3.162 \(\int x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx\)

Optimal. Leaf size=154 \[ \frac{i x \text{PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{\text{PolyLog}\left (4,i c e^{2 i a+2 i b x}\right )}{8 b^3}+\frac{x^2 \text{PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac{b x^4}{12} \]

[Out]

(b*x^4)/12 + (x^3*ArcCot[c + (1 + I*c)*Tan[a + b*x]])/3 + (I/6)*x^3*Log[1 - I*c*E^((2*I)*a + (2*I)*b*x)] + (x^
2*PolyLog[2, I*c*E^((2*I)*a + (2*I)*b*x)])/(4*b) + ((I/4)*x*PolyLog[3, I*c*E^((2*I)*a + (2*I)*b*x)])/b^2 - Pol
yLog[4, I*c*E^((2*I)*a + (2*I)*b*x)]/(8*b^3)

________________________________________________________________________________________

Rubi [A]  time = 0.252764, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5172, 2184, 2190, 2531, 6609, 2282, 6589} \[ \frac{i x \text{PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{\text{PolyLog}\left (4,i c e^{2 i a+2 i b x}\right )}{8 b^3}+\frac{x^2 \text{PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac{b x^4}{12} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCot[c + (1 + I*c)*Tan[a + b*x]],x]

[Out]

(b*x^4)/12 + (x^3*ArcCot[c + (1 + I*c)*Tan[a + b*x]])/3 + (I/6)*x^3*Log[1 - I*c*E^((2*I)*a + (2*I)*b*x)] + (x^
2*PolyLog[2, I*c*E^((2*I)*a + (2*I)*b*x)])/(4*b) + ((I/4)*x*PolyLog[3, I*c*E^((2*I)*a + (2*I)*b*x)])/b^2 - Pol
yLog[4, I*c*E^((2*I)*a + (2*I)*b*x)]/(8*b^3)

Rule 5172

Int[ArcCot[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcCot[c + d*Tan[a + b*x]])/(f*(m + 1)), x] + Dist[(I*b)/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c + I*d + c*
E^(2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c + I*d)^2, -1]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && 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^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \, dx &=\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac{1}{3} (i b) \int \frac{x^3}{i (1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))-\frac{1}{3} (b c) \int \frac{e^{2 i a+2 i b x} x^3}{i (1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac{1}{2} i \int x^2 \log \left (1+\frac{c e^{2 i a+2 i b x}}{i (1+i c)+c}\right ) \, dx\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{x^2 \text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{\int x \text{Li}_2\left (-\frac{c e^{2 i a+2 i b x}}{i (1+i c)+c}\right ) \, dx}{2 b}\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{x^2 \text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{i x \text{Li}_3\left (i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{i \int \text{Li}_3\left (-\frac{c e^{2 i a+2 i b x}}{i (1+i c)+c}\right ) \, dx}{4 b^2}\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{x^2 \text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{i x \text{Li}_3\left (i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_3(i c x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{x^2 \text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{i x \text{Li}_3\left (i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{\text{Li}_4\left (i c e^{2 i a+2 i b x}\right )}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.202893, size = 136, normalized size = 0.88 \[ \frac{1}{24} \left (\frac{6 i x \text{PolyLog}\left (3,-\frac{i e^{-2 i (a+b x)}}{c}\right )}{b^2}+\frac{3 \text{PolyLog}\left (4,-\frac{i e^{-2 i (a+b x)}}{c}\right )}{b^3}-\frac{6 x^2 \text{PolyLog}\left (2,-\frac{i e^{-2 i (a+b x)}}{c}\right )}{b}+4 i x^3 \log \left (1+\frac{i e^{-2 i (a+b x)}}{c}\right )+8 x^3 \cot ^{-1}(c+(1+i c) \tan (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCot[c + (1 + I*c)*Tan[a + b*x]],x]

[Out]

(8*x^3*ArcCot[c + (1 + I*c)*Tan[a + b*x]] + (4*I)*x^3*Log[1 + I/(c*E^((2*I)*(a + b*x)))] - (6*x^2*PolyLog[2, (
-I)/(c*E^((2*I)*(a + b*x)))])/b + ((6*I)*x*PolyLog[3, (-I)/(c*E^((2*I)*(a + b*x)))])/b^2 + (3*PolyLog[4, (-I)/
(c*E^((2*I)*(a + b*x)))])/b^3)/24

