3.2.13 \(\int \frac {x^2 (a+b \text {ArcTan}(c x))^2}{(d+i c d x)^3} \, dx\) [113]

Optimal. Leaf size=304 \[ -\frac {i b^2}{16 c^3 d^3 (i-c x)^2}+\frac {13 b^2}{16 c^3 d^3 (i-c x)}-\frac {13 b^2 \text {ArcTan}(c x)}{16 c^3 d^3}+\frac {b (a+b \text {ArcTan}(c x))}{4 c^3 d^3 (i-c x)^2}+\frac {7 i b (a+b \text {ArcTan}(c x))}{4 c^3 d^3 (i-c x)}-\frac {7 i (a+b \text {ArcTan}(c x))^2}{8 c^3 d^3}+\frac {i (a+b \text {ArcTan}(c x))^2}{2 c^3 d^3 (i-c x)^2}-\frac {2 (a+b \text {ArcTan}(c x))^2}{c^3 d^3 (i-c x)}-\frac {i (a+b \text {ArcTan}(c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^3}+\frac {b (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3 d^3}-\frac {i b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^3 d^3} \]

[Out]

-1/16*I*b^2/c^3/d^3/(I-c*x)^2+13/16*b^2/c^3/d^3/(I-c*x)-13/16*b^2*arctan(c*x)/c^3/d^3+1/4*b*(a+b*arctan(c*x))/
c^3/d^3/(I-c*x)^2+7/4*I*b*(a+b*arctan(c*x))/c^3/d^3/(I-c*x)-7/8*I*(a+b*arctan(c*x))^2/c^3/d^3+1/2*I*(a+b*arcta
n(c*x))^2/c^3/d^3/(I-c*x)^2-2*(a+b*arctan(c*x))^2/c^3/d^3/(I-c*x)-I*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/c^3/d^
3+b*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))/c^3/d^3-1/2*I*b^2*polylog(3,1-2/(1+I*c*x))/c^3/d^3

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Rubi [A]
time = 0.40, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4996, 4974, 4972, 641, 46, 209, 5004, 4964, 5114, 6745} \begin {gather*} \frac {b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))}{c^3 d^3}+\frac {7 i b (a+b \text {ArcTan}(c x))}{4 c^3 d^3 (-c x+i)}+\frac {b (a+b \text {ArcTan}(c x))}{4 c^3 d^3 (-c x+i)^2}-\frac {2 (a+b \text {ArcTan}(c x))^2}{c^3 d^3 (-c x+i)}+\frac {i (a+b \text {ArcTan}(c x))^2}{2 c^3 d^3 (-c x+i)^2}-\frac {7 i (a+b \text {ArcTan}(c x))^2}{8 c^3 d^3}-\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2}{c^3 d^3}-\frac {13 b^2 \text {ArcTan}(c x)}{16 c^3 d^3}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c^3 d^3}+\frac {13 b^2}{16 c^3 d^3 (-c x+i)}-\frac {i b^2}{16 c^3 d^3 (-c x+i)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x)^3,x]

[Out]

((-1/16*I)*b^2)/(c^3*d^3*(I - c*x)^2) + (13*b^2)/(16*c^3*d^3*(I - c*x)) - (13*b^2*ArcTan[c*x])/(16*c^3*d^3) +
(b*(a + b*ArcTan[c*x]))/(4*c^3*d^3*(I - c*x)^2) + (((7*I)/4)*b*(a + b*ArcTan[c*x]))/(c^3*d^3*(I - c*x)) - (((7
*I)/8)*(a + b*ArcTan[c*x])^2)/(c^3*d^3) + ((I/2)*(a + b*ArcTan[c*x])^2)/(c^3*d^3*(I - c*x)^2) - (2*(a + b*ArcT
an[c*x])^2)/(c^3*d^3*(I - c*x)) - (I*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^3*d^3) + (b*(a + b*ArcTan[c*
x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^3*d^3) - ((I/2)*b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(c^3*d^3)

