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

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

[Out]

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

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Rubi [A]  time = 0.0705051, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5360, 2218} \[ \frac{e^{-a} (e x)^{m+1} \left (b x^n\right )^{-\frac{m+1}{n}} \text{Gamma}\left (\frac{m+1}{n},b x^n\right )}{2 e n}-\frac{e^a (e x)^{m+1} \left (-b x^n\right )^{-\frac{m+1}{n}} \text{Gamma}\left (\frac{m+1}{n},-b x^n\right )}{2 e n} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b*x^n],x]

[Out]

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

Rule 5360

Int[((e_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)], 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 (e x)^m \sinh \left (a+b x^n\right ) \, dx &=-\left (\frac{1}{2} \int e^{-a-b x^n} (e x)^m \, dx\right )+\frac{1}{2} \int e^{a+b x^n} (e x)^m \, dx\\ &=-\frac{e^a (e x)^{1+m} \left (-b x^n\right )^{-\frac{1+m}{n}} \Gamma \left (\frac{1+m}{n},-b x^n\right )}{2 e n}+\frac{e^{-a} (e x)^{1+m} \left (b x^n\right )^{-\frac{1+m}{n}} \Gamma \left (\frac{1+m}{n},b x^n\right )}{2 e n}\\ \end{align*}

Mathematica [A]  time = 0.170415, size = 102, normalized size = 1.03 \[ -\frac{x (e x)^m \left (-b^2 x^{2 n}\right )^{-\frac{m+1}{n}} \left ((\sinh (a)+\cosh (a)) \left (b x^n\right )^{\frac{m+1}{n}} \text{Gamma}\left (\frac{m+1}{n},-b x^n\right )-(\cosh (a)-\sinh (a)) \left (-b x^n\right )^{\frac{m+1}{n}} \text{Gamma}\left (\frac{m+1}{n},b x^n\right )\right )}{2 n} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sinh[a + b*x^n],x]

[Out]

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

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Maple [C]  time = 0.22, size = 115, normalized size = 1.2 \begin{align*}{\frac{ \left ( ex \right ) ^{m}x\sinh \left ( a \right ) }{1+m}{\mbox{$_1$F$_2$}({\frac{m}{2\,n}}+{\frac{1}{2\,n}};\,{\frac{1}{2}},1+{\frac{m}{2\,n}}+{\frac{1}{2\,n}};\,{\frac{{x}^{2\,n}{b}^{2}}{4}})}}+{\frac{ \left ( ex \right ) ^{m}{x}^{n+1}b\cosh \left ( a \right ) }{m+n+1}{\mbox{$_1$F$_2$}({\frac{1}{2}}+{\frac{m}{2\,n}}+{\frac{1}{2\,n}};\,{\frac{3}{2}},{\frac{3}{2}}+{\frac{m}{2\,n}}+{\frac{1}{2\,n}};\,{\frac{{x}^{2\,n}{b}^{2}}{4}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x)^m/(1+m)*x*hypergeom([1/2/n*m+1/2/n],[1/2,1+1/2/n*m+1/2/n],1/4*x^(2*n)*b^2)*sinh(a)+(e*x)^m/(m+n+1)*x^(n+
1)*b*hypergeom([1/2+1/2/n*m+1/2/n],[3/2,3/2+1/2/n*m+1/2/n],1/4*x^(2*n)*b^2)*cosh(a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*sinh(b*x^n + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x)^m*sinh(b*x^n + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*sinh(b*x^n + a), x)