3.55 \(\int (e x)^m \sinh (a+\frac{b}{x^2}) \, dx\)

Optimal. Leaf size=87 \[ \frac{1}{4} e^a x \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b}{x^2}\right )-\frac{1}{4} e^{-a} x \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{b}{x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0869012, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5350, 5328, 2218} \[ \frac{1}{4} e^a x \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b}{x^2}\right )-\frac{1}{4} e^{-a} x \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} (e x)^m \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{b}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b/x^2],x]

[Out]

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

Rule 5350

Int[((e_.)*(x_))^(m_)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Dist[(e*x)^m*(x^(-1))
^m, Subst[Int[(a + b*Sinh[c + d/x^n])^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IntegerQ
[p] && ILtQ[n, 0] &&  !RationalQ[m]

Rule 5328

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}, x] && IGtQ[n, 0]

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

Mathematica [A]  time = 0.139264, size = 84, normalized size = 0.97 \[ \frac{1}{4} x (e x)^m \left ((\sinh (a)+\cosh (a)) \left (-\frac{b}{x^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b}{x^2}\right )-(\cosh (a)-\sinh (a)) \left (\frac{b}{x^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),\frac{b}{x^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sinh[a + b/x^2],x]

[Out]

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

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Maple [C]  time = 0.034, size = 77, normalized size = 0.9 \begin{align*}{\frac{ \left ( ex \right ) ^{m}b\cosh \left ( a \right ) }{ \left ( -1+m \right ) x}{\mbox{$_1$F$_2$}({\frac{1}{4}}-{\frac{m}{4}};\,{\frac{3}{2}},{\frac{5}{4}}-{\frac{m}{4}};\,{\frac{{b}^{2}}{4\,{x}^{4}}})}}+{\frac{ \left ( ex \right ) ^{m}x\sinh \left ( a \right ) }{1+m}{\mbox{$_1$F$_2$}(-{\frac{1}{4}}-{\frac{m}{4}};\,{\frac{1}{2}},{\frac{3}{4}}-{\frac{m}{4}};\,{\frac{{b}^{2}}{4\,{x}^{4}}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x)^m*b/(-1+m)/x*hypergeom([1/4-1/4*m],[3/2,5/4-1/4*m],1/4/x^4*b^2)*cosh(a)+(e*x)^m/(1+m)*x*hypergeom([-1/4-
1/4*m],[1/2,3/4-1/4*m],1/4/x^4*b^2)*sinh(a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*sinh(a + b/x^2), 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 (\frac{a x^{2} + b}{x^{2}}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x)^m*sinh((a*x^2 + b)/x^2), 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 + \frac{b}{x^{2}} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*sinh(a + b/x^2), x)