\(\int \frac {\sinh (a+b x^n)}{x^3} \, dx\) [63]

Optimal result

Integrand size = 12, antiderivative size = 75 \[ \int \frac {\sinh \left (a+b x^n\right )}{x^3} \, dx=-\frac {e^a \left (-b x^n\right )^{2/n} \Gamma \left (-\frac {2}{n},-b x^n\right )}{2 n x^2}+\frac {e^{-a} \left (b x^n\right )^{2/n} \Gamma \left (-\frac {2}{n},b x^n\right )}{2 n x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5468, 2250} \[ \int \frac {\sinh \left (a+b x^n\right )}{x^3} \, dx=\frac {e^{-a} \left (b x^n\right )^{2/n} \Gamma \left (-\frac {2}{n},b x^n\right )}{2 n x^2}-\frac {e^a \left (-b x^n\right )^{2/n} \Gamma \left (-\frac {2}{n},-b x^n\right )}{2 n x^2} \]

[In]

Int[Sinh[a + b*x^n]/x^3,x]

[Out]

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

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 5468

Int[((e_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)], 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]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \int \frac {e^{-a-b x^n}}{x^3} \, dx\right )+\frac {1}{2} \int \frac {e^{a+b x^n}}{x^3} \, dx \\ & = -\frac {e^a \left (-b x^n\right )^{2/n} \Gamma \left (-\frac {2}{n},-b x^n\right )}{2 n x^2}+\frac {e^{-a} \left (b x^n\right )^{2/n} \Gamma \left (-\frac {2}{n},b x^n\right )}{2 n x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.91 \[ \int \frac {\sinh \left (a+b x^n\right )}{x^3} \, dx=-\frac {e^a \left (-b x^n\right )^{2/n} \Gamma \left (-\frac {2}{n},-b x^n\right )-e^{-a} \left (b x^n\right )^{2/n} \Gamma \left (-\frac {2}{n},b x^n\right )}{2 n x^2} \]

[In]

Integrate[Sinh[a + b*x^n]/x^3,x]

[Out]

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

Maple [C] (verified)

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

Time = 0.59 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03

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

[In]

int(sinh(a+b*x^n)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/x^2*hypergeom([-1/n],[1/2,1-1/n],1/4*x^(2*n)*b^2)*sinh(a)+1/(-2+n)*x^(-2+n)*b*hypergeom([1/2-1/n],[3/2,3/
2-1/n],1/4*x^(2*n)*b^2)*cosh(a)

Fricas [F]

\[ \int \frac {\sinh \left (a+b x^n\right )}{x^3} \, dx=\int { \frac {\sinh \left (b x^{n} + a\right )}{x^{3}} \,d x } \]

[In]

integrate(sinh(a+b*x^n)/x^3,x, algorithm="fricas")

[Out]

integral(sinh(b*x^n + a)/x^3, x)

Sympy [F]

\[ \int \frac {\sinh \left (a+b x^n\right )}{x^3} \, dx=\int \frac {\sinh {\left (a + b x^{n} \right )}}{x^{3}}\, dx \]

[In]

integrate(sinh(a+b*x**n)/x**3,x)

[Out]

Integral(sinh(a + b*x**n)/x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {\sinh \left (a+b x^n\right )}{x^3} \, dx=\frac {\left (b x^{n}\right )^{\frac {2}{n}} e^{\left (-a\right )} \Gamma \left (-\frac {2}{n}, b x^{n}\right )}{2 \, n x^{2}} - \frac {\left (-b x^{n}\right )^{\frac {2}{n}} e^{a} \Gamma \left (-\frac {2}{n}, -b x^{n}\right )}{2 \, n x^{2}} \]

[In]

integrate(sinh(a+b*x^n)/x^3,x, algorithm="maxima")

[Out]

1/2*(b*x^n)^(2/n)*e^(-a)*gamma(-2/n, b*x^n)/(n*x^2) - 1/2*(-b*x^n)^(2/n)*e^a*gamma(-2/n, -b*x^n)/(n*x^2)

Giac [F]

\[ \int \frac {\sinh \left (a+b x^n\right )}{x^3} \, dx=\int { \frac {\sinh \left (b x^{n} + a\right )}{x^{3}} \,d x } \]

[In]

integrate(sinh(a+b*x^n)/x^3,x, algorithm="giac")

[Out]

integrate(sinh(b*x^n + a)/x^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh \left (a+b x^n\right )}{x^3} \, dx=\int \frac {\mathrm {sinh}\left (a+b\,x^n\right )}{x^3} \,d x \]

[In]

int(sinh(a + b*x^n)/x^3,x)

[Out]

int(sinh(a + b*x^n)/x^3, x)