\(\int (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}) \, dx\) [71]

Optimal result

Integrand size = 22, antiderivative size = 58 \[ \int \left (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}\right ) \, dx=-\frac {2 x^2 \cosh (x)}{\sqrt {\sinh (x)}}+8 x \sqrt {\sinh (x)}-\frac {16 i E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {\sinh (x)}}{\sqrt {i \sinh (x)}} \]

[Out]

-2*x^2*cosh(x)/sinh(x)^(1/2)+8*x*sinh(x)^(1/2)-16*I*(sin(1/4*Pi+1/2*I*x)^2)^(1/2)/sin(1/4*Pi+1/2*I*x)*Elliptic
E(cos(1/4*Pi+1/2*I*x),2^(1/2))*sinh(x)^(1/2)/(I*sinh(x))^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3397, 2721, 2719} \[ \int \left (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}\right ) \, dx=-\frac {2 x^2 \cosh (x)}{\sqrt {\sinh (x)}}+8 x \sqrt {\sinh (x)}-\frac {16 i \sqrt {\sinh (x)} E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{\sqrt {i \sinh (x)}} \]

[In]

Int[x^2/Sinh[x]^(3/2) - x^2*Sqrt[Sinh[x]],x]

[Out]

(-2*x^2*Cosh[x])/Sqrt[Sinh[x]] + 8*x*Sqrt[Sinh[x]] - ((16*I)*EllipticE[Pi/4 - (I/2)*x, 2]*Sqrt[Sinh[x]])/Sqrt[
I*Sinh[x]]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 3397

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(c + d*x)^m*Cos[e + f*x
]*((b*Sin[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + (Dist[(n + 2)/(b^2*(n + 1)), Int[(c + d*x)^m*(b*Sin[e + f*x])
^(n + 2), x], x] + Dist[d^2*m*((m - 1)/(b^2*f^2*(n + 1)*(n + 2))), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^(n +
2), x], x] - Simp[d*m*(c + d*x)^(m - 1)*((b*Sin[e + f*x])^(n + 2)/(b^2*f^2*(n + 1)*(n + 2))), x]) /; FreeQ[{b
, c, d, e, f}, x] && LtQ[n, -1] && NeQ[n, -2] && GtQ[m, 1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{\sinh ^{\frac {3}{2}}(x)} \, dx-\int x^2 \sqrt {\sinh (x)} \, dx \\ & = -\frac {2 x^2 \cosh (x)}{\sqrt {\sinh (x)}}+8 x \sqrt {\sinh (x)}-8 \int \sqrt {\sinh (x)} \, dx \\ & = -\frac {2 x^2 \cosh (x)}{\sqrt {\sinh (x)}}+8 x \sqrt {\sinh (x)}-\frac {\left (8 \sqrt {\sinh (x)}\right ) \int \sqrt {i \sinh (x)} \, dx}{\sqrt {i \sinh (x)}} \\ & = -\frac {2 x^2 \cosh (x)}{\sqrt {\sinh (x)}}+8 x \sqrt {\sinh (x)}-\frac {16 i E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {\sinh (x)}}{\sqrt {i \sinh (x)}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 1.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.17 \[ \int \left (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}\right ) \, dx=-\frac {2 \left (x^2 \cosh (x)-4 (-2+x) \sinh (x)-8 \sqrt {2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},\cosh (2 x)+\sinh (2 x)\right ) (-\cosh (x)+\sinh (x)) \sqrt {-\sinh (x) (\cosh (x)+\sinh (x))}\right )}{\sqrt {\sinh (x)}} \]

[In]

Integrate[x^2/Sinh[x]^(3/2) - x^2*Sqrt[Sinh[x]],x]

[Out]

(-2*(x^2*Cosh[x] - 4*(-2 + x)*Sinh[x] - 8*Sqrt[2]*Hypergeometric2F1[-1/4, 1/2, 3/4, Cosh[2*x] + Sinh[2*x]]*(-C
osh[x] + Sinh[x])*Sqrt[-(Sinh[x]*(Cosh[x] + Sinh[x]))]))/Sqrt[Sinh[x]]

Maple [F]

\[\int \left (\frac {x^{2}}{\sinh \left (x \right )^{\frac {3}{2}}}-x^{2} \sqrt {\sinh \left (x \right )}\right )d x\]

[In]

int(x^2/sinh(x)^(3/2)-x^2*sinh(x)^(1/2),x)

[Out]

int(x^2/sinh(x)^(3/2)-x^2*sinh(x)^(1/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \left (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}\right ) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^2/sinh(x)^(3/2)-x^2*sinh(x)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int \left (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}\right ) \, dx=- \int \left (- \frac {x^{2}}{\sinh ^{\frac {3}{2}}{\left (x \right )}}\right )\, dx - \int x^{2} \sqrt {\sinh {\left (x \right )}}\, dx \]

[In]

integrate(x**2/sinh(x)**(3/2)-x**2*sinh(x)**(1/2),x)

[Out]

-Integral(-x**2/sinh(x)**(3/2), x) - Integral(x**2*sqrt(sinh(x)), x)

Maxima [F]

\[ \int \left (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}\right ) \, dx=\int { -x^{2} \sqrt {\sinh \left (x\right )} + \frac {x^{2}}{\sinh \left (x\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^2/sinh(x)^(3/2)-x^2*sinh(x)^(1/2),x, algorithm="maxima")

[Out]

integrate(-x^2*sqrt(sinh(x)) + x^2/sinh(x)^(3/2), x)

Giac [F]

\[ \int \left (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}\right ) \, dx=\int { -x^{2} \sqrt {\sinh \left (x\right )} + \frac {x^{2}}{\sinh \left (x\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^2/sinh(x)^(3/2)-x^2*sinh(x)^(1/2),x, algorithm="giac")

[Out]

integrate(-x^2*sqrt(sinh(x)) + x^2/sinh(x)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \left (\frac {x^2}{\sinh ^{\frac {3}{2}}(x)}-x^2 \sqrt {\sinh (x)}\right ) \, dx=-\int x^2\,\sqrt {\mathrm {sinh}\left (x\right )}-\frac {x^2}{{\mathrm {sinh}\left (x\right )}^{3/2}} \,d x \]

[In]

int(x^2/sinh(x)^(3/2) - x^2*sinh(x)^(1/2),x)

[Out]

-int(x^2*sinh(x)^(1/2) - x^2/sinh(x)^(3/2), x)