\(\int \frac {1}{(a \sin ^4(x))^{5/2}} \, dx\) [18]

Optimal result

Integrand size = 10, antiderivative size = 118 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {4 \cos ^2(x) \cot (x)}{3 a^2 \sqrt {a \sin ^4(x)}}-\frac {6 \cos ^2(x) \cot ^3(x)}{5 a^2 \sqrt {a \sin ^4(x)}}-\frac {4 \cos ^2(x) \cot ^5(x)}{7 a^2 \sqrt {a \sin ^4(x)}}-\frac {\cos ^2(x) \cot ^7(x)}{9 a^2 \sqrt {a \sin ^4(x)}}-\frac {\cos (x) \sin (x)}{a^2 \sqrt {a \sin ^4(x)}} \]

[Out]

-4/3*cos(x)^2*cot(x)/a^2/(a*sin(x)^4)^(1/2)-6/5*cos(x)^2*cot(x)^3/a^2/(a*sin(x)^4)^(1/2)-4/7*cos(x)^2*cot(x)^5
/a^2/(a*sin(x)^4)^(1/2)-1/9*cos(x)^2*cot(x)^7/a^2/(a*sin(x)^4)^(1/2)-cos(x)*sin(x)/a^2/(a*sin(x)^4)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3286, 3852} \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {\sin (x) \cos (x)}{a^2 \sqrt {a \sin ^4(x)}}-\frac {\cos ^2(x) \cot ^7(x)}{9 a^2 \sqrt {a \sin ^4(x)}}-\frac {4 \cos ^2(x) \cot ^5(x)}{7 a^2 \sqrt {a \sin ^4(x)}}-\frac {6 \cos ^2(x) \cot ^3(x)}{5 a^2 \sqrt {a \sin ^4(x)}}-\frac {4 \cos ^2(x) \cot (x)}{3 a^2 \sqrt {a \sin ^4(x)}} \]

[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 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
]])

Rule 3852

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*} \text {integral}& = \frac {\sin ^2(x) \int \csc ^{10}(x) \, dx}{a^2 \sqrt {a \sin ^4(x)}} \\ & = -\frac {\sin ^2(x) \text {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] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {\cos (x) \left (128+64 \csc ^2(x)+48 \csc ^4(x)+40 \csc ^6(x)+35 \csc ^8(x)\right ) \sin (x)}{315 a^2 \sqrt {a \sin ^4(x)}} \]

[In]

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

[Out]

-1/315*(Cos[x]*(128 + 64*Csc[x]^2 + 48*Csc[x]^4 + 40*Csc[x]^6 + 35*Csc[x]^8)*Sin[x])/(a^2*Sqrt[a*Sin[x]^4])

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42

method	result	size
default	\(-\frac {\cot \left (x \right ) \left (\csc ^{6}\left (x \right )\right ) \left (128 \left (\cos ^{8}\left (x \right )\right )-576 \left (\cos ^{6}\left (x \right )\right )+1008 \left (\cos ^{4}\left (x \right )\right )-840 \left (\cos ^{2}\left (x \right )\right )+315\right ) \sqrt {16}}{1260 \sqrt {a \left (\sin ^{4}\left (x \right )\right )}\, a^{2}}\)	\(49\)
risch	\(\frac {256 i \left (126 \,{\mathrm e}^{6 i x}-84 \,{\mathrm e}^{4 i x}-9+37 \cos \left (2 x \right )+35 i \sin \left (2 x \right )\right )}{315 a^{2} \left ({\mathrm e}^{2 i x}-1\right )^{7} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}\)	\(63\)

[In]

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

[Out]

-1/1260*cot(x)*csc(x)^6*(128*cos(x)^8-576*cos(x)^6+1008*cos(x)^4-840*cos(x)^2+315)/(a*sin(x)^4)^(1/2)/a^2*16^(
1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=\frac {{\left (128 \, \cos \left (x\right )^{9} - 576 \, \cos \left (x\right )^{7} + 1008 \, \cos \left (x\right )^{5} - 840 \, \cos \left (x\right )^{3} + 315 \, \cos \left (x\right )\right )} \sqrt {a \cos \left (x\right )^{4} - 2 \, a \cos \left (x\right )^{2} + a}}{315 \, {\left (a^{3} \cos \left (x\right )^{10} - 5 \, a^{3} \cos \left (x\right )^{8} + 10 \, a^{3} \cos \left (x\right )^{6} - 10 \, a^{3} \cos \left (x\right )^{4} + 5 \, a^{3} \cos \left (x\right )^{2} - a^{3}\right )} \sin \left (x\right )} \]

[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))

Sympy [F]

\[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a \sin ^{4}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral((a*sin(x)**4)**(-5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {315 \, \tan \left (x\right )^{8} + 420 \, \tan \left (x\right )^{6} + 378 \, \tan \left (x\right )^{4} + 180 \, \tan \left (x\right )^{2} + 35}{315 \, a^{\frac {5}{2}} \tan \left (x\right )^{9}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {315 \, \tan \left (x\right )^{8} + 420 \, \tan \left (x\right )^{6} + 378 \, \tan \left (x\right )^{4} + 180 \, \tan \left (x\right )^{2} + 35}{315 \, a^{\frac {5}{2}} \tan \left (x\right )^{9}} \]

[In]

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

[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)

Mupad [B] (verification not implemented)

Time = 15.91 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=\frac {256\,\left ({\mathrm {e}}^{x\,46{}\mathrm {i}}\,1{}\mathrm {i}-{\mathrm {e}}^{x\,48{}\mathrm {i}}\,9{}\mathrm {i}+{\mathrm {e}}^{x\,50{}\mathrm {i}}\,36{}\mathrm {i}-{\mathrm {e}}^{x\,52{}\mathrm {i}}\,84{}\mathrm {i}+{\mathrm {e}}^{x\,54{}\mathrm {i}}\,126{}\mathrm {i}\right )}{315\,a^{5/2}\,\left ({\mathrm {e}}^{x\,46{}\mathrm {i}}-9\,{\mathrm {e}}^{x\,48{}\mathrm {i}}+36\,{\mathrm {e}}^{x\,50{}\mathrm {i}}-84\,{\mathrm {e}}^{x\,52{}\mathrm {i}}+126\,{\mathrm {e}}^{x\,54{}\mathrm {i}}-126\,{\mathrm {e}}^{x\,56{}\mathrm {i}}+84\,{\mathrm {e}}^{x\,58{}\mathrm {i}}-36\,{\mathrm {e}}^{x\,60{}\mathrm {i}}+9\,{\mathrm {e}}^{x\,62{}\mathrm {i}}-{\mathrm {e}}^{x\,64{}\mathrm {i}}\right )} \]

[In]

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

[Out]

(256*(exp(x*46i)*1i - exp(x*48i)*9i + exp(x*50i)*36i - exp(x*52i)*84i + exp(x*54i)*126i))/(315*a^(5/2)*(exp(x*
46i) - 9*exp(x*48i) + 36*exp(x*50i) - 84*exp(x*52i) + 126*exp(x*54i) - 126*exp(x*56i) + 84*exp(x*58i) - 36*exp
(x*60i) + 9*exp(x*62i) - exp(x*64i)))