3.1.12 \(\int \frac {1}{(a \sin ^3(x))^{5/2}} \, dx\) [12]

Optimal. Leaf size=123 \[ -\frac {154 \cot (x)}{585 a^2 \sqrt {a \sin ^3(x)}}-\frac {22 \cot (x) \csc ^2(x)}{117 a^2 \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc ^4(x)}{13 a^2 \sqrt {a \sin ^3(x)}}-\frac {154 \cos (x) \sin (x)}{195 a^2 \sqrt {a \sin ^3(x)}}+\frac {154 E\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sin ^{\frac {3}{2}}(x)}{195 a^2 \sqrt {a \sin ^3(x)}} \]

[Out]

-154/585*cot(x)/a^2/(a*sin(x)^3)^(1/2)-22/117*cot(x)*csc(x)^2/a^2/(a*sin(x)^3)^(1/2)-2/13*cot(x)*csc(x)^4/a^2/
(a*sin(x)^3)^(1/2)-154/195*cos(x)*sin(x)/a^2/(a*sin(x)^3)^(1/2)+154/195*(sin(1/4*Pi+1/2*x)^2)^(1/2)/sin(1/4*Pi
+1/2*x)*EllipticE(cos(1/4*Pi+1/2*x),2^(1/2))*sin(x)^(3/2)/a^2/(a*sin(x)^3)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3286, 2716, 2719} \begin {gather*} -\frac {154 \sin (x) \cos (x)}{195 a^2 \sqrt {a \sin ^3(x)}}-\frac {154 \cot (x)}{585 a^2 \sqrt {a \sin ^3(x)}}+\frac {154 \sin ^{\frac {3}{2}}(x) E\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right )}{195 a^2 \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc ^4(x)}{13 a^2 \sqrt {a \sin ^3(x)}}-\frac {22 \cot (x) \csc ^2(x)}{117 a^2 \sqrt {a \sin ^3(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*Sin[x]^3)^(-5/2),x]

[Out]

(-154*Cot[x])/(585*a^2*Sqrt[a*Sin[x]^3]) - (22*Cot[x]*Csc[x]^2)/(117*a^2*Sqrt[a*Sin[x]^3]) - (2*Cot[x]*Csc[x]^
4)/(13*a^2*Sqrt[a*Sin[x]^3]) - (154*Cos[x]*Sin[x])/(195*a^2*Sqrt[a*Sin[x]^3]) + (154*EllipticE[Pi/4 - x/2, 2]*
Sin[x]^(3/2))/(195*a^2*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 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 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 )^{5/2}} \, dx &=\frac {\sin ^{\frac {3}{2}}(x) \int \frac {1}{\sin ^{\frac {15}{2}}(x)} \, dx}{a^2 \sqrt {a \sin ^3(x)}}\\ &=-\frac {2 \cot (x) \csc ^4(x)}{13 a^2 \sqrt {a \sin ^3(x)}}+\frac {\left (11 \sin ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sin ^{\frac {11}{2}}(x)} \, dx}{13 a^2 \sqrt {a \sin ^3(x)}}\\ &=-\frac {22 \cot (x) \csc ^2(x)}{117 a^2 \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc ^4(x)}{13 a^2 \sqrt {a \sin ^3(x)}}+\frac {\left (77 \sin ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sin ^{\frac {7}{2}}(x)} \, dx}{117 a^2 \sqrt {a \sin ^3(x)}}\\ &=-\frac {154 \cot (x)}{585 a^2 \sqrt {a \sin ^3(x)}}-\frac {22 \cot (x) \csc ^2(x)}{117 a^2 \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc ^4(x)}{13 a^2 \sqrt {a \sin ^3(x)}}+\frac {\left (77 \sin ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sin ^{\frac {3}{2}}(x)} \, dx}{195 a^2 \sqrt {a \sin ^3(x)}}\\ &=-\frac {154 \cot (x)}{585 a^2 \sqrt {a \sin ^3(x)}}-\frac {22 \cot (x) \csc ^2(x)}{117 a^2 \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc ^4(x)}{13 a^2 \sqrt {a \sin ^3(x)}}-\frac {154 \cos (x) \sin (x)}{195 a^2 \sqrt {a \sin ^3(x)}}-\frac {\left (77 \sin ^{\frac {3}{2}}(x)\right ) \int \sqrt {\sin (x)} \, dx}{195 a^2 \sqrt {a \sin ^3(x)}}\\ &=-\frac {154 \cot (x)}{585 a^2 \sqrt {a \sin ^3(x)}}-\frac {22 \cot (x) \csc ^2(x)}{117 a^2 \sqrt {a \sin ^3(x)}}-\frac {2 \cot (x) \csc ^4(x)}{13 a^2 \sqrt {a \sin ^3(x)}}-\frac {154 \cos (x) \sin (x)}{195 a^2 \sqrt {a \sin ^3(x)}}+\frac {154 E\left (\left .\frac {\pi }{4}-\frac {x}{2}\right |2\right ) \sin ^{\frac {3}{2}}(x)}{195 a^2 \sqrt {a \sin ^3(x)}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 60, normalized size = 0.49 \begin {gather*} -\frac {2 \left (\cot (x) \left (77+55 \csc ^2(x)+45 \csc ^4(x)\right )+231 \cos (x) \sin (x)-231 E\left (\left .\frac {1}{4} (\pi -2 x)\right |2\right ) \sin ^{\frac {3}{2}}(x)\right )}{585 a^2 \sqrt {a \sin ^3(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[x]^3)^(-5/2),x]

[Out]

