3.1.64 \(\int x^2 \sinh ^2(a+b x^n) \, dx\) [64]

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

[Out]

-1/6*x^3-2^(-2-3/n)*exp(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)/exp(
2*a)/n/((b*x^n)^(3/n))

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Rubi [A]
time = 0.09, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 = {5470, 5469, 2250} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sinh[a + b*x^n]^2,x]

[Out]

-1/6*x^3 - (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 2250

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

Rule 5469

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 5470

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]

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*}

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Mathematica [A]
time = 1.09, size = 89, normalized size = 0.90 \begin {gather*} -\frac {x^3 \left (2 n+3\ 8^{-1/n} e^{2 a} \left (-b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},-2 b x^n\right )+3\ 8^{-1/n} e^{-2 a} \left (b x^n\right )^{-3/n} \Gamma \left (\frac {3}{n},2 b x^n\right )\right )}{12 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sinh[a + b*x^n]^2,x]

[Out]

-1/12*(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))))/n

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Maple [F]
time = 1.65, size = 0, normalized size = 0.00 \[\int x^{2} \left (\sinh ^{2}\left (a +b \,x^{n}\right )\right )\, dx\]

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)

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Maxima [A]
time = 0.09, size = 82, normalized size = 0.83 \begin {gather*} -\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 {gather*}

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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sinh ^{2}{\left (a + b x^{n} \right )}\, dx \end {gather*}

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\mathrm {sinh}\left (a+b\,x^n\right )}^2 \,d x \end {gather*}

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)

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