________________________________________________________________________________________

Maple [C]  time = 23.174, size = 1526, normalized size = 9.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccot(c+(1+I*c)*tan(b*x+a)),x)

[Out]

1/2*I/b^2*a^2*ln(1+I*exp(I*(b*x+a))*(-I*c)^(1/2))*x-1/2*I/b^2*ln(1-I*c*exp(2*I*(b*x+a)))*x*a^2+1/2*I/b^2*a^2*l
n(1-I*exp(I*(b*x+a))*(-I*c)^(1/2))*x+1/12*x^3*Pi*csgn(I*exp(2*I*(b*x+a)))^3-1/12*x^3*Pi*csgn(I/(exp(2*I*(b*x+a
))+1))*csgn(I*(c*exp(2*I*(b*x+a))+I))*csgn(I*(c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))-1/12*x^3*Pi*csgn(I*e
xp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))^2+1/12*x^3*Pi*cs
gn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2+1/12*x^3*Pi*csgn(I*(c*exp(2*I
*(b*x+a))+I))*csgn(I*(c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2-1/12*x^3*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(
I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))^2+1/12*x^3*Pi*csgn(I*(c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1
))*csgn((c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2-1/12*x^3*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(c-I)/(
exp(2*I*(b*x+a))+1))^2-1/12*x^3*Pi*csgn(I*(c-I))*csgn(I*(c-I)/(exp(2*I*(b*x+a))+1))^2-1/12*x^3*Pi*csgn(I*(c-I)
/(exp(2*I*(b*x+a))+1))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))^2+1/12*b*x^4+1/12*x^3*Pi*csgn(I*exp
(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))+1/2*I/b^3*a^3*ln(1
-I*exp(I*(b*x+a))*(-I*c)^(1/2))-1/12*x^3*Pi*csgn(I*(c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^3+1/2/b^3*a^2*
dilog(1-I*exp(I*(b*x+a))*(-I*c)^(1/2))+1/2/b^3*a^2*dilog(1+I*exp(I*(b*x+a))*(-I*c)^(1/2))+1/4*x^2*polylog(2,I*
c*exp(2*I*(b*x+a)))/b-1/12*x^3*Pi*csgn(I*(c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))*csgn((c*exp(2*I*(b*x+a))
+I)/(exp(2*I*(b*x+a))+1))+1/6*I*x^3*ln(c-I)+1/12*x^3*Pi*csgn(I*(c-I)/(exp(2*I*(b*x+a))+1))^3+1/3*I*x^3*ln(exp(
I*(b*x+a)))+1/12*x^3*Pi*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(c-I))*csgn(I*(c-I)/(exp(2*I*(b*x+a))+1))+1/12*x^3
*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*(c-I)/(exp(2*I*(b*x+a))+1))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a)
)+1))+1/2*I/b^3*a^3*ln(1+I*exp(I*(b*x+a))*(-I*c)^(1/2))-1/3*I/b^3*ln(1-I*c*exp(2*I*(b*x+a)))*a^3-1/6*I/b^3*a^3
*ln(c*exp(2*I*(b*x+a))+I)+1/4*I*x*polylog(3,I*c*exp(2*I*(b*x+a)))/b^2-1/12*x^3*Pi*csgn(exp(2*I*(b*x+a))*(c-I)/
(exp(2*I*(b*x+a))+1))^3+1/12*x^3*Pi*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))^2+1/6*I*x^3*ln(1-I*c*exp
(2*I*(b*x+a)))+1/12*x^3*Pi*csgn(I*exp(I*(b*x+a)))^2*csgn(I*exp(2*I*(b*x+a)))-1/6*I*x^3*ln(c*exp(2*I*(b*x+a))+I
)-1/4/b^3*polylog(2,I*c*exp(2*I*(b*x+a)))*a^2+1/12*x^3*Pi*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))+1))^
3-1/8*polylog(4,I*c*exp(2*I*(b*x+a)))/b^3-1/12*x^3*Pi*csgn((c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^3+1/12
*x^3*Pi*csgn((c*exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2-1/6*x^3*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I*exp(2*I*(
b*x+a)))^2

