\(\int \frac {1}{(a \csc ^4(x))^{3/2}} \, dx\) [66]

Optimal result

Integrand size = 10, antiderivative size = 86 \[ \int \frac {1}{\left (a \csc ^4(x)\right )^{3/2}} \, dx=-\frac {5 \cot (x)}{16 a \sqrt {a \csc ^4(x)}}+\frac {5 x \csc ^2(x)}{16 a \sqrt {a \csc ^4(x)}}-\frac {5 \cos (x) \sin (x)}{24 a \sqrt {a \csc ^4(x)}}-\frac {\cos (x) \sin ^3(x)}{6 a \sqrt {a \csc ^4(x)}} \]

[Out]

-5/16*cot(x)/a/(a*csc(x)^4)^(1/2)+5/16*x*csc(x)^2/a/(a*csc(x)^4)^(1/2)-5/24*cos(x)*sin(x)/a/(a*csc(x)^4)^(1/2)
-1/6*cos(x)*sin(x)^3/a/(a*csc(x)^4)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4208, 2715, 8} \[ \int \frac {1}{\left (a \csc ^4(x)\right )^{3/2}} \, dx=\frac {5 x \csc ^2(x)}{16 a \sqrt {a \csc ^4(x)}}-\frac {5 \cot (x)}{16 a \sqrt {a \csc ^4(x)}}-\frac {\sin ^3(x) \cos (x)}{6 a \sqrt {a \csc ^4(x)}}-\frac {5 \sin (x) \cos (x)}{24 a \sqrt {a \csc ^4(x)}} \]

[In]

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

[Out]

(-5*Cot[x])/(16*a*Sqrt[a*Csc[x]^4]) + (5*x*Csc[x]^2)/(16*a*Sqrt[a*Csc[x]^4]) - (5*Cos[x]*Sin[x])/(24*a*Sqrt[a*
Csc[x]^4]) - (Cos[x]*Sin[x]^3)/(6*a*Sqrt[a*Csc[x]^4])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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}& = \frac {\csc ^2(x) \int \sin ^6(x) \, dx}{a \sqrt {a \csc ^4(x)}} \\ & = -\frac {\cos (x) \sin ^3(x)}{6 a \sqrt {a \csc ^4(x)}}+\frac {\left (5 \csc ^2(x)\right ) \int \sin ^4(x) \, dx}{6 a \sqrt {a \csc ^4(x)}} \\ & = -\frac {5 \cos (x) \sin (x)}{24 a \sqrt {a \csc ^4(x)}}-\frac {\cos (x) \sin ^3(x)}{6 a \sqrt {a \csc ^4(x)}}+\frac {\left (5 \csc ^2(x)\right ) \int \sin ^2(x) \, dx}{8 a \sqrt {a \csc ^4(x)}} \\ & = -\frac {5 \cot (x)}{16 a \sqrt {a \csc ^4(x)}}-\frac {5 \cos (x) \sin (x)}{24 a \sqrt {a \csc ^4(x)}}-\frac {\cos (x) \sin ^3(x)}{6 a \sqrt {a \csc ^4(x)}}+\frac {\left (5 \csc ^2(x)\right ) \int 1 \, dx}{16 a \sqrt {a \csc ^4(x)}} \\ & = -\frac {5 \cot (x)}{16 a \sqrt {a \csc ^4(x)}}+\frac {5 x \csc ^2(x)}{16 a \sqrt {a \csc ^4(x)}}-\frac {5 \cos (x) \sin (x)}{24 a \sqrt {a \csc ^4(x)}}-\frac {\cos (x) \sin ^3(x)}{6 a \sqrt {a \csc ^4(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (a \csc ^4(x)\right )^{3/2}} \, dx=-\frac {\csc ^6(x) (-60 x+45 \sin (2 x)-9 \sin (4 x)+\sin (6 x))}{192 \left (a \csc ^4(x)\right )^{3/2}} \]

[In]

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

[Out]

-1/192*(Csc[x]^6*(-60*x + 45*Sin[2*x] - 9*Sin[4*x] + Sin[6*x]))/(a*Csc[x]^4)^(3/2)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.52

method	result	size
default	\(-\frac {\left (8 \cos \left (x \right )^{4} \cot \left (x \right )-26 \cos \left (x \right )^{2} \cot \left (x \right )+33 \cot \left (x \right )-15 \csc \left (x \right )^{2} x \right ) \sqrt {16}}{192 \sqrt {a \csc \left (x \right )^{4}}\, a}\)	\(45\)
risch	\(-\frac {5 \,{\mathrm e}^{2 i x} x}{16 a \left ({\mathrm e}^{2 i x}-1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{4}}}}-\frac {i {\mathrm e}^{8 i x}}{384 a \left ({\mathrm e}^{2 i x}-1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{4}}}}+\frac {3 i {\mathrm e}^{6 i x}}{128 a \left ({\mathrm e}^{2 i x}-1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{4}}}}-\frac {15 i {\mathrm e}^{4 i x}}{128 a \left ({\mathrm e}^{2 i x}-1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{4}}}}+\frac {15 i}{128 a \left ({\mathrm e}^{2 i x}-1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{4}}}}-\frac {3 i {\mathrm e}^{-2 i x}}{128 a \left ({\mathrm e}^{2 i x}-1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{4}}}}+\frac {i {\mathrm e}^{-4 i x}}{384 a \left ({\mathrm e}^{2 i x}-1\right )^{2} \sqrt {\frac {a \,{\mathrm e}^{4 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{4}}}}\)	\(263\)

[In]

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

[Out]

-1/192/(a*csc(x)^4)^(1/2)/a*(8*cos(x)^4*cot(x)-26*cos(x)^2*cot(x)+33*cot(x)-15*csc(x)^2*x)*16^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (a \csc ^4(x)\right )^{3/2}} \, dx=-\frac {{\left (15 \, x \cos \left (x\right )^{2} - {\left (8 \, \cos \left (x\right )^{7} - 34 \, \cos \left (x\right )^{5} + 59 \, \cos \left (x\right )^{3} - 33 \, \cos \left (x\right )\right )} \sin \left (x\right ) - 15 \, x\right )} \sqrt {\frac {a}{\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1}}}{48 \, a^{2}} \]

[In]

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

[Out]

-1/48*(15*x*cos(x)^2 - (8*cos(x)^7 - 34*cos(x)^5 + 59*cos(x)^3 - 33*cos(x))*sin(x) - 15*x)*sqrt(a/(cos(x)^4 -
2*cos(x)^2 + 1))/a^2

Sympy [F]

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

[In]

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

[Out]

Integral((a*csc(x)**4)**(-3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\left (a \csc ^4(x)\right )^{3/2}} \, dx=-\frac {33 \, \tan \left (x\right )^{5} + 40 \, \tan \left (x\right )^{3} + 15 \, \tan \left (x\right )}{48 \, {\left (a^{\frac {3}{2}} \tan \left (x\right )^{6} + 3 \, a^{\frac {3}{2}} \tan \left (x\right )^{4} + 3 \, a^{\frac {3}{2}} \tan \left (x\right )^{2} + a^{\frac {3}{2}}\right )}} + \frac {5 \, x}{16 \, a^{\frac {3}{2}}} \]

[In]

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

[Out]

-1/48*(33*tan(x)^5 + 40*tan(x)^3 + 15*tan(x))/(a^(3/2)*tan(x)^6 + 3*a^(3/2)*tan(x)^4 + 3*a^(3/2)*tan(x)^2 + a^
(3/2)) + 5/16*x/a^(3/2)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a \csc ^4(x)\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

int(1/(a/sin(x)^4)^(3/2), x)