3.2.6 \(\int \frac {x (a+b \text {ArcTan}(c x))^2}{(d+i c d x)^2} \, dx\) [106]

Optimal. Leaf size=216 \[ \frac {i b^2}{2 c^2 d^2 (i-c x)}-\frac {i b^2 \text {ArcTan}(c x)}{2 c^2 d^2}-\frac {b (a+b \text {ArcTan}(c x))}{c^2 d^2 (i-c x)}+\frac {(a+b \text {ArcTan}(c x))^2}{2 c^2 d^2}-\frac {i (a+b \text {ArcTan}(c x))^2}{c^2 d^2 (i-c x)}+\frac {(a+b \text {ArcTan}(c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^2 d^2}+\frac {i b (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^2 d^2}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^2 d^2} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4996, 4974, 4972, 641, 46, 209, 5004, 4964, 5114, 6745} \begin {gather*} \frac {i b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))}{c^2 d^2}-\frac {b (a+b \text {ArcTan}(c x))}{c^2 d^2 (-c x+i)}-\frac {i (a+b \text {ArcTan}(c x))^2}{c^2 d^2 (-c x+i)}+\frac {(a+b \text {ArcTan}(c x))^2}{2 c^2 d^2}+\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2}{c^2 d^2}-\frac {i b^2 \text {ArcTan}(c x)}{2 c^2 d^2}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c^2 d^2}+\frac {i b^2}{2 c^2 d^2 (-c x+i)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*b^2)/(c^2*d^2*(I - c*x)) - ((I/2)*b^2*ArcTan[c*x])/(c^2*d^2) - (b*(a + b*ArcTan[c*x]))/(c^2*d^2*(I - c*
x)) + (a + b*ArcTan[c*x])^2/(2*c^2*d^2) - (I*(a + b*ArcTan[c*x])^2)/(c^2*d^2*(I - c*x)) + ((a + b*ArcTan[c*x])
^2*Log[2/(1 + I*c*x)])/(c^2*d^2) + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^2*d^2) + (b^2*Po
lyLog[3, 1 - 2/(1 + I*c*x)])/(2*c^2*d^2)

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

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Mathematica [A]
time = 0.48, size = 300, normalized size = 1.39 \begin {gather*} \frac {\frac {12 i a^2}{-i+c x}-12 i a^2 \text {ArcTan}(c x)-6 a^2 \log \left (1+c^2 x^2\right )-6 i a b \left (4 \text {ArcTan}(c x)^2-\cos (2 \text {ArcTan}(c x))+2 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )-2 i \text {ArcTan}(c x) \left (\cos (2 \text {ArcTan}(c x))-2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )-i \sin (2 \text {ArcTan}(c x))\right )+i \sin (2 \text {ArcTan}(c x))\right )+b^2 \left (-8 i \text {ArcTan}(c x)^3+3 \cos (2 \text {ArcTan}(c x))+6 i \text {ArcTan}(c x) \cos (2 \text {ArcTan}(c x))-6 \text {ArcTan}(c x)^2 \cos (2 \text {ArcTan}(c x))+12 \text {ArcTan}(c x)^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )-12 i \text {ArcTan}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )+6 \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )-3 i \sin (2 \text {ArcTan}(c x))+6 \text {ArcTan}(c x) \sin (2 \text {ArcTan}(c x))+6 i \text {ArcTan}(c x)^2 \sin (2 \text {ArcTan}(c x))\right )}{12 c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((12*I)*a^2)/(-I + c*x) - (12*I)*a^2*ArcTan[c*x] - 6*a^2*Log[1 + c^2*x^2] - (6*I)*a*b*(4*ArcTan[c*x]^2 - Cos[
2*ArcTan[c*x]] + 2*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - (2*I)*ArcTan[c*x]*(Cos[2*ArcTan[c*x]] - 2*Log[1 + E^((
2*I)*ArcTan[c*x])] - I*Sin[2*ArcTan[c*x]]) + I*Sin[2*ArcTan[c*x]]) + b^2*((-8*I)*ArcTan[c*x]^3 + 3*Cos[2*ArcTa
n[c*x]] + (6*I)*ArcTan[c*x]*Cos[2*ArcTan[c*x]] - 6*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] + 12*ArcTan[c*x]^2*Log[1 +
E^((2*I)*ArcTan[c*x])] - (12*I)*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 6*PolyLog[3, -E^((2*I)*ArcTa
n[c*x])] - (3*I)*Sin[2*ArcTan[c*x]] + 6*ArcTan[c*x]*Sin[2*ArcTan[c*x]] + (6*I)*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]
]))/(12*c^2*d^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.41, size = 969, normalized size = 4.49 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

a^2*(I/(c^3*d^2*x - I*c^2*d^2) - log(c*x - I)/(c^2*d^2)) - 1/32*(-8*I*b^2*arctan(c*x)^2 - 8*(-I*b^2*c*x - b^2)
*arctan(c*x)^3 - (b^2*c*x - I*b^2)*log(c^2*x^2 + 1)^3 - 2*(-I*b^2 + (-I*b^2*c*x - b^2)*arctan(c*x))*log(c^2*x^
2 + 1)^2 - (2*b^2*c^3*(c^2/(c^9*d^2*x^2 + c^7*d^2) + log(c^2*x^2 + 1)/(c^7*d^2*x^2 + c^5*d^2)) - 640*b^2*c^3*i
ntegrate(1/16*x^3*arctan(c*x)^2/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) - 96*b^2*c^3*integrate(1/16*x^3*log(
c^2*x^2 + 1)^2/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) - 1024*a*b*c^3*integrate(1/16*x^3*arctan(c*x)/(c^5*d^
2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) - 256*b^2*c^2*integrate(1/16*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^5*d^2*x^4
+ 2*c^3*d^2*x^2 + c*d^2), x) + 256*b^2*c^2*integrate(1/16*x^2*arctan(c*x)/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2
), x) + 16*(c*(x/(c^5*d^2*x^2 + c^3*d^2) + arctan(c*x)/(c^4*d^2)) - 2*arctan(c*x)/(c^5*d^2*x^2 + c^3*d^2))*a*b
*c + 128*b^2*c*integrate(1/16*x*arctan(c*x)^2/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) + b^2*c*log(c^2*x^2 +
1)^2/(c^5*d^2*x^2 + c^3*d^2) + 256*b^2*integrate(1/16*arctan(c*x)/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x))*(
c^3*d^2*x - I*c^2*d^2) - 32*(I*c^3*d^2*x + c^2*d^2)*integrate(-1/8*(32*a*b*c^2*x^2*arctan(c*x) - b^2*log(c^2*x
^2 + 1)^2 + 4*(2*b^2*c^2*x^2 - b^2)*arctan(c*x)^2 - 2*(b^2*c^2*x^2 + b^2 + (b^2*c^3*x^3 - b^2*c*x)*arctan(c*x)
)*log(c^2*x^2 + 1))/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) + 4*(2*b^2*arctan(c*x) - (b^2*c*x - I*b^2)*arcta
n(c*x)^2)*log(c^2*x^2 + 1))/(c^3*d^2*x - I*c^2*d^2)

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

[Out]

integral(1/4*(b^2*x*log(-(c*x + I)/(c*x - I))^2 - 4*I*a*b*x*log(-(c*x + I)/(c*x - I)) - 4*a^2*x)/(c^2*d^2*x^2
- 2*I*c*d^2*x - d^2), 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*(a+b*atan(c*x))**2/(d+I*c*d*x)**2,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*(a+b*arctan(c*x))^2/(d+I*c*d*x)^2,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\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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