\(\int (a \csc ^4(x))^{5/2} \, dx\) [62]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.03 (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 = {4208, 3852} \[ \int \left (a \csc ^4(x)\right )^{5/2} \, dx=-\frac {1}{9} a^2 \cos ^2(x) \cot ^7(x) \sqrt {a \csc ^4(x)}-\frac {4}{7} a^2 \cos ^2(x) \cot ^5(x) \sqrt {a \csc ^4(x)}-\frac {6}{5} a^2 \cos ^2(x) \cot ^3(x) \sqrt {a \csc ^4(x)}-\frac {4}{3} a^2 \cos ^2(x) \cot (x) \sqrt {a \csc ^4(x)}-a^2 \sin (x) \cos (x) \sqrt {a \csc ^4(x)} \]

[In]

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

[Out]

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

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]

Rule 4208

Int[((b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Sec[e + f*x])^n)^
FracPart[p]/(c*Sec[e + f*x])^(n*FracPart[p])), Int[(c*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{b, c, e, f, n, p},
x] &&  !IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.40 \[ \int \left (a \csc ^4(x)\right )^{5/2} \, dx=-\frac {1}{315} a^2 \cos (x) \sqrt {a \csc ^4(x)} \left (128+64 \csc ^2(x)+48 \csc ^4(x)+40 \csc ^6(x)+35 \csc ^8(x)\right ) \sin (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75 \[ \int \left (a \csc ^4(x)\right )^{5/2} \, dx=\frac {{\left (128 \, a^{2} \cos \left (x\right )^{9} - 576 \, a^{2} \cos \left (x\right )^{7} + 1008 \, a^{2} \cos \left (x\right )^{5} - 840 \, a^{2} \cos \left (x\right )^{3} + 315 \, a^{2} \cos \left (x\right )\right )} \sqrt {\frac {a}{\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1}}}{315 \, {\left (\cos \left (x\right )^{6} - 3 \, \cos \left (x\right )^{4} + 3 \, \cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate((a*csc(x)**4)**(5/2),x)

[Out]

Integral((a*csc(x)**4)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.41 \[ \int \left (a \csc ^4(x)\right )^{5/2} \, dx=-\frac {315 \, a^{\frac {5}{2}} \tan \left (x\right )^{8} + 420 \, a^{\frac {5}{2}} \tan \left (x\right )^{6} + 378 \, a^{\frac {5}{2}} \tan \left (x\right )^{4} + 180 \, a^{\frac {5}{2}} \tan \left (x\right )^{2} + 35 \, a^{\frac {5}{2}}}{315 \, \tan \left (x\right )^{9}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.43 \[ \int \left (a \csc ^4(x)\right )^{5/2} \, dx=-\frac {{\left (315 \, a^{2} \tan \left (x\right )^{8} + 420 \, a^{2} \tan \left (x\right )^{6} + 378 \, a^{2} \tan \left (x\right )^{4} + 180 \, a^{2} \tan \left (x\right )^{2} + 35 \, a^{2}\right )} \sqrt {a}}{315 \, \tan \left (x\right )^{9}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 19.51 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03 \[ \int \left (a \csc ^4(x)\right )^{5/2} \, dx=\frac {128\,a^{5/2}\,\left ({\mathrm {e}}^{x\,46{}\mathrm {i}}\,1{}\mathrm {i}-{\mathrm {e}}^{x\,48{}\mathrm {i}}\,9{}\mathrm {i}+{\mathrm {e}}^{x\,50{}\mathrm {i}}\,36{}\mathrm {i}-{\mathrm {e}}^{x\,52{}\mathrm {i}}\,84{}\mathrm {i}+{\mathrm {e}}^{x\,54{}\mathrm {i}}\,126{}\mathrm {i}\right )}{315\,\left (\frac {{\mathrm {e}}^{-x\,2{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{x\,2{}\mathrm {i}}}{2}-1\right )\,\left ({\mathrm {e}}^{x\,48{}\mathrm {i}}-7\,{\mathrm {e}}^{x\,50{}\mathrm {i}}+21\,{\mathrm {e}}^{x\,52{}\mathrm {i}}-35\,{\mathrm {e}}^{x\,54{}\mathrm {i}}+35\,{\mathrm {e}}^{x\,56{}\mathrm {i}}-21\,{\mathrm {e}}^{x\,58{}\mathrm {i}}+7\,{\mathrm {e}}^{x\,60{}\mathrm {i}}-{\mathrm {e}}^{x\,62{}\mathrm {i}}\right )} \]

[In]

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

[Out]

(128*a^(5/2)*(exp(x*46i)*1i - exp(x*48i)*9i + exp(x*50i)*36i - exp(x*52i)*84i + exp(x*54i)*126i))/(315*(exp(-x
*2i)/2 + exp(x*2i)/2 - 1)*(exp(x*48i) - 7*exp(x*50i) + 21*exp(x*52i) - 35*exp(x*54i) + 35*exp(x*56i) - 21*exp(
x*58i) + 7*exp(x*60i) - exp(x*62i)))