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

Optimal. Leaf size=95 \[ \frac {e^{-a} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},b x^2\right )}{4 e}-\frac {e^a \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-b x^2\right )}{4 e} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 = {5328, 2218} \[ \frac {e^{-a} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},b x^2\right )}{4 e}-\frac {e^a \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},-b x^2\right )}{4 e} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b*x^2],x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 98, normalized size = 1.03 \[ -\frac {1}{4} x \left (-b^2 x^4\right )^{\frac {1}{2} (-m-1)} (e x)^m \left ((\sinh (a)+\cosh (a)) \left (b x^2\right )^{\frac {m+1}{2}} \Gamma \left (\frac {m+1}{2},-b x^2\right )-(\cosh (a)-\sinh (a)) \left (-b x^2\right )^{\frac {m+1}{2}} \Gamma \left (\frac {m+1}{2},b x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sinh[a + b*x^2],x]

[Out]

-1/4*(x*(e*x)^m*(-(b^2*x^4))^((-1 - m)/2)*(-((-(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])))

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fricas [A]  time = 0.45, size = 124, normalized size = 1.31 \[ \frac {e \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {b}{e^{2}}\right ) + a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, b x^{2}\right ) + e \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {b}{e^{2}}\right ) - a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2}\right ) - e \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {b}{e^{2}}\right ) + a\right ) - e \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {b}{e^{2}}\right ) - a\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(e*cosh(1/2*(m - 1)*log(b/e^2) + a)*gamma(1/2*m + 1/2, b*x^2) + e*cosh(1/2*(m - 1)*log(-b/e^2) - a)*gamma(
1/2*m + 1/2, -b*x^2) - e*gamma(1/2*m + 1/2, b*x^2)*sinh(1/2*(m - 1)*log(b/e^2) + a) - e*gamma(1/2*m + 1/2, -b*
x^2)*sinh(1/2*(m - 1)*log(-b/e^2) - a))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*sinh(b*x^2 + a), x)

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maple [C]  time = 0.10, size = 77, normalized size = 0.81 \[ \frac {\left (e x \right )^{m} x \hypergeom \left (\left [\frac {m}{4}+\frac {1}{4}\right ], \left [\frac {1}{2}, \frac {5}{4}+\frac {m}{4}\right ], \frac {x^{4} b^{2}}{4}\right ) \sinh \relax (a )}{1+m}+\frac {\left (e x \right )^{m} b \,x^{3} \hypergeom \left (\left [\frac {3}{4}+\frac {m}{4}\right ], \left [\frac {3}{2}, \frac {7}{4}+\frac {m}{4}\right ], \frac {x^{4} b^{2}}{4}\right ) \cosh \relax (a )}{3+m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x)^m/(1+m)*x*hypergeom([1/4*m+1/4],[1/2,5/4+1/4*m],1/4*x^4*b^2)*sinh(a)+(e*x)^m*b/(3+m)*x^3*hypergeom([3/4+
1/4*m],[3/2,7/4+1/4*m],1/4*x^4*b^2)*cosh(a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x)^m*sinh(b*x^2 + a), x)

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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