3.1.52 \(\int x^2 \text {ArcTan}(c+(1+i c) \tan (a+b x)) \, dx\) [52]

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

[Out]

-1/12*b*x^4+1/3*x^3*arctan(c+(1+I*c)*tan(b*x+a))-1/6*I*x^3*ln(1-I*c*exp(2*I*a+2*I*b*x))-1/4*x^2*polylog(2,I*c*
exp(2*I*a+2*I*b*x))/b-1/4*I*x*polylog(3,I*c*exp(2*I*a+2*I*b*x))/b^2+1/8*polylog(4,I*c*exp(2*I*a+2*I*b*x))/b^3

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Rubi [A]
time = 0.18, antiderivative size = 154, normalized size of antiderivative = 1.00, 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 = {5279, 2215, 2221, 2611, 6744, 2320, 6724} \begin {gather*} \frac {1}{3} x^3 \text {ArcTan}(c+(1+i c) \tan (a+b x))+\frac {\text {Li}_4\left (i c e^{2 i a+2 i b x}\right )}{8 b^3}-\frac {i x \text {Li}_3\left (i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {x^2 \text {Li}_2\left (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 {b x^4}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b*x^4) + (x^3*ArcTan[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 +
PolyLog[4, I*c*E^((2*I)*a + (2*I)*b*x)]/(8*b^3)

Rule 2215

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 5279

Int[ArcTan[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m +
1)*(ArcTan[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 6724

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]

Rule 6744

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]

Rubi steps

\begin {align*} \int x^2 \tan ^{-1}(c+(1+i c) \tan (a+b x)) \, dx &=\frac {1}{3} x^3 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 {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 \tan ^{-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*}

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Mathematica [A]
time = 0.34, size = 140, normalized size = 0.91 \begin {gather*} \frac {1}{3} x^3 \text {ArcTan}(c+(1+i c) \tan (a+b x))-\frac {4 i b^3 x^3 \log \left (1+\frac {i e^{-2 i (a+b x)}}{c}\right )-6 b^2 x^2 \text {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{c}\right )+6 i b x \text {PolyLog}\left (3,-\frac {i e^{-2 i (a+b x)}}{c}\right )+3 \text {PolyLog}\left (4,-\frac {i e^{-2 i (a+b x)}}{c}\right )}{24 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.81, size = 1532, normalized size = 9.95

method	result	size
risch	\(\text {Expression too large to display}\)	\(1532\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctan(c+(1+I*c)*tan(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (107) = 214\).
time = 0.28, size = 310, normalized size = 2.01 \begin {gather*} \frac {\frac {4 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \arctan \left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )}{b^{2}} - \frac {{\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} - 2 \, {\left (4 i \, {\left (b x + a\right )}^{3} - 9 i \, {\left (b x + a\right )}^{2} a + 9 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 ) - 3 \, {\left (4 i \, {\left (b x + a\right )}^{2} - 6 i \, {\left (b x + a\right )} a + 3 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 (c - i\right )}}}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arctan((I*c + 1)*tan(b*x + a) + c)/b^2 - (-3*I*(b*x
+ a)^4 + 12*I*(b*x + a)^3*a - 18*I*(b*x + a)^2*a^2 - 2*(4*I*(b*x + a)^3 - 9*I*(b*x + a)^2*a + 9*I*(b*x + a)*a^
2)*arctan2(c*cos(2*b*x + 2*a), c*sin(2*b*x + 2*a) + 1) - 3*(4*I*(b*x + a)^2 - 6*I*(b*x + a)*a + 3*I*a^2)*dilog
(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*po
lylog(4, I*c*e^(2*I*b*x + 2*I*a)))*(I*c + 1)/(b^2*(c - I)))/b

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (107) = 214\).
time = 0.89, size = 322, normalized size = 2.09 \begin {gather*} -\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 ) + 2 \, {\left (i \, b^{3} x^{3} + i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) + 2 \, {\left (i \, b^{3} x^{3} + 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(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*
polylog(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 + I*a^3)*log(1/2*sqrt(4*I*c)*e^(I*b*x + I*a) + 1) + 2*(I*b^3*x^3 + 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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: CoercionFailed >> Cannot convert _t0**2 + exp(2*I*a) of type <class 'sympy.core.add.Add'> to
QQ_I[x,b,_t0,exp(I*a)]

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\mathrm {atan}\left (c+\mathrm {tan}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atan(c + tan(a + b*x)*(c*1i + 1)),x)

[Out]

int(x^2*atan(c + tan(a + b*x)*(c*1i + 1)), x)

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