Integrand size = 11, antiderivative size = 72 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx=-x+\frac {7 x}{4 \sqrt {2}}-\frac {7 \arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{4 \sqrt {2}}-\frac {\tan ^3(x)}{2 \left (2+\tan ^2(x)\right )^2}+\frac {\tan (x)}{4 \left (2+\tan ^2(x)\right )} \] Output:
-x+7/8*x*2^(1/2)-7/8*2^(1/2)*arctan(cos(x)*sin(x)/(1+2^(1/2)+cos(x)^2))-1/ 2*tan(x)^3/(2+tan(x)^2)^2+tan(x)/(8+4*tan(x)^2)
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx=\frac {-76 x-48 x \cos (2 x)+7 \sqrt {2} \arctan \left (\frac {\tan (x)}{\sqrt {2}}\right ) (3+\cos (2 x))^2-4 x \cos (4 x)+2 \sin (2 x)+3 \sin (4 x)}{8 (3+\cos (2 x))^2} \] Input:
Integrate[(Cot[x]^2 + Csc[x]^2)^(-3),x]
Output:
(-76*x - 48*x*Cos[2*x] + 7*Sqrt[2]*ArcTan[Tan[x]/Sqrt[2]]*(3 + Cos[2*x])^2 - 4*x*Cos[4*x] + 2*Sin[2*x] + 3*Sin[4*x])/(8*(3 + Cos[2*x])^2)
Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3042, 4889, 372, 27, 440, 397, 216}
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.
\(\displaystyle \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\cot (x)^2+\csc (x)^2\right )^3}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {\tan ^6(x)}{\left (\tan ^2(x)+1\right ) \left (\tan ^2(x)+2\right )^3}d\tan (x)\) |
\(\Big \downarrow \) 372 |
\(\displaystyle \frac {1}{4} \int \frac {2 \tan ^2(x) \left (\tan ^2(x)+3\right )}{\left (\tan ^2(x)+1\right ) \left (\tan ^2(x)+2\right )^2}d\tan (x)-\frac {\tan ^3(x)}{2 \left (\tan ^2(x)+2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {\tan ^2(x) \left (\tan ^2(x)+3\right )}{\left (\tan ^2(x)+1\right ) \left (\tan ^2(x)+2\right )^2}d\tan (x)-\frac {\tan ^3(x)}{2 \left (\tan ^2(x)+2\right )^2}\) |
\(\Big \downarrow \) 440 |
\(\displaystyle \frac {1}{2} \left (\frac {\tan (x)}{2 \left (\tan ^2(x)+2\right )}-\frac {1}{2} \int \frac {1-3 \tan ^2(x)}{\left (\tan ^2(x)+1\right ) \left (\tan ^2(x)+2\right )}d\tan (x)\right )-\frac {\tan ^3(x)}{2 \left (\tan ^2(x)+2\right )^2}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (7 \int \frac {1}{\tan ^2(x)+2}d\tan (x)-4 \int \frac {1}{\tan ^2(x)+1}d\tan (x)\right )+\frac {\tan (x)}{2 \left (\tan ^2(x)+2\right )}\right )-\frac {\tan ^3(x)}{2 \left (\tan ^2(x)+2\right )^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {7 \arctan \left (\frac {\tan (x)}{\sqrt {2}}\right )}{\sqrt {2}}-4 \arctan (\tan (x))\right )+\frac {\tan (x)}{2 \left (\tan ^2(x)+2\right )}\right )-\frac {\tan ^3(x)}{2 \left (\tan ^2(x)+2\right )^2}\) |
Input:
Int[(Cot[x]^2 + Csc[x]^2)^(-3),x]
Output:
-1/2*Tan[x]^3/(2 + Tan[x]^2)^2 + ((-4*ArcTan[Tan[x]] + (7*ArcTan[Tan[x]/Sq rt[2]])/Sqrt[2])/2 + Tan[x]/(2*(2 + Tan[x]^2)))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 1]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Time = 12.86 (sec) , antiderivative size = 2, normalized size of antiderivative = 0.03
method | result | size |
parallelrisch | \(0\) | \(2\) |
default | \(\frac {-\frac {\tan \left (x \right )^{3}}{4}+\frac {\tan \left (x \right )}{2}}{\left (2+\tan \left (x \right )^{2}\right )^{2}}+\frac {7 \sqrt {2}\, \arctan \left (\frac {\tan \left (x \right ) \sqrt {2}}{2}\right )}{8}-\arctan \left (\tan \left (x \right )\right )\) | \(41\) |
