3.268 \(\int \sinh ^3(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=149 \[ \frac{x \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}+\frac{6 b^2 n^2 x \sinh \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4-10 b^2 n^2+1}-\frac{6 b^3 n^3 x \cosh \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4-10 b^2 n^2+1}-\frac{3 b n x \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2} \]

[Out]

(-6*b^3*n^3*x*Cosh[a + b*Log[c*x^n]])/(1 - 10*b^2*n^2 + 9*b^4*n^4) + (6*b^2*n^2*x*Sinh[a + b*Log[c*x^n]])/(1 -
10*b^2*n^2 + 9*b^4*n^4) - (3*b*n*x*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]]^2)/(1 - 9*b^2*n^2) + (x*Sinh
[a + b*Log[c*x^n]]^3)/(1 - 9*b^2*n^2)

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Rubi [A]  time = 0.0405997, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5519, 5517} \[ \frac{x \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}+\frac{6 b^2 n^2 x \sinh \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4-10 b^2 n^2+1}-\frac{6 b^3 n^3 x \cosh \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4-10 b^2 n^2+1}-\frac{3 b n x \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*Log[c*x^n]]^3,x]

[Out]

(-6*b^3*n^3*x*Cosh[a + b*Log[c*x^n]])/(1 - 10*b^2*n^2 + 9*b^4*n^4) + (6*b^2*n^2*x*Sinh[a + b*Log[c*x^n]])/(1 -
10*b^2*n^2 + 9*b^4*n^4) - (3*b*n*x*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]]^2)/(1 - 9*b^2*n^2) + (x*Sinh
[a + b*Log[c*x^n]]^3)/(1 - 9*b^2*n^2)

Rule 5519

Int[Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> -Simp[(x*Sinh[d*(a + b*Log[c*x^n])]^p
)/(b^2*d^2*n^2*p^2 - 1), x] + (-Dist[(b^2*d^2*n^2*p*(p - 1))/(b^2*d^2*n^2*p^2 - 1), Int[Sinh[d*(a + b*Log[c*x^
n])]^(p - 2), x], x] + Simp[(b*d*n*p*x*Cosh[d*(a + b*Log[c*x^n])]*Sinh[d*(a + b*Log[c*x^n])]^(p - 1))/(b^2*d^2
*n^2*p^2 - 1), x]) /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 - 1, 0]

Rule 5517

Int[Sinh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_Symbol] :> -Simp[(x*Sinh[d*(a + b*Log[c*x^n])])/(b^2*
d^2*n^2 - 1), x] + Simp[(b*d*n*x*Cosh[d*(a + b*Log[c*x^n])])/(b^2*d^2*n^2 - 1), x] /; FreeQ[{a, b, c, d, n}, x
] && NeQ[b^2*d^2*n^2 - 1, 0]

Rubi steps

\begin{align*} \int \sinh ^3\left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac{3 b n x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}+\frac{x \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}+\frac{\left (6 b^2 n^2\right ) \int \sinh \left (a+b \log \left (c x^n\right )\right ) \, dx}{1-9 b^2 n^2}\\ &=-\frac{6 b^3 n^3 x \cosh \left (a+b \log \left (c x^n\right )\right )}{1-10 b^2 n^2+9 b^4 n^4}+\frac{6 b^2 n^2 x \sinh \left (a+b \log \left (c x^n\right )\right )}{1-10 b^2 n^2+9 b^4 n^4}-\frac{3 b n x \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}+\frac{x \sinh ^3\left (a+b \log \left (c x^n\right )\right )}{1-9 b^2 n^2}\\ \end{align*}

Mathematica [A]  time = 0.502005, size = 120, normalized size = 0.81 \[ \frac{x \left (-3 b n \left (9 b^2 n^2-1\right ) \cosh \left (a+b \log \left (c x^n\right )\right )+3 b n \left (b^2 n^2-1\right ) \cosh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )-2 \sinh \left (a+b \log \left (c x^n\right )\right ) \left (\left (b^2 n^2-1\right ) \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-13 b^2 n^2+1\right )\right )}{36 b^4 n^4-40 b^2 n^2+4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*Log[c*x^n]]^3,x]

[Out]

(x*(-3*b*n*(-1 + 9*b^2*n^2)*Cosh[a + b*Log[c*x^n]] + 3*b*n*(-1 + b^2*n^2)*Cosh[3*(a + b*Log[c*x^n])] - 2*(1 -
13*b^2*n^2 + (-1 + b^2*n^2)*Cosh[2*(a + b*Log[c*x^n])])*Sinh[a + b*Log[c*x^n]]))/(4 - 40*b^2*n^2 + 36*b^4*n^4)

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Maple [F]  time = 0.108, size = 0, normalized size = 0. \begin{align*} \int \left ( \sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b*ln(c*x^n))^3,x)

[Out]

int(sinh(a+b*ln(c*x^n))^3,x)

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Maxima [A]  time = 1.21326, size = 154, normalized size = 1.03 \begin{align*} \frac{c^{3 \, b} x e^{\left (3 \, b \log \left (x^{n}\right ) + 3 \, a\right )}}{8 \,{\left (3 \, b n + 1\right )}} - \frac{3 \, c^{b} x e^{\left (b \log \left (x^{n}\right ) + a\right )}}{8 \,{\left (b n + 1\right )}} - \frac{3 \, x e^{\left (-b \log \left (x^{n}\right ) - a\right )}}{8 \,{\left (b c^{b} n - c^{b}\right )}} + \frac{x e^{\left (-3 \, b \log \left (x^{n}\right ) - 3 \, a\right )}}{8 \,{\left (3 \, b c^{3 \, b} n - c^{3 \, b}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^3,x, algorithm="maxima")