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 + c^2*x
^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 4972

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a + b*
ArcTan[c*x])/(e*(q + 1))), x] - Dist[b*(c/(e*(q + 1))), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 4974

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a
+ b*ArcTan[c*x])^p/(e*(q + 1))), x] - Dist[b*c*(p/(e*(q + 1))), Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p -
1), (d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && N
eQ[q, -1]

Rule 4996

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 5004

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x] + Dist[b*p*(I/2), Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 -
u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - 2
*(I/(I - c*x)))^2, 0]

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{(d+i c d x)^3} \, dx &=\int \left (-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^3 (-i+c x)^3}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^3 (-i+c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^3 (-i+c x)}\right ) \, dx\\ &=-\frac {i \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^3} \, dx}{c^2 d^3}+\frac {i \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{-i+c x} \, dx}{c^2 d^3}-\frac {2 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{c^2 d^3}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^3}-\frac {(i b) \int \left (-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^3}+\frac {a+b \tan ^{-1}(c x)}{4 (-i+c x)^2}-\frac {a+b \tan ^{-1}(c x)}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}+\frac {(2 i b) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d^3}-\frac {(4 b) \int \left (-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d^3}-\frac {(i b) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{4 c^2 d^3}+\frac {(i b) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{4 c^2 d^3}+\frac {(2 i b) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^2 d^3}-\frac {(2 i b) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^2 d^3}-\frac {b \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{2 c^2 d^3}-\frac {b^2 \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d^3}\\ &=\frac {b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac {7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac {7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d^3}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d^3}-\frac {\left (i b^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{4 c^2 d^3}+\frac {\left (2 i b^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^2 d^3}-\frac {b^2 \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{4 c^2 d^3}\\ &=\frac {b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac {7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac {7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d^3}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d^3}-\frac {\left (i b^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{4 c^2 d^3}+\frac {\left (2 i b^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{c^2 d^3}-\frac {b^2 \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{4 c^2 d^3}\\ &=\frac {b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac {7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac {7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d^3}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d^3}-\frac {\left (i b^2\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{4 c^2 d^3}+\frac {\left (2 i b^2\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}-\frac {b^2 \int \left (-\frac {i}{2 (-i+c x)^3}+\frac {1}{4 (-i+c x)^2}-\frac {1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{4 c^2 d^3}\\ &=-\frac {i b^2}{16 c^3 d^3 (i-c x)^2}+\frac {13 b^2}{16 c^3 d^3 (i-c x)}+\frac {b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac {7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac {7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d^3}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d^3}+\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{16 c^2 d^3}+\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{8 c^2 d^3}-\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{c^2 d^3}\\ &=-\frac {i b^2}{16 c^3 d^3 (i-c x)^2}+\frac {13 b^2}{16 c^3 d^3 (i-c x)}-\frac {13 b^2 \tan ^{-1}(c x)}{16 c^3 d^3}+\frac {b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac {7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac {7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d^3}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.72, size = 431, normalized size = 1.42 \begin {gather*} \frac {\frac {96 i a^2}{(-i+c x)^2}+\frac {384 a^2}{-i+c x}-192 a^2 \text {ArcTan}(c x)+96 i a^2 \log \left (1+c^2 x^2\right )-b^2 \left (128 \text {ArcTan}(c x)^3+72 i \cos (2 \text {ArcTan}(c x))-144 \text {ArcTan}(c x) \cos (2 \text {ArcTan}(c x))-144 i \text {ArcTan}(c x)^2 \cos (2 \text {ArcTan}(c x))-3 i \cos (4 \text {ArcTan}(c x))+12 \text {ArcTan}(c x) \cos (4 \text {ArcTan}(c x))+24 i \text {ArcTan}(c x)^2 \cos (4 \text {ArcTan}(c x))+192 i \text {ArcTan}(c x)^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )+192 \text {ArcTan}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )+96 i \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )+72 \sin (2 \text {ArcTan}(c x))+144 i \text {ArcTan}(c x) \sin (2 \text {ArcTan}(c x))-144 \text {ArcTan}(c x)^2 \sin (2 \text {ArcTan}(c x))-3 \sin (4 \text {ArcTan}(c x))-12 i \text {ArcTan}(c x) \sin (4 \text {ArcTan}(c x))+24 \text {ArcTan}(c x)^2 \sin (4 \text {ArcTan}(c x))\right )-12 a b \left (32 \text {ArcTan}(c x)^2-12 \cos (2 \text {ArcTan}(c x))+\cos (4 \text {ArcTan}(c x))+16 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )+12 i \sin (2 \text {ArcTan}(c x))-i \sin (4 \text {ArcTan}(c x))+4 \text {ArcTan}(c x) \left (-6 i \cos (2 \text {ArcTan}(c x))+i \cos (4 \text {ArcTan}(c x))+8 i \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )-6 \sin (2 \text {ArcTan}(c x))+\sin (4 \text {ArcTan}(c x))\right )\right )}{192 c^3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x)^3,x]