risch | \(-x +\frac {i \left (17 \,{\mathrm e}^{6 i x}+57 \,{\mathrm e}^{4 i x}+19 \,{\mathrm e}^{2 i x}+3\right )}{2 \left ({\mathrm e}^{4 i x}+6 \,{\mathrm e}^{2 i x}+1\right )^{2}}+\frac {7 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}+3\right )}{16}-\frac {7 i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}+3\right )}{16}\) | \(85\) |
Input:
int(1/(csc(x)^2+cot(x)^2)^3,x,method=_RETURNVERBOSE)
Output:
0
Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx=-\frac {16 \, x \cos \left (x\right )^{4} + 32 \, x \cos \left (x\right )^{2} + 7 \, {\left (\sqrt {2} \cos \left (x\right )^{4} + 2 \, \sqrt {2} \cos \left (x\right )^{2} + \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) - 4 \, {\left (3 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) + 16 \, x}{16 \, {\left (\cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:
integrate(1/(cot(x)^2+csc(x)^2)^3,x, algorithm="fricas")
Output:
-1/16*(16*x*cos(x)^4 + 32*x*cos(x)^2 + 7*(sqrt(2)*cos(x)^4 + 2*sqrt(2)*cos (x)^2 + sqrt(2))*arctan(1/4*(3*sqrt(2)*cos(x)^2 - sqrt(2))/(cos(x)*sin(x)) ) - 4*(3*cos(x)^3 - cos(x))*sin(x) + 16*x)/(cos(x)^4 + 2*cos(x)^2 + 1)
Timed out. \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(cot(x)**2+csc(x)**2)**3,x)
Output:
Timed out
Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx=\frac {7}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \tan \left (x\right )\right ) - x - \frac {\tan \left (x\right )^{3} - 2 \, \tan \left (x\right )}{4 \, {\left (\tan \left (x\right )^{4} + 4 \, \tan \left (x\right )^{2} + 4\right )}} \] Input:
integrate(1/(cot(x)^2+csc(x)^2)^3,x, algorithm="maxima")
Output:
7/8*sqrt(2)*arctan(1/2*sqrt(2)*tan(x)) - x - 1/4*(tan(x)^3 - 2*tan(x))/(ta n(x)^4 + 4*tan(x)^2 + 4)
Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx=\frac {7}{8} \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - \cos \left (2 \, x\right ) + 1}\right )\right )} - x - \frac {\tan \left (x\right )^{3} - 2 \, \tan \left (x\right )}{4 \, {\left (\tan \left (x\right )^{2} + 2\right )}^{2}} \] Input:
integrate(1/(cot(x)^2+csc(x)^2)^3,x, algorithm="giac")
Output:
7/8*sqrt(2)*(x + arctan(-(sqrt(2)*sin(2*x) - sin(2*x))/(sqrt(2)*cos(2*x) + sqrt(2) - cos(2*x) + 1))) - x - 1/4*(tan(x)^3 - 2*tan(x))/(tan(x)^2 + 2)^ 2
Time = 16.60 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx=\frac {\frac {\mathrm {tan}\left (x\right )}{2}-\frac {{\mathrm {tan}\left (x\right )}^3}{4}}{{\mathrm {tan}\left (x\right )}^4+4\,{\mathrm {tan}\left (x\right )}^2+4}-x+\frac {7\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{2}\right )}{8} \] Input:
int(1/(cot(x)^2 + 1/sin(x)^2)^3,x)
Output:
(tan(x)/2 - tan(x)^3/4)/(4*tan(x)^2 + tan(x)^4 + 4) - x + (7*2^(1/2)*atan( (2^(1/2)*tan(x))/2))/8
\[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx=\int \frac {1}{\cot \left (x \right )^{6}+3 \cot \left (x \right )^{4} \csc \left (x \right )^{2}+3 \cot \left (x \right )^{2} \csc \left (x \right )^{4}+\csc \left (x \right )^{6}}d x \] Input:
int(1/(cot(x)^2+csc(x)^2)^3,x)
Output:
int(1/(cot(x)**6 + 3*cot(x)**4*csc(x)**2 + 3*cot(x)**2*csc(x)**4 + csc(x)* *6),x)