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

Optimal. Leaf size=430 \[ \frac{i b d^2 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^3}-\frac{i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^3}-\frac{b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 e^3}+\frac{b^2 d^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^3}-\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c e^2}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 e}-\frac{d^2 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e^3}+\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^3}-\frac{d x \left (a+b \tan ^{-1}(c x)\right )^2}{e^2}-\frac{i d \left (a+b \tan ^{-1}(c x)\right )^2}{c e^2}-\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c e^2}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}-\frac{a b x}{c e}+\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^2 e}-\frac{b^2 x \tan ^{-1}(c x)}{c e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.424568, antiderivative size = 430, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.524, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4852, 4916, 260, 4884, 4858} \[ \frac{i b d^2 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^3}-\frac{i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^3}-\frac{b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 e^3}+\frac{b^2 d^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^3}-\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c e^2}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 e}-\frac{d^2 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e^3}+\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^3}-\frac{d x \left (a+b \tan ^{-1}(c x)\right )^2}{e^2}-\frac{i d \left (a+b \tan ^{-1}(c x)\right )^2}{c e^2}-\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c e^2}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}-\frac{a b x}{c e}+\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^2 e}-\frac{b^2 x \tan ^{-1}(c x)}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 4858

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^2*Log[2/
(1 - I*c*x)])/e, x] + (Simp[((a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] + Sim
p[(I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e, x] - Simp[(I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 -
(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] - Simp[(b^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*e), x] + Simp
[(b^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && Ne
Q[c^2*d^2 + e^2, 0]

Rubi steps

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

Mathematica [F]  time = 122.316, size = 0, normalized size = 0. \[ \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d+e x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [C]  time = 8.652, size = 1784, normalized size = 4.2 \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/(e*x+d),x)

[Out]

