3.2.8 \(\int \frac {(a+b \text {ArcTan}(c x))^2}{x (d+i c d x)^2} \, dx\) [108]

Optimal. Leaf size=221 \[ -\frac {i b^2}{2 d^2 (i-c x)}+\frac {i b^2 \text {ArcTan}(c x)}{2 d^2}+\frac {b (a+b \text {ArcTan}(c x))}{d^2 (i-c x)}-\frac {(a+b \text {ArcTan}(c x))^2}{2 d^2}+\frac {i (a+b \text {ArcTan}(c x))^2}{d^2 (i-c x)}+\frac {2 (a+b \text {ArcTan}(c x))^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^2}+\frac {(a+b \text {ArcTan}(c x))^2 \log \left (\frac {2}{1+i c x}\right )}{d^2}+\frac {i b (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d^2}+\frac {b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d^2} \]

[Out]

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

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Rubi [A]
time = 0.45, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {4996, 4942, 5108, 5004, 5114, 6745, 4974, 4972, 641, 46, 209, 4964} \begin {gather*} \frac {i b \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))}{d^2}+\frac {b (a+b \text {ArcTan}(c x))}{d^2 (-c x+i)}+\frac {i (a+b \text {ArcTan}(c x))^2}{d^2 (-c x+i)}-\frac {(a+b \text {ArcTan}(c x))^2}{2 d^2}+\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2}{d^2}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2}{d^2}+\frac {i b^2 \text {ArcTan}(c x)}{2 d^2}+\frac {b^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )}{2 d^2}-\frac {i b^2}{2 d^2 (-c x+i)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*b^2)/(d^2*(I - c*x)) + ((I/2)*b^2*ArcTan[c*x])/d^2 + (b*(a + b*ArcTan[c*x]))/(d^2*(I - c*x)) - (a +
b*ArcTan[c*x])^2/(2*d^2) + (I*(a + b*ArcTan[c*x])^2)/(d^2*(I - c*x)) + (2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/
(1 + I*c*x)])/d^2 + ((a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/d^2 + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, -1 +
2/(1 + I*c*x)])/d^2 + (b^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/(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 4942

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

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 5108

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

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

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Mathematica [A]
time = 0.55, size = 299, normalized size = 1.35 \begin {gather*} \frac {-\frac {24 i a^2}{-i+c x}-24 i a^2 \text {ArcTan}(c x)+24 a^2 \log (c x)-12 a^2 \log \left (1+c^2 x^2\right )-12 a b \left (4 i \text {ArcTan}(c x)^2+i \cos (2 \text {ArcTan}(c x))+2 i \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )-2 \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 )+\sin (2 \text {ArcTan}(c x))\right )+b^2 \left (-i \pi ^3-6 \cos (2 \text {ArcTan}(c x))-12 i \text {ArcTan}(c x) \cos (2 \text {ArcTan}(c x))+12 \text {ArcTan}(c x)^2 \cos (2 \text {ArcTan}(c x))+24 \text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )+24 i \text {ArcTan}(c x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )+12 \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )+6 i \sin (2 \text {ArcTan}(c x))-12 \text {ArcTan}(c x) \sin (2 \text {ArcTan}(c x))-12 i \text {ArcTan}(c x)^2 \sin (2 \text {ArcTan}(c x))\right )}{24 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-24*I)*a^2)/(-I + c*x) - (24*I)*a^2*ArcTan[c*x] + 24*a^2*Log[c*x] - 12*a^2*Log[1 + c^2*x^2] - 12*a*b*((4*I)
*ArcTan[c*x]^2 + I*Cos[2*ArcTan[c*x]] + (2*I)*PolyLog[2, E^((2*I)*ArcTan[c*x])] - 2*ArcTan[c*x]*(Cos[2*ArcTan[
c*x]] + 2*Log[1 - E^((2*I)*ArcTan[c*x])] - I*Sin[2*ArcTan[c*x]]) + Sin[2*ArcTan[c*x]]) + b^2*((-I)*Pi^3 - 6*Co
s[2*ArcTan[c*x]] - (12*I)*ArcTan[c*x]*Cos[2*ArcTan[c*x]] + 12*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] + 24*ArcTan[c*x
]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + 12*PolyLog[3, E^
((-2*I)*ArcTan[c*x])] + (6*I)*Sin[2*ArcTan[c*x]] - 12*ArcTan[c*x]*Sin[2*ArcTan[c*x]] - (12*I)*ArcTan[c*x]^2*Si
n[2*ArcTan[c*x]]))/(24*d^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.94, size = 1921, normalized size = 8.69

