3.18 \(\int \frac{1}{(a \sin ^4(x))^{5/2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac{\sin (x) \cos (x)}{a^2 \sqrt{a \sin ^4(x)}}-\frac{\cos ^2(x) \cot ^7(x)}{9 a^2 \sqrt{a \sin ^4(x)}}-\frac{4 \cos ^2(x) \cot ^5(x)}{7 a^2 \sqrt{a \sin ^4(x)}}-\frac{6 \cos ^2(x) \cot ^3(x)}{5 a^2 \sqrt{a \sin ^4(x)}}-\frac{4 \cos ^2(x) \cot (x)}{3 a^2 \sqrt{a \sin ^4(x)}} \]

[Out]

(-4*Cos[x]^2*Cot[x])/(3*a^2*Sqrt[a*Sin[x]^4]) - (6*Cos[x]^2*Cot[x]^3)/(5*a^2*Sqrt[a*Sin[x]^4]) - (4*Cos[x]^2*C
ot[x]^5)/(7*a^2*Sqrt[a*Sin[x]^4]) - (Cos[x]^2*Cot[x]^7)/(9*a^2*Sqrt[a*Sin[x]^4]) - (Cos[x]*Sin[x])/(a^2*Sqrt[a
*Sin[x]^4])

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Rubi [A]  time = 0.0282477, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3207, 3767} \[ -\frac{\sin (x) \cos (x)}{a^2 \sqrt{a \sin ^4(x)}}-\frac{\cos ^2(x) \cot ^7(x)}{9 a^2 \sqrt{a \sin ^4(x)}}-\frac{4 \cos ^2(x) \cot ^5(x)}{7 a^2 \sqrt{a \sin ^4(x)}}-\frac{6 \cos ^2(x) \cot ^3(x)}{5 a^2 \sqrt{a \sin ^4(x)}}-\frac{4 \cos ^2(x) \cot (x)}{3 a^2 \sqrt{a \sin ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[x]^4)^(-5/2),x]

[Out]

(-4*Cos[x]^2*Cot[x])/(3*a^2*Sqrt[a*Sin[x]^4]) - (6*Cos[x]^2*Cot[x]^3)/(5*a^2*Sqrt[a*Sin[x]^4]) - (4*Cos[x]^2*C
ot[x]^5)/(7*a^2*Sqrt[a*Sin[x]^4]) - (Cos[x]^2*Cot[x]^7)/(9*a^2*Sqrt[a*Sin[x]^4]) - (Cos[x]*Sin[x])/(a^2*Sqrt[a
*Sin[x]^4])

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
]])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx &=\frac{\sin ^2(x) \int \csc ^{10}(x) \, dx}{a^2 \sqrt{a \sin ^4(x)}}\\ &=-\frac{\sin ^2(x) \operatorname{Subst}\left (\int \left (1+4 x^2+6 x^4+4 x^6+x^8\right ) \, dx,x,\cot (x)\right )}{a^2 \sqrt{a \sin ^4(x)}}\\ &=-\frac{4 \cos ^2(x) \cot (x)}{3 a^2 \sqrt{a \sin ^4(x)}}-\frac{6 \cos ^2(x) \cot ^3(x)}{5 a^2 \sqrt{a \sin ^4(x)}}-\frac{4 \cos ^2(x) \cot ^5(x)}{7 a^2 \sqrt{a \sin ^4(x)}}-\frac{\cos ^2(x) \cot ^7(x)}{9 a^2 \sqrt{a \sin ^4(x)}}-\frac{\cos (x) \sin (x)}{a^2 \sqrt{a \sin ^4(x)}}\\ \end{align*}

Mathematica [A]  time = 0.049783, size = 47, normalized size = 0.4 \[ -\frac{\sin (x) \cos (x) \left (35 \csc ^8(x)+40 \csc ^6(x)+48 \csc ^4(x)+64 \csc ^2(x)+128\right )}{315 a^2 \sqrt{a \sin ^4(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sin[x]^4)^(-5/2),x]

[Out]

-(Cos[x]*(128 + 64*Csc[x]^2 + 48*Csc[x]^4 + 40*Csc[x]^6 + 35*Csc[x]^8)*Sin[x])/(315*a^2*Sqrt[a*Sin[x]^4])

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Maple [A]  time = 0.16, size = 41, normalized size = 0.4 \begin{align*} -{\frac{ \left ( 128\, \left ( \cos \left ( x \right ) \right ) ^{8}-576\, \left ( \cos \left ( x \right ) \right ) ^{6}+1008\, \left ( \cos \left ( x \right ) \right ) ^{4}-840\, \left ( \cos \left ( x \right ) \right ) ^{2}+315 \right ) \sin \left ( x \right ) \cos \left ( x \right ) }{315} \left ( a \left ( \sin \left ( x \right ) \right ) ^{4} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sin(x)^4)^(5/2),x)

[Out]

-1/315*(128*cos(x)^8-576*cos(x)^6+1008*cos(x)^4-840*cos(x)^2+315)*sin(x)*cos(x)/(a*sin(x)^4)^(5/2)

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Maxima [A]  time = 1.44887, size = 47, normalized size = 0.4 \begin{align*} -\frac{315 \, \tan \left (x\right )^{8} + 420 \, \tan \left (x\right )^{6} + 378 \, \tan \left (x\right )^{4} + 180 \, \tan \left (x\right )^{2} + 35}{315 \, a^{\frac{5}{2}} \tan \left (x\right )^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)^4)^(5/2),x, algorithm="maxima")

[Out]

-1/315*(315*tan(x)^8 + 420*tan(x)^6 + 378*tan(x)^4 + 180*tan(x)^2 + 35)/(a^(5/2)*tan(x)^9)

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Fricas [A]  time = 1.62853, size = 294, normalized size = 2.49 \begin{align*} \frac{{\left (128 \, \cos \left (x\right )^{9} - 576 \, \cos \left (x\right )^{7} + 1008 \, \cos \left (x\right )^{5} - 840 \, \cos \left (x\right )^{3} + 315 \, \cos \left (x\right )\right )} \sqrt{a \cos \left (x\right )^{4} - 2 \, a \cos \left (x\right )^{2} + a}}{315 \,{\left (a^{3} \cos \left (x\right )^{10} - 5 \, a^{3} \cos \left (x\right )^{8} + 10 \, a^{3} \cos \left (x\right )^{6} - 10 \, a^{3} \cos \left (x\right )^{4} + 5 \, a^{3} \cos \left (x\right )^{2} - a^{3}\right )} \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)^4)^(5/2),x, algorithm="fricas")

[Out]

1/315*(128*cos(x)^9 - 576*cos(x)^7 + 1008*cos(x)^5 - 840*cos(x)^3 + 315*cos(x))*sqrt(a*cos(x)^4 - 2*a*cos(x)^2
+ a)/((a^3*cos(x)^10 - 5*a^3*cos(x)^8 + 10*a^3*cos(x)^6 - 10*a^3*cos(x)^4 + 5*a^3*cos(x)^2 - a^3)*sin(x))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)**4)**(5/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(x)^4)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError