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

Optimal. Leaf size=74 \[ \frac {\text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}-\frac {i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}+\frac {x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3717, 2190, 2531, 2282, 6589} \[ -\frac {i x \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}+\frac {\text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}+\frac {x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cot[a + b*x],x]

[Out]

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

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

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Mathematica [A]  time = 0.46, size = 136, normalized size = 1.84 \[ \frac {3 b^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+3 b^2 x^2 \log \left (1+e^{-i (a+b x)}\right )+6 i b x \text {Li}_2\left (-e^{-i (a+b x)}\right )+6 i b x \text {Li}_2\left (e^{-i (a+b x)}\right )+6 \text {Li}_3\left (-e^{-i (a+b x)}\right )+6 \text {Li}_3\left (e^{-i (a+b x)}\right )+i b^3 x^3}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cot[a + b*x],x]

[Out]

(I*b^3*x^3 + 3*b^2*x^2*Log[1 - E^((-I)*(a + b*x))] + 3*b^2*x^2*Log[1 + E^((-I)*(a + b*x))] + (6*I)*b*x*PolyLog
[2, -E^((-I)*(a + b*x))] + (6*I)*b*x*PolyLog[2, E^((-I)*(a + b*x))] + 6*PolyLog[3, -E^((-I)*(a + b*x))] + 6*Po
lyLog[3, E^((-I)*(a + b*x))])/(3*b^3)

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fricas [C]  time = 0.57, size = 244, normalized size = 3.30 \[ \frac {-2 i \, b x {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 i \, b x {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right )}{4 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cot \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*cot(b*x + a), x)

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maple [B]  time = 0.80, size = 198, normalized size = 2.68 \[ -\frac {i x^{3}}{3}+\frac {2 i a^{2} x}{b^{2}}-\frac {\ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}+\frac {4 i a^{3}}{3 b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}-\frac {2 i \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {2 i \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.49, size = 257, normalized size = 3.47 \[ -\frac {2 i \, {\left (b x + a\right )}^{3} - 6 i \, {\left (b x + a\right )}^{2} a + 12 i \, b x {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + 12 i \, b x {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) - 6 \, a^{2} \log \left (\sin \left (b x + a\right )\right ) - {\left (6 i \, {\left (b x + a\right )}^{2} - 12 i \, {\left (b x + a\right )} a\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - {\left (-6 i \, {\left (b x + a\right )}^{2} + 12 i \, {\left (b x + a\right )} a\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - 12 \, {\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) - 12 \, {\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )})}{6 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {cot}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cot(a + b*x),x)

[Out]

int(x^2*cot(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cot {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*cot(a + b*x), x)

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