3.1.11 \(\int \frac {1}{(a \sin ^3(x))^{3/2}} \, dx\) [11]

Optimal. Leaf size=77 \[ -\frac {10 \cos (x)}{21 a \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc (x)}{7 a \sqrt {a \sin ^3(x)}}-\frac {10 F\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sin ^{\frac {3}{2}}(x)}{21 a \sqrt {a \sin ^3(x)}} \]

[Out]

-10/21*cos(x)/a/(a*sin(x)^3)^(1/2)-2/7*cot(x)*csc(x)/a/(a*sin(x)^3)^(1/2)-10/21*(sin(1/4*Pi+1/2*x)^2)^(1/2)/si
n(1/4*Pi+1/2*x)*EllipticF(cos(1/4*Pi+1/2*x),2^(1/2))*sin(x)^(3/2)/a/(a*sin(x)^3)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3286, 2716, 2720} \begin {gather*} -\frac {10 \cos (x)}{21 a \sqrt {a \sin ^3(x)}}-\frac {10 \sin ^{\frac {3}{2}}(x) F\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right )}{21 a \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc (x)}{7 a \sqrt {a \sin ^3(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Cos[x])/(21*a*Sqrt[a*Sin[x]^3]) - (2*Cot[x]*Csc[x])/(7*a*Sqrt[a*Sin[x]^3]) - (10*EllipticF[Pi/4 - x/2, 2]
*Sin[x]^(3/2))/(21*a*Sqrt[a*Sin[x]^3])

Rule 2716

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

Rule 2720

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

Rule 3286

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 \frac {1}{\left (a \sin ^3(x)\right )^{3/2}} \, dx &=\frac {\sin ^{\frac {3}{2}}(x) \int \frac {1}{\sin ^{\frac {9}{2}}(x)} \, dx}{a \sqrt {a \sin ^3(x)}}\\ &=-\frac {2 \cot (x) \csc (x)}{7 a \sqrt {a \sin ^3(x)}}+\frac {\left (5 \sin ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sin ^{\frac {5}{2}}(x)} \, dx}{7 a \sqrt {a \sin ^3(x)}}\\ &=-\frac {10 \cos (x)}{21 a \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc (x)}{7 a \sqrt {a \sin ^3(x)}}+\frac {\left (5 \sin ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sqrt {\sin (x)}} \, dx}{21 a \sqrt {a \sin ^3(x)}}\\ &=-\frac {10 \cos (x)}{21 a \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc (x)}{7 a \sqrt {a \sin ^3(x)}}-\frac {10 F\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sin ^{\frac {3}{2}}(x)}{21 a \sqrt {a \sin ^3(x)}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 48, normalized size = 0.62 \begin {gather*} -\frac {2 \sin ^2(x) \left (3 \cot (x)+5 \cos (x) \sin (x)+5 F\left (\left .\frac {1}{4} (\pi -2 x)\right |2\right ) \sin ^{\frac {5}{2}}(x)\right )}{21 \left (a \sin ^3(x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sin[x]^2*(3*Cot[x] + 5*Cos[x]*Sin[x] + 5*EllipticF[(Pi - 2*x)/4, 2]*Sin[x]^(5/2)))/(21*(a*Sin[x]^3)^(3/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.36, size = 372, normalized size = 4.83

