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

Optimal. Leaf size=101 \[ -\frac{3 i x^2 \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i \text{PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4}+\frac{x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{i x^4}{4} \]

[Out]

(-I/4)*x^4 + (x^3*Log[1 - E^((2*I)*(a + b*x))])/b - (((3*I)/2)*x^2*PolyLog[2, E^((2*I)*(a + b*x))])/b^2 + (3*x
*PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^3) + (((3*I)/4)*PolyLog[4, E^((2*I)*(a + b*x))])/b^4

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Rubi [A]  time = 0.170506, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3717, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 i x^2 \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i \text{PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4}+\frac{x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{i x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Cot[a + b*x],x]

[Out]

(-I/4)*x^4 + (x^3*Log[1 - E^((2*I)*(a + b*x))])/b - (((3*I)/2)*x^2*PolyLog[2, E^((2*I)*(a + b*x))])/b^2 + (3*x
*PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^3) + (((3*I)/4)*PolyLog[4, E^((2*I)*(a + b*x))])/b^4

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x^3 \cot (a+b x) \, dx &=-\frac{i x^4}{4}-2 i \int \frac{e^{2 i (a+b x)} x^3}{1-e^{2 i (a+b x)}} \, dx\\ &=-\frac{i x^4}{4}+\frac{x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 \int x^2 \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=-\frac{i x^4}{4}+\frac{x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 i x^2 \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac{(3 i) \int x \text{Li}_2\left (e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{i x^4}{4}+\frac{x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 i x^2 \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 \int \text{Li}_3\left (e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=-\frac{i x^4}{4}+\frac{x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 i x^2 \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{4 b^4}\\ &=-\frac{i x^4}{4}+\frac{x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 i x^2 \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i \text{Li}_4\left (e^{2 i (a+b x)}\right )}{4 b^4}\\ \end{align*}

Mathematica [A]  time = 0.679236, size = 184, normalized size = 1.82 \[ \frac{12 i b^2 x^2 \text{PolyLog}\left (2,-e^{-i (a+b x)}\right )+12 i b^2 x^2 \text{PolyLog}\left (2,e^{-i (a+b x)}\right )+24 b x \text{PolyLog}\left (3,-e^{-i (a+b x)}\right )+24 b x \text{PolyLog}\left (3,e^{-i (a+b x)}\right )-24 i \text{PolyLog}\left (4,-e^{-i (a+b x)}\right )-24 i \text{PolyLog}\left (4,e^{-i (a+b x)}\right )+4 b^3 x^3 \log \left (1-e^{-i (a+b x)}\right )+4 b^3 x^3 \log \left (1+e^{-i (a+b x)}\right )+i b^4 x^4}{4 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cot[a + b*x],x]

[Out]

(I*b^4*x^4 + 4*b^3*x^3*Log[1 - E^((-I)*(a + b*x))] + 4*b^3*x^3*Log[1 + E^((-I)*(a + b*x))] + (12*I)*b^2*x^2*Po
lyLog[2, -E^((-I)*(a + b*x))] + (12*I)*b^2*x^2*PolyLog[2, E^((-I)*(a + b*x))] + 24*b*x*PolyLog[3, -E^((-I)*(a
+ b*x))] + 24*b*x*PolyLog[3, E^((-I)*(a + b*x))] - (24*I)*PolyLog[4, -E^((-I)*(a + b*x))] - (24*I)*PolyLog[4,
E^((-I)*(a + b*x))])/(4*b^4)

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Maple [B]  time = 0.258, size = 240, normalized size = 2.4 \begin{align*} -{\frac{i}{4}}{x}^{4}+{\frac{\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{3}}{b}}+{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ){x}^{3}}{b}}-{\frac{2\,i{a}^{3}x}{{b}^{3}}}-{\frac{3\,i{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{{b}^{2}}}-{\frac{3\,i{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{{b}^{2}}}+{\frac{6\,i{\it polylog} \left ( 4,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}+{\frac{\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){a}^{3}}{{b}^{4}}}+{\frac{6\,i{\it polylog} \left ( 4,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-{\frac{{\frac{3\,i}{2}}{a}^{4}}{{b}^{4}}}+2\,{\frac{{a}^{3}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-{\frac{{a}^{3}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{4}}}+6\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{3}}}+6\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cot(b*x+a),x)

[Out]

-1/4*I*x^4+1/b*ln(1-exp(I*(b*x+a)))*x^3+1/b*ln(exp(I*(b*x+a))+1)*x^3-2*I/b^3*a^3*x-3*I/b^2*polylog(2,-exp(I*(b
*x+a)))*x^2-3*I/b^2*polylog(2,exp(I*(b*x+a)))*x^2+6*I/b^4*polylog(4,exp(I*(b*x+a)))+1/b^4*ln(1-exp(I*(b*x+a)))
*a^3+6*I/b^4*polylog(4,-exp(I*(b*x+a)))-3/2*I/b^4*a^4+2/b^4*a^3*ln(exp(I*(b*x+a)))-1/b^4*a^3*ln(exp(I*(b*x+a))
-1)+6/b^3*polylog(3,exp(I*(b*x+a)))*x+6/b^3*polylog(3,-exp(I*(b*x+a)))*x

