3.1258 \(\int \frac{(d+e x^2)^2 (a+b \tan ^{-1}(c x))^2}{x} \, dx\)

Optimal. Leaf size=355 \[ -i b d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d^2 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )+\frac{d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}+\frac{a b e^2 x}{2 c^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+2 d^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{2 a b d e x}{c}+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{b^2 d e \log \left (c^2 x^2+1\right )}{c^2}+\frac{b^2 e^2 x^2}{12 c^2}-\frac{b^2 e^2 \log \left (c^2 x^2+1\right )}{3 c^4}+\frac{b^2 e^2 x \tan ^{-1}(c x)}{2 c^3}-\frac{2 b^2 d e x \tan ^{-1}(c x)}{c} \]

[Out]

(-2*a*b*d*e*x)/c + (a*b*e^2*x)/(2*c^3) + (b^2*e^2*x^2)/(12*c^2) - (2*b^2*d*e*x*ArcTan[c*x])/c + (b^2*e^2*x*Arc
Tan[c*x])/(2*c^3) - (b*e^2*x^3*(a + b*ArcTan[c*x]))/(6*c) + (d*e*(a + b*ArcTan[c*x])^2)/c^2 - (e^2*(a + b*ArcT
an[c*x])^2)/(4*c^4) + d*e*x^2*(a + b*ArcTan[c*x])^2 + (e^2*x^4*(a + b*ArcTan[c*x])^2)/4 + 2*d^2*(a + b*ArcTan[
c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + (b^2*d*e*Log[1 + c^2*x^2])/c^2 - (b^2*e^2*Log[1 + c^2*x^2])/(3*c^4) - I*b
*d^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*d^2*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*
c*x)] - (b^2*d^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/2 + (b^2*d^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/2

________________________________________________________________________________________

Rubi [A]  time = 0.689985, antiderivative size = 355, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 12, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.522, Rules used = {4980, 4850, 4988, 4884, 4994, 6610, 4852, 4916, 4846, 260, 266, 43} \[ -i b d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d^2 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )+\frac{d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}+\frac{a b e^2 x}{2 c^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+2 d^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{2 a b d e x}{c}+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{b^2 d e \log \left (c^2 x^2+1\right )}{c^2}+\frac{b^2 e^2 x^2}{12 c^2}-\frac{b^2 e^2 \log \left (c^2 x^2+1\right )}{3 c^4}+\frac{b^2 e^2 x \tan ^{-1}(c x)}{2 c^3}-\frac{2 b^2 d e x \tan ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b*d*e*x)/c + (a*b*e^2*x)/(2*c^3) + (b^2*e^2*x^2)/(12*c^2) - (2*b^2*d*e*x*ArcTan[c*x])/c + (b^2*e^2*x*Arc
Tan[c*x])/(2*c^3) - (b*e^2*x^3*(a + b*ArcTan[c*x]))/(6*c) + (d*e*(a + b*ArcTan[c*x])^2)/c^2 - (e^2*(a + b*ArcT
an[c*x])^2)/(4*c^4) + d*e*x^2*(a + b*ArcTan[c*x])^2 + (e^2*x^4*(a + b*ArcTan[c*x])^2)/4 + 2*d^2*(a + b*ArcTan[
c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + (b^2*d*e*Log[1 + c^2*x^2])/c^2 - (b^2*e^2*Log[1 + c^2*x^2])/(3*c^4) - I*b
*d^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*d^2*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*
c*x)] - (b^2*d^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/2 + (b^2*d^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/2

Rule 4980

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

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 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 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 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]

