}^{-c-d\,x}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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