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

Optimal. Leaf size=58 \[ -\frac {2 x^2 \cosh (x)}{\sqrt {\sinh (x)}}+8 x \sqrt {\sinh (x)}-\frac {16 i \sqrt {\sinh (x)} E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{\sqrt {i \sinh (x)}} \]

[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)

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Rubi [A]  time = 0.12, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3316, 2640, 2639} \[ -\frac {2 x^2 \cosh (x)}{\sqrt {\sinh (x)}}+8 x \sqrt {\sinh (x)}-\frac {16 i \sqrt {\sinh (x)} E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right )}{\sqrt {i \sinh (x)}} \]

Antiderivative was successfully verified.

[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 2639

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 3316

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

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Mathematica [C]  time = 1.23, size = 68, normalized size = 1.17 \[ -\frac {2 \left (-8 \sqrt {2} (\sinh (x)-\cosh (x)) \sqrt {-\sinh (x) (\sinh (x)+\cosh (x))} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};\cosh (2 x)+\sinh (2 x)\right )+x^2 \cosh (x)-4 (x-2) \sinh (x)\right )}{\sqrt {\sinh (x)}} \]

Antiderivative was successfully verified.

[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]]

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -x^{2} \sqrt {\sinh \relax (x)} + \frac {x^{2}}{\sinh \relax (x)^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sinh \relax (x )^{\frac {3}{2}}}-x^{2} \left (\sqrt {\sinh }\relax (x )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -x^{2} \sqrt {\sinh \relax (x)} + \frac {x^{2}}{\sinh \relax (x)^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int x^2\,\sqrt {\mathrm {sinh}\relax (x)}-\frac {x^2}{{\mathrm {sinh}\relax (x)}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {x^{2}}{\sinh ^{\frac {3}{2}}{\relax (x )}}\right )\, dx - \int x^{2} \sqrt {\sinh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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