\(\int \frac {1}{(\cot ^2(x)+\csc ^2(x))^2} \, dx\) [491]

Optimal result

Integrand size = 11, antiderivative size = 47 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=x-\frac {x}{\sqrt {2}}+\frac {\arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{\sqrt {2}}-\frac {\tan (x)}{2+\tan ^2(x)} \]

[Out]

x-1/2*x*2^(1/2)+1/2*arctan(cos(x)*sin(x)/(1+cos(x)^2+2^(1/2)))*2^(1/2)-tan(x)/(2+tan(x)^2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {481, 12, 400, 209} \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=\frac {\arctan \left (\frac {\sin (x) \cos (x)}{\cos ^2(x)+\sqrt {2}+1}\right )}{\sqrt {2}}-\frac {x}{\sqrt {2}}+x-\frac {\tan (x)}{\tan ^2(x)+2} \]

[In]

Int[(Cot[x]^2 + Csc[x]^2)^(-2),x]

[Out]

x - x/Sqrt[2] + ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Cos[x]^2)]/Sqrt[2] - Tan[x]/(2 + Tan[x]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 400

Int[1/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^n),
x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0]

Rule 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right ) \left (2+x^2\right )^2} \, dx,x,\tan (x)\right ) \\ & = -\frac {\tan (x)}{2+\tan ^2(x)}+\frac {1}{2} \text {Subst}\left (\int \frac {2}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx,x,\tan (x)\right ) \\ & = -\frac {\tan (x)}{2+\tan ^2(x)}+\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx,x,\tan (x)\right ) \\ & = -\frac {\tan (x)}{2+\tan ^2(x)}+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (x)\right )-\text {Subst}\left (\int \frac {1}{2+x^2} \, dx,x,\tan (x)\right ) \\ & = 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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=\frac {(3+\cos (2 x)) \csc ^4(x) \left (6 x+2 x \cos (2 x)-\sqrt {2} \arctan \left (\frac {\tan (x)}{\sqrt {2}}\right ) (3+\cos (2 x))-2 \sin (2 x)\right )}{8 \left (\cot ^2(x)+\csc ^2(x)\right )^2} \]

[In]

Integrate[(Cot[x]^2 + Csc[x]^2)^(-2),x]

[Out]

((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)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.60

\[-\frac {\tan \left (x \right )}{2+\tan \left (x \right )^{2}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \tan \left (x \right )}{2}\right )}{2}+x\]

[In]

int(1/(cot(x)^2+csc(x)^2)^2,x)

[Out]

-tan(x)/(2+tan(x)^2)-1/2*2^(1/2)*arctan(1/2*2^(1/2)*tan(x))+x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=\frac {4 \, x \cos \left (x\right )^{2} + {\left (\sqrt {2} \cos \left (x\right )^{2} + \sqrt {2}\right )} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) - 4 \, \cos \left (x\right ) \sin \left (x\right ) + 4 \, x}{4 \, {\left (\cos \left (x\right )^{2} + 1\right )}} \]

[In]

integrate(1/(cot(x)^2+csc(x)^2)^2,x, algorithm="fricas")

[Out]

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)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=\text {Timed out} \]

[In]

integrate(1/(cot(x)**2+csc(x)**2)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \tan \left (x\right )\right ) + x - \frac {\tan \left (x\right )}{\tan \left (x\right )^{2} + 2} \]

[In]

integrate(1/(cot(x)^2+csc(x)^2)^2,x, algorithm="maxima")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*tan(x)) + x - tan(x)/(tan(x)^2 + 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=-\frac {1}{2} \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - \cos \left (2 \, x\right ) + 1}\right )\right )} + x - \frac {\tan \left (x\right )}{\tan \left (x\right )^{2} + 2} \]

[In]

integrate(1/(cot(x)^2+csc(x)^2)^2,x, algorithm="giac")

[Out]

-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 - ta
n(x)/(tan(x)^2 + 2)

Mupad [B] (verification not implemented)

Time = 27.91 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (\cot ^2(x)+\csc ^2(x)\right )^2} \, dx=x-\frac {\mathrm {tan}\left (x\right )}{{\mathrm {tan}\left (x\right )}^2+2}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\left (x\right )}{2}\right )}{2} \]

[In]

int(1/(cot(x)^2 + 1/sin(x)^2)^2,x)

[Out]

x - tan(x)/(tan(x)^2 + 2) - (2^(1/2)*atan((2^(1/2)*tan(x))/2))/2