________________________________________________________________________________________

Maxima [B]  time = 1.13204, size = 417, normalized size = 2.71 \begin{align*} \frac{\frac{{\left ({\left (b x + a\right )}^{3} - 3 \,{\left (b x + a\right )}^{2} a + 3 \,{\left (b x + a\right )} a^{2}\right )} \operatorname{arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )}{b^{2}} - \frac{3 \,{\left (-3 i \,{\left (b x + a\right )}^{4} + 12 i \,{\left (b x + a\right )}^{3} a - 18 i \,{\left (b x + a\right )}^{2} a^{2} +{\left (-8 i \,{\left (b x + a\right )}^{3} + 18 i \,{\left (b x + a\right )}^{2} a - 18 i \,{\left (b x + a\right )} a^{2}\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) +{\left (-12 i \,{\left (b x + a\right )}^{2} + 18 i \,{\left (b x + a\right )} a - 9 i \, a^{2}\right )}{\rm Li}_2\left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) +{\left (4 \,{\left (b x + a\right )}^{3} - 9 \,{\left (b x + a\right )}^{2} a + 9 \,{\left (b x + a\right )} a^{2}\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \,{\left (4 \, b x + a\right )}{\rm Li}_{3}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \,{\rm Li}_{4}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )}{\left (-i \, c - 1\right )}}{b^{2}{\left (12 \, c - 12 i\right )}}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="maxima")

[Out]

1/3*(((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arccot((I*c + 1)*tan(b*x + a) + c)/b^2 - 3*(-3*I*(b*x +
a)^4 + 12*I*(b*x + a)^3*a - 18*I*(b*x + a)^2*a^2 + (-8*I*(b*x + a)^3 + 18*I*(b*x + a)^2*a - 18*I*(b*x + a)*a^
2)*arctan2(c*cos(2*b*x + 2*a), c*sin(2*b*x + 2*a) + 1) + (-12*I*(b*x + a)^2 + 18*I*(b*x + a)*a - 9*I*a^2)*dilo
g(I*c*e^(2*I*b*x + 2*I*a)) + (4*(b*x + a)^3 - 9*(b*x + a)^2*a + 9*(b*x + a)*a^2)*log(c^2*cos(2*b*x + 2*a)^2 +
c^2*sin(2*b*x + 2*a)^2 + 2*c*sin(2*b*x + 2*a) + 1) + 3*(4*b*x + a)*polylog(3, I*c*e^(2*I*b*x + 2*I*a)) + 6*I*p
olylog(4, I*c*e^(2*I*b*x + 2*I*a)))*(-I*c - 1)/(b^2*(12*c - 12*I)))/b

________________________________________________________________________________________

Fricas [C]  time = 2.68018, size = 902, normalized size = 5.86 \begin{align*} \frac{b^{4} x^{4} - 2 i \, b^{3} x^{3} \log \left (\frac{{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - a^{4} - 2 i \, a^{3} \log \left (\frac{2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt{4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac{2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt{4 i \, c}}{2 \, c}\right ) + 12 i \, b x{\rm polylog}\left (3, \frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 12 i \, b x{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) +{\left (2 i \, b^{3} x^{3} + 2 i \, a^{3}\right )} \log \left (\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) +{\left (2 i \, b^{3} x^{3} + 2 i \, a^{3}\right )} \log \left (-\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 \,{\rm polylog}\left (4, \frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \,{\rm polylog}\left (4, -\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="fricas")

[Out]

1/12*(b^4*x^4 - 2*I*b^3*x^3*log((c*e^(2*I*b*x + 2*I*a) + I)*e^(-2*I*b*x - 2*I*a)/(c - I)) + 6*b^2*x^2*dilog(1/
2*sqrt(4*I*c)*e^(I*b*x + I*a)) + 6*b^2*x^2*dilog(-1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) - a^4 - 2*I*a^3*log(1/2*(2*
c*e^(I*b*x + I*a) + I*sqrt(4*I*c))/c) - 2*I*a^3*log(1/2*(2*c*e^(I*b*x + I*a) - I*sqrt(4*I*c))/c) + 12*I*b*x*po
lylog(3, 1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) + 12*I*b*x*polylog(3, -1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) + (2*I*b^3*x
^3 + 2*I*a^3)*log(1/2*sqrt(4*I*c)*e^(I*b*x + I*a) + 1) + (2*I*b^3*x^3 + 2*I*a^3)*log(-1/2*sqrt(4*I*c)*e^(I*b*x
+ I*a) + 1) - 12*polylog(4, 1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) - 12*polylog(4, -1/2*sqrt(4*I*c)*e^(I*b*x + I*a)
))/b^3

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acot(c+(1+I*c)*tan(b*x+a)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^2*arccot((I*c + 1)*tan(b*x + a) + c), x)