\(\int \frac {1}{(a \sin ^4(x))^{3/2}} \, dx\) [17]

Optimal result

Integrand size = 10, antiderivative size = 68 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{3/2}} \, dx=-\frac {2 \cos ^2(x) \cot (x)}{3 a \sqrt {a \sin ^4(x)}}-\frac {\cos ^2(x) \cot ^3(x)}{5 a \sqrt {a \sin ^4(x)}}-\frac {\cos (x) \sin (x)}{a \sqrt {a \sin ^4(x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 68, 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} \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{3/2}} \, dx=-\frac {\sin (x) \cos (x)}{a \sqrt {a \sin ^4(x)}}-\frac {\cos ^2(x) \cot ^3(x)}{5 a \sqrt {a \sin ^4(x)}}-\frac {2 \cos ^2(x) \cot (x)}{3 a \sqrt {a \sin ^4(x)}} \]

[In]

Int[(a*Sin[x]^4)^(-3/2),x]

[Out]

(-2*Cos[x]^2*Cot[x])/(3*a*Sqrt[a*Sin[x]^4]) - (Cos[x]^2*Cot[x]^3)/(5*a*Sqrt[a*Sin[x]^4]) - (Cos[x]*Sin[x])/(a*
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 ^6(x) \, dx}{a \sqrt {a \sin ^4(x)}} \\ & = -\frac {\sin ^2(x) \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (x)\right )}{a \sqrt {a \sin ^4(x)}} \\ & = -\frac {2 \cos ^2(x) \cot (x)}{3 a \sqrt {a \sin ^4(x)}}-\frac {\cos ^2(x) \cot ^3(x)}{5 a \sqrt {a \sin ^4(x)}}-\frac {\cos (x) \sin (x)}{a \sqrt {a \sin ^4(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{3/2}} \, dx=-\frac {\cos (x) \left (8+4 \csc ^2(x)+3 \csc ^4(x)\right ) \sin ^5(x)}{15 \left (a \sin ^4(x)\right )^{3/2}} \]

[In]

Integrate[(a*Sin[x]^4)^(-3/2),x]

[Out]

-1/15*(Cos[x]*(8 + 4*Csc[x]^2 + 3*Csc[x]^4)*Sin[x]^5)/(a*Sin[x]^4)^(3/2)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.54

method	result	size
default	\(-\frac {\left (\csc ^{2}\left (x \right )\right ) \cot \left (x \right ) \left (8 \left (\cos ^{4}\left (x \right )\right )-20 \left (\cos ^{2}\left (x \right )\right )+15\right ) \sqrt {16}}{60 \sqrt {a \left (\sin ^{4}\left (x \right )\right )}\, a}\)	\(37\)
risch	\(\frac {16 i \left (-5+11 \cos \left (2 x \right )+9 i \sin \left (2 x \right )\right )}{15 a \left ({\mathrm e}^{2 i x}-1\right )^{3} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}\)	\(49\)

[In]

int(1/(a*sin(x)^4)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/60*csc(x)^2*cot(x)*(8*cos(x)^4-20*cos(x)^2+15)/(a*sin(x)^4)^(1/2)/a*16^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{3/2}} \, dx=\frac {\sqrt {a \cos \left (x\right )^{4} - 2 \, a \cos \left (x\right )^{2} + a} {\left (8 \, \cos \left (x\right )^{5} - 20 \, \cos \left (x\right )^{3} + 15 \, \cos \left (x\right )\right )}}{15 \, {\left (a^{2} \cos \left (x\right )^{6} - 3 \, a^{2} \cos \left (x\right )^{4} + 3 \, a^{2} \cos \left (x\right )^{2} - a^{2}\right )} \sin \left (x\right )} \]

[In]

integrate(1/(a*sin(x)^4)^(3/2),x, algorithm="fricas")

[Out]

1/15*sqrt(a*cos(x)^4 - 2*a*cos(x)^2 + a)*(8*cos(x)^5 - 20*cos(x)^3 + 15*cos(x))/((a^2*cos(x)^6 - 3*a^2*cos(x)^
4 + 3*a^2*cos(x)^2 - a^2)*sin(x))

Sympy [F]

\[ \int \frac {1}{\left (a \sin ^4(x)\right )^{3/2}} \, dx=\int \frac {1}{\left (a \sin ^{4}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(a*sin(x)**4)**(3/2),x)

[Out]

Integral((a*sin(x)**4)**(-3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{3/2}} \, dx=-\frac {15 \, \tan \left (x\right )^{4} + 10 \, \tan \left (x\right )^{2} + 3}{15 \, a^{\frac {3}{2}} \tan \left (x\right )^{5}} \]

[In]

integrate(1/(a*sin(x)^4)^(3/2),x, algorithm="maxima")

[Out]

-1/15*(15*tan(x)^4 + 10*tan(x)^2 + 3)/(a^(3/2)*tan(x)^5)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{3/2}} \, dx=-\frac {15 \, \tan \left (x\right )^{4} + 10 \, \tan \left (x\right )^{2} + 3}{15 \, a^{\frac {3}{2}} \tan \left (x\right )^{5}} \]

[In]

integrate(1/(a*sin(x)^4)^(3/2),x, algorithm="giac")

[Out]

-1/15*(15*tan(x)^4 + 10*tan(x)^2 + 3)/(a^(3/2)*tan(x)^5)

Mupad [B] (verification not implemented)

Time = 13.79 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{3/2}} \, dx=\frac {\frac {8{}\mathrm {i}}{15\,a^{3/2}}-\frac {4\,\left (2\,{\sin \left (2\,x\right )}^3-9\,\sin \left (2\,x\right )+3\,\sin \left (4\,x\right )+2{}\mathrm {i}\right )}{15\,a^{3/2}}}{{\left (\cos \left (2\,x\right )-1\right )}^3} \]

[In]

int(1/(a*sin(x)^4)^(3/2),x)

[Out]

(8i/(15*a^(3/2)) - (4*(3*sin(4*x) - 9*sin(2*x) + 2*sin(2*x)^3 + 2i))/(15*a^(3/2)))/(cos(2*x) - 1)^3