3.113 \(\int \frac {x^2 (a+b \tan ^{-1}(c x))^2}{(d+i c d x)^3} \, dx\)

Optimal. Leaf size=304 \[ \frac {b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3}+\frac {7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (-c x+i)}+\frac {b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (-c x+i)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (-c x+i)}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (-c x+i)^2}-\frac {7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac {i b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c^3 d^3}+\frac {13 b^2}{16 c^3 d^3 (-c x+i)}-\frac {i b^2}{16 c^3 d^3 (-c x+i)^2}-\frac {13 b^2 \tan ^{-1}(c x)}{16 c^3 d^3} \]

[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 = 0.56, 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 = {4876, 4864, 4862, 627, 44, 203, 4884, 4854, 4994, 6610} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3}-\frac {i b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^3 d^3}+\frac {7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (-c x+i)}+\frac {b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (-c x+i)^2}-\frac {2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (-c x+i)}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (-c x+i)^2}-\frac {7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {13 b^2}{16 c^3 d^3 (-c x+i)}-\frac {i b^2}{16 c^3 d^3 (-c x+i)^2}-\frac {13 b^2 \tan ^{-1}(c x)}{16 c^3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/16)*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*ArcTan
[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 44

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] && L
tQ[m + n + 2, 0])

Rule 203

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[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 4862

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 4864

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 4876

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 4884

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 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[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 6610

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 = 1.26, size = 431, normalized size = 1.42 \[ \frac {96 i a^2 \log \left (c^2 x^2+1\right )+\frac {384 a^2}{c x-i}+\frac {96 i a^2}{(c x-i)^2}-192 a^2 \tan ^{-1}(c x)-12 a b \left (16 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+32 \tan ^{-1}(c x)^2+12 i \sin \left (2 \tan ^{-1}(c x)\right )-i \sin \left (4 \tan ^{-1}(c x)\right )-12 \cos \left (2 \tan ^{-1}(c x)\right )+\cos \left (4 \tan ^{-1}(c x)\right )+4 \tan ^{-1}(c x) \left (8 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-6 \sin \left (2 \tan ^{-1}(c x)\right )+\sin \left (4 \tan ^{-1}(c x)\right )-6 i \cos \left (2 \tan ^{-1}(c x)\right )+i \cos \left (4 \tan ^{-1}(c x)\right )\right )\right )-b^2 \left (192 \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+96 i \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+128 \tan ^{-1}(c x)^3+192 i \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-144 \tan ^{-1}(c x)^2 \sin \left (2 \tan ^{-1}(c x)\right )+24 \tan ^{-1}(c x)^2 \sin \left (4 \tan ^{-1}(c x)\right )+144 i \tan ^{-1}(c x) \sin \left (2 \tan ^{-1}(c x)\right )-12 i \tan ^{-1}(c x) \sin \left (4 \tan ^{-1}(c x)\right )+72 \sin \left (2 \tan ^{-1}(c x)\right )-3 \sin \left (4 \tan ^{-1}(c x)\right )-144 i \tan ^{-1}(c x)^2 \cos \left (2 \tan ^{-1}(c x)\right )+24 i \tan ^{-1}(c x)^2 \cos \left (4 \tan ^{-1}(c x)\right )-144 \tan ^{-1}(c x) \cos \left (2 \tan ^{-1}(c x)\right )+12 \tan ^{-1}(c x) \cos \left (4 \tan ^{-1}(c x)\right )+72 i \cos \left (2 \tan ^{-1}(c x)\right )-3 i \cos \left (4 \tan ^{-1}(c x)\right )\right )}{192 c^3 d^3} \]

Warning: Unable to verify antiderivative.

[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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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {-i \, b^{2} x^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 4 \, a b x^{2} \log \left (-\frac {c x + i}{c x - i}\right ) + 4 i \, a^{2} x^{2}}{4 \, c^{3} d^{3} x^{3} - 12 i \, c^{2} d^{3} x^{2} - 12 \, c d^{3} x + 4 i \, d^{3}}, x\right ) \]

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((-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)/(4*c^3*d
^3*x^3 - 12*I*c^2*d^3*x^2 - 12*c*d^3*x + 4*I*d^3), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [C]  time = 0.47, size = 1276, normalized size = 4.20 \[ \text {result too large to display} \]

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)

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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