\(\int (e x)^m \sinh ^2(a+b x^2) \, dx\) [25]

Optimal result

Integrand size = 16, antiderivative size = 135 \[ \int (e x)^m \sinh ^2\left (a+b x^2\right ) \, dx=-\frac {(e x)^{1+m}}{2 e (1+m)}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{2 a} (e x)^{1+m} \left (-b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 b x^2\right )}{e}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{-2 a} (e x)^{1+m} \left (b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 b x^2\right )}{e} \]

[Out]

-1/2*(e*x)^(1+m)/e/(1+m)-2^(-7/2-1/2*m)*exp(2*a)*(e*x)^(1+m)*(-b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,-2*b*x^2)/e
-2^(-7/2-1/2*m)*(e*x)^(1+m)*(b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,2*b*x^2)/e/exp(2*a)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5448, 5437, 2250} \[ \int (e x)^m \sinh ^2\left (a+b x^2\right ) \, dx=-\frac {e^{2 a} 2^{-\frac {m}{2}-\frac {7}{2}} \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-2 b x^2\right )}{e}-\frac {e^{-2 a} 2^{-\frac {m}{2}-\frac {7}{2}} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},2 b x^2\right )}{e}-\frac {(e x)^{m+1}}{2 e (m+1)} \]

[In]

Int[(e*x)^m*Sinh[a + b*x^2]^2,x]

[Out]

-1/2*(e*x)^(1 + m)/(e*(1 + m)) - (2^(-7/2 - m/2)*E^(2*a)*(e*x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2
, -2*b*x^2])/e - (2^(-7/2 - m/2)*(e*x)^(1 + m)*(b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, 2*b*x^2])/(e*E^(2*a))

Rule 2250

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

Rule 5437

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

Rule 5448

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2} (e x)^m+\frac {1}{2} (e x)^m \cosh \left (2 a+2 b x^2\right )\right ) \, dx \\ & = -\frac {(e x)^{1+m}}{2 e (1+m)}+\frac {1}{2} \int (e x)^m \cosh \left (2 a+2 b x^2\right ) \, dx \\ & = -\frac {(e x)^{1+m}}{2 e (1+m)}+\frac {1}{4} \int e^{-2 a-2 b x^2} (e x)^m \, dx+\frac {1}{4} \int e^{2 a+2 b x^2} (e x)^m \, dx \\ & = -\frac {(e x)^{1+m}}{2 e (1+m)}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{2 a} (e x)^{1+m} \left (-b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 b x^2\right )}{e}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{-2 a} (e x)^{1+m} \left (b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 b x^2\right )}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.84 \[ \int (e x)^m \sinh ^2\left (a+b x^2\right ) \, dx=\frac {1}{8} x (e x)^m \left (-\frac {4}{1+m}-2^{-\frac {1}{2}-\frac {m}{2}} e^{2 a} \left (-b x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},-2 b x^2\right )-2^{-\frac {1}{2}-\frac {m}{2}} e^{-2 a} \left (b x^2\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {1+m}{2},2 b x^2\right )\right ) \]

[In]

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

[Out]

(x*(e*x)^m*(-4/(1 + m) - 2^(-1/2 - m/2)*E^(2*a)*(-(b*x^2))^(-1/2 - m/2)*Gamma[(1 + m)/2, -2*b*x^2] - (2^(-1/2
- m/2)*(b*x^2)^(-1/2 - m/2)*Gamma[(1 + m)/2, 2*b*x^2])/E^(2*a)))/8

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.29 \[ \int (e x)^m \sinh ^2\left (a+b x^2\right ) \, dx=-\frac {8 \, b x \cosh \left (m \log \left (e x\right )\right ) + {\left (e m + e\right )} \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 \, b}{e^{2}}\right ) + 2 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 \, b x^{2}\right ) - {\left (e m + e\right )} \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 \, b}{e^{2}}\right ) - 2 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 \, b x^{2}\right ) + 8 \, b x \sinh \left (m \log \left (e x\right )\right ) - {\left (e m + e\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 \, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 \, b}{e^{2}}\right ) + 2 \, a\right ) + {\left (e m + e\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 \, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 \, b}{e^{2}}\right ) - 2 \, a\right )}{16 \, {\left (b m + b\right )}} \]

[In]

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

[Out]

-1/16*(8*b*x*cosh(m*log(e*x)) + (e*m + e)*cosh(1/2*(m - 1)*log(2*b/e^2) + 2*a)*gamma(1/2*m + 1/2, 2*b*x^2) - (
e*m + e)*cosh(1/2*(m - 1)*log(-2*b/e^2) - 2*a)*gamma(1/2*m + 1/2, -2*b*x^2) + 8*b*x*sinh(m*log(e*x)) - (e*m +
e)*gamma(1/2*m + 1/2, 2*b*x^2)*sinh(1/2*(m - 1)*log(2*b/e^2) + 2*a) + (e*m + e)*gamma(1/2*m + 1/2, -2*b*x^2)*s
inh(1/2*(m - 1)*log(-2*b/e^2) - 2*a))/(b*m + b)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

1/4*e^m*integrate(e^(2*b*x^2 + m*log(x) + 2*a), x) + 1/4*e^m*integrate(e^(-2*b*x^2 + m*log(x) - 2*a), x) - 1/2
*(e*x)^(m + 1)/(e*(m + 1))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(sinh(a + b*x^2)^2*(e*x)^m,x)

[Out]

int(sinh(a + b*x^2)^2*(e*x)^m, x)