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

Optimal result

Integrand size = 14, antiderivative size = 87 \[ \int (e x)^m \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {1}{4} e^a \left (-\frac {b}{x^2}\right )^{\frac {1+m}{2}} x (e x)^m \Gamma \left (\frac {1}{2} (-1-m),-\frac {b}{x^2}\right )-\frac {1}{4} e^{-a} \left (\frac {b}{x^2}\right )^{\frac {1+m}{2}} x (e x)^m \Gamma \left (\frac {1}{2} (-1-m),\frac {b}{x^2}\right ) \] Output:

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)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int (e x)^m \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {1}{4} e^{-a} x (e x)^m \left (e^{2 a} \left (-\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),-\frac {b}{x^2}\right )-\left (\frac {b}{x^2}\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1}{2} (-1-m),\frac {b}{x^2}\right )\right ) \] Input:

Integrate[(e*x)^m*Sinh[a + b/x^2],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5873, 5851, 2648}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5873

\(\displaystyle -\left (\frac {1}{x}\right )^m (e x)^m \int \left (\frac {1}{x}\right )^{-m-2} \sinh \left (a+\frac {b}{x^2}\right )d\frac {1}{x}\)

\(\Big \downarrow \) 5851

\(\displaystyle -\left (\frac {1}{x}\right )^m (e x)^m \left (\frac {1}{2} \int e^{a+\frac {b}{x^2}} \left (\frac {1}{x}\right )^{-m-2}d\frac {1}{x}-\frac {1}{2} \int e^{-a-\frac {b}{x^2}} \left (\frac {1}{x}\right )^{-m-2}d\frac {1}{x}\right )\)

\(\Big \downarrow \) 2648

\(\displaystyle -\left (\frac {1}{x}\right )^m (e x)^m \left (\frac {1}{4} e^{-a} \left (\frac {1}{x}\right )^{-m-1} \left (\frac {b}{x^2}\right )^{\frac {m+1}{2}} \Gamma \left (\frac {1}{2} (-m-1),\frac {b}{x^2}\right )-\frac {1}{4} e^a \left (\frac {1}{x}\right )^{-m-1} \left (-\frac {b}{x^2}\right )^{\frac {m+1}{2}} \Gamma \left (\frac {1}{2} (-m-1),-\frac {b}{x^2}\right )\right )\)

Input:

Int[(e*x)^m*Sinh[a + b/x^2],x]
 

Output:

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

Defintions of rubi rules used

rule 2648

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] /; FreeQ[{F 
, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]
 

rule 5851

Int[((e_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[1/2 
   Int[(e*x)^m*E^(c + d*x^n), x], x] - Simp[1/2   Int[(e*x)^m*E^(-c - d*x^n 
), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]
 

rule 5873

Int[((e_.)*(x_))^(m_)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), 
x_Symbol] :> Simp[(-(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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89

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

Input:

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

Output:

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

Fricas [F]

\[ \int (e x)^m \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right ) \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int (e x)^m \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int \left (e x\right )^{m} \sinh {\left (a + \frac {b}{x^{2}} \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int (e x)^m \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right ) \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int (e x)^m \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (a + \frac {b}{x^{2}}\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int \mathrm {sinh}\left (a+\frac {b}{x^2}\right )\,{\left (e\,x\right )}^m \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (e x)^m \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {e^{m} \left (2 x^{m} \cosh \left (\frac {a \,x^{2}+b}{x^{2}}\right ) b +x^{m} \sinh \left (\frac {a \,x^{2}+b}{x^{2}}\right ) m \,x^{2}-x^{m} \sinh \left (\frac {a \,x^{2}+b}{x^{2}}\right ) x^{2}+4 \left (\int \frac {x^{m} \sinh \left (\frac {a \,x^{2}+b}{x^{2}}\right )}{x^{4}}d x \right ) b^{2} x \right )}{x \left (m^{2}-1\right )} \] Input:

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

Output:

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