Integrand size = 10, antiderivative size = 118 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {4 \cos ^2(x) \cot (x)}{3 a^2 \sqrt {a \sin ^4(x)}}-\frac {6 \cos ^2(x) \cot ^3(x)}{5 a^2 \sqrt {a \sin ^4(x)}}-\frac {4 \cos ^2(x) \cot ^5(x)}{7 a^2 \sqrt {a \sin ^4(x)}}-\frac {\cos ^2(x) \cot ^7(x)}{9 a^2 \sqrt {a \sin ^4(x)}}-\frac {\cos (x) \sin (x)}{a^2 \sqrt {a \sin ^4(x)}} \] Output:
-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*s in(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*c ot(x)^7/a^2/(a*sin(x)^4)^(1/2)-cos(x)*sin(x)/a^2/(a*sin(x)^4)^(1/2)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {\cos (x) \left (128+64 \csc ^2(x)+48 \csc ^4(x)+40 \csc ^6(x)+35 \csc ^8(x)\right ) \sin (x)}{315 a^2 \sqrt {a \sin ^4(x)}} \] Input:
Integrate[(a*Sin[x]^4)^(-5/2),x]
Output:
-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])
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.46, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3686, 3042, 4254, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \sin (x)^4\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {\sin ^2(x) \int \csc ^{10}(x)dx}{a^2 \sqrt {a \sin ^4(x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin ^2(x) \int \csc (x)^{10}dx}{a^2 \sqrt {a \sin ^4(x)}}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle -\frac {\sin ^2(x) \int \left (\cot ^8(x)+4 \cot ^6(x)+6 \cot ^4(x)+4 \cot ^2(x)+1\right )d\cot (x)}{a^2 \sqrt {a \sin ^4(x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sin ^2(x) \left (\frac {\cot ^9(x)}{9}+\frac {4 \cot ^7(x)}{7}+\frac {6 \cot ^5(x)}{5}+\frac {4 \cot ^3(x)}{3}+\cot (x)\right )}{a^2 \sqrt {a \sin ^4(x)}}\) |
Input:
Int[(a*Sin[x]^4)^(-5/2),x]
Output:
-(((Cot[x] + (4*Cot[x]^3)/3 + (6*Cot[x]^5)/5 + (4*Cot[x]^7)/7 + Cot[x]^9/9 )*Sin[x]^2)/(a^2*Sqrt[a*Sin[x]^4]))
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[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]])
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 0.48 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42
method | result | size |
default | \(-\frac {\cot \left (x \right ) \csc \left (x \right )^{6} \left (128 \cos \left (x \right )^{8}-576 \cos \left (x \right )^{6}+1008 \cos \left (x \right )^{4}-840 \cos \left (x \right )^{2}+315\right ) \sqrt {16}}{1260 \sqrt {a \sin \left (x \right )^{4}}\, a^{2}}\) | \(49\) |
risch | \(\frac {256 i \left (126 \,{\mathrm e}^{6 i x}-84 \,{\mathrm e}^{4 i x}-9+37 \cos \left (2 x \right )+35 i \sin \left (2 x \right )\right )}{315 a^{2} \left ({\mathrm e}^{2 i x}-1\right )^{7} \sqrt {a \left ({\mathrm e}^{2 i x}-1\right )^{4} {\mathrm e}^{-4 i x}}}\) | \(63\) |
Input:
int(1/(a*sin(x)^4)^(5/2),x,method=_RETURNVERBOSE)
Output:
-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)
Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=\frac {{\left (128 \, \cos \left (x\right )^{9} - 576 \, \cos \left (x\right )^{7} + 1008 \, \cos \left (x\right )^{5} - 840 \, \cos \left (x\right )^{3} + 315 \, \cos \left (x\right )\right )} \sqrt {a \cos \left (x\right )^{4} - 2 \, a \cos \left (x\right )^{2} + a}}{315 \, {\left (a^{3} \cos \left (x\right )^{10} - 5 \, a^{3} \cos \left (x\right )^{8} + 10 \, a^{3} \cos \left (x\right )^{6} - 10 \, a^{3} \cos \left (x\right )^{4} + 5 \, a^{3} \cos \left (x\right )^{2} - a^{3}\right )} \sin \left (x\right )} \] Input:
integrate(1/(a*sin(x)^4)^(5/2),x, algorithm="fricas")
Output:
1/315*(128*cos(x)^9 - 576*cos(x)^7 + 1008*cos(x)^5 - 840*cos(x)^3 + 315*co s(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))
\[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a \sin ^{4}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a*sin(x)**4)**(5/2),x)
Output:
Integral((a*sin(x)**4)**(-5/2), x)
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {315 \, \tan \left (x\right )^{8} + 420 \, \tan \left (x\right )^{6} + 378 \, \tan \left (x\right )^{4} + 180 \, \tan \left (x\right )^{2} + 35}{315 \, a^{\frac {5}{2}} \tan \left (x\right )^{9}} \] Input:
integrate(1/(a*sin(x)^4)^(5/2),x, algorithm="maxima")
Output:
-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)
Time = 0.45 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=-\frac {315 \, \tan \left (x\right )^{8} + 420 \, \tan \left (x\right )^{6} + 378 \, \tan \left (x\right )^{4} + 180 \, \tan \left (x\right )^{2} + 35}{315 \, a^{\frac {5}{2}} \tan \left (x\right )^{9}} \] Input:
integrate(1/(a*sin(x)^4)^(5/2),x, algorithm="giac")
Output:
-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)
Time = 41.72 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=\frac {256\,\left ({\mathrm {e}}^{x\,46{}\mathrm {i}}\,1{}\mathrm {i}-{\mathrm {e}}^{x\,48{}\mathrm {i}}\,9{}\mathrm {i}+{\mathrm {e}}^{x\,50{}\mathrm {i}}\,36{}\mathrm {i}-{\mathrm {e}}^{x\,52{}\mathrm {i}}\,84{}\mathrm {i}+{\mathrm {e}}^{x\,54{}\mathrm {i}}\,126{}\mathrm {i}\right )}{315\,a^{5/2}\,\left ({\mathrm {e}}^{x\,46{}\mathrm {i}}-9\,{\mathrm {e}}^{x\,48{}\mathrm {i}}+36\,{\mathrm {e}}^{x\,50{}\mathrm {i}}-84\,{\mathrm {e}}^{x\,52{}\mathrm {i}}+126\,{\mathrm {e}}^{x\,54{}\mathrm {i}}-126\,{\mathrm {e}}^{x\,56{}\mathrm {i}}+84\,{\mathrm {e}}^{x\,58{}\mathrm {i}}-36\,{\mathrm {e}}^{x\,60{}\mathrm {i}}+9\,{\mathrm {e}}^{x\,62{}\mathrm {i}}-{\mathrm {e}}^{x\,64{}\mathrm {i}}\right )} \] Input:
int(1/(a*sin(x)^4)^(5/2),x)
Output:
(256*(exp(x*46i)*1i - exp(x*48i)*9i + exp(x*50i)*36i - exp(x*52i)*84i + ex p(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)))
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\left (a \sin ^4(x)\right )^{5/2}} \, dx=\frac {\sqrt {a}\, \cos \left (x \right ) \left (-128 \sin \left (x \right )^{8}-64 \sin \left (x \right )^{6}-48 \sin \left (x \right )^{4}-40 \sin \left (x \right )^{2}-35\right )}{315 \sin \left (x \right )^{9} a^{3}} \] Input:
int(1/(a*sin(x)^4)^(5/2),x)
Output:
(sqrt(a)*cos(x)*( - 128*sin(x)**8 - 64*sin(x)**6 - 48*sin(x)**4 - 40*sin(x )**2 - 35))/(315*sin(x)**9*a**3)