[Out]

1/8*c^(3*b)*x*e^(3*b*log(x^n) + 3*a)/(3*b*n + 1) - 3/8*c^b*x*e^(b*log(x^n) + a)/(b*n + 1) - 3/8*x*e^(-b*log(x^
n) - a)/(b*c^b*n - c^b) + 1/8*x*e^(-3*b*log(x^n) - 3*a)/(3*b*c^(3*b)*n - c^(3*b))

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Fricas [A]  time = 2.14739, size = 532, normalized size = 3.57 \begin{align*} \frac{3 \,{\left (b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} + 9 \,{\left (b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} -{\left (b^{2} n^{2} - 1\right )} x \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 3 \,{\left (9 \, b^{3} n^{3} - b n\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - 3 \,{\left ({\left (b^{2} n^{2} - 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} -{\left (9 \, b^{2} n^{2} - 1\right )} x\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{4 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^3,x, algorithm="fricas")

[Out]

1/4*(3*(b^3*n^3 - b*n)*x*cosh(b*n*log(x) + b*log(c) + a)^3 + 9*(b^3*n^3 - b*n)*x*cosh(b*n*log(x) + b*log(c) +
a)*sinh(b*n*log(x) + b*log(c) + a)^2 - (b^2*n^2 - 1)*x*sinh(b*n*log(x) + b*log(c) + a)^3 - 3*(9*b^3*n^3 - b*n)
*x*cosh(b*n*log(x) + b*log(c) + a) - 3*((b^2*n^2 - 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2 - (9*b^2*n^2 - 1)*x)
*sinh(b*n*log(x) + b*log(c) + a))/(9*b^4*n^4 - 10*b^2*n^2 + 1)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*ln(c*x**n))**3,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.27506, size = 898, normalized size = 6.03 \begin{align*} \frac{3 \, b^{3} c^{3 \, b} n^{3} x x^{3 \, b n} e^{\left (3 \, a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac{27 \, b^{3} c^{b} n^{3} x x^{b n} e^{a}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac{b^{2} c^{3 \, b} n^{2} x x^{3 \, b n} e^{\left (3 \, a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} + \frac{27 \, b^{2} c^{b} n^{2} x x^{b n} e^{a}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac{3 \, b c^{3 \, b} n x x^{3 \, b n} e^{\left (3 \, a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac{27 \, b^{3} n^{3} x e^{\left (-a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{b} x^{b n}} + \frac{3 \, b^{3} n^{3} x e^{\left (-3 \, a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{3 \, b} x^{3 \, b n}} + \frac{3 \, b c^{b} n x x^{b n} e^{a}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} + \frac{c^{3 \, b} x x^{3 \, b n} e^{\left (3 \, a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} - \frac{27 \, b^{2} n^{2} x e^{\left (-a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{b} x^{b n}} + \frac{b^{2} n^{2} x e^{\left (-3 \, a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{3 \, b} x^{3 \, b n}} - \frac{3 \, c^{b} x x^{b n} e^{a}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )}} + \frac{3 \, b n x e^{\left (-a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{b} x^{b n}} - \frac{3 \, b n x e^{\left (-3 \, a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{3 \, b} x^{3 \, b n}} + \frac{3 \, x e^{\left (-a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{b} x^{b n}} - \frac{x e^{\left (-3 \, a\right )}}{8 \,{\left (9 \, b^{4} n^{4} - 10 \, b^{2} n^{2} + 1\right )} c^{3 \, b} x^{3 \, b n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^3,x, algorithm="giac")

[Out]

3/8*b^3*c^(3*b)*n^3*x*x^(3*b*n)*e^(3*a)/(9*b^4*n^4 - 10*b^2*n^2 + 1) - 27/8*b^3*c^b*n^3*x*x^(b*n)*e^a/(9*b^4*n
^4 - 10*b^2*n^2 + 1) - 1/8*b^2*c^(3*b)*n^2*x*x^(3*b*n)*e^(3*a)/(9*b^4*n^4 - 10*b^2*n^2 + 1) + 27/8*b^2*c^b*n^2
*x*x^(b*n)*e^a/(9*b^4*n^4 - 10*b^2*n^2 + 1) - 3/8*b*c^(3*b)*n*x*x^(3*b*n)*e^(3*a)/(9*b^4*n^4 - 10*b^2*n^2 + 1)
- 27/8*b^3*n^3*x*e^(-a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^b*x^(b*n)) + 3/8*b^3*n^3*x*e^(-3*a)/((9*b^4*n^4 - 10*
b^2*n^2 + 1)*c^(3*b)*x^(3*b*n)) + 3/8*b*c^b*n*x*x^(b*n)*e^a/(9*b^4*n^4 - 10*b^2*n^2 + 1) + 1/8*c^(3*b)*x*x^(3*
b*n)*e^(3*a)/(9*b^4*n^4 - 10*b^2*n^2 + 1) - 27/8*b^2*n^2*x*e^(-a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^b*x^(b*n)) +
1/8*b^2*n^2*x*e^(-3*a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^(3*b)*x^(3*b*n)) - 3/8*c^b*x*x^(b*n)*e^a/(9*b^4*n^4 -
10*b^2*n^2 + 1) + 3/8*b*n*x*e^(-a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^b*x^(b*n)) - 3/8*b*n*x*e^(-3*a)/((9*b^4*n^4
- 10*b^2*n^2 + 1)*c^(3*b)*x^(3*b*n)) + 3/8*x*e^(-a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^b*x^(b*n)) - 1/8*x*e^(-3*
a)/((9*b^4*n^4 - 10*b^2*n^2 + 1)*c^(3*b)*x^(3*b*n))