(-2*(Cot[x]*(77 + 55*Csc[x]^2 + 45*Csc[x]^4) + 231*Cos[x]*Sin[x] - 231*EllipticE[(Pi - 2*x)/4, 2]*Sin[x]^(3/2)
))/(585*a^2*Sqrt[a*Sin[x]^3])

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Maple [C] Result contains complex when optimal does not.
time = 0.34, size = 1349, normalized size = 10.97

method	result	size
default	\(\text {Expression too large to display}\)	\(1349\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sin(x)^3)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/585*(462*cos(x)^7*2^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^(1/2)*EllipticE(
((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(x))/sin(x))^(1/2)-231*cos(x)^7*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+cos(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)
+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))+462*cos(x)^6*2^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(
x)-I)/sin(x))^(1/2)*EllipticE(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(x))/sin(x))^(1/2)-23
1*cos(x)^6*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+cos(x))/sin(
x))^(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))-1386*cos(x)^5*2^(1/2)*((I*cos(x)+sin(x)-I)
/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^(1/2)*EllipticE(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))*(
-I*(-1+cos(x))/sin(x))^(1/2)+693*cos(x)^5*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+cos(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))-1386*cos(x)
^4*2^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^(1/2)*EllipticE(((I*cos(x)+sin(x)-
I)/sin(x))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(x))/sin(x))^(1/2)+693*cos(x)^4*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+cos(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))
^(1/2),1/2*2^(1/2))+1386*cos(x)^3*2^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^(1/
2)*EllipticE(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(x))/sin(x))^(1/2)-693*cos(x)^3*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+cos(x))/sin(x))^(1/2)*Ellipti
cF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))+1386*cos(x)^2*2^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-
(I*cos(x)-sin(x)-I)/sin(x))^(1/2)*EllipticE(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(x))/si
n(x))^(1/2)-693*cos(x)^2*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+cos(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))-462*cos(x)^6-462*cos(x)*2^(1
/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^(1/2)*EllipticE(((I*cos(x)+sin(x)-I)/sin(
x))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(x))/sin(x))^(1/2)+231*cos(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+cos(x))/sin(x))^(1/2)*EllipticF(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/
2*2^(1/2))+154*cos(x)^5-462*2^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-(I*cos(x)-sin(x)-I)/sin(x))^(1/2)*Ell
ipticE(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(x))/sin(x))^(1/2)+231*2^(1/2)*((I*cos(x)+si
n(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*cos(x)+s
in(x)-I)/sin(x))^(1/2),1/2*2^(1/2))+1386*cos(x)^4-418*cos(x)^3-1386*cos(x)^2+354*cos(x)+462)*sin(x)/(a*sin(x)^
3)^(5/2)

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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/(a*sin(x)^3)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(x)^3)^(-5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 209, normalized size = 1.70 \begin {gather*} -\frac {231 \, {\left (i \, \sqrt {2} \cos \left (x\right )^{8} - 4 i \, \sqrt {2} \cos \left (x\right )^{6} + 6 i \, \sqrt {2} \cos \left (x\right )^{4} - 4 i \, \sqrt {2} \cos \left (x\right )^{2} + i \, \sqrt {2}\right )} \sqrt {-i \, a} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) + i \, \sin \left (x\right )\right )\right ) + 231 \, {\left (-i \, \sqrt {2} \cos \left (x\right )^{8} + 4 i \, \sqrt {2} \cos \left (x\right )^{6} - 6 i \, \sqrt {2} \cos \left (x\right )^{4} + 4 i \, \sqrt {2} \cos \left (x\right )^{2} - i \, \sqrt {2}\right )} \sqrt {i \, a} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (x\right ) - i \, \sin \left (x\right )\right )\right ) - 2 \, {\left (231 \, \cos \left (x\right )^{7} - 770 \, \cos \left (x\right )^{5} + 902 \, \cos \left (x\right )^{3} - 408 \, \cos \left (x\right )\right )} \sqrt {-{\left (a \cos \left (x\right )^{2} - a\right )} \sin \left (x\right )}}{585 \, {\left (a^{3} \cos \left (x\right )^{8} - 4 \, a^{3} \cos \left (x\right )^{6} + 6 \, a^{3} \cos \left (x\right )^{4} - 4 \, a^{3} \cos \left (x\right )^{2} + a^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)^3)^(5/2),x, algorithm="fricas")

[Out]

-1/585*(231*(I*sqrt(2)*cos(x)^8 - 4*I*sqrt(2)*cos(x)^6 + 6*I*sqrt(2)*cos(x)^4 - 4*I*sqrt(2)*cos(x)^2 + I*sqrt(
2))*sqrt(-I*a)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(x) + I*sin(x))) + 231*(-I*sqrt(2)*cos(x)^8
+ 4*I*sqrt(2)*cos(x)^6 - 6*I*sqrt(2)*cos(x)^4 + 4*I*sqrt(2)*cos(x)^2 - I*sqrt(2))*sqrt(I*a)*weierstrassZeta(4,
0, weierstrassPInverse(4, 0, cos(x) - I*sin(x))) - 2*(231*cos(x)^7 - 770*cos(x)^5 + 902*cos(x)^3 - 408*cos(x)
)*sqrt(-(a*cos(x)^2 - a)*sin(x)))/(a^3*cos(x)^8 - 4*a^3*cos(x)^6 + 6*a^3*cos(x)^4 - 4*a^3*cos(x)^2 + a^3)

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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 {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)**3)**(5/2),x)

[Out]

Integral((a*sin(x)**3)**(-5/2), 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)^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(x)^3)^(-5/2), 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 )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sin(x)^3)^(5/2),x)

[Out]

int(1/(a*sin(x)^3)^(5/2), x)

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