3.7 \(\int x^2 \cot ^2(a+b x) \, dx\)

Optimal. Leaf size=74 \[ -\frac{i \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac{2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac{x^2 \cot (a+b x)}{b}-\frac{i x^2}{b}-\frac{x^3}{3} \]

[Out]

((-I)*x^2)/b - x^3/3 - (x^2*Cot[a + b*x])/b + (2*x*Log[1 - E^((2*I)*(a + b*x))])/b^2 - (I*PolyLog[2, E^((2*I)*
(a + b*x))])/b^3

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Rubi [A]  time = 0.119014, antiderivative size = 74, 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 = {3720, 3717, 2190, 2279, 2391, 30} \[ -\frac{i \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac{2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac{x^2 \cot (a+b x)}{b}-\frac{i x^2}{b}-\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cot[a + b*x]^2,x]

[Out]

((-I)*x^2)/b - x^3/3 - (x^2*Cot[a + b*x])/b + (2*x*Log[1 - E^((2*I)*(a + b*x))])/b^2 - (I*PolyLog[2, E^((2*I)*
(a + b*x))])/b^3

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^2 \cot ^2(a+b x) \, dx &=-\frac{x^2 \cot (a+b x)}{b}+\frac{2 \int x \cot (a+b x) \, dx}{b}-\int x^2 \, dx\\ &=-\frac{i x^2}{b}-\frac{x^3}{3}-\frac{x^2 \cot (a+b x)}{b}-\frac{(4 i) \int \frac{e^{2 i (a+b x)} x}{1-e^{2 i (a+b x)}} \, dx}{b}\\ &=-\frac{i x^2}{b}-\frac{x^3}{3}-\frac{x^2 \cot (a+b x)}{b}+\frac{2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac{2 \int \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{i x^2}{b}-\frac{x^3}{3}-\frac{x^2 \cot (a+b x)}{b}+\frac{2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{b^3}\\ &=-\frac{i x^2}{b}-\frac{x^3}{3}-\frac{x^2 \cot (a+b x)}{b}+\frac{2 x \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac{i \text{Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}\\ \end{align*}

Mathematica [B]  time = 4.59087, size = 153, normalized size = 2.07 \[ \frac{-i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-b^2 x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt{\sec ^2(a)}+i b x \left (\pi -2 \tan ^{-1}(\tan (a))\right )+2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )+\pi \log \left (1+e^{-2 i b x}\right )-\pi \log (\cos (b x))}{b^3}+\frac{x^2 \csc (a) \sin (b x) \csc (a+b x)}{b}-\frac{x^3}{3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*Cot[a + b*x]^2,x]

[Out]

-x^3/3 + (I*b*x*(Pi - 2*ArcTan[Tan[a]]) + Pi*Log[1 + E^((-2*I)*b*x)] + 2*(b*x + ArcTan[Tan[a]])*Log[1 - E^((2*
I)*(b*x + ArcTan[Tan[a]]))] - Pi*Log[Cos[b*x]] - 2*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] - I*PolyLog[2
, E^((2*I)*(b*x + ArcTan[Tan[a]]))] - b^2*E^(I*ArcTan[Tan[a]])*x^2*Cot[a]*Sqrt[Sec[a]^2])/b^3 + (x^2*Csc[a]*Cs
c[a + b*x]*Sin[b*x])/b

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Maple [B]  time = 0.134, size = 183, normalized size = 2.5 \begin{align*} -{\frac{{x}^{3}}{3}}-{\frac{2\,i{x}^{2}}{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) }}-{\frac{2\,i{x}^{2}}{b}}-{\frac{4\,iax}{{b}^{2}}}-{\frac{2\,i{a}^{2}}{{b}^{3}}}+2\,{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{2}}}-{\frac{2\,i{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{3}}}-{\frac{2\,i{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-2\,{\frac{a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{3}}}+4\,{\frac{a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cot(b*x+a)^2,x)

[Out]

-1/3*x^3-2*I*x^2/b/(exp(2*I*(b*x+a))-1)-2*I*x^2/b-4*I/b^2*a*x-2*I/b^3*a^2+2/b^2*ln(exp(I*(b*x+a))+1)*x-2*I/b^3
*polylog(2,-exp(I*(b*x+a)))+2/b^2*ln(1-exp(I*(b*x+a)))*x+2/b^3*ln(1-exp(I*(b*x+a)))*a-2*I/b^3*polylog(2,exp(I*
(b*x+a)))-2/b^3*a*ln(exp(I*(b*x+a))-1)+4/b^3*a*ln(exp(I*(b*x+a)))

