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

Optimal. Leaf size=90 \[ -e^{2 a} b 2^{m-1} \left (-\frac{b}{x}\right )^m (e x)^m \text{Gamma}\left (-m-1,-\frac{2 b}{x}\right )+e^{-2 a} b 2^{m-1} \left (\frac{b}{x}\right )^m (e x)^m \text{Gamma}\left (-m-1,\frac{2 b}{x}\right )-\frac{x (e x)^m}{2 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.157582, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5350, 3312, 3307, 2181} \[ -e^{2 a} b 2^{m-1} \left (-\frac{b}{x}\right )^m (e x)^m \text{Gamma}\left (-m-1,-\frac{2 b}{x}\right )+e^{-2 a} b 2^{m-1} \left (\frac{b}{x}\right )^m (e x)^m \text{Gamma}\left (-m-1,\frac{2 b}{x}\right )-\frac{x (e x)^m}{2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b/x]^2,x]

[Out]

-(x*(e*x)^m)/(2*(1 + m)) - 2^(-1 + m)*b*E^(2*a)*(-(b/x))^m*(e*x)^m*Gamma[-1 - m, (-2*b)/x] + (2^(-1 + m)*b*(b/
x)^m*(e*x)^m*Gamma[-1 - m, (2*b)/x])/E^(2*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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

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

Mathematica [A]  time = 0.25195, size = 88, normalized size = 0.98 \[ -\frac{(e x)^m \left (b 2^m (m+1) (\sinh (a)+\cosh (a))^2 \left (-\frac{b}{x}\right )^m \text{Gamma}\left (-m-1,-\frac{2 b}{x}\right )-b 2^m (m+1) (\cosh (a)-\sinh (a))^2 \left (\frac{b}{x}\right )^m \text{Gamma}\left (-m-1,\frac{2 b}{x}\right )+x\right )}{2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sinh[a + b/x]^2,x]

[Out]

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

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Maple [F]  time = 0.058, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( \sinh \left ( a+{\frac{b}{x}} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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]

Exception raised: ValueError

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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 + b}{x}\right )^{2}, 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 + 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 ^{2}{\left (a + \frac{b}{x} \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}\right )^{2}\,{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)