\(\int \frac {\sinh ^2(a+b x^n)}{x^2} \, dx\) [68]

Optimal result

Integrand size = 14, antiderivative size = 91 \[ \int \frac {\sinh ^2\left (a+b x^n\right )}{x^2} \, dx=\frac {1}{2 x}-\frac {2^{-2+\frac {1}{n}} e^{2 a} \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-2 b x^n\right )}{n x}-\frac {2^{-2+\frac {1}{n}} e^{-2 a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},2 b x^n\right )}{n x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5470, 5469, 2250} \[ \int \frac {\sinh ^2\left (a+b x^n\right )}{x^2} \, dx=-\frac {e^{2 a} 2^{\frac {1}{n}-2} \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-2 b x^n\right )}{n x}-\frac {e^{-2 a} 2^{\frac {1}{n}-2} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},2 b x^n\right )}{n x}+\frac {1}{2 x} \]

[In]

Int[Sinh[a + b*x^n]^2/x^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.87 \[ \int \frac {\sinh ^2\left (a+b x^n\right )}{x^2} \, dx=-\frac {-2 n+2^{\frac {1}{n}} e^{2 a} \left (-b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},-2 b x^n\right )+2^{\frac {1}{n}} e^{-2 a} \left (b x^n\right )^{\frac {1}{n}} \Gamma \left (-\frac {1}{n},2 b x^n\right )}{4 n x} \]

[In]

Integrate[Sinh[a + b*x^n]^2/x^2,x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(sinh(b*x^n + a)^2/x^2, x)

Sympy [F]

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

[In]

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

[Out]

Integral(sinh(a + b*x**n)**2/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.81 \[ \int \frac {\sinh ^2\left (a+b x^n\right )}{x^2} \, dx=-\frac {\left (2 \, b x^{n}\right )^{\left (\frac {1}{n}\right )} e^{\left (-2 \, a\right )} \Gamma \left (-\frac {1}{n}, 2 \, b x^{n}\right )}{4 \, n x} - \frac {\left (-2 \, b x^{n}\right )^{\left (\frac {1}{n}\right )} e^{\left (2 \, a\right )} \Gamma \left (-\frac {1}{n}, -2 \, b x^{n}\right )}{4 \, n x} + \frac {1}{2 \, x} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(sinh(b*x^n + a)^2/x^2, x)

Mupad [F(-1)]

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

[In]

int(sinh(a + b*x^n)^2/x^2,x)

[Out]

int(sinh(a + b*x^n)^2/x^2, x)