3.1.37 \(\int (e x)^m \sinh ^3(a+\frac {b}{x}) \, dx\) [37]

Optimal. Leaf size=146 \[ -\frac {1}{8} 3^{1+m} b e^{3 a} \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,-\frac {3 b}{x}\right )+\frac {3}{8} b e^a \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,-\frac {b}{x}\right )+\frac {3}{8} b e^{-a} \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,\frac {b}{x}\right )-\frac {1}{8} 3^{1+m} b e^{-3 a} \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,\frac {3 b}{x}\right ) \]

[Out]

-1/8*3^(1+m)*b*exp(3*a)*(-b/x)^m*(e*x)^m*GAMMA(-1-m,-3*b/x)+3/8*b*exp(a)*(-b/x)^m*(e*x)^m*GAMMA(-1-m,-b/x)+3/8
*b*(b/x)^m*(e*x)^m*GAMMA(-1-m,b/x)/exp(a)-1/8*3^(1+m)*b*(b/x)^m*(e*x)^m*GAMMA(-1-m,3*b/x)/exp(3*a)

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5458, 3393, 3389, 2212} \begin {gather*} -\frac {1}{8} e^{3 a} b 3^{m+1} \left (-\frac {b}{x}\right )^m (e x)^m \text {Gamma}\left (-m-1,-\frac {3 b}{x}\right )+\frac {3}{8} e^a b \left (-\frac {b}{x}\right )^m (e x)^m \text {Gamma}\left (-m-1,-\frac {b}{x}\right )+\frac {3}{8} e^{-a} b \left (\frac {b}{x}\right )^m (e x)^m \text {Gamma}\left (-m-1,\frac {b}{x}\right )-\frac {1}{8} e^{-3 a} b 3^{m+1} \left (\frac {b}{x}\right )^m (e x)^m \text {Gamma}\left (-m-1,\frac {3 b}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b/x]^3,x]

[Out]

-1/8*(3^(1 + m)*b*E^(3*a)*(-(b/x))^m*(e*x)^m*Gamma[-1 - m, (-3*b)/x]) + (3*b*E^a*(-(b/x))^m*(e*x)^m*Gamma[-1 -
m, -(b/x)])/8 + (3*b*(b/x)^m*(e*x)^m*Gamma[-1 - m, b/x])/(8*E^a) - (3^(1 + m)*b*(b/x)^m*(e*x)^m*Gamma[-1 - m,
(3*b)/x])/(8*E^(3*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 3389

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

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 ^3\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 ^3(a+b x) \, dx,x,\frac {1}{x}\right )\right )\\ &=-\left (\left (i \left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int \left (\frac {3}{4} i x^{-2-m} \sinh (a+b x)-\frac {1}{4} i x^{-2-m} \sinh (3 a+3 b x)\right ) \, dx,x,\frac {1}{x}\right )\right )\\ &=-\left (\frac {1}{4} \left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int x^{-2-m} \sinh (3 a+3 b x) \, dx,x,\frac {1}{x}\right )\right )+\frac {1}{4} \left (3 \left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int x^{-2-m} \sinh (a+b x) \, dx,x,\frac {1}{x}\right )\\ &=-\left (\frac {1}{8} \left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int e^{-i (3 i a+3 i b x)} x^{-2-m} \, dx,x,\frac {1}{x}\right )\right )+\frac {1}{8} \left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int e^{i (3 i a+3 i b x)} x^{-2-m} \, dx,x,\frac {1}{x}\right )+\frac {1}{8} \left (3 \left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int e^{-i (i a+i b x)} x^{-2-m} \, dx,x,\frac {1}{x}\right )-\frac {1}{8} \left (3 \left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int e^{i (i a+i b x)} x^{-2-m} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {1}{8} 3^{1+m} b e^{3 a} \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,-\frac {3 b}{x}\right )+\frac {3}{8} b e^a \left (-\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,-\frac {b}{x}\right )+\frac {3}{8} b e^{-a} \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,\frac {b}{x}\right )-\frac {1}{8} 3^{1+m} b e^{-3 a} \left (\frac {b}{x}\right )^m (e x)^m \Gamma \left (-1-m,\frac {3 b}{x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]
time = 180.00, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(e*x)^m*Sinh[a + b/x]^3,x]

[Out]

$Aborted

________________________________________________________________________________________

Maple [F]
time = 2.00, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\sinh ^{3}\left (a +\frac {b}{x}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*sinh(a+b/x)^3,x)

[Out]

int((e*x)^m*sinh(a+b/x)^3,x)

________________________________________________________________________________________

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)^3,x, algorithm="maxima")

[Out]

integrate((x*e)^m*sinh(a + b/x)^3, x)

________________________________________________________________________________________

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)^3,x, algorithm="fricas")

[Out]

integral((x*e)^m*sinh((a*x + b)/x)^3, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e x\right )^{m} \sinh ^{3}{\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)**3,x)

[Out]

Integral((e*x)**m*sinh(a + b/x)**3, x)

________________________________________________________________________________________

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)^3,x, algorithm="giac")

[Out]

integrate((e*x)^m*sinh(a + b/x)^3, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {sinh}\left (a+\frac {b}{x}\right )}^3\,{\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)^3*(e*x)^m,x)

[Out]

int(sinh(a + b/x)^3*(e*x)^m, x)

________________________________________________________________________________________