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

Optimal. Leaf size=317 \[ -c^2 d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+2 b c d^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 b c d^2 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )-2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-2 b c d^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 b c d^2 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+4 i c d^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-i b^2 c d^2 \text {Li}_2\left (\frac {2}{1-i c x}-1\right )-i b^2 c d^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )-i b^2 c d^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+i b^2 c d^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \]

[Out]

-2*I*c*d^2*(a+b*arctan(c*x))^2-d^2*(a+b*arctan(c*x))^2/x-c^2*d^2*x*(a+b*arctan(c*x))^2-4*I*c*d^2*(a+b*arctan(c
*x))^2*arctanh(-1+2/(1+I*c*x))-2*b*c*d^2*(a+b*arctan(c*x))*ln(2/(1+I*c*x))+2*b*c*d^2*(a+b*arctan(c*x))*ln(2-2/
(1-I*c*x))-I*b^2*c*d^2*polylog(2,-1+2/(1-I*c*x))-I*b^2*c*d^2*polylog(2,1-2/(1+I*c*x))+2*b*c*d^2*(a+b*arctan(c*
x))*polylog(2,1-2/(1+I*c*x))-2*b*c*d^2*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))-I*b^2*c*d^2*polylog(3,1-2/(
1+I*c*x))+I*b^2*c*d^2*polylog(3,-1+2/(1+I*c*x))

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Rubi [A]  time = 0.62, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4852, 4924, 4868, 2447, 4850, 4988, 4884, 4994, 6610} \[ 2 b c d^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 b c d^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-i b^2 c d^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )-i b^2 c d^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-i b^2 c d^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+i b^2 c d^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )-c^2 d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-2 b c d^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 b c d^2 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+4 i c d^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*I)*c*d^2*(a + b*ArcTan[c*x])^2 - (d^2*(a + b*ArcTan[c*x])^2)/x - c^2*d^2*x*(a + b*ArcTan[c*x])^2 + (4*I)*c
*d^2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] - 2*b*c*d^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)] + 2*b
*c*d^2*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)] - I*b^2*c*d^2*PolyLog[2, -1 + 2/(1 - I*c*x)] - I*b^2*c*d^2*P
olyLog[2, 1 - 2/(1 + I*c*x)] + 2*b*c*d^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] - 2*b*c*d^2*(a + b*
ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] - I*b^2*c*d^2*PolyLog[3, 1 - 2/(1 + I*c*x)] + I*b^2*c*d^2*PolyLog[
3, -1 + 2/(1 + I*c*x)]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4846

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

Rule 4850

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 4852

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

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 4868

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTan[c*x]
)^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2 - 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 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 4920

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

Rule 4924

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

Rule 4988

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[(
Log[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 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 {(d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx &=\int \left (-c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x^2}+\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d^2 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx+\left (2 i c d^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx-\left (c^2 d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-c^2 d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+4 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\left (2 b c d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx-\left (8 i b c^2 d^2\right ) \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+\left (2 b c^3 d^2\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-c^2 d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+4 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\left (2 i b c d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx+\left (4 i b c^2 d^2\right ) \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-\left (4 i b c^2 d^2\right ) \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-\left (2 b c^2 d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx\\ &=-2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-c^2 d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+4 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )+2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\left (2 b^2 c^2 d^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 b^2 c^2 d^2\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx-\left (2 b^2 c^2 d^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (2 b^2 c^2 d^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-c^2 d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+4 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )+2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d^2 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )+2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-i b^2 c d^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+i b^2 c d^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )-\left (2 i b^2 c d^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )\\ &=-2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-c^2 d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+4 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )+2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d^2 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )-i b^2 c d^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-2 b c d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-i b^2 c d^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+i b^2 c d^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.68, size = 378, normalized size = 1.19 \[ -\frac {d^2 \left (12 a^2 c^2 x^2-24 i a^2 c x \log (c x)+12 a^2+24 a b c^2 x^2 \tan ^{-1}(c x)+24 a b c x \text {Li}_2(-i c x)-24 a b c x \text {Li}_2(i c x)-24 a b c x \log (c x)+24 a b \tan ^{-1}(c x)+12 b^2 c^2 x^2 \tan ^{-1}(c x)^2+24 b^2 c x \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+12 b^2 c x \left (2 \tan ^{-1}(c x)-i\right ) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+12 i b^2 c x \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-12 i b^2 c x \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )+12 i b^2 c x \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )-\pi ^3 b^2 c x+16 b^2 c x \tan ^{-1}(c x)^3+12 b^2 \tan ^{-1}(c x)^2-24 i b^2 c x \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+24 i b^2 c x \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-24 b^2 c x \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+24 b^2 c x \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{12 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/12*(d^2*(12*a^2 - b^2*c*Pi^3*x + 12*a^2*c^2*x^2 + 24*a*b*ArcTan[c*x] + 24*a*b*c^2*x^2*ArcTan[c*x] + 12*b^2*
ArcTan[c*x]^2 + 12*b^2*c^2*x^2*ArcTan[c*x]^2 + 16*b^2*c*x*ArcTan[c*x]^3 - (24*I)*b^2*c*x*ArcTan[c*x]^2*Log[1 -
E^((-2*I)*ArcTan[c*x])] - 24*b^2*c*x*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + 24*b^2*c*x*ArcTan[c*x]*Log[
1 + E^((2*I)*ArcTan[c*x])] + (24*I)*b^2*c*x*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - (24*I)*a^2*c*x*Log[
c*x] - 24*a*b*c*x*Log[c*x] + 24*b^2*c*x*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + 12*b^2*c*x*(-I + 2*Ar
cTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + (12*I)*b^2*c*x*PolyLog[2, E^((2*I)*ArcTan[c*x])] + 24*a*b*c*x*
PolyLog[2, (-I)*c*x] - 24*a*b*c*x*PolyLog[2, I*c*x] - (12*I)*b^2*c*x*PolyLog[3, E^((-2*I)*ArcTan[c*x])] + (12*
I)*b^2*c*x*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/x

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {4 \, a^{2} c^{2} d^{2} x^{2} - 8 i \, a^{2} c d^{2} x - 4 \, a^{2} d^{2} - {\left (b^{2} c^{2} d^{2} x^{2} - 2 i \, b^{2} c d^{2} x - b^{2} d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - {\left (-4 i \, a b c^{2} d^{2} x^{2} - 8 \, a b c d^{2} x + 4 i \, a b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{4 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-1/4*(4*a^2*c^2*d^2*x^2 - 8*I*a^2*c*d^2*x - 4*a^2*d^2 - (b^2*c^2*d^2*x^2 - 2*I*b^2*c*d^2*x - b^2*d^2)
*log(-(c*x + I)/(c*x - I))^2 - (-4*I*a*b*c^2*d^2*x^2 - 8*a*b*c*d^2*x + 4*I*a*b*d^2)*log(-(c*x + I)/(c*x - I)))
/x^2, 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((d+I*c*d*x)^2*(a+b*arctan(c*x))^2/x^2,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 2.58, size = 11959, normalized size = 37.73 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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