3.1.55 \(\int (e x)^m \sinh (a+\frac {b}{x^2}) \, dx\) [55]

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

[Out]

1/4*exp(a)*(-b/x^2)^(1/2+1/2*m)*x*(e*x)^m*GAMMA(-1/2-1/2*m,-b/x^2)-1/4*(b/x^2)^(1/2+1/2*m)*x*(e*x)^m*GAMMA(-1/
2-1/2*m,b/x^2)/exp(a)

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Rubi [A]
time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 = {5458, 5436, 2250} \begin {gather*} \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 ) \end {gather*}

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 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 5436

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 5458

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] && Intege
rQ[p] && ILtQ[n, 0] &&  !RationalQ[m]

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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.11, size = 84, normalized size = 0.97 \begin {gather*} \frac {1}{4} x (e x)^m \left (-\left (\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),\frac {b}{x^2}\right ) (\cosh (a)-\sinh (a))+\left (-\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),-\frac {b}{x^2}\right ) (\cosh (a)+\sinh (a))\right ) \end {gather*}

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] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.45, size = 77, normalized size = 0.89

method	result	size
meijerg	\(\frac {\left (e x \right )^{m} b \hypergeom \left (\left [\frac {1}{4}-\frac {m}{4}\right ], \left [\frac {3}{2}, \frac {5}{4}-\frac {m}{4}\right ], \frac {b^{2}}{4 x^{4}}\right ) \cosh \left (a \right )}{\left (-1+m \right ) x}+\frac {\left (e x \right )^{m} x \hypergeom \left (\left [-\frac {1}{4}-\frac {m}{4}\right ], \left [\frac {1}{2}, \frac {3}{4}-\frac {m}{4}\right ], \frac {b^{2}}{4 x^{4}}\right ) \sinh \left (a \right )}{1+m}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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((x*e)^m*sinh(a + b/x^2), x)

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

[Out]

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

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

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 = 0.00, 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((e*x)^m*sinh(a+b/x^2),x, algorithm="giac")

[Out]

integrate((e*x)^m*sinh(a + b/x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x^2)*(e*x)^m,x)

[Out]

int(sinh(a + b/x^2)*(e*x)^m, x)

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