3.267 \(\int \sinh ^2(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=88 \[ \frac{x \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}-\frac{2 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}+\frac{2 b^2 n^2 x}{1-4 b^2 n^2} \]

[Out]

(2*b^2*n^2*x)/(1 - 4*b^2*n^2) - (2*b*n*x*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]])/(1 - 4*b^2*n^2) + (x*S
inh[a + b*Log[c*x^n]]^2)/(1 - 4*b^2*n^2)

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Rubi [A]  time = 0.0206036, antiderivative size = 88, 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, 8} \[ \frac{x \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}-\frac{2 b n x \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{1-4 b^2 n^2}+\frac{2 b^2 n^2 x}{1-4 b^2 n^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*Log[c*x^n]]^2,x]

[Out]

(2*b^2*n^2*x)/(1 - 4*b^2*n^2) - (2*b*n*x*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*Log[c*x^n]])/(1 - 4*b^2*n^2) + (x*S
inh[a + b*Log[c*x^n]]^2)/(1 - 4*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 8

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

Rubi steps

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

Mathematica [A]  time = 0.100934, size = 55, normalized size = 0.62 \[ -\frac{x \left (-2 b n \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+4 b^2 n^2-1\right )}{8 b^2 n^2-2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*Log[c*x^n]]^2,x]

[Out]

-((x*(-1 + 4*b^2*n^2 + Cosh[2*(a + b*Log[c*x^n])] - 2*b*n*Sinh[2*(a + b*Log[c*x^n])]))/(-2 + 8*b^2*n^2))

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Maple [F]  time = 0.095, size = 0, normalized size = 0. \begin{align*} \int \left ( \sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b*ln(c*x^n))^2,x)

[Out]

int(sinh(a+b*ln(c*x^n))^2,x)

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Maxima [A]  time = 1.16068, size = 90, normalized size = 1.02 \begin{align*} \frac{c^{2 \, b} x e^{\left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )}}{4 \,{\left (2 \, b n + 1\right )}} - \frac{1}{2} \, x - \frac{x e^{\left (-2 \, a\right )}}{4 \,{\left (2 \, b c^{2 \, b} n - c^{2 \, b}\right )}{\left (x^{n}\right )}^{2 \, b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

1/4*c^(2*b)*x*e^(2*b*log(x^n) + 2*a)/(2*b*n + 1) - 1/2*x - 1/4*x*e^(-2*a)/((2*b*c^(2*b)*n - c^(2*b))*(x^n)^(2*
b))

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Fricas [A]  time = 2.02648, size = 258, normalized size = 2.93 \begin{align*} \frac{4 \, b n 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 ) - x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - x \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} -{\left (4 \, b^{2} n^{2} - 1\right )} x}{2 \,{\left (4 \, 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))^2,x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.20731, size = 228, normalized size = 2.59 \begin{align*} \frac{b c^{2 \, b} n x x^{2 \, b n} e^{\left (2 \, a\right )}}{2 \,{\left (4 \, b^{2} n^{2} - 1\right )}} - \frac{2 \, b^{2} n^{2} x}{4 \, b^{2} n^{2} - 1} - \frac{c^{2 \, b} x x^{2 \, b n} e^{\left (2 \, a\right )}}{4 \,{\left (4 \, b^{2} n^{2} - 1\right )}} - \frac{b n x e^{\left (-2 \, a\right )}}{2 \,{\left (4 \, b^{2} n^{2} - 1\right )} c^{2 \, b} x^{2 \, b n}} + \frac{x}{2 \,{\left (4 \, b^{2} n^{2} - 1\right )}} - \frac{x e^{\left (-2 \, a\right )}}{4 \,{\left (4 \, b^{2} n^{2} - 1\right )} c^{2 \, b} x^{2 \, b n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

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