3.147 \(\int (a \sinh ^3(x))^{3/2} \, dx\)

Optimal. Leaf size=83 \[ -\frac {14}{45} a \cosh (x) \sqrt {a \sinh ^3(x)}+\frac {2}{9} a \sinh ^2(x) \cosh (x) \sqrt {a \sinh ^3(x)}+\frac {14 i a \text {csch}(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {a \sinh ^3(x)}}{15 \sqrt {i \sinh (x)}} \]

[Out]

-14/45*a*cosh(x)*(a*sinh(x)^3)^(1/2)+2/9*a*cosh(x)*sinh(x)^2*(a*sinh(x)^3)^(1/2)+14/15*I*a*csch(x)*(sin(1/4*Pi
+1/2*I*x)^2)^(1/2)/sin(1/4*Pi+1/2*I*x)*EllipticE(cos(1/4*Pi+1/2*I*x),2^(1/2))*(a*sinh(x)^3)^(1/2)/(I*sinh(x))^
(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3207, 2635, 2640, 2639} \[ \frac {2}{9} a \sinh ^2(x) \cosh (x) \sqrt {a \sinh ^3(x)}-\frac {14}{45} a \cosh (x) \sqrt {a \sinh ^3(x)}+\frac {14 i a \text {csch}(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {a \sinh ^3(x)}}{15 \sqrt {i \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*a*Cosh[x]*Sqrt[a*Sinh[x]^3])/45 + (((14*I)/15)*a*Csch[x]*EllipticE[Pi/4 - (I/2)*x, 2]*Sqrt[a*Sinh[x]^3])/
Sqrt[I*Sinh[x]] + (2*a*Cosh[x]*Sinh[x]^2*Sqrt[a*Sinh[x]^3])/9

Rule 2635

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

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 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \left (a \sinh ^3(x)\right )^{3/2} \, dx &=\frac {\left (a \sqrt {a \sinh ^3(x)}\right ) \int \sinh ^{\frac {9}{2}}(x) \, dx}{\sinh ^{\frac {3}{2}}(x)}\\ &=\frac {2}{9} a \cosh (x) \sinh ^2(x) \sqrt {a \sinh ^3(x)}-\frac {\left (7 a \sqrt {a \sinh ^3(x)}\right ) \int \sinh ^{\frac {5}{2}}(x) \, dx}{9 \sinh ^{\frac {3}{2}}(x)}\\ &=-\frac {14}{45} a \cosh (x) \sqrt {a \sinh ^3(x)}+\frac {2}{9} a \cosh (x) \sinh ^2(x) \sqrt {a \sinh ^3(x)}+\frac {\left (7 a \sqrt {a \sinh ^3(x)}\right ) \int \sqrt {\sinh (x)} \, dx}{15 \sinh ^{\frac {3}{2}}(x)}\\ &=-\frac {14}{45} a \cosh (x) \sqrt {a \sinh ^3(x)}+\frac {2}{9} a \cosh (x) \sinh ^2(x) \sqrt {a \sinh ^3(x)}+\frac {\left (7 a \text {csch}(x) \sqrt {a \sinh ^3(x)}\right ) \int \sqrt {i \sinh (x)} \, dx}{15 \sqrt {i \sinh (x)}}\\ &=-\frac {14}{45} a \cosh (x) \sqrt {a \sinh ^3(x)}+\frac {14 i a \text {csch}(x) E\left (\left .\frac {\pi }{4}-\frac {i x}{2}\right |2\right ) \sqrt {a \sinh ^3(x)}}{15 \sqrt {i \sinh (x)}}+\frac {2}{9} a \cosh (x) \sinh ^2(x) \sqrt {a \sinh ^3(x)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 57, normalized size = 0.69 \[ \frac {1}{180} a \text {csch}(x) \sqrt {a \sinh ^3(x)} \left (-38 \sinh (2 x)+5 \sinh (4 x)+168 \sqrt {i \sinh (x)} \text {csch}(x) E\left (\left .\frac {1}{4} (\pi -2 i x)\right |2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Csch[x]*Sqrt[a*Sinh[x]^3]*(168*Csch[x]*EllipticE[(Pi - (2*I)*x)/4, 2]*Sqrt[I*Sinh[x]] - 38*Sinh[2*x] + 5*Si
nh[4*x]))/180

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a \sinh \relax (x)^{3}} a \sinh \relax (x)^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*sinh(x)^3)*a*sinh(x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sinh(x)^3)^(3/2), x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sinh ^{3}\relax (x )\right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sinh(x)^3)^(3/2),x)

[Out]

int((a*sinh(x)^3)^(3/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sinh(x)^3)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,{\mathrm {sinh}\relax (x)}^3\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sinh(x)^3)^(3/2),x)

[Out]

int((a*sinh(x)^3)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sinh(x)**3)**(3/2),x)

[Out]

Integral((a*sinh(x)**3)**(3/2), x)

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