\(\int (e x)^m \sinh ^2(a+b x^n) \, dx\) [81]

Optimal result

Integrand size = 16, antiderivative size = 143 \[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=-\frac {(e x)^{1+m}}{2 e (1+m)}-\frac {2^{-\frac {1+m+2 n}{n}} e^{2 a} (e x)^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 b x^n\right )}{e n}-\frac {2^{-\frac {1+m+2 n}{n}} e^{-2 a} (e x)^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 b x^n\right )}{e n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 143, 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 = {5470, 5469, 2250} \[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=-\frac {e^{2 a} 2^{-\frac {m+2 n+1}{n}} (e x)^{m+1} \left (-b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},-2 b x^n\right )}{e n}-\frac {e^{-2 a} 2^{-\frac {m+2 n+1}{n}} (e x)^{m+1} \left (b x^n\right )^{-\frac {m+1}{n}} \Gamma \left (\frac {m+1}{n},2 b x^n\right )}{e n}-\frac {(e x)^{m+1}}{2 e (m+1)} \]

[In]

Int[(e*x)^m*Sinh[a + b*x^n]^2,x]

[Out]

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

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 5469

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, n}, x]

Rule 5470

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, n}, x] && IGtQ[p, 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^n\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^n\right ) \, dx \\ & = -\frac {(e x)^{1+m}}{2 e (1+m)}+\frac {1}{4} \int e^{-2 a-2 b x^n} (e x)^m \, dx+\frac {1}{4} \int e^{2 a+2 b x^n} (e x)^m \, dx \\ & = -\frac {(e x)^{1+m}}{2 e (1+m)}-\frac {2^{-\frac {1+m+2 n}{n}} e^{2 a} (e x)^{1+m} \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 b x^n\right )}{e n}-\frac {2^{-\frac {1+m+2 n}{n}} e^{-2 a} (e x)^{1+m} \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 b x^n\right )}{e n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.82 \[ \int (e x)^m \sinh ^2\left (a+b x^n\right ) \, dx=-\frac {x (e x)^m \left (2 n+2^{-\frac {1+m}{n}} e^{2 a} (1+m) \left (-b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},-2 b x^n\right )+2^{-\frac {1+m}{n}} e^{-2 a} (1+m) \left (b x^n\right )^{-\frac {1+m}{n}} \Gamma \left (\frac {1+m}{n},2 b x^n\right )\right )}{4 (1+m) n} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(sinh(a + b*x^n)^2*(e*x)^m,x)

[Out]

int(sinh(a + b*x^n)^2*(e*x)^m, x)