\(\int (e x)^m \sinh ^3(a+b x^2) \, dx\) [24]

Optimal result

Integrand size = 16, antiderivative size = 214 \[ \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx=-\frac {3^{-\frac {1}{2}-\frac {m}{2}} e^{3 a} (e x)^{1+m} \left (-b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-3 b x^2\right )}{16 e}+\frac {3 e^a (e x)^{1+m} \left (-b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-b x^2\right )}{16 e}-\frac {3 e^{-a} (e x)^{1+m} \left (b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},b x^2\right )}{16 e}+\frac {3^{-\frac {1}{2}-\frac {m}{2}} e^{-3 a} (e x)^{1+m} \left (b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},3 b x^2\right )}{16 e} \] Output:

-1/16*3^(-1/2-1/2*m)*exp(3*a)*(e*x)^(1+m)*(-b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+ 
1/2*m,-3*b*x^2)/e+3/16*exp(a)*(e*x)^(1+m)*(-b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+ 
1/2*m,-b*x^2)/e-3/16*(e*x)^(1+m)*(b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,b*x^ 
2)/e/exp(a)+1/16*3^(-1/2-1/2*m)*(e*x)^(1+m)*(b*x^2)^(-1/2-1/2*m)*GAMMA(1/2 
+1/2*m,3*b*x^2)/e/exp(3*a)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.84 \[ \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx=\frac {1}{16} 3^{\frac {1}{2} (-1-m)} e^{-3 a} x (e x)^m \left (-b^2 x^4\right )^{\frac {1}{2} (-1-m)} \left (-e^{6 a} \left (b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-3 b x^2\right )+3^{\frac {3+m}{2}} e^{4 a} \left (b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-b x^2\right )+\left (-b x^2\right )^{\frac {1+m}{2}} \left (-3^{\frac {3+m}{2}} e^{2 a} \Gamma \left (\frac {1+m}{2},b x^2\right )+\Gamma \left (\frac {1+m}{2},3 b x^2\right )\right )\right ) \] Input:

Integrate[(e*x)^m*Sinh[a + b*x^2]^3,x]
 

Output:

(3^((-1 - m)/2)*x*(e*x)^m*(-(b^2*x^4))^((-1 - m)/2)*(-(E^(6*a)*(b*x^2)^((1 
 + m)/2)*Gamma[(1 + m)/2, -3*b*x^2]) + 3^((3 + m)/2)*E^(4*a)*(b*x^2)^((1 + 
 m)/2)*Gamma[(1 + m)/2, -(b*x^2)] + (-(b*x^2))^((1 + m)/2)*(-(3^((3 + m)/2 
)*E^(2*a)*Gamma[(1 + m)/2, b*x^2]) + Gamma[(1 + m)/2, 3*b*x^2])))/(16*E^(3 
*a))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5863, 2009}

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.

\(\displaystyle \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx\)

\(\Big \downarrow \) 5863

\(\displaystyle \int \left (\frac {1}{4} (e x)^m \sinh \left (3 a+3 b x^2\right )-\frac {3}{4} (e x)^m \sinh \left (a+b x^2\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {e^{3 a} 3^{-\frac {m}{2}-\frac {1}{2}} \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-3 b x^2\right )}{16 e}+\frac {3 e^a \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-b x^2\right )}{16 e}-\frac {3 e^{-a} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},b x^2\right )}{16 e}+\frac {e^{-3 a} 3^{-\frac {m}{2}-\frac {1}{2}} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},3 b x^2\right )}{16 e}\)

Input:

Int[(e*x)^m*Sinh[a + b*x^2]^3,x]
 

Output:

-1/16*(3^(-1/2 - m/2)*E^(3*a)*(e*x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[ 
(1 + m)/2, -3*b*x^2])/e + (3*E^a*(e*x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gam 
ma[(1 + m)/2, -(b*x^2)])/(16*e) - (3*(e*x)^(1 + m)*(b*x^2)^((-1 - m)/2)*Ga 
mma[(1 + m)/2, b*x^2])/(16*e*E^a) + (3^(-1/2 - m/2)*(e*x)^(1 + m)*(b*x^2)^ 
((-1 - m)/2)*Gamma[(1 + m)/2, 3*b*x^2])/(16*e*E^(3*a))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5863

