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

Optimal. Leaf size=277 \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}+\frac{i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^3 d}-\frac{i b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^3 d}+\frac{i a b x}{c^2 d}+\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac{2 b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}-\frac{i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}-\frac{i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}-\frac{i b^2 \log \left (c^2 x^2+1\right )}{2 c^3 d}+\frac{i b^2 x \tan ^{-1}(c x)}{c^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.513029, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {4866, 4852, 4916, 4846, 260, 4884, 4920, 4854, 2402, 2315, 4994, 6610} \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}+\frac{i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^3 d}-\frac{i b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^3 d}+\frac{i a b x}{c^2 d}+\frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac{2 b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}-\frac{i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}-\frac{i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}-\frac{i b^2 \log \left (c^2 x^2+1\right )}{2 c^3 d}+\frac{i b^2 x \tan ^{-1}(c x)}{c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4866

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

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 4916

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

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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

Mathematica [A]  time = 0.554076, size = 330, normalized size = 1.19 \[ -\frac{i \left (6 b \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right ) \left (-i a-i b \tan ^{-1}(c x)+b\right )+3 b^2 \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )+3 a^2 c^2 x^2-3 a^2 \log \left (c^2 x^2+1\right )+6 i a^2 c x-6 i a^2 \tan ^{-1}(c x)-6 i a b \log \left (c^2 x^2+1\right )+6 a b c^2 x^2 \tan ^{-1}(c x)-6 a b c x-12 i a b \tan ^{-1}(c x)^2+6 a b \tan ^{-1}(c x)+12 i a b c x \tan ^{-1}(c x)+12 a b \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+3 b^2 \log \left (c^2 x^2+1\right )+3 b^2 c^2 x^2 \tan ^{-1}(c x)^2-4 i b^2 \tan ^{-1}(c x)^3+9 b^2 \tan ^{-1}(c x)^2+6 i b^2 c x \tan ^{-1}(c x)^2-6 b^2 c x \tan ^{-1}(c x)+6 b^2 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+12 i b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{6 c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I/6)*((6*I)*a^2*c*x - 6*a*b*c*x + 3*a^2*c^2*x^2 - (6*I)*a^2*ArcTan[c*x] + 6*a*b*ArcTan[c*x] + (12*I)*a*b*c*
x*ArcTan[c*x] - 6*b^2*c*x*ArcTan[c*x] + 6*a*b*c^2*x^2*ArcTan[c*x] - (12*I)*a*b*ArcTan[c*x]^2 + 9*b^2*ArcTan[c*
x]^2 + (6*I)*b^2*c*x*ArcTan[c*x]^2 + 3*b^2*c^2*x^2*ArcTan[c*x]^2 - (4*I)*b^2*ArcTan[c*x]^3 + 12*a*b*ArcTan[c*x
]*Log[1 + E^((2*I)*ArcTan[c*x])] + (12*I)*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 6*b^2*ArcTan[c*x]^2
*Log[1 + E^((2*I)*ArcTan[c*x])] - 3*a^2*Log[1 + c^2*x^2] - (6*I)*a*b*Log[1 + c^2*x^2] + 3*b^2*Log[1 + c^2*x^2]
 + 6*b*((-I)*a + b - I*b*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 3*b^2*PolyLog[3, -E^((2*I)*ArcTan[c
*x])]))/(c^3*d)

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Maple [C]  time = 1.643, size = 1212, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

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),x)

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 d x - 4 i \, d}, x\right ) \end{align*}

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),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*d*x
- 4*I*d), x)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{i \, c d x + d}\,{d x} \end{align*}

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),x, algorithm="giac")

[Out]

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