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1921\)
default	\(\text {Expression too large to display}\)	\(1921\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*b*a/d^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-1/2*I*b^2/d^2*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2
+1)+1))^2*arctan(c*x)^2-1/2*a^2/d^2*ln(c^2*x^2+1)-1/2*b^2/d^2*arctan(c*x)^2+a^2/d^2*ln(c*x)-2*b*a/d^2*arctan(c
*x)*ln(c*x-I)-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))*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))^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(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^
2*x^2+1)+1))^3*arctan(c*x)^2+1/2*I*b^2/d^2*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*
arctan(c*x)^2+I*b*a/d^2*ln(c*x)*ln(1+I*c*x)-I*b*a/d^2*ln(c*x)*ln(1-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))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+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))^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+3/2*I*b^2/d^2*Pi*arctan(c*x)^2-2*
I*b^2/d^2*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*b^2/d^2*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2
*x^2+1)^(1/2))+2*b*a/d^2*arctan(c*x)*ln(c*x)+I*b*a/d^2*dilog(1+I*c*x)-I*b*a/d^2*dilog(1-I*c*x)+1/2*I*b^2/d^2*P
i*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-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+b^2/d^2*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^2/d^2*arct
an(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-b^2/d^2*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+b^2/d^2*arctan
(c*x)^2*ln(c*x)-I*a^2/d^2/(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*I*b^2/d^2*arctan(c*x)/(4*c*x-4*I)*c*x-1/2*I*b^2/d^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*
((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-2*b^2/d^2*arctan(c*x)/(4*c*x-4*I)-2/3
*I*b^2/d^2*arctan(c*x)^3-2*I*b*a/d^2*arctan(c*x)/(c*x-I)-1/2*I*b^2/d^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(
(1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-1/2*
I*b^2/d^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-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(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I
*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+2*b^2/d^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1
/2))+2*b^2/d^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-I*a^2/d^2*arctan(c*x)+1/4*I*b^2/d^2/(c*x-I)-b*a/d^2/(c*
x-I)-b*a/d^2*arctan(c*x)+I*b*a/d^2*dilog(-1/2*I*(c*x+I))-1/2*I*b*a/d^2*ln(c*x-I)^2+1/4*b^2/d^2/(c*x-I)*c*x-I*b
^2/d^2*arctan(c*x)^2/(c*x-I)

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

[Out]

a^2*(-I/(c*d^2*x - I*d^2) - log(c*x - I)/d^2 + log(x)/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 - (6*b^2*c^4*(c^2/(c^8*d^2*x^2 + c^6*d^2) + log(c^2*x^2 + 1)/(c^6*d^2*x^2 + c^4*d^2)) - 256*b^2*c^4*i
ntegrate(1/16*x^4*arctan(c*x)^2/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x) - 64*b^2*c^4*integrate(1/16*x^4*log(
c^2*x^2 + 1)^2/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x) - 256*b^2*c^3*integrate(1/16*x^3*arctan(c*x)/(c^4*d^2
*x^5 + 2*c^2*d^2*x^3 + d^2*x), x) - 16*(c*(x/(c^4*d^2*x^2 + c^2*d^2) + arctan(c*x)/(c^3*d^2)) - 2*arctan(c*x)/
(c^4*d^2*x^2 + c^2*d^2))*a*b*c^2 - 640*b^2*c^2*integrate(1/16*x^2*arctan(c*x)^2/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 +
d^2*x), x) + 3*b^2*c^2*log(c^2*x^2 + 1)^2/(c^4*d^2*x^2 + c^2*d^2) - 256*b^2*c*integrate(1/16*x*arctan(c*x)*lo
g(c^2*x^2 + 1)/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x) - 256*b^2*c*integrate(1/16*x*arctan(c*x)/(c^4*d^2*x^5
+ 2*c^2*d^2*x^3 + d^2*x), x) + 384*b^2*integrate(1/16*arctan(c*x)^2/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x)
+ 32*b^2*integrate(1/16*log(c^2*x^2 + 1)^2/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x) + 1024*a*b*integrate(1/1
6*arctan(c*x)/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x))*(c*d^2*x - I*d^2) - 32*(I*c*d^2*x + d^2)*integrate(1/
8*(b^2*c^3*x^3*log(c^2*x^2 + 1)^2 - 32*a*b*c*x*arctan(c*x) + 4*(b^2*c^3*x^3 - 2*b^2*c*x)*arctan(c*x)^2 - 2*(b^
2*c^3*x^3 + b^2*c*x - (b^2*c^2*x^2 - b^2)*arctan(c*x))*log(c^2*x^2 + 1))/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x)
, x) - 4*(2*b^2*arctan(c*x) + (b^2*c*x - I*b^2)*arctan(c*x)^2)*log(c^2*x^2 + 1))/(c*d^2*x - I*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((a+b*arctan(c*x))^2/x/(d+I*c*d*x)^2,x, algorithm="fricas")

[Out]

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

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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((a+b*atan(c*x))**2/x/(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((a+b*arctan(c*x))^2/x/(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 {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x\,{\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((a + b*atan(c*x))^2/(x*(d + c*d*x*1i)^2),x)

[Out]

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

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