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

Optimal. Leaf size=293 \[ \frac{2 b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^2}-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{17 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^2}-\frac{4 i b d^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^2}+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{6} b c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{2}{3} i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{3 a b d^2 x}{2 c}+\frac{5 b^2 d^2 \log \left (c^2 x^2+1\right )}{6 c^2}-\frac{2 i b^2 d^2 \tan ^{-1}(c x)}{3 c^2}+\frac{2 i b^2 d^2 x}{3 c}-\frac{3 b^2 d^2 x \tan ^{-1}(c x)}{2 c}-\frac{1}{12} b^2 d^2 x^2 \]

[Out]

(-3*a*b*d^2*x)/(2*c) + (((2*I)/3)*b^2*d^2*x)/c - (b^2*d^2*x^2)/12 - (((2*I)/3)*b^2*d^2*ArcTan[c*x])/c^2 - (3*b
^2*d^2*x*ArcTan[c*x])/(2*c) - ((2*I)/3)*b*d^2*x^2*(a + b*ArcTan[c*x]) + (b*c*d^2*x^3*(a + b*ArcTan[c*x]))/6 +
(17*d^2*(a + b*ArcTan[c*x])^2)/(12*c^2) + (d^2*x^2*(a + b*ArcTan[c*x])^2)/2 + ((2*I)/3)*c*d^2*x^3*(a + b*ArcTa
n[c*x])^2 - (c^2*d^2*x^4*(a + b*ArcTan[c*x])^2)/4 - (((4*I)/3)*b*d^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c
^2 + (5*b^2*d^2*Log[1 + c^2*x^2])/(6*c^2) + (2*b^2*d^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(3*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.621398, antiderivative size = 293, normalized size of antiderivative = 1., number of steps used = 28, number of rules used = 14, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.609, Rules used = {4876, 4852, 4916, 4846, 260, 4884, 321, 203, 4920, 4854, 2402, 2315, 266, 43} \[ \frac{2 b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^2}-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{17 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^2}-\frac{4 i b d^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^2}+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{6} b c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{2}{3} i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{3 a b d^2 x}{2 c}+\frac{5 b^2 d^2 \log \left (c^2 x^2+1\right )}{6 c^2}-\frac{2 i b^2 d^2 \tan ^{-1}(c x)}{3 c^2}+\frac{2 i b^2 d^2 x}{3 c}-\frac{3 b^2 d^2 x \tan ^{-1}(c x)}{2 c}-\frac{1}{12} b^2 d^2 x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*b*d^2*x)/(2*c) + (((2*I)/3)*b^2*d^2*x)/c - (b^2*d^2*x^2)/12 - (((2*I)/3)*b^2*d^2*ArcTan[c*x])/c^2 - (3*b
^2*d^2*x*ArcTan[c*x])/(2*c) - ((2*I)/3)*b*d^2*x^2*(a + b*ArcTan[c*x]) + (b*c*d^2*x^3*(a + b*ArcTan[c*x]))/6 +
(17*d^2*(a + b*ArcTan[c*x])^2)/(12*c^2) + (d^2*x^2*(a + b*ArcTan[c*x])^2)/2 + ((2*I)/3)*c*d^2*x^3*(a + b*ArcTa
n[c*x])^2 - (c^2*d^2*x^4*(a + b*ArcTan[c*x])^2)/4 - (((4*I)/3)*b*d^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c
^2 + (5*b^2*d^2*Log[1 + c^2*x^2])/(6*c^2) + (2*b^2*d^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(3*c^2)

