3.65 \(\int x \sinh ^2(a+b x^n) \, dx\)

Optimal. Leaf size=99 \[ -\frac{e^{2 a} 4^{-\frac{1}{n}-1} x^2 \left (-b x^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-2 b x^n\right )}{n}-\frac{e^{-2 a} 4^{-\frac{1}{n}-1} x^2 \left (b x^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},2 b x^n\right )}{n}-\frac{x^2}{4} \]

[Out]

-x^2/4 - (4^(-1 - n^(-1))*E^(2*a)*x^2*Gamma[2/n, -2*b*x^n])/(n*(-(b*x^n))^(2/n)) - (4^(-1 - n^(-1))*x^2*Gamma[
2/n, 2*b*x^n])/(E^(2*a)*n*(b*x^n)^(2/n))

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Rubi [A]  time = 0.107395, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5362, 5361, 2218} \[ -\frac{e^{2 a} 4^{-\frac{1}{n}-1} x^2 \left (-b x^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-2 b x^n\right )}{n}-\frac{e^{-2 a} 4^{-\frac{1}{n}-1} x^2 \left (b x^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},2 b x^n\right )}{n}-\frac{x^2}{4} \]

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b*x^n]^2,x]

[Out]

-x^2/4 - (4^(-1 - n^(-1))*E^(2*a)*x^2*Gamma[2/n, -2*b*x^n])/(n*(-(b*x^n))^(2/n)) - (4^(-1 - n^(-1))*x^2*Gamma[
2/n, 2*b*x^n])/(E^(2*a)*n*(b*x^n)^(2/n))

Rule 5362

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

Rule 5361

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^(c + d*x^n), x], x]
+ Dist[1/2, Int[(e*x)^m*E^(-c - d*x^n), x], x] /; FreeQ[{c, d, e, m, n}, x]

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 1.24474, size = 85, normalized size = 0.86 \[ -\frac{x^2 \left (e^{2 a} 4^{-1/n} \left (-b x^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},-2 b x^n\right )+e^{-2 a} 4^{-1/n} \left (b x^n\right )^{-2/n} \text{Gamma}\left (\frac{2}{n},2 b x^n\right )+n\right )}{4 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[a + b*x^n]^2,x]

[Out]

-(x^2*(n + (E^(2*a)*Gamma[2/n, -2*b*x^n])/(4^n^(-1)*(-(b*x^n))^(2/n)) + Gamma[2/n, 2*b*x^n]/(4^n^(-1)*E^(2*a)*
(b*x^n)^(2/n))))/(4*n)

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Maple [F]  time = 0.098, size = 0, normalized size = 0. \begin{align*} \int x \left ( \sinh \left ( a+b{x}^{n} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.18564, size = 111, normalized size = 1.12 \begin{align*} -\frac{1}{4} \, x^{2} - \frac{x^{2} e^{\left (-2 \, a\right )} \Gamma \left (\frac{2}{n}, 2 \, b x^{n}\right )}{4 \, \left (2 \, b x^{n}\right )^{\frac{2}{n}} n} - \frac{x^{2} e^{\left (2 \, a\right )} \Gamma \left (\frac{2}{n}, -2 \, b x^{n}\right )}{4 \, \left (-2 \, b x^{n}\right )^{\frac{2}{n}} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^2 - 1/4*x^2*e^(-2*a)*gamma(2/n, 2*b*x^n)/((2*b*x^n)^(2/n)*n) - 1/4*x^2*e^(2*a)*gamma(2/n, -2*b*x^n)/((-
2*b*x^n)^(2/n)*n)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x \sinh \left (b x^{n} + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*sinh(b*x^n + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(a+b*x**n)**2,x)

[Out]

Integral(x*sinh(a + b*x**n)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sinh(b*x^n + a)^2, x)