3.13 \(\int x \cot ^3(a+b x) \, dx\)

Optimal. Leaf size=91 \[ \frac{i \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac{\cot (a+b x)}{2 b^2}-\frac{x \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{x \cot ^2(a+b x)}{2 b}-\frac{x}{2 b}+\frac{i x^2}{2} \]

[Out]

-x/(2*b) + (I/2)*x^2 - Cot[a + b*x]/(2*b^2) - (x*Cot[a + b*x]^2)/(2*b) - (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 = 0.106975, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {3720, 3473, 8, 3717, 2190, 2279, 2391} \[ \frac{i \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac{\cot (a+b x)}{2 b^2}-\frac{x \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{x \cot ^2(a+b x)}{2 b}-\frac{x}{2 b}+\frac{i x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*Cot[a + b*x]^3,x]

[Out]

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

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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 ^3(a+b x) \, dx &=-\frac{x \cot ^2(a+b x)}{2 b}+\frac{\int \cot ^2(a+b x) \, dx}{2 b}-\int x \cot (a+b x) \, dx\\ &=\frac{i x^2}{2}-\frac{\cot (a+b x)}{2 b^2}-\frac{x \cot ^2(a+b x)}{2 b}+2 i \int \frac{e^{2 i (a+b x)} x}{1-e^{2 i (a+b x)}} \, dx-\frac{\int 1 \, dx}{2 b}\\ &=-\frac{x}{2 b}+\frac{i x^2}{2}-\frac{\cot (a+b x)}{2 b^2}-\frac{x \cot ^2(a+b x)}{2 b}-\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{x}{2 b}+\frac{i x^2}{2}-\frac{\cot (a+b x)}{2 b^2}-\frac{x \cot ^2(a+b x)}{2 b}-\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{x}{2 b}+\frac{i x^2}{2}-\frac{\cot (a+b x)}{2 b^2}-\frac{x \cot ^2(a+b x)}{2 b}-\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 [A]  time = 4.08289, size = 179, normalized size = 1.97 \[ \frac{i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-b^2 x^2 \cot (a)+b^2 x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt{\sec ^2(a)}-b x \csc ^2(a+b x)-2 b x \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+\csc (a) \sin (b x) \csc (a+b x)+2 \tan ^{-1}(\tan (a)) \left (-\log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+\log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )+i b x\right )-i \pi b x-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))}{2 b^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*Cot[a + b*x]^3,x]

[Out]

((-I)*b*Pi*x - b^2*x^2*Cot[a] - b*x*Csc[a + b*x]^2 - Pi*Log[1 + E^((-2*I)*b*x)] - 2*b*x*Log[1 - E^((2*I)*(b*x
+ ArcTan[Tan[a]]))] + Pi*Log[Cos[b*x]] + 2*ArcTan[Tan[a]]*(I*b*x - Log[1 - E^((2*I)*(b*x + 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] + Csc[a]*Csc[a + b*x]*Sin[b*x])/(2*b^2)

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Maple [B]  time = 0.145, size = 197, normalized size = 2.2 \begin{align*}{\frac{i}{2}}{x}^{2}+{\frac{2\,bx{{\rm e}^{2\,i \left ( bx+a \right ) }}-i{{\rm e}^{2\,i \left ( bx+a \right ) }}+i}{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{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)^3,x)

[Out]

1/2*I*x^2+(2*b*x*exp(2*I*(b*x+a))-I*exp(2*I*(b*x+a))+I)/b^2/(exp(2*I*(b*x+a))-1)^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(exp(I*(b*x+a)))

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Maxima [B]  time = 1.53682, size = 798, normalized size = 8.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cot(b*x+a)^3,x, algorithm="maxima")

[Out]

(b^2*x^2*cos(4*b*x + 4*a) + I*b^2*x^2*sin(4*b*x + 4*a) + b^2*x^2 - (2*b*x*cos(4*b*x + 4*a) - 4*b*x*cos(2*b*x +
2*a) + 2*I*b*x*sin(4*b*x + 4*a) - 4*I*b*x*sin(2*b*x + 2*a) + 2*b*x)*arctan2(sin(b*x + a), cos(b*x + a) + 1) +
(2*b*x*cos(4*b*x + 4*a) - 4*b*x*cos(2*b*x + 2*a) + 2*I*b*x*sin(4*b*x + 4*a) - 4*I*b*x*sin(2*b*x + 2*a) + 2*b*
x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 2*(b^2*x^2 + 2*I*b*x + 1)*cos(2*b*x + 2*a) + (2*cos(4*b*x + 4*a)
- 4*cos(2*b*x + 2*a) + 2*I*sin(4*b*x + 4*a) - 4*I*sin(2*b*x + 2*a) + 2)*dilog(-e^(I*b*x + I*a)) + (2*cos(4*b*
x + 4*a) - 4*cos(2*b*x + 2*a) + 2*I*sin(4*b*x + 4*a) - 4*I*sin(2*b*x + 2*a) + 2)*dilog(e^(I*b*x + I*a)) - (-I*
b*x*cos(4*b*x + 4*a) + 2*I*b*x*cos(2*b*x + 2*a) + b*x*sin(4*b*x + 4*a) - 2*b*x*sin(2*b*x + 2*a) - I*b*x)*log(c
os(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (-I*b*x*cos(4*b*x + 4*a) + 2*I*b*x*cos(2*b*x + 2*a) + b
*x*sin(4*b*x + 4*a) - 2*b*x*sin(2*b*x + 2*a) - I*b*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1
) - (2*I*b^2*x^2 - 4*b*x + 2*I)*sin(2*b*x + 2*a) + 2)/(-2*I*b^2*cos(4*b*x + 4*a) + 4*I*b^2*cos(2*b*x + 2*a) +
2*b^2*sin(4*b*x + 4*a) - 4*b^2*sin(2*b*x + 2*a) - 2*I*b^2)

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Fricas [B]  time = 1.77028, size = 761, normalized size = 8.36 \begin{align*} \frac{4 \, b x +{\left (i \, \cos \left (2 \, b x + 2 \, a\right ) - i\right )}{\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) +{\left (-i \, \cos \left (2 \, b x + 2 \, a\right ) + i\right )}{\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 \,{\left (a \cos \left (2 \, b x + 2 \, a\right ) - a\right )} \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 (a \cos \left (2 \, b x + 2 \, a\right ) - a\right )} \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 -{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + 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 -{\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + a\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, \sin \left (2 \, b x + 2 \, a\right )}{4 \,{\left (b^{2} \cos \left (2 \, b x + 2 \, a\right ) - b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cot(b*x+a)^3,x, algorithm="fricas")

[Out]

1/4*(4*b*x + (I*cos(2*b*x + 2*a) - I)*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) + (-I*cos(2*b*x + 2*a) + I)
*dilog(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)) + 2*(a*cos(2*b*x + 2*a) - a)*log(-1/2*cos(2*b*x + 2*a) + 1/2*I*s
in(2*b*x + 2*a) + 1/2) + 2*(a*cos(2*b*x + 2*a) - a)*log(-1/2*cos(2*b*x + 2*a) - 1/2*I*sin(2*b*x + 2*a) + 1/2)
+ 2*(b*x - (b*x + a)*cos(2*b*x + 2*a) + a)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1) + 2*(b*x - (b*x + a
)*cos(2*b*x + 2*a) + a)*log(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1) + 2*sin(2*b*x + 2*a))/(b^2*cos(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 ^{3}{\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)**3,x)

[Out]

Integral(x*cot(a + b*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \cot \left (b x + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cot(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x*cot(b*x + a)^3, x)