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 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x (d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+2 i c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-c^2 d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^2 \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+\left (2 i c d^2\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx-\left (c^2 d^2\right ) \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\left (b c d^2\right ) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac{1}{3} \left (4 i b c^2 d^2\right ) \int \frac{x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx+\frac{1}{2} \left (b c^3 d^2\right ) \int \frac{x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{3} \left (4 i b d^2\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac{1}{3} \left (4 i b d^2\right ) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac{\left (b d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c}+\frac{\left (b d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c}+\frac{1}{2} \left (b c d^2\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx-\frac{1}{2} \left (b c d^2\right ) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-\frac{a b d^2 x}{c}-\frac{2}{3} i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} b c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{7 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^2}+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{\left (4 i b d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{3 c}-\frac{\left (b d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}+\frac{\left (b d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c}-\frac{\left (b^2 d^2\right ) \int \tan ^{-1}(c x) \, dx}{c}+\frac{1}{3} \left (2 i b^2 c d^2\right ) \int \frac{x^2}{1+c^2 x^2} \, dx-\frac{1}{6} \left (b^2 c^2 d^2\right ) \int \frac{x^3}{1+c^2 x^2} \, dx\\ &=-\frac{3 a b d^2 x}{2 c}+\frac{2 i b^2 d^2 x}{3 c}-\frac{b^2 d^2 x \tan ^{-1}(c x)}{c}-\frac{2}{3} i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} b c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{17 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^2}+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^2}+\left (b^2 d^2\right ) \int \frac{x}{1+c^2 x^2} \, dx-\frac{\left (2 i b^2 d^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{3 c}+\frac{\left (4 i b^2 d^2\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{3 c}-\frac{\left (b^2 d^2\right ) \int \tan ^{-1}(c x) \, dx}{2 c}-\frac{1}{12} \left (b^2 c^2 d^2\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{3 a b d^2 x}{2 c}+\frac{2 i b^2 d^2 x}{3 c}-\frac{2 i b^2 d^2 \tan ^{-1}(c x)}{3 c^2}-\frac{3 b^2 d^2 x \tan ^{-1}(c x)}{2 c}-\frac{2}{3} i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} b c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{17 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^2}+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^2}+\frac{b^2 d^2 \log \left (1+c^2 x^2\right )}{2 c^2}+\frac{1}{2} \left (b^2 d^2\right ) \int \frac{x}{1+c^2 x^2} \, dx+\frac{\left (4 b^2 d^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{3 c^2}-\frac{1}{12} \left (b^2 c^2 d^2\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{3 a b d^2 x}{2 c}+\frac{2 i b^2 d^2 x}{3 c}-\frac{1}{12} b^2 d^2 x^2-\frac{2 i b^2 d^2 \tan ^{-1}(c x)}{3 c^2}-\frac{3 b^2 d^2 x \tan ^{-1}(c x)}{2 c}-\frac{2}{3} i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{6} b c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{17 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^2}+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^2}+\frac{5 b^2 d^2 \log \left (1+c^2 x^2\right )}{6 c^2}+\frac{2 b^2 d^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{3 c^2}\\ \end{align*}

Mathematica [A]  time = 0.775093, size = 257, normalized size = 0.88 \[ -\frac{d^2 \left (8 b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+3 a^2 c^4 x^4-8 i a^2 c^3 x^3-6 a^2 c^2 x^2-2 a b c^3 x^3+8 i a b c^2 x^2-8 i a b \log \left (c^2 x^2+1\right )+2 b \tan ^{-1}(c x) \left (a \left (3 c^4 x^4-8 i c^3 x^3-6 c^2 x^2-9\right )+b \left (-c^3 x^3+4 i c^2 x^2+9 c x+4 i\right )+8 i b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+18 a b c x+b^2 c^2 x^2-10 b^2 \log \left (c^2 x^2+1\right )-8 i b^2 c x+b^2 (c x-i)^3 (3 c x+i) \tan ^{-1}(c x)^2+b^2\right )}{12 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(d^2*(b^2 + 18*a*b*c*x - (8*I)*b^2*c*x - 6*a^2*c^2*x^2 + (8*I)*a*b*c^2*x^2 + b^2*c^2*x^2 - (8*I)*a^2*c^3*x^3
- 2*a*b*c^3*x^3 + 3*a^2*c^4*x^4 + b^2*(-I + c*x)^3*(I + 3*c*x)*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]*(b*(4*I + 9*c*x
+ (4*I)*c^2*x^2 - c^3*x^3) + a*(-9 - 6*c^2*x^2 - (8*I)*c^3*x^3 + 3*c^4*x^4) + (8*I)*b*Log[1 + E^((2*I)*ArcTan
[c*x])]) - (8*I)*a*b*Log[1 + c^2*x^2] - 10*b^2*Log[1 + c^2*x^2] + 8*b^2*PolyLog[2, -E^((2*I)*ArcTan[c*x])]))/(
12*c^2)