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Maxima [B]  time = 1.41269, size = 528, normalized size = 5.23 \begin{align*} -\frac{i \,{\left (b x + a\right )}^{4} - 4 i \,{\left (b x + a\right )}^{3} a + 6 i \,{\left (b x + a\right )}^{2} a^{2} + 4 \, a^{3} \log \left (\sin \left (b x + a\right )\right ) - 24 \, b x{\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) - 24 \, b x{\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) -{\left (4 i \,{\left (b x + a\right )}^{3} - 12 i \,{\left (b x + a\right )}^{2} a + 12 i \,{\left (b x + a\right )} a^{2}\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) -{\left (-4 i \,{\left (b x + a\right )}^{3} + 12 i \,{\left (b x + a\right )}^{2} a - 12 i \,{\left (b x + a\right )} a^{2}\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) -{\left (-12 i \,{\left (b x + a\right )}^{2} + 24 i \,{\left (b x + a\right )} a - 12 i \, a^{2}\right )}{\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) -{\left (-12 i \,{\left (b x + a\right )}^{2} + 24 i \,{\left (b x + a\right )} a - 12 i \, a^{2}\right )}{\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) - 2 \,{\left ({\left (b x + a\right )}^{3} - 3 \,{\left (b x + a\right )}^{2} a + 3 \,{\left (b x + a\right )} a^{2}\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 ) - 2 \,{\left ({\left (b x + a\right )}^{3} - 3 \,{\left (b x + a\right )}^{2} a + 3 \,{\left (b x + a\right )} a^{2}\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 ) - 24 i \,{\rm Li}_{4}(-e^{\left (i \, b x + i \, a\right )}) - 24 i \,{\rm Li}_{4}(e^{\left (i \, b x + i \, a\right )})}{4 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(I*(b*x + a)^4 - 4*I*(b*x + a)^3*a + 6*I*(b*x + a)^2*a^2 + 4*a^3*log(sin(b*x + a)) - 24*b*x*polylog(3, -e
^(I*b*x + I*a)) - 24*b*x*polylog(3, e^(I*b*x + I*a)) - (4*I*(b*x + a)^3 - 12*I*(b*x + a)^2*a + 12*I*(b*x + a)*
a^2)*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (-4*I*(b*x + a)^3 + 12*I*(b*x + a)^2*a - 12*I*(b*x + a)*a^2)*ar
ctan2(sin(b*x + a), -cos(b*x + a) + 1) - (-12*I*(b*x + a)^2 + 24*I*(b*x + a)*a - 12*I*a^2)*dilog(-e^(I*b*x + I
*a)) - (-12*I*(b*x + a)^2 + 24*I*(b*x + a)*a - 12*I*a^2)*dilog(e^(I*b*x + I*a)) - 2*((b*x + a)^3 - 3*(b*x + a)
^2*a + 3*(b*x + a)*a^2)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - 2*((b*x + a)^3 - 3*(b*x +
a)^2*a + 3*(b*x + a)*a^2)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - 24*I*polylog(4, -e^(I*b*
x + I*a)) - 24*I*polylog(4, e^(I*b*x + I*a)))/b^4

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Fricas [C]  time = 1.8394, size = 842, normalized size = 8.34 \begin{align*} \frac{-6 i \, b^{2} x^{2}{\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 6 i \, b^{2} x^{2}{\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - 4 \, a^{3} \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 ) - 4 \, a^{3} \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 ) + 6 \, b x{\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 6 \, b x{\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 4 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 4 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 i \,{\rm polylog}\left (4, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - 3 i \,{\rm polylog}\left (4, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right )}{8 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(-6*I*b^2*x^2*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) + 6*I*b^2*x^2*dilog(cos(2*b*x + 2*a) - I*sin(2*
b*x + 2*a)) - 4*a^3*log(-1/2*cos(2*b*x + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2) - 4*a^3*log(-1/2*cos(2*b*x + 2*a
) - 1/2*I*sin(2*b*x + 2*a) + 1/2) + 6*b*x*polylog(3, cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) + 6*b*x*polylog(3,
cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)) + 4*(b^3*x^3 + a^3)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1) +
4*(b^3*x^3 + a^3)*log(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1) + 3*I*polylog(4, cos(2*b*x + 2*a) + I*sin(2*
b*x + 2*a)) - 3*I*polylog(4, cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cot(b*x+a),x)

[Out]

Integral(x**3*cot(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*cot(b*x + a), x)