3.6 \(\int x^3 \cot ^2(a+b x) \, dx\)

Optimal. Leaf size=97 \[ \frac {3 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^4}-\frac {3 i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {3 x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {x^3 \cot (a+b x)}{b}-\frac {i x^3}{b}-\frac {x^4}{4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3720, 3717, 2190, 2531, 2282, 6589, 30} \[ -\frac {3 i x \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac {3 \text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^4}+\frac {3 x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {x^3 \cot (a+b x)}{b}-\frac {i x^3}{b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Cot[a + b*x]^2,x]

[Out]

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

Rule 30

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

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 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 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 ^2(a+b x) \, dx &=-\frac {x^3 \cot (a+b x)}{b}+\frac {3 \int x^2 \cot (a+b x) \, dx}{b}-\int x^3 \, dx\\ &=-\frac {i x^3}{b}-\frac {x^4}{4}-\frac {x^3 \cot (a+b x)}{b}-\frac {(6 i) \int \frac {e^{2 i (a+b x)} x^2}{1-e^{2 i (a+b x)}} \, dx}{b}\\ &=-\frac {i x^3}{b}-\frac {x^4}{4}-\frac {x^3 \cot (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {6 \int x \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {i x^3}{b}-\frac {x^4}{4}-\frac {x^3 \cot (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {(3 i) \int \text {Li}_2\left (e^{2 i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {i x^3}{b}-\frac {x^4}{4}-\frac {x^3 \cot (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^4}\\ &=-\frac {i x^3}{b}-\frac {x^4}{4}-\frac {x^3 \cot (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {3 \text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.92, size = 171, normalized size = 1.76 \[ \frac {-\frac {2 i b^3 x^3}{-1+e^{2 i a}}+3 b^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+3 b^2 x^2 \log \left (1+e^{-i (a+b x)}\right )+6 i b x \text {Li}_2\left (-e^{-i (a+b x)}\right )+6 i b x \text {Li}_2\left (e^{-i (a+b x)}\right )+6 \text {Li}_3\left (-e^{-i (a+b x)}\right )+6 \text {Li}_3\left (e^{-i (a+b x)}\right )}{b^4}+\frac {x^3 \csc (a) \sin (b x) \csc (a+b x)}{b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cot[a + b*x]^2,x]

[Out]

-1/4*x^4 + (((-2*I)*b^3*x^3)/(-1 + E^((2*I)*a)) + 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*P
olyLog[3, -E^((-I)*(a + b*x))] + 6*PolyLog[3, E^((-I)*(a + b*x))])/b^4 + (x^3*Csc[a]*Csc[a + b*x]*Sin[b*x])/b

________________________________________________________________________________________

fricas [C]  time = 0.69, size = 372, normalized size = 3.84 \[ -\frac {b^{4} x^{4} \sin \left (2 \, b x + 2 \, a\right ) + 4 \, b^{3} x^{3} \cos \left (2 \, b x + 2 \, a\right ) + 4 \, b^{3} x^{3} + 6 i \, b x {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 i \, b x {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) - 3 \, {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 3 \, {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right )}{4 \, b^{4} \sin \left (2 \, b x + 2 \, a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cot(b*x+a)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \cot \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cot(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^3*cot(b*x + a)^2, x)

________________________________________________________________________________________

maple [B]  time = 0.95, size = 231, normalized size = 2.38 \[ -\frac {x^{4}}{4}-\frac {2 i x^{3}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {6 i a^{2} x}{b^{3}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b^{2}}-\frac {6 i \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {6 \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {3 \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{4}}-\frac {6 i \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {6 \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{4}}-\frac {6 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {2 i x^{3}}{b}+\frac {4 i a^{3}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cot(b*x+a)^2,x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.59, size = 953, normalized size = 9.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cot(b*x+a)^2,x, algorithm="maxima")

[Out]

1/2*(2*(b*x + a + 1/tan(b*x + a))*a^3 - 3*((b*x + a)^2*cos(2*b*x + 2*a)^2 + (b*x + a)^2*sin(2*b*x + 2*a)^2 - 2
*(b*x + a)^2*cos(2*b*x + 2*a) + (b*x + a)^2 - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) +
1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*co
s(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 4*(b*x + a)*sin(2*b*x + 2*a))*
a^2/(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1) + 2*(-I*(b*x + a)^4 + 4*I*(b*x + a)^3*a
- (12*(b*x + a)^2 - 24*(b*x + a)*a - 12*((b*x + a)^2 - 2*(b*x + a)*a)*cos(2*b*x + 2*a) - (12*I*(b*x + a)^2 -
24*I*(b*x + a)*a)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (12*(b*x + a)^2 - 24*(b*x + a)*a
- 12*((b*x + a)^2 - 2*(b*x + a)*a)*cos(2*b*x + 2*a) + (-12*I*(b*x + a)^2 + 24*I*(b*x + a)*a)*sin(2*b*x + 2*a)
)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (I*(b*x + a)^4 - 4*(b*x + a)^3*(I*a + 2) + 24*(b*x + a)^2*a)*cos(
2*b*x + 2*a) - 24*(b*x*cos(2*b*x + 2*a) + I*b*x*sin(2*b*x + 2*a) - b*x)*dilog(-e^(I*b*x + I*a)) - 24*(b*x*cos(
2*b*x + 2*a) + I*b*x*sin(2*b*x + 2*a) - b*x)*dilog(e^(I*b*x + I*a)) + (6*I*(b*x + a)^2 - 12*I*(b*x + a)*a + (-
6*I*(b*x + a)^2 + 12*I*(b*x + a)*a)*cos(2*b*x + 2*a) + 6*((b*x + a)^2 - 2*(b*x + a)*a)*sin(2*b*x + 2*a))*log(c
os(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (6*I*(b*x + a)^2 - 12*I*(b*x + a)*a + (-6*I*(b*x + a)^2
+ 12*I*(b*x + a)*a)*cos(2*b*x + 2*a) + 6*((b*x + a)^2 - 2*(b*x + a)*a)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 +
sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + (-24*I*cos(2*b*x + 2*a) + 24*sin(2*b*x + 2*a) + 24*I)*polylog(3, -e^(I
*b*x + I*a)) + (-24*I*cos(2*b*x + 2*a) + 24*sin(2*b*x + 2*a) + 24*I)*polylog(3, e^(I*b*x + I*a)) - ((b*x + a)^
4 - (b*x + a)^3*(4*a - 8*I) - 24*I*(b*x + a)^2*a)*sin(2*b*x + 2*a))/(-4*I*cos(2*b*x + 2*a) + 4*sin(2*b*x + 2*a
) + 4*I))/b^4

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\mathrm {cot}\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cot(a + b*x)^2,x)

[Out]

int(x^3*cot(a + b*x)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \cot ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cot(b*x+a)**2,x)

[Out]

Integral(x**3*cot(a + b*x)**2, x)

________________________________________________________________________________________