3.38 \(\int (1-\cot ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=54 \[ \frac{1}{2} \cot (x) \sqrt{1-\cot ^2(x)}-2 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \cot (x)}{\sqrt{1-\cot ^2(x)}}\right )+\frac{5}{2} \sin ^{-1}(\cot (x)) \]

[Out]

(5*ArcSin[Cot[x]])/2 - 2*Sqrt[2]*ArcTan[(Sqrt[2]*Cot[x])/Sqrt[1 - Cot[x]^2]] + (Cot[x]*Sqrt[1 - Cot[x]^2])/2

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Rubi [A]  time = 0.0459119, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3661, 416, 523, 216, 377, 203} \[ \frac{1}{2} \cot (x) \sqrt{1-\cot ^2(x)}-2 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \cot (x)}{\sqrt{1-\cot ^2(x)}}\right )+\frac{5}{2} \sin ^{-1}(\cot (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cot[x]^2)^(3/2),x]

[Out]

(5*ArcSin[Cot[x]])/2 - 2*Sqrt[2]*ArcTan[(Sqrt[2]*Cot[x])/Sqrt[1 - Cot[x]^2]] + (Cot[x]*Sqrt[1 - Cot[x]^2])/2

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

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

Rule 216

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

Rule 377

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

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 \left (1-\cot ^2(x)\right )^{3/2} \, dx &=-\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{3/2}}{1+x^2} \, dx,x,\cot (x)\right )\\ &=\frac{1}{2} \cot (x) \sqrt{1-\cot ^2(x)}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{3-5 x^2}{\sqrt{1-x^2} \left (1+x^2\right )} \, dx,x,\cot (x)\right )\\ &=\frac{1}{2} \cot (x) \sqrt{1-\cot ^2(x)}+\frac{5}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2}} \, dx,x,\cot (x)\right )-4 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \left (1+x^2\right )} \, dx,x,\cot (x)\right )\\ &=\frac{5}{2} \sin ^{-1}(\cot (x))+\frac{1}{2} \cot (x) \sqrt{1-\cot ^2(x)}-4 \operatorname{Subst}\left (\int \frac{1}{1+2 x^2} \, dx,x,\frac{\cot (x)}{\sqrt{1-\cot ^2(x)}}\right )\\ &=\frac{5}{2} \sin ^{-1}(\cot (x))-2 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \cot (x)}{\sqrt{1-\cot ^2(x)}}\right )+\frac{1}{2} \cot (x) \sqrt{1-\cot ^2(x)}\\ \end{align*}

Mathematica [B]  time = 0.390845, size = 123, normalized size = 2.28 \[ \frac{1}{2} \left (1-\cot ^2(x)\right )^{3/2} \sec ^2(2 x) \left (-\frac{1}{4} \sin (4 x)-4 \sqrt{2} \sin ^3(x) \sqrt{\cos (2 x)} \log \left (\sqrt{2} \cos (x)+\sqrt{\cos (2 x)}\right )+\sin ^3(x) \sqrt{-\cos (2 x)} \tan ^{-1}\left (\frac{\cos (x)}{\sqrt{-\cos (2 x)}}\right )+4 \sin ^3(x) \sqrt{\cos (2 x)} \tanh ^{-1}\left (\frac{\cos (x)}{\sqrt{\cos (2 x)}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cot[x]^2)^(3/2),x]

[Out]

((1 - Cot[x]^2)^(3/2)*Sec[2*x]^2*(ArcTan[Cos[x]/Sqrt[-Cos[2*x]]]*Sqrt[-Cos[2*x]]*Sin[x]^3 + 4*ArcTanh[Cos[x]/S
qrt[Cos[2*x]]]*Sqrt[Cos[2*x]]*Sin[x]^3 - 4*Sqrt[2]*Sqrt[Cos[2*x]]*Log[Sqrt[2]*Cos[x] + Sqrt[Cos[2*x]]]*Sin[x]^
3 - Sin[4*x]/4))/2