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_), 
x_Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sinh[c + d*x^n])^p, x], x 
] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]
 

Maple [F]

\[\int \left (e x \right )^{m} \sinh \left (b \,x^{2}+a \right )^{3}d x\]

Input:

int((e*x)^m*sinh(b*x^2+a)^3,x)
 

Output:

int((e*x)^m*sinh(b*x^2+a)^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.18 \[ \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx=\frac {e \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {3 \, b}{e^{2}}\right ) + 3 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 3 \, b x^{2}\right ) - 9 \, e \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {b}{e^{2}}\right ) + a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, b x^{2}\right ) - 9 \, e \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {b}{e^{2}}\right ) - a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2}\right ) + e \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {3 \, b}{e^{2}}\right ) - 3 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -3 \, b x^{2}\right ) - e \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 3 \, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {3 \, b}{e^{2}}\right ) + 3 \, a\right ) + 9 \, e \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {b}{e^{2}}\right ) + a\right ) + 9 \, e \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {b}{e^{2}}\right ) - a\right ) - e \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -3 \, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {3 \, b}{e^{2}}\right ) - 3 \, a\right )}{48 \, b} \] Input:

integrate((e*x)^m*sinh(b*x^2+a)^3,x, algorithm="fricas")
 

Output:

1/48*(e*cosh(1/2*(m - 1)*log(3*b/e^2) + 3*a)*gamma(1/2*m + 1/2, 3*b*x^2) - 
 9*e*cosh(1/2*(m - 1)*log(b/e^2) + a)*gamma(1/2*m + 1/2, b*x^2) - 9*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(-3*b/e^2) - 3*a)*gamma(1/2*m + 1/2, -3*b*x^2) - e*gamma(1/2*m + 1/ 
2, 3*b*x^2)*sinh(1/2*(m - 1)*log(3*b/e^2) + 3*a) + 9*e*gamma(1/2*m + 1/2, 
b*x^2)*sinh(1/2*(m - 1)*log(b/e^2) + a) + 9*e*gamma(1/2*m + 1/2, -b*x^2)*s 
inh(1/2*(m - 1)*log(-b/e^2) - a) - e*gamma(1/2*m + 1/2, -3*b*x^2)*sinh(1/2 
*(m - 1)*log(-3*b/e^2) - 3*a))/b
 

Sympy [F]

\[ \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx=\int \left (e x\right )^{m} \sinh ^{3}{\left (a + b x^{2} \right )}\, dx \] Input:

integrate((e*x)**m*sinh(b*x**2+a)**3,x)
 

Output:

Integral((e*x)**m*sinh(a + b*x**2)**3, x)
 

Maxima [F]

\[ \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{2} + a\right )^{3} \,d x } \] Input:

integrate((e*x)^m*sinh(b*x^2+a)^3,x, algorithm="maxima")
 

Output:

integrate((e*x)^m*sinh(b*x^2 + a)^3, x)
 

Giac [F]

\[ \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx=\int { \left (e x\right )^{m} \sinh \left (b x^{2} + a\right )^{3} \,d x } \] Input:

integrate((e*x)^m*sinh(b*x^2+a)^3,x, algorithm="giac")
 

Output:

integrate((e*x)^m*sinh(b*x^2 + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx=\int {\mathrm {sinh}\left (b\,x^2+a\right )}^3\,{\left (e\,x\right )}^m \,d x \] Input:

int(sinh(a + b*x^2)^3*(e*x)^m,x)
 

Output:

int(sinh(a + b*x^2)^3*(e*x)^m, x)
 

Reduce [F]

\[ \int (e x)^m \sinh ^3\left (a+b x^2\right ) \, dx=e^{m} \left (\int x^{m} \sinh \left (b \,x^{2}+a \right )^{3}d x \right ) \] Input:

int((e*x)^m*sinh(b*x^2+a)^3,x)
 

Output:

e**m*int(x**m*sinh(a + b*x**2)**3,x)