method result size
default \(-\frac {\left (\cos \left (x \right )+1\right )^{2} \left (-1+\cos \left (x \right )\right )^{2} \left (5 i \left (\cos ^{3}\left (x \right )\right ) \sin \left (x \right ) \sqrt {2}\, \sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \cos \left (x \right )-\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (x \right )\right )}{\sin \left (x \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}, \frac {\sqrt {2}}{2}\right )+5 i \left (\cos ^{2}\left (x \right )\right ) \sin \left (x \right ) \sqrt {2}\, \sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \cos \left (x \right )-\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (x \right )\right )}{\sin \left (x \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}, \frac {\sqrt {2}}{2}\right )-5 i \cos \left (x \right ) \sin \left (x \right ) \sqrt {2}\, \sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \cos \left (x \right )-\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (x \right )\right )}{\sin \left (x \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}, \frac {\sqrt {2}}{2}\right )-5 i \sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {2}\, \sqrt {-\frac {i \cos \left (x \right )-\sin \left (x \right )-i}{\sin \left (x \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (x \right )\right )}{\sin \left (x \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (x \right )+\sin \left (x \right )-i}{\sin \left (x \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (x \right )-10 \left (\cos ^{3}\left (x \right )\right )+16 \cos \left (x \right )\right )}{21 \left (a \left (\sin ^{3}\left (x \right )\right )\right )^{\frac {3}{2}} \sin \left (x \right )^{3}}\) \(372\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sin(x)^3)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/21*(cos(x)+1)^2*(-1+cos(x))^2*(5*I*cos(x)^3*sin(x)*2^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-s
in(x)-I)/sin(x))^(1/2)*(-I*(-1+cos(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))
+5*I*cos(x)^2*sin(x)*2^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^(1/2)*(-I*(-1+co
s(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))-5*I*cos(x)*sin(x)*2^(1/2)*((I*co
s(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^(1/2)*(-I*(-1+cos(x))/sin(x))^(1/2)*EllipticF(((I*c
os(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))-5*I*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*2^(1/2)*(-(I*cos(x)-sin(x)-I
)/sin(x))^(1/2)*(-I*(-1+cos(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))*sin(x)
-10*cos(x)^3+16*cos(x))/(a*sin(x)^3)^(3/2)/sin(x)^3

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

[Out]

integrate((a*sin(x)^3)^(-3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 139, normalized size = 1.81 \begin {gather*} \frac {5 \, {\left (\sqrt {2} \cos \left (x\right )^{4} - 2 \, \sqrt {2} \cos \left (x\right )^{2} + \sqrt {2}\right )} \sqrt {-i \, a} \sin \left (x\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) + i \, \sin \left (x\right )\right ) + 5 \, {\left (\sqrt {2} \cos \left (x\right )^{4} - 2 \, \sqrt {2} \cos \left (x\right )^{2} + \sqrt {2}\right )} \sqrt {i \, a} \sin \left (x\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) - i \, \sin \left (x\right )\right ) + 2 \, {\left (5 \, \cos \left (x\right )^{3} - 8 \, \cos \left (x\right )\right )} \sqrt {-{\left (a \cos \left (x\right )^{2} - a\right )} \sin \left (x\right )}}{21 \, {\left (a^{2} \cos \left (x\right )^{4} - 2 \, a^{2} \cos \left (x\right )^{2} + a^{2}\right )} \sin \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)^3)^(3/2),x, algorithm="fricas")

[Out]

1/21*(5*(sqrt(2)*cos(x)^4 - 2*sqrt(2)*cos(x)^2 + sqrt(2))*sqrt(-I*a)*sin(x)*weierstrassPInverse(4, 0, cos(x) +
 I*sin(x)) + 5*(sqrt(2)*cos(x)^4 - 2*sqrt(2)*cos(x)^2 + sqrt(2))*sqrt(I*a)*sin(x)*weierstrassPInverse(4, 0, co
s(x) - I*sin(x)) + 2*(5*cos(x)^3 - 8*cos(x))*sqrt(-(a*cos(x)^2 - a)*sin(x)))/((a^2*cos(x)^4 - 2*a^2*cos(x)^2 +
 a^2)*sin(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)**3)**(3/2),x)

[Out]

Integral((a*sin(x)**3)**(-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/(a*sin(x)^3)^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(x)^3)^(-3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a\,{\sin \left (x\right )}^3\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sin(x)^3)^(3/2),x)

[Out]

int(1/(a*sin(x)^3)^(3/2), x)

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