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Maple [A]  time = 0.041, size = 51, normalized size = 0.9 \begin{align*}{\frac{\cot \left ( x \right ) }{2}\sqrt{1- \left ( \cot \left ( x \right ) \right ) ^{2}}}+{\frac{5\,\arcsin \left ( \cot \left ( x \right ) \right ) }{2}}+2\,\sqrt{2}\arctan \left ({\frac{\sqrt{2}\sqrt{1- \left ( \cot \left ( x \right ) \right ) ^{2}}\cot \left ( x \right ) }{-1+ \left ( \cot \left ( x \right ) \right ) ^{2}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-cot(x)^2)^(3/2),x)

[Out]

1/2*cot(x)*(1-cot(x)^2)^(1/2)+5/2*arcsin(cot(x))+2*2^(1/2)*arctan(2^(1/2)*(1-cot(x)^2)^(1/2)/(-1+cot(x)^2)*cot
(x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-\cot \left (x\right )^{2} + 1\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-cot(x)^2 + 1)^(3/2), x)

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Fricas [B]  time = 1.98102, size = 317, normalized size = 5.87 \begin{align*} \frac{4 \, \sqrt{2} \arctan \left (\frac{\sqrt{\frac{\cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) - 1}} \sin \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 1}\right ) \sin \left (2 \, x\right ) + \sqrt{2} \sqrt{\frac{\cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) - 1}}{\left (\cos \left (2 \, x\right ) + 1\right )} - 5 \, \arctan \left (\frac{\sqrt{2} \sqrt{\frac{\cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) - 1}} \sin \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 1}\right ) \sin \left (2 \, x\right )}{2 \, \sin \left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*sqrt(2)*arctan(sqrt(cos(2*x)/(cos(2*x) - 1))*sin(2*x)/(cos(2*x) + 1))*sin(2*x) + sqrt(2)*sqrt(cos(2*x)/
(cos(2*x) - 1))*(cos(2*x) + 1) - 5*arctan(sqrt(2)*sqrt(cos(2*x)/(cos(2*x) - 1))*sin(2*x)/(cos(2*x) + 1))*sin(2
*x))/sin(2*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (1 - \cot ^{2}{\left (x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cot(x)**2)**(3/2),x)

[Out]

Integral((1 - cot(x)**2)**(3/2), x)

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Giac [B]  time = 1.39459, size = 347, normalized size = 6.43 \begin{align*} \frac{1}{4} \,{\left (5 \, \pi \mathrm{sgn}\left (\cos \left (x\right )\right ) - 4 \, \sqrt{2}{\left (\pi \mathrm{sgn}\left (\cos \left (x\right )\right ) + 2 \, \arctan \left (-\frac{{\left (\frac{{\left (\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}\right )}^{2}}{\cos \left (x\right )^{2}} - 4\right )} \cos \left (x\right )}{4 \,{\left (\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}\right )}}\right )\right )} + \frac{4 \, \sqrt{2}{\left (\frac{\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}}{\cos \left (x\right )} - \frac{4 \, \cos \left (x\right )}{\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}}\right )}}{{\left (\frac{\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}}{\cos \left (x\right )} - \frac{4 \, \cos \left (x\right )}{\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}}\right )}^{2} + 8} + 10 \, \arctan \left (-\frac{\sqrt{2}{\left (\frac{{\left (\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}\right )}^{2}}{\cos \left (x\right )^{2}} - 4\right )} \cos \left (x\right )}{4 \,{\left (\sqrt{2} \sqrt{-2 \, \cos \left (x\right )^{2} + 1} - \sqrt{2}\right )}}\right )\right )} \mathrm{sgn}\left (\sin \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(5*pi*sgn(cos(x)) - 4*sqrt(2)*(pi*sgn(cos(x)) + 2*arctan(-1/4*((sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2))^2
/cos(x)^2 - 4)*cos(x)/(sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2)))) + 4*sqrt(2)*((sqrt(2)*sqrt(-2*cos(x)^2 + 1)
- sqrt(2))/cos(x) - 4*cos(x)/(sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2)))/(((sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqr
t(2))/cos(x) - 4*cos(x)/(sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2)))^2 + 8) + 10*arctan(-1/4*sqrt(2)*((sqrt(2)*s
qrt(-2*cos(x)^2 + 1) - sqrt(2))^2/cos(x)^2 - 4)*cos(x)/(sqrt(2)*sqrt(-2*cos(x)^2 + 1) - sqrt(2))))*sgn(sin(x))