3.492 \(\int \frac{1}{(\cot ^2(x)+\csc ^2(x))^3} \, dx\)

Optimal. Leaf size=72 \[ \frac{7 x}{4 \sqrt{2}}-x-\frac{\tan ^3(x)}{2 \left (\tan ^2(x)+2\right )^2}+\frac{\tan (x)}{4 \left (\tan ^2(x)+2\right )}-\frac{7 \tan ^{-1}\left (\frac{\sin (x) \cos (x)}{\cos ^2(x)+\sqrt{2}+1}\right )}{4 \sqrt{2}} \]

[Out]

-x + (7*x)/(4*Sqrt[2]) - (7*ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Cos[x]^2)])/(4*Sqrt[2]) - Tan[x]^3/(2*(2 + T
an[x]^2)^2) + Tan[x]/(4*(2 + Tan[x]^2))

________________________________________________________________________________________

Rubi [A]  time = 0.0755278, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {470, 578, 522, 203} \[ \frac{7 x}{4 \sqrt{2}}-x-\frac{\tan ^3(x)}{2 \left (\tan ^2(x)+2\right )^2}+\frac{\tan (x)}{4 \left (\tan ^2(x)+2\right )}-\frac{7 \tan ^{-1}\left (\frac{\sin (x) \cos (x)}{\cos ^2(x)+\sqrt{2}+1}\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[x]^2 + Csc[x]^2)^(-3),x]

[Out]

-x + (7*x)/(4*Sqrt[2]) - (7*ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Cos[x]^2)])/(4*Sqrt[2]) - Tan[x]^3/(2*(2 + T
an[x]^2)^2) + Tan[x]/(4*(2 + Tan[x]^2))

Rule 470

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]

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 522

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{1}{\left (\cot ^2(x)+\csc ^2(x)\right )^3} \, dx &=\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right ) \left (2+x^2\right )^3} \, dx,x,\tan (x)\right )\\ &=-\frac{\tan ^3(x)}{2 \left (2+\tan ^2(x)\right )^2}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{x^2 \left (6+2 x^2\right )}{\left (1+x^2\right ) \left (2+x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=-\frac{\tan ^3(x)}{2 \left (2+\tan ^2(x)\right )^2}+\frac{\tan (x)}{4 \left (2+\tan ^2(x)\right )}-\frac{1}{8} \operatorname{Subst}\left (\int \frac{2-6 x^2}{\left (1+x^2\right ) \left (2+x^2\right )} \, dx,x,\tan (x)\right )\\ &=-\frac{\tan ^3(x)}{2 \left (2+\tan ^2(x)\right )^2}+\frac{\tan (x)}{4 \left (2+\tan ^2(x)\right )}+\frac{7}{4} \operatorname{Subst}\left (\int \frac{1}{2+x^2} \, dx,x,\tan (x)\right )-\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-x+\frac{7 x}{4 \sqrt{2}}-\frac{7 \tan ^{-1}\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 )}\\ \end{align*}

Mathematica [A]  time = 0.16371, size = 66, normalized size = 0.92 \[ \frac{-76 x+2 \sin (2 x)+3 \sin (4 x)-48 x \cos (2 x)-4 x \cos (4 x)+7 \sqrt{2} (\cos (2 x)+3)^2 \tan ^{-1}\left (\frac{\tan (x)}{\sqrt{2}}\right )}{8 (\cos (2 x)+3)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[x]^2 + Csc[x]^2)^(-3),x]

[Out]

(-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)

________________________________________________________________________________________

Maple [A]  time = 0.121, size = 39, normalized size = 0.5 \begin{align*} 2\,{\frac{-1/8\, \left ( \tan \left ( x \right ) \right ) ^{3}+1/4\,\tan \left ( x \right ) }{ \left ( 2+ \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{7\,\sqrt{2}}{8}\arctan \left ({\frac{\tan \left ( x \right ) \sqrt{2}}{2}} \right ) }-x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cot(x)^2+csc(x)^2)^3,x)

[Out]

2*(-1/8*tan(x)^3+1/4*tan(x))/(2+tan(x)^2)^2+7/8*2^(1/2)*arctan(1/2*tan(x)*2^(1/2))-x

________________________________________________________________________________________

Maxima [A]  time = 1.51483, size = 57, normalized size = 0.79 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cot(x)^2+csc(x)^2)^3,x, algorithm="maxima")

[Out]

7/8*sqrt(2)*arctan(1/2*sqrt(2)*tan(x)) - x - 1/4*(tan(x)^3 - 2*tan(x))/(tan(x)^4 + 4*tan(x)^2 + 4)

________________________________________________________________________________________

Fricas [A]  time = 1.85431, size = 297, normalized size = 4.12 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cot(x)^2+csc(x)^2)^3,x, algorithm="fricas")

[Out]

-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)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cot(x)**2+csc(x)**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.13138, size = 93, normalized size = 1.29 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cot(x)^2+csc(x)^2)^3,x, algorithm="giac")

[Out]

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