3.1.38 \(\int (e x)^m \sinh ^2(a+\frac {b}{x}) \, dx\) [38]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5458, 3393, 3388, 2212} \begin {gather*} -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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b/x]^2,x]

[Out]

-1/2*(x*(e*x)^m)/(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 2212

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

Rule 3388

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 3393

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 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 ^2\left (a+\frac {b}{x}\right ) \, dx &=-\left (\left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.18, size = 88, normalized size = 0.98 \begin {gather*} -\frac {(e x)^m \left (x-2^m b (1+m) \left (\frac {b}{x}\right )^m \Gamma \left (-1-m,\frac {2 b}{x}\right ) (\cosh (a)-\sinh (a))^2+2^m b (1+m) \left (-\frac {b}{x}\right )^m \Gamma \left (-1-m,-\frac {2 b}{x}\right ) (\cosh (a)+\sinh (a))^2\right )}{2 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sinh[a + b/x]^2,x]

[Out]

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

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Maple [F]
time = 0.86, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\sinh ^{2}\left (a +\frac {b}{x}\right )\right )\, dx\]

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]
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]

-1/2*(x*e)^(m + 1)*e^(-1)/(m + 1) + 1/4*integrate(e^(m*log(x) + 2*a + m + 2*b/x), x) + 1/4*integrate(e^(m*log(
x) - 2*a + m - 2*b/x), 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 + 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 ^{2}{\left (a + \frac {b}{x} \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}\right )}^2\,{\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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