[Out]

(((96*I)*a^2)/(-I + c*x)^2 + (384*a^2)/(-I + c*x) - 192*a^2*ArcTan[c*x] + (96*I)*a^2*Log[1 + c^2*x^2] - b^2*(1
28*ArcTan[c*x]^3 + (72*I)*Cos[2*ArcTan[c*x]] - 144*ArcTan[c*x]*Cos[2*ArcTan[c*x]] - (144*I)*ArcTan[c*x]^2*Cos[
2*ArcTan[c*x]] - (3*I)*Cos[4*ArcTan[c*x]] + 12*ArcTan[c*x]*Cos[4*ArcTan[c*x]] + (24*I)*ArcTan[c*x]^2*Cos[4*Arc
Tan[c*x]] + (192*I)*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + 192*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan
[c*x])] + (96*I)*PolyLog[3, -E^((2*I)*ArcTan[c*x])] + 72*Sin[2*ArcTan[c*x]] + (144*I)*ArcTan[c*x]*Sin[2*ArcTan
[c*x]] - 144*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]] - 3*Sin[4*ArcTan[c*x]] - (12*I)*ArcTan[c*x]*Sin[4*ArcTan[c*x]] +
24*ArcTan[c*x]^2*Sin[4*ArcTan[c*x]]) - 12*a*b*(32*ArcTan[c*x]^2 - 12*Cos[2*ArcTan[c*x]] + Cos[4*ArcTan[c*x]]
+ 16*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + (12*I)*Sin[2*ArcTan[c*x]] - I*Sin[4*ArcTan[c*x]] + 4*ArcTan[c*x]*((-
6*I)*Cos[2*ArcTan[c*x]] + I*Cos[4*ArcTan[c*x]] + (8*I)*Log[1 + E^((2*I)*ArcTan[c*x])] - 6*Sin[2*ArcTan[c*x]] +
Sin[4*ArcTan[c*x]])))/(192*c^3*d^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.61, size = 1164, normalized size = 3.83