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Maxima [B]  time = 2.16097, size = 518, normalized size = 7. \begin{align*} \frac{-i \, b^{3} x^{3} + 6 \,{\left (b x \cos \left (2 \, b x + 2 \, a\right ) + i \, b x \sin \left (2 \, b x + 2 \, a\right ) - b x\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 6 \,{\left (b x \cos \left (2 \, b x + 2 \, a\right ) + i \, b x \sin \left (2 \, b x + 2 \, a\right ) - b x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) +{\left (i \, b^{3} x^{3} - 6 \, b^{2} x^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) - 6 \,{\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) - 1\right )}{\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 6 \,{\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) - 1\right )}{\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) +{\left (-3 i \, b x \cos \left (2 \, b x + 2 \, a\right ) + 3 \, b x \sin \left (2 \, b x + 2 \, a\right ) + 3 i \, b x\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) +{\left (-3 i \, b x \cos \left (2 \, b x + 2 \, a\right ) + 3 \, b x \sin \left (2 \, b x + 2 \, a\right ) + 3 i \, b x\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) -{\left (b^{3} x^{3} + 6 i \, b^{2} x^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )}{-3 i \, b^{3} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, b^{3} \sin \left (2 \, b x + 2 \, a\right ) + 3 i \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cot(b*x+a)^2,x, algorithm="maxima")

[Out]

(-I*b^3*x^3 + 6*(b*x*cos(2*b*x + 2*a) + I*b*x*sin(2*b*x + 2*a) - b*x)*arctan2(sin(b*x + a), cos(b*x + a) + 1)
- 6*(b*x*cos(2*b*x + 2*a) + I*b*x*sin(2*b*x + 2*a) - b*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (I*b^3*x^
3 - 6*b^2*x^2)*cos(2*b*x + 2*a) - 6*(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) - 1)*dilog(-e^(I*b*x + I*a)) - 6*(c
os(2*b*x + 2*a) + I*sin(2*b*x + 2*a) - 1)*dilog(e^(I*b*x + I*a)) + (-3*I*b*x*cos(2*b*x + 2*a) + 3*b*x*sin(2*b*
x + 2*a) + 3*I*b*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (-3*I*b*x*cos(2*b*x + 2*a) + 3
*b*x*sin(2*b*x + 2*a) + 3*I*b*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - (b^3*x^3 + 6*I*b^
2*x^2)*sin(2*b*x + 2*a))/(-3*I*b^3*cos(2*b*x + 2*a) + 3*b^3*sin(2*b*x + 2*a) + 3*I*b^3)

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Fricas [B]  time = 1.78521, size = 749, normalized size = 10.12 \begin{align*} -\frac{2 \, b^{3} x^{3} \sin \left (2 \, b x + 2 \, a\right ) + 6 \, b^{2} x^{2} \cos \left (2 \, b x + 2 \, a\right ) + 6 \, b^{2} x^{2} + 6 \, a \log \left (-\frac{1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac{1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac{1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) + 6 \, a \log \left (-\frac{1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac{1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac{1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \,{\left (b x + a\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \,{\left (b x + a\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) + 3 i \,{\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 3 i \,{\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right )}{6 \, b^{3} \sin \left (2 \, b x + 2 \, a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cot(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/6*(2*b^3*x^3*sin(2*b*x + 2*a) + 6*b^2*x^2*cos(2*b*x + 2*a) + 6*b^2*x^2 + 6*a*log(-1/2*cos(2*b*x + 2*a) + 1/
2*I*sin(2*b*x + 2*a) + 1/2)*sin(2*b*x + 2*a) + 6*a*log(-1/2*cos(2*b*x + 2*a) - 1/2*I*sin(2*b*x + 2*a) + 1/2)*s
in(2*b*x + 2*a) - 6*(b*x + a)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1)*sin(2*b*x + 2*a) - 6*(b*x + a)*l
og(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1)*sin(2*b*x + 2*a) + 3*I*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2
*a))*sin(2*b*x + 2*a) - 3*I*dilog(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a))*sin(2*b*x + 2*a))/(b^3*sin(2*b*x + 2*
a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cot(b*x+a)**2,x)

[Out]

Integral(x**2*cot(a + b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \cot \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cot(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^2*cot(b*x + a)^2, x)