3.81 \(\int (e x)^m \sinh ^2(a+b x^n) \, dx\)

Optimal. Leaf size=143 \[ -\frac {e^{2 a} 2^{-\frac {m+2 n+1}{n}} (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-2 b x^n\right )}{e n}-\frac {e^{-2 a} 2^{-\frac {m+2 n+1}{n}} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},2 b x^n\right )}{e n}-\frac {(e x)^{m+1}}{2 e (m+1)} \]

[Out]

-1/2*(e*x)^(1+m)/e/(1+m)-exp(2*a)*(e*x)^(1+m)*GAMMA((1+m)/n,-2*b*x^n)/(2^((1+m+2*n)/n))/e/n/((-b*x^n)^((1+m)/n
))-(e*x)^(1+m)*GAMMA((1+m)/n,2*b*x^n)/(2^((1+m+2*n)/n))/e/exp(2*a)/n/((b*x^n)^((1+m)/n))

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Rubi [A]  time = 0.17, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5362, 5361, 2218} \[ -\frac {e^{2 a} 2^{-\frac {m+2 n+1}{n}} (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},-2 b x^n\right )}{e n}-\frac {e^{-2 a} 2^{-\frac {m+2 n+1}{n}} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \text {Gamma}\left (\frac {m+1}{n},2 b x^n\right )}{e n}-\frac {(e x)^{m+1}}{2 e (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b*x^n]^2,x]

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 2.00, size = 117, normalized size = 0.82 \[ -\frac {x (e x)^m \left (e^{2 a} (m+1) 2^{-\frac {m+1}{n}} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-2 b x^n\right )+e^{-2 a} (m+1) 2^{-\frac {m+1}{n}} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},2 b x^n\right )+2 n\right )}{4 (m+1) n} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sinh[a + b*x^n]^2,x]

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x)^m*sinh(b*x^n + a)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{m} \sinh \left (b x^{n} + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\sinh ^{2}\left (a +b \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, e^{m} \int e^{\left (2 \, b x^{n} + m \log \relax (x) + 2 \, a\right )}\,{d x} + \frac {1}{4} \, e^{m} \int e^{\left (-2 \, b x^{n} + m \log \relax (x) - 2 \, a\right )}\,{d x} - \frac {\left (e x\right )^{m + 1}}{2 \, e {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {sinh}\left (a+b\,x^n\right )}^2\,{\left (e\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^n)^2*(e*x)^m,x)

[Out]

int(sinh(a + b*x^n)^2*(e*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*x)**m*sinh(a + b*x**n)**2, x)

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