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1164\)
default	\(\text {Expression too large to display}\)	\(1164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/c^3*(1/64*I*b^2/d^3/(c*x-I)^2*c^2*x^2-3*b^2/d^3/(8*c*x-8*I)-2/3*b^2/d^3*arctan(c*x)^3+1/4*b*a/d^3/(c*x-I)^2-
7/16*b*a/d^3*ln(c^2*x^2+1)+7/32*b*a/d^3*ln(c^4*x^4+10*c^2*x^2+9)-1/2*b*a/d^3*ln(c*x-I)^2-a^2/d^3*arctan(c*x)+2
*a^2/d^3/(c*x-I)-1/2*b^2/d^3*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x
^2+1)+1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-1/8*I*b^2/d^3*arctan(c*x)/(c*x-I)^2*c*x+3*I*b^2/d^
3/(8*c*x-8*I)*c*x+I*b*a/d^3*arctan(c*x)/(c*x-I)^2+2*I*b*a/d^3*arctan(c*x)*ln(c*x-I)+I*b^2/d^3*arctan(c*x)^2*ln
(c*x-I)+1/2*I*b^2/d^3*arctan(c*x)^2/(c*x-I)^2-3/4*I*b^2/d^3*arctan(c*x)/(c*x-I)-I*b^2/d^3*arctan(c*x)^2*ln(2*I
*(1+I*c*x)^2/(c^2*x^2+1))-7/4*I*b*a/d^3/(c*x-I)-7/8*I*b*a/d^3*arctan(c*x)+7/16*I*b*a/d^3*arctan(1/2*c*x)-7/16*
I*b*a/d^3*arctan(1/6*c^3*x^3+7/6*c*x)-7/8*I*b*a/d^3*arctan(1/2*c*x-1/2*I)-7/8*I*b^2/d^3*arctan(c*x)^2-1/64*I*b
^2/d^3/(c*x-I)^2-1/2*I*b^2/d^3*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+1/2*I*a^2/d^3*ln(c^2*x^2+1)+1/2*I*a^2/d^3/(
c*x-I)^2+b*a/d^3*dilog(-1/2*I*(c*x+I))+b^2/d^3*Pi*arctan(c*x)^2-b^2/d^3*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^
2*x^2+1))+2*b^2/d^3*arctan(c*x)^2/(c*x-I)+1/16*b^2/d^3*arctan(c*x)/(c*x-I)^2-1/2*b^2/d^3*Pi*csgn((1+I*c*x)^2/(
c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*b^2/d^3*Pi*csgn((1+I
*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-3/4*b^2/d
^3*arctan(c*x)/(c*x-I)*c*x-1/16*b^2/d^3*arctan(c*x)/(c*x-I)^2*c^2*x^2+4*b*a/d^3*arctan(c*x)/(c*x-I)+b*a/d^3*ln
(c*x-I)*ln(-1/2*I*(c*x+I))-1/32*b^2/d^3/(c*x-I)^2*c*x-1/2*b^2/d^3*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2
/(c^2*x^2+1)+1))^3*arctan(c*x)^2-b^2/d^3*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan
(c*x)^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="maxima")

[Out]

