3.3 \(\int x \cot (a+b x) \, dx\)

Optimal. Leaf size=53 \[ -\frac{i \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}+\frac{x \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{i x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.0849125, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3717, 2190, 2279, 2391} \[ -\frac{i \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}+\frac{x \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{i x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*Cot[a + b*x],x]

[Out]

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

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]

Rubi steps

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

Mathematica [B]  time = 3.555, size = 135, normalized size = 2.55 \[ \frac{1}{2} \left (-\frac{i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-i b x \left (\pi -2 \tan ^{-1}(\tan (a))\right )-2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))}{b^2}+x^2 \cot (a)-x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt{\sec ^2(a)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*Cot[a + b*x],x]

[Out]

(x^2*Cot[a] - ((-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*P
olyLog[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))])/b^2 - E^(I*ArcTan[Tan[a]])*x^2*Cot[a]*Sqrt[Sec[a]^2])/2

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Maple [B]  time = 0.231, size = 150, normalized size = 2.8 \begin{align*} -{\frac{i}{2}}{x}^{2}-{\frac{2\,iax}{b}}-{\frac{i{a}^{2}}{{b}^{2}}}+{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}-{\frac{i{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+{\frac{\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-{\frac{i{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{2}}}+2\,{\frac{a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.48665, size = 189, normalized size = 3.57 \begin{align*} \frac{-i \, b^{2} x^{2} + 2 i \, b x \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 2 i \, b x \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + b x \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 ) + b x \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 ) - 2 i \,{\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 2 i \,{\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.76379, size = 486, normalized size = 9.17 \begin{align*} -\frac{2 \, 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 ) + 2 \, 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 ) - 2 \,{\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 ) - 2 \,{\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 ) + i \,{\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - i \,{\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a*log(-1/2*cos(2*b*x + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2) + 2*a*log(-1/2*cos(2*b*x + 2*a) - 1/2*I*si
n(2*b*x + 2*a) + 1/2) - 2*(b*x + a)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1) - 2*(b*x + a)*log(-cos(2*b
*x + 2*a) - I*sin(2*b*x + 2*a) + 1) + I*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) - I*dilog(cos(2*b*x + 2*a
) - I*sin(2*b*x + 2*a)))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*cot(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*cot(b*x + a), x)