1/2*I*b^2/e^3*d^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/
(c^2*x^2+1)+I*e+d*c))*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(
c^2*x^2+1)+1))*arctan(c*x)^2-a*b*x/c/e-b^2*x*arctan(c*x)/c/e-1/2*I*b^2/e^3*d^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2
+1)+1))*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))
^2*arctan(c*x)^2-1/2*I*b^2/e^3*d^2*Pi*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c
))*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*ar
ctan(c*x)^2-I*c*b^2*d^3/e^3/(d*c-I*e)*arctan(c*x)*polylog(2,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))-1/c*a
*b*d/e^2-a^2*d/e^2*x+1/2*b^2*arctan(c*x)^2*x^2/e+a^2*d^2/e^3*ln(c*e*x+c*d)-1/2*b^2*d^2/e^3*polylog(3,-(1+I*c*x
)^2/(c^2*x^2+1))+1/2/c^2*b^2*arctan(c*x)^2/e-1/c^2*b^2/e*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+I*a*b/e^3*d^2*ln(c*e*x+
c*d)*ln((I*e-e*c*x)/(d*c+I*e))+c*b^2*d^3/e^3/(d*c-I*e)*arctan(c*x)^2*ln(1-(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2
*x^2+1))-I*b^2*d^2/e^2/(d*c-I*e)*arctan(c*x)^2*ln(1-(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+1/2*I*b^2/e^3
*d^2*Pi*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))
^3*arctan(c*x)^2-I*a*b/e^3*d^2*ln(c*e*x+c*d)*ln((I*e+e*c*x)/(I*e-d*c))+1/c*a*b/e^2*d*ln(c^2*d^2-2*(c*e*x+c*d)*
c*d+(c*e*x+c*d)^2+e^2)+1/2*c*b^2*d^3/e^3/(d*c-I*e)*polylog(3,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+I/c*
b^2/e^2*d*arctan(c*x)^2-2/c*b^2/e^2*d*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-2/c*b^2/e^2*d*arctan(c*x
)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*a*b/e^3*d^2*dilog((I*e-e*c*x)/(d*c+I*e))-2*a*b*arctan(c*x)*d/e^2*x-b^2
*d^2/e^2/(d*c-I*e)*arctan(c*x)*polylog(2,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+I*b^2*d^2/e^3*arctan(c*x
)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+2*a*b*arctan(c*x)*d^2/e^3*ln(c*e*x+c*d)+2*I/c*b^2/e^2*d*dilog(1+I*(1+I*c
*x)/(c^2*x^2+1)^(1/2))+2*I/c*b^2/e^2*d*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*I*b^2*d^2/e^2/(d*c-I*e)*poly
log(3,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))-I*a*b/e^3*d^2*dilog((I*e+e*c*x)/(I*e-d*c))+I/c^2*b^2*arctan
(c*x)/e+1/c^2*a*b/e*arctan(c*x)+a*b*arctan(c*x)*x^2/e-b^2*arctan(c*x)^2*d/e^2*x+b^2*arctan(c*x)^2*d^2/e^3*ln(c
*e*x+c*d)-b^2*d^2/e^3*arctan(c*x)^2*ln(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)+1/2*a
^2*x^2/e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{2}{\left (\frac{2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac{e x^{2} - 2 \, d x}{e^{2}}\right )} + \frac{4 \,{\left (b^{2} e x^{2} - 2 \, b^{2} d x\right )} \arctan \left (c x\right )^{2} + 2 \, e^{2} \int \frac{12 \,{\left (b^{2} c^{2} e^{2} x^{4} + b^{2} e^{2} x^{2}\right )} \arctan \left (c x\right )^{2} +{\left (b^{2} c^{2} e^{2} x^{4} + b^{2} e^{2} x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + 4 \,{\left (8 \, a b c^{2} e^{2} x^{4} - b^{2} c e^{2} x^{3} + 2 \, b^{2} c d^{2} x +{\left (b^{2} c d e + 8 \, a b e^{2}\right )} x^{2}\right )} \arctan \left (c x\right ) + 2 \,{\left (b^{2} c^{2} e^{2} x^{4} - b^{2} c^{2} d e x^{3} - 2 \, b^{2} c^{2} d^{2} x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{c^{2} e^{3} x^{3} + c^{2} d e^{2} x^{2} + e^{3} x + d e^{2}}\,{d x} -{\left (b^{2} e x^{2} - 2 \, b^{2} d x\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{32 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + 1/32*(4*(b^2*e*x^2 - 2*b^2*d*x)*arctan(c*x)^2 + 32*e^
2*integrate(1/16*(12*(b^2*c^2*e^2*x^4 + b^2*e^2*x^2)*arctan(c*x)^2 + (b^2*c^2*e^2*x^4 + b^2*e^2*x^2)*log(c^2*x
^2 + 1)^2 + 4*(8*a*b*c^2*e^2*x^4 - b^2*c*e^2*x^3 + 2*b^2*c*d^2*x + (b^2*c*d*e + 8*a*b*e^2)*x^2)*arctan(c*x) +
2*(b^2*c^2*e^2*x^4 - b^2*c^2*d*e*x^3 - 2*b^2*c^2*d^2*x^2)*log(c^2*x^2 + 1))/(c^2*e^3*x^3 + c^2*d*e^2*x^2 + e^3
*x + d*e^2), x) - (b^2*e*x^2 - 2*b^2*d*x)*log(c^2*x^2 + 1)^2)/e^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} x^{2} \arctan \left (c x\right )^{2} + 2 \, a b x^{2} \arctan \left (c x\right ) + a^{2} x^{2}}{e x + 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/(e*x+d),x, algorithm="fricas")

[Out]

integral((b^2*x^2*arctan(c*x)^2 + 2*a*b*x^2*arctan(c*x) + a^2*x^2)/(e*x + d), x)

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Sympy [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*atan(c*x))**2/(e*x+d),x)

[Out]

Timed out

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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}}{e 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/(e*x+d),x, algorithm="giac")

[Out]

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