3.1.95 \(\int (1-\sinh ^2(x))^{3/2} \, dx\) [95]

Optimal. Leaf size=45 \[ -2 i E(i x|-1)+\frac {2}{3} i F(i x|-1)-\frac {1}{3} \cosh (x) \sinh (x) \sqrt {1-\sinh ^2(x)} \]

[Out]

-2*I*(cosh(x)^2)^(1/2)/cosh(x)*EllipticE(I*sinh(x),I)+2/3*I*(cosh(x)^2)^(1/2)/cosh(x)*EllipticF(I*sinh(x),I)-1
/3*cosh(x)*sinh(x)*(1-sinh(x)^2)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3259, 3251, 3256, 3261} \begin {gather*} \frac {2}{3} i F(i x|-1)-2 i E(i x|-1)-\frac {1}{3} \sinh (x) \sqrt {1-\sinh ^2(x)} \cosh (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - Sinh[x]^2)^(3/2),x]

[Out]

(-2*I)*EllipticE[I*x, -1] + ((2*I)/3)*EllipticF[I*x, -1] - (Cosh[x]*Sinh[x]*Sqrt[1 - Sinh[x]^2])/3

Rule 3251

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[
B/b, Int[Sqrt[a + b*Sin[e + f*x]^2], x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /;
FreeQ[{a, b, e, f, A, B}, x]

Rule 3256

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]/f)*EllipticE[e + f*x, -b/a], x] /
; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3259

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p - 1)/(2*f*p)), x] + Dist[1/(2*p), Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(
2*a + b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && GtQ[p, 1]

Rule 3261

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(Sqrt[a]*f))*EllipticF[e + f*x, -b/a]
, x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rubi steps

\begin {align*} \int \left (1-\sinh ^2(x)\right )^{3/2} \, dx &=-\frac {1}{3} \cosh (x) \sinh (x) \sqrt {1-\sinh ^2(x)}+\frac {1}{3} \int \frac {4-6 \sinh ^2(x)}{\sqrt {1-\sinh ^2(x)}} \, dx\\ &=-\frac {1}{3} \cosh (x) \sinh (x) \sqrt {1-\sinh ^2(x)}-\frac {2}{3} \int \frac {1}{\sqrt {1-\sinh ^2(x)}} \, dx+2 \int \sqrt {1-\sinh ^2(x)} \, dx\\ &=-2 i E(i x|-1)+\frac {2}{3} i F(i x|-1)-\frac {1}{3} \cosh (x) \sinh (x) \sqrt {1-\sinh ^2(x)}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 45, normalized size = 1.00 \begin {gather*} \frac {1}{12} \left (-24 i E(i x|-1)+8 i F(i x|-1)-\sqrt {6-2 \cosh (2 x)} \sinh (2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - Sinh[x]^2)^(3/2),x]

[Out]

((-24*I)*EllipticE[I*x, -1] + (8*I)*EllipticF[I*x, -1] - Sqrt[6 - 2*Cosh[2*x]]*Sinh[2*x])/12

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Maple [A]
time = 0.98, size = 103, normalized size = 2.29

method	result	size
default	\(\frac {\sqrt {-\left (-1+\sinh ^{2}\left (x \right )\right ) \left (\cosh ^{2}\left (x \right )\right )}\, \left (\left (\cosh ^{4}\left (x \right )\right ) \sinh \left (x \right )+10 \sqrt {-\left (\cosh ^{2}\left (x \right )\right )+2}\, \sqrt {\frac {1}{2}+\frac {\cosh \left (2 x \right )}{2}}\, \EllipticF \left (\sinh \left (x \right ), i\right )-6 \sqrt {-\left (\cosh ^{2}\left (x \right )\right )+2}\, \sqrt {\frac {1}{2}+\frac {\cosh \left (2 x \right )}{2}}\, \EllipticE \left (\sinh \left (x \right ), i\right )-2 \left (\cosh ^{2}\left (x \right )\right ) \sinh \left (x \right )\right )}{3 \sqrt {1-\left (\sinh ^{4}\left (x \right )\right )}\, \cosh \left (x \right ) \sqrt {1-\left (\sinh ^{2}\left (x \right )\right )}}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-sinh(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(-(-1+sinh(x)^2)*cosh(x)^2)^(1/2)*(cosh(x)^4*sinh(x)+10*(-cosh(x)^2+2)^(1/2)*(cosh(x)^2)^(1/2)*EllipticF(s
inh(x),I)-6*(-cosh(x)^2+2)^(1/2)*(cosh(x)^2)^(1/2)*EllipticE(sinh(x),I)-2*cosh(x)^2*sinh(x))/(1-sinh(x)^4)^(1/
2)/cosh(x)/(1-sinh(x)^2)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-sinh(x)^2 + 1)^(3/2), x)

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Fricas [F]
time = 0.08, size = 12, normalized size = 0.27 \begin {gather*} {\rm integral}\left ({\left (-\sinh \left (x\right )^{2} + 1\right )}^{\frac {3}{2}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-sinh(x)^2 + 1)^(3/2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (1 - \sinh ^{2}{\left (x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((1 - sinh(x)**2)**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-sinh(x)^2 + 1)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\left (1-{\mathrm {sinh}\left (x\right )}^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - sinh(x)^2)^(3/2),x)

[Out]

int((1 - sinh(x)^2)^(3/2), x)

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