1/128*(144*a^2*c^2*x^2*arctan2(1, c*x) - 32*a^2*c*x*(9*I*arctan2(1, c*x) - 8) - 32*(b^2*c^2*x^2 - 2*I*b^2*c*x
- b^2)*arctan(c*x)^3 + 4*(-I*b^2*c^2*x^2 - 2*b^2*c*x + I*b^2)*log(c^2*x^2 + 1)^3 - 48*a^2*(3*arctan2(1, c*x) +
4*I) + 16*(4*b^2*c*x - 3*I*b^2)*arctan(c*x)^2 - 4*(4*b^2*c*x - 3*I*b^2 + 2*(b^2*c^2*x^2 - 2*I*b^2*c*x - b^2)*
arctan(c*x))*log(c^2*x^2 + 1)^2 + 6*(I*b^2*c^6*d^3*x^2 + 2*b^2*c^5*d^3*x - I*b^2*c^4*d^3)*(((8*c^2*x^2 + 7)*c^
2/(c^14*d^3*x^4 + 2*c^12*d^3*x^2 + c^10*d^3) + 2*(4*c^2*x^2 + 3)*log(c^2*x^2 + 1)/(c^12*d^3*x^4 + 2*c^10*d^3*x
^2 + c^8*d^3))*c^4 + 2*(2*c^2*x^2 + 1)*c^2*log(c^2*x^2 + 1)^2/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3) - c^2*(
c^2/(c^12*d^3*x^4 + 2*c^10*d^3*x^2 + c^8*d^3) + 2*log(c^2*x^2 + 1)/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3)) -
512*c^2*integrate(1/16*x^3*arctan(c*x)^2/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x) - 2*log(
c^2*x^2 + 1)^2/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3) + 512*integrate(1/16*x*arctan(c*x)^2/(c^8*d^3*x^6 + 3*c
^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x)) - 4*(-I*b^2*c^8*d^3*x^2 - 2*b^2*c^7*d^3*x + I*b^2*c^6*d^3)*(((8*c^2
*x^2 + 7)*c^2/(c^14*d^3*x^4 + 2*c^12*d^3*x^2 + c^10*d^3) + 2*(4*c^2*x^2 + 3)*log(c^2*x^2 + 1)/(c^12*d^3*x^4 +
2*c^10*d^3*x^2 + c^8*d^3))*c^2 + 512*c^2*integrate(1/16*x^5*arctan(c*x)^2/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4
*d^3*x^2 + c^2*d^3), x) + 128*c^2*integrate(1/16*x^5*log(c^2*x^2 + 1)^2/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d
^3*x^2 + c^2*d^3), x) + 2*(2*c^2*x^2 + 1)*log(c^2*x^2 + 1)^2/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3) - 512*in
tegrate(1/16*x^3*arctan(c*x)^2/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x)) + 8*(-I*b^2*c^7*d^
3*x^2 - 2*b^2*c^6*d^3*x + I*b^2*c^5*d^3)*(((8*c^2*x^2 + 7)*c^2/(c^13*d^3*x^4 + 2*c^11*d^3*x^2 + c^9*d^3) + 2*(
4*c^2*x^2 + 3)*log(c^2*x^2 + 1)/(c^11*d^3*x^4 + 2*c^9*d^3*x^2 + c^7*d^3))*c^2 + 2*(2*c^2*x^2 + 1)*log(c^2*x^2
+ 1)^2/(c^9*d^3*x^4 + 2*c^7*d^3*x^2 + c^5*d^3) - 256*integrate(1/8*x^3*arctan(c*x)^2/(c^7*d^3*x^6 + 3*c^5*d^3*
x^4 + 3*c^3*d^3*x^2 + c*d^3), x)) + 14*(I*b^2*c^6*d^3*x^2 + 2*b^2*c^5*d^3*x - I*b^2*c^4*d^3)*(((4*c^2*x^2 + 3)
*c^2/(c^12*d^3*x^4 + 2*c^10*d^3*x^2 + c^8*d^3) + 2*(2*c^2*x^2 + 1)*log(c^2*x^2 + 1)/(c^10*d^3*x^4 + 2*c^8*d^3*
x^2 + c^6*d^3))*c^2 - 256*c*integrate(1/8*x^2*arctan(c*x)/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d
^3), x) - c^2/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3) - 2*log(c^2*x^2 + 1)/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4
*d^3)) - 7*(b^2*c^6*d^3*x^2 - 2*I*b^2*c^5*d^3*x - b^2*c^4*d^3)*((c*((5*c^2*x^3 + 3*x)/(c^10*d^3*x^4 + 2*c^8*d^
3*x^2 + c^6*d^3) + 5*arctan(c*x)/(c^7*d^3)) - 8*(2*c^2*x^2 + 1)*arctan(c*x)/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^
6*d^3))*c^2 - c*((3*c^2*x^3 + 5*x)/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3) + 3*arctan(c*x)/(c^5*d^3)) - 128*c*
integrate(1/4*x^2*log(c^2*x^2 + 1)/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x) + 8*arctan(c*x)
