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

Optimal. Leaf size=99 \[ -\frac{e^{2 a} 2^{-\frac{3}{n}-2} x^3 \left (-b x^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-2 b x^n\right )}{n}-\frac{e^{-2 a} 2^{-\frac{3}{n}-2} x^3 \left (b x^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},2 b x^n\right )}{n}-\frac{x^3}{6} \]

[Out]

-x^3/6 - (2^(-2 - 3/n)*E^(2*a)*x^3*Gamma[3/n, -2*b*x^n])/(n*(-(b*x^n))^(3/n)) - (2^(-2 - 3/n)*x^3*Gamma[3/n, 2
*b*x^n])/(E^(2*a)*n*(b*x^n)^(3/n))

________________________________________________________________________________________

Rubi [A]  time = 0.136861, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5362, 5361, 2218} \[ -\frac{e^{2 a} 2^{-\frac{3}{n}-2} x^3 \left (-b x^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-2 b x^n\right )}{n}-\frac{e^{-2 a} 2^{-\frac{3}{n}-2} x^3 \left (b x^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},2 b x^n\right )}{n}-\frac{x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sinh[a + b*x^n]^2,x]

[Out]

-x^3/6 - (2^(-2 - 3/n)*E^(2*a)*x^3*Gamma[3/n, -2*b*x^n])/(n*(-(b*x^n))^(3/n)) - (2^(-2 - 3/n)*x^3*Gamma[3/n, 2
*b*x^n])/(E^(2*a)*n*(b*x^n)^(3/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^2 \sinh ^2\left (a+b x^n\right ) \, dx &=\int \left (-\frac{x^2}{2}+\frac{1}{2} x^2 \cosh \left (2 a+2 b x^n\right )\right ) \, dx\\ &=-\frac{x^3}{6}+\frac{1}{2} \int x^2 \cosh \left (2 a+2 b x^n\right ) \, dx\\ &=-\frac{x^3}{6}+\frac{1}{4} \int e^{-2 a-2 b x^n} x^2 \, dx+\frac{1}{4} \int e^{2 a+2 b x^n} x^2 \, dx\\ &=-\frac{x^3}{6}-\frac{2^{-2-\frac{3}{n}} e^{2 a} x^3 \left (-b x^n\right )^{-3/n} \Gamma \left (\frac{3}{n},-2 b x^n\right )}{n}-\frac{2^{-2-\frac{3}{n}} e^{-2 a} x^3 \left (b x^n\right )^{-3/n} \Gamma \left (\frac{3}{n},2 b x^n\right )}{n}\\ \end{align*}

Mathematica [A]  time = 1.43503, size = 89, normalized size = 0.9 \[ -\frac{x^3 \left (3 e^{2 a} 8^{-1/n} \left (-b x^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},-2 b x^n\right )+3 e^{-2 a} 8^{-1/n} \left (b x^n\right )^{-3/n} \text{Gamma}\left (\frac{3}{n},2 b x^n\right )+2 n\right )}{12 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sinh[a + b*x^n]^2,x]

[Out]

-(x^3*(2*n + (3*E^(2*a)*Gamma[3/n, -2*b*x^n])/(8^n^(-1)*(-(b*x^n))^(3/n)) + (3*Gamma[3/n, 2*b*x^n])/(8^n^(-1)*
E^(2*a)*(b*x^n)^(3/n))))/(12*n)

________________________________________________________________________________________

Maple [F]  time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \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^2*sinh(a+b*x^n)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.22367, size = 111, normalized size = 1.12 \begin{align*} -\frac{1}{6} \, x^{3} - \frac{x^{3} e^{\left (-2 \, a\right )} \Gamma \left (\frac{3}{n}, 2 \, b x^{n}\right )}{4 \, \left (2 \, b x^{n}\right )^{\frac{3}{n}} n} - \frac{x^{3} e^{\left (2 \, a\right )} \Gamma \left (\frac{3}{n}, -2 \, b x^{n}\right )}{4 \, \left (-2 \, b x^{n}\right )^{\frac{3}{n}} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*x^3 - 1/4*x^3*e^(-2*a)*gamma(3/n, 2*b*x^n)/((2*b*x^n)^(3/n)*n) - 1/4*x^3*e^(2*a)*gamma(3/n, -2*b*x^n)/((-
2*b*x^n)^(3/n)*n)

________________________________________________________________________________________

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

[Out]

integral(x^2*sinh(b*x^n + a)^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sinh(a + b*x**n)**2, x)

________________________________________________________________________________________

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

[Out]

integrate(x^2*sinh(b*x^n + a)^2, x)