________________________________________________________________________________________

Maple [B]  time = 0.097, size = 556, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c^2*d^2*a^2*x^4+1/3/c^2*d^2*b^2*dilog(-1/2*I*(c*x+I))+3/4/c^2*d^2*b^2*arctan(c*x)^2-1/6/c^2*d^2*b^2*ln(c*
x+I)^2+1/6/c^2*d^2*b^2*ln(c*x-I)^2-1/3/c^2*d^2*b^2*dilog(1/2*I*(c*x-I))+1/2*d^2*b^2*arctan(c*x)^2*x^2+4/3*I*c*
d^2*a*b*arctan(c*x)*x^3+2/3*I*c*d^2*a^2*x^3+d^2*a*b*arctan(c*x)*x^2-2/3*I*d^2*a*b*x^2+1/3/c^2*d^2*b^2*ln(c*x+I
)*ln(c^2*x^2+1)+1/3/c^2*d^2*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-2/3*I*d^2*b^2*arctan(c*x)*x^2-1/4*c^2*d^2*b^2*arc
tan(c*x)^2*x^4+3/2/c^2*d^2*a*b*arctan(c*x)-1/3/c^2*d^2*b^2*ln(c*x-I)*ln(c^2*x^2+1)+1/6*c*d^2*b^2*arctan(c*x)*x
^3-1/3/c^2*d^2*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))+1/6*c*d^2*a*b*x^3-1/12*b^2*d^2*x^2+2/3*I/c^2*d^2*b^2*arctan(c*x
)*ln(c^2*x^2+1)+2/3*I/c^2*d^2*a*b*ln(c^2*x^2+1)+2/3*I*c*d^2*b^2*arctan(c*x)^2*x^3-1/2*c^2*d^2*a*b*arctan(c*x)*
x^4+1/2*d^2*a^2*x^2+2/3*I*b^2*d^2*x/c-2/3*I*b^2*d^2*arctan(c*x)/c^2-3/2*a*b*d^2*x/c-3/2*b^2*d^2*x*arctan(c*x)/
c+5/6*b^2*d^2*ln(c^2*x^2+1)/c^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*c^2*d^2*x^4 + 2/3*I*a^2*c*d^2*x^3 + 1/2*b^2*d^2*x^2*arctan(c*x)^2 - 1/6*(3*x^4*arctan(c*x) - c*((c^2*
x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*a*b*c^2*d^2 + 2/3*I*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^
4))*a*b*c*d^2 + 1/2*a^2*d^2*x^2 + (x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a*b*d^2 - 1/2*(2*c*(x/c^2 -
arctan(c*x)/c^3)*arctan(c*x) + (arctan(c*x)^2 - log(c^2*x^2 + 1))/c^2)*b^2*d^2 - 1/48*(3*b^2*c^2*d^2*x^4 - 8*I
*b^2*c*d^2*x^3)*arctan(c*x)^2 - 1/192*(12*I*b^2*c^2*d^2*x^4 + 32*b^2*c*d^2*x^3)*arctan(c*x)*log(c^2*x^2 + 1) +
1/192*(3*b^2*c^2*d^2*x^4 - 8*I*b^2*c*d^2*x^3)*log(c^2*x^2 + 1)^2 - integrate(-1/48*(22*b^2*c^3*d^2*x^4*arctan
(c*x) - 36*(b^2*c^4*d^2*x^5 + b^2*c^2*d^2*x^3)*arctan(c*x)^2 - 3*(b^2*c^4*d^2*x^5 + b^2*c^2*d^2*x^3)*log(c^2*x
^2 + 1)^2 - (3*b^2*c^4*d^2*x^5 - 8*b^2*c^2*d^2*x^3 - 24*(b^2*c^3*d^2*x^4 + b^2*c*d^2*x^2)*arctan(c*x))*log(c^2
*x^2 + 1))/(c^2*x^2 + 1), x) + I*integrate(1/48*(72*(b^2*c^3*d^2*x^4 + b^2*c*d^2*x^2)*arctan(c*x)^2 + 6*(b^2*c
^3*d^2*x^4 + b^2*c*d^2*x^2)*log(c^2*x^2 + 1)^2 + 2*(3*b^2*c^4*d^2*x^5 - 8*b^2*c^2*d^2*x^3)*arctan(c*x) + (11*b
^2*c^3*d^2*x^4 + 12*(b^2*c^4*d^2*x^5 + b^2*c^2*d^2*x^3)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^2 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{48} \,{\left (3 \, b^{2} c^{2} d^{2} x^{4} - 8 i \, b^{2} c d^{2} x^{3} - 6 \, b^{2} d^{2} x^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} +{\rm integral}\left (-\frac{12 \, a^{2} c^{4} d^{2} x^{5} - 24 i \, a^{2} c^{3} d^{2} x^{4} - 24 i \, a^{2} c d^{2} x^{2} - 12 \, a^{2} d^{2} x -{\left (-12 i \, a b c^{4} d^{2} x^{5} - 3 \,{\left (8 \, a b - i \, b^{2}\right )} c^{3} d^{2} x^{4} + 8 \, b^{2} c^{2} d^{2} x^{3} - 6 \,{\left (4 \, a b + i \, b^{2}\right )} c d^{2} x^{2} + 12 i \, a b d^{2} x\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{12 \,{\left (c^{2} x^{2} + 1\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*b^2*c^2*d^2*x^4 - 8*I*b^2*c*d^2*x^3 - 6*b^2*d^2*x^2)*log(-(c*x + I)/(c*x - I))^2 + integral(-1/12*(12*
a^2*c^4*d^2*x^5 - 24*I*a^2*c^3*d^2*x^4 - 24*I*a^2*c*d^2*x^2 - 12*a^2*d^2*x - (-12*I*a*b*c^4*d^2*x^5 - 3*(8*a*b
- I*b^2)*c^3*d^2*x^4 + 8*b^2*c^2*d^2*x^3 - 6*(4*a*b + I*b^2)*c*d^2*x^2 + 12*I*a*b*d^2*x)*log(-(c*x + I)/(c*x
- I)))/(c^2*x^2 + 1), 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*(d+I*c*d*x)**2*(a+b*atan(c*x))**2,x)

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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