/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3)) + 4*(I*b^2*c^5*d^3*x^2 + 2*b^2*c^4*d^3*x - I*b^2*c^3*d^3)*(c^2*(c^2/
(c^11*d^3*x^4 + 2*c^9*d^3*x^2 + c^7*d^3) + 2*log(c^2*x^2 + 1)/(c^9*d^3*x^4 + 2*c^7*d^3*x^2 + c^5*d^3)) + 2*log
(c^2*x^2 + 1)^2/(c^7*d^3*x^4 + 2*c^5*d^3*x^2 + c^3*d^3) - 256*integrate(1/8*x*arctan(c*x)^2/(c^7*d^3*x^6 + 3*c
^5*d^3*x^4 + 3*c^3*d^3*x^2 + c*d^3), x)) - 4*(-I*a*b*c^8*d^3*x^2 - 2*a*b*c^7*d^3*x + I*a*b*c^6*d^3)*(128*c^2*i
ntegrate(1/4*x^5*arctan(c*x)/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x) - c*((5*c^2*x^3 + 3*x
)/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3) + 5*arctan(c*x)/(c^7*d^3)) + 128*c*integrate(1/4*x^4*log(c^2*x^2 +
1)/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x) + 8*(2*c^2*x^2 + 1)*arctan(c*x)/(c^10*d^3*x^4 +
2*c^8*d^3*x^2 + c^6*d^3)) + 4*(I*a*b*c^8*d^3*x^2 + 2*a*b*c^7*d^3*x - I*a*b*c^6*d^3)*(128*c^2*integrate(1/4*x^
5*arctan(c*x)/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x) - c*((5*c^2*x^3 + 3*x)/(c^10*d^3*x^4
+ 2*c^8*d^3*x^2 + c^6*d^3) + 5*arctan(c*x)/(c^7*d^3)) - 128*c*integrate(1/4*x^4*log(c^2*x^2 + 1)/(c^8*d^3*x^6
+ 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x) + 8*(2*c^2*x^2 + 1)*arctan(c*x)/(c^10*d^3*x^4 + 2*c^8*d^3*x^2
+ c^6*d^3)) - 8*(a*b*c^8*d^3*x^2 - 2*I*a*b*c^7*d^3*x - a*b*c^6*d^3)*(64*c^2*integrate(1/8*x^5*log(c^2*x^2 + 1)
/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x) + (4*c^2*x^2 + 3)*c^2/(c^12*d^3*x^4 + 2*c^10*d^3*
x^2 + c^8*d^3) + 256*c*integrate(1/8*x^4*arctan(c*x)/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3),
x) + 2*(2*c^2*x^2 + 1)*log(c^2*x^2 + 1)/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3)) + 8*(a*b*c^8*d^3*x^2 - 2*I*a
*b*c^7*d^3*x - a*b*c^6*d^3)*(64*c^2*integrate(1/8*x^5*log(c^2*x^2 + 1)/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^
3*x^2 + c^2*d^3), x) + (4*c^2*x^2 + 3)*c^2/(c^12*d^3*x^4 + 2*c^10*d^3*x^2 + c^8*d^3) - 256*c*integrate(1/8*x^4
*arctan(c*x)/(c^8*d^3*x^6 + 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 + c^2*d^3), x) + 2*(2*c^2*x^2 + 1)*log(c^2*x^2 + 1)/
(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3)) - 16*((a*b + I*b^2)*c^7*d^3*x^2 - 2*(I*a*b - b^2)*c^6*d^3*x - (a*b +
I*b^2)*c^5*d^3)*(32*c^2*integrate(1/4*x^4*arct...

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Fricas [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*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="fricas")

[Out]

integral(1/4*(-I*b^2*x^2*log(-(c*x + I)/(c*x - I))^2 - 4*a*b*x^2*log(-(c*x + I)/(c*x - I)) + 4*I*a^2*x^2)/(c^3
*d^3*x^3 - 3*I*c^2*d^3*x^2 - 3*c*d^3*x + I*d^3), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*atan(c*x))**2/(d+I*c*d*x)**3,x)

[Out]

Timed out

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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*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*atan(c*x))^2)/(d + c*d*x*1i)^3,x)

[Out]

int((x^2*(a + b*atan(c*x))^2)/(d + c*d*x*1i)^3, x)

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