Integrand size = 11, antiderivative size = 47 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=x-\frac {x}{\sqrt {2}}+\frac {\arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{\sqrt {2}}-\frac {\tan (x)}{2+\tan ^2(x)} \] Output:
x-1/2*x*2^(1/2)+1/2*2^(1/2)*arctan(cos(x)*sin(x)/(1+2^(1/2)+cos(x)^2))-tan (x)/(2+tan(x)^2)
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=\frac {(3+\cos (2 x)) \csc ^4(x) \left (6 x+2 x \cos (2 x)-\sqrt {2} \arctan \left (\frac {\tan (x)}{\sqrt {2}}\right ) (3+\cos (2 x))-2 \sin (2 x)\right )}{8 \left (\cot ^2(x)+\csc ^2(x)\right )^2} \] Input:
Integrate[(Cot[x]^2 + Csc[x]^2)^(-2),x]
Output:
((3 + Cos[2*x])*Csc[x]^4*(6*x + 2*x*Cos[2*x] - Sqrt[2]*ArcTan[Tan[x]/Sqrt[ 2]]*(3 + Cos[2*x]) - 2*Sin[2*x]))/(8*(Cot[x]^2 + Csc[x]^2)^2)
Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.68, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4889, 372, 27, 303, 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.
\(\displaystyle \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\cot (x)^2+\csc (x)^2\right )^2}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {\tan ^4(x)}{\left (\tan ^2(x)+1\right ) \left (\tan ^2(x)+2\right )^2}d\tan (x)\) |
\(\Big \downarrow \) 372 |
\(\displaystyle \frac {1}{2} \int \frac {2}{\left (\tan ^2(x)+1\right ) \left (\tan ^2(x)+2\right )}d\tan (x)-\frac {\tan (x)}{\tan ^2(x)+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{\left (\tan ^2(x)+1\right ) \left (\tan ^2(x)+2\right )}d\tan (x)-\frac {\tan (x)}{\tan ^2(x)+2}\) |
\(\Big \downarrow \) 303 |
\(\displaystyle \int \frac {1}{\tan ^2(x)+1}d\tan (x)-\int \frac {1}{\tan ^2(x)+2}d\tan (x)-\frac {\tan (x)}{\tan ^2(x)+2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \arctan (\tan (x))-\frac {\arctan \left (\frac {\tan (x)}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\tan (x)}{\tan ^2(x)+2}\) |
Input:
Int[(Cot[x]^2 + Csc[x]^2)^(-2),x]
Output:
ArcTan[Tan[x]] - ArcTan[Tan[x]/Sqrt[2]]/Sqrt[2] - Tan[x]/(2 + Tan[x]^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[1/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[b/(b *c - a*d) Int[1/(a + b*x^2), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x ^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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[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 = 98.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.64
method | result | size |
default | \(-\frac {\tan \left (x \right )}{2+\tan \left (x \right )^{2}}-\frac {\sqrt {2}\, \arctan \left (\frac {\tan \left (x \right ) \sqrt {2}}{2}\right )}{2}+\arctan \left (\tan \left (x \right )\right )\) | \(30\) |
risch | \(x -\frac {2 i \left (3 \,{\mathrm e}^{2 i x}+1\right )}{{\mathrm e}^{4 i x}+6 \,{\mathrm e}^{2 i x}+1}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}+3\right )}{4}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}+3\right )}{4}\) | \(69\) |
Input:
int(1/(csc(x)^2+cot(x)^2)^2,x,method=_RETURNVERBOSE)
Output:
-tan(x)/(2+tan(x)^2)-1/2*2^(1/2)*arctan(1/2*tan(x)*2^(1/2))+arctan(tan(x))
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=\frac {4 \, x \cos \left (x\right )^{2} + {\left (\sqrt {2} \cos \left (x\right )^{2} + \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) - 4 \, \cos \left (x\right ) \sin \left (x\right ) + 4 \, x}{4 \, {\left (\cos \left (x\right )^{2} + 1\right )}} \] Input:
integrate(1/(cot(x)^2+csc(x)^2)^2,x, algorithm="fricas")
Output:
1/4*(4*x*cos(x)^2 + (sqrt(2)*cos(x)^2 + sqrt(2))*arctan(1/4*(3*sqrt(2)*cos (x)^2 - sqrt(2))/(cos(x)*sin(x))) - 4*cos(x)*sin(x) + 4*x)/(cos(x)^2 + 1)
\[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=\int \frac {1}{\left (\cot ^{2}{\left (x \right )} + \csc ^{2}{\left (x \right )}\right )^{2}}\, dx \] Input:
integrate(1/(cot(x)**2+csc(x)**2)**2,x)
Output:
Integral((cot(x)**2 + csc(x)**2)**(-2), x)
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \tan \left (x\right )\right ) + x - \frac {\tan \left (x\right )}{\tan \left (x\right )^{2} + 2} \] Input:
integrate(1/(cot(x)^2+csc(x)^2)^2,x, algorithm="maxima")
Output:
-1/2*sqrt(2)*arctan(1/2*sqrt(2)*tan(x)) + x - tan(x)/(tan(x)^2 + 2)
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=-\frac {1}{2} \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - \cos \left (2 \, x\right ) + 1}\right )\right )} + x - \frac {\tan \left (x\right )}{\tan \left (x\right )^{2} + 2} \] Input:
integrate(1/(cot(x)^2+csc(x)^2)^2,x, algorithm="giac")
Output:
-1/2*sqrt(2)*(x + arctan(-(sqrt(2)*sin(2*x) - sin(2*x))/(sqrt(2)*cos(2*x) + sqrt(2) - cos(2*x) + 1))) + x - tan(x)/(tan(x)^2 + 2)
Time = 16.80 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=x-\frac {\mathrm {tan}\left (x\right )}{{\mathrm {tan}\left (x\right )}^2+2}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{2}\right )}{2} \] Input:
int(1/(cot(x)^2 + 1/sin(x)^2)^2,x)
Output:
x - tan(x)/(tan(x)^2 + 2) - (2^(1/2)*atan((2^(1/2)*tan(x))/2))/2
\[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=\int \frac {1}{\cot \left (x \right )^{4}+2 \cot \left (x \right )^{2} \csc \left (x \right )^{2}+\csc \left (x \right )^{4}}d x \] Input:
int(1/(cot(x)^2+csc(x)^2)^2,x)
Output:
int(1/(cot(x)**4 + 2*cot(x)**2*csc(x)**2 + csc(x)**4),x)