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 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 \frac{\left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx &=\int \left (\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 d e x \left (a+b \tan ^{-1}(c x)\right )^2+e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^2 \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx+(2 d e) \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+e^2 \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-\left (4 b c 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-(2 b c d e) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac{1}{2} \left (b c e^2\right ) \int \frac{x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 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{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 b c 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-\frac{(2 b d e) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c}+\frac{(2 b d e) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c}-\frac{\left (b e^2\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}+\frac{\left (b e^2\right ) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c}\\ &=-\frac{2 a b d e x}{c}-\frac{b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )+\left (i b^2 c d^2\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c d^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac{\left (2 b^2 d e\right ) \int \tan ^{-1}(c x) \, dx}{c}+\frac{1}{6} \left (b^2 e^2\right ) \int \frac{x^3}{1+c^2 x^2} \, dx+\frac{\left (b e^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac{\left (b e^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^3}\\ &=-\frac{2 a b d e x}{c}+\frac{a b e^2 x}{2 c^3}-\frac{2 b^2 d e x \tan ^{-1}(c x)}{c}-\frac{b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )+\left (2 b^2 d e\right ) \int \frac{x}{1+c^2 x^2} \, dx+\frac{1}{12} \left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )+\frac{\left (b^2 e^2\right ) \int \tan ^{-1}(c x) \, dx}{2 c^3}\\ &=-\frac{2 a b d e x}{c}+\frac{a b e^2 x}{2 c^3}-\frac{2 b^2 d e x \tan ^{-1}(c x)}{c}+\frac{b^2 e^2 x \tan ^{-1}(c x)}{2 c^3}-\frac{b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+\frac{b^2 d e \log \left (1+c^2 x^2\right )}{c^2}-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )+\frac{1}{12} \left (b^2 e^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{\left (b^2 e^2\right ) \int \frac{x}{1+c^2 x^2} \, dx}{2 c^2}\\ &=-\frac{2 a b d e x}{c}+\frac{a b e^2 x}{2 c^3}+\frac{b^2 e^2 x^2}{12 c^2}-\frac{2 b^2 d e x \tan ^{-1}(c x)}{c}+\frac{b^2 e^2 x \tan ^{-1}(c x)}{2 c^3}-\frac{b e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{d e \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+\frac{b^2 d e \log \left (1+c^2 x^2\right )}{c^2}-\frac{b^2 e^2 \log \left (1+c^2 x^2\right )}{3 c^4}-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.668017, size = 389, normalized size = 1.1 \[ i a b d^2 (\text{PolyLog}(2,-i c x)-\text{PolyLog}(2,i c x))+b^2 d^2 \left (i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+i \tan ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )+\frac{2}{3} i \tan ^{-1}(c x)^3+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-\frac{i \pi ^3}{24}\right )+a^2 d^2 \log (x)+a^2 d e x^2+\frac{1}{4} a^2 e^2 x^4+\frac{2 a b d e \left (\left (c^2 x^2+1\right ) \tan ^{-1}(c x)-c x\right )}{c^2}+\frac{a b e^2 \left (-c^3 x^3+3 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)+3 c x\right )}{6 c^4}+\frac{b^2 d e \left (\log \left (c^2 x^2+1\right )+\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2-2 c x \tan ^{-1}(c x)\right )}{c^2}+\frac{b^2 e^2 \left (c^2 x^2-4 \log \left (c^2 x^2+1\right )+3 \left (c^4 x^4-1\right ) \tan ^{-1}(c x)^2+\left (6 c x-2 c^3 x^3\right ) \tan ^{-1}(c x)+1\right )}{12 c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a^2*d*e*x^2 + (a^2*e^2*x^4)/4 + (2*a*b*d*e*(-(c*x) + (1 + c^2*x^2)*ArcTan[c*x]))/c^2 + (a*b*e^2*(3*c*x - c^3*x
^3 + 3*(-1 + c^4*x^4)*ArcTan[c*x]))/(6*c^4) + a^2*d^2*Log[x] + (b^2*e^2*(1 + c^2*x^2 + (6*c*x - 2*c^3*x^3)*Arc
Tan[c*x] + 3*(-1 + c^4*x^4)*ArcTan[c*x]^2 - 4*Log[1 + c^2*x^2]))/(12*c^4) + (b^2*d*e*(-2*c*x*ArcTan[c*x] + (1
+ c^2*x^2)*ArcTan[c*x]^2 + Log[1 + c^2*x^2]))/c^2 + I*a*b*d^2*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x]) + b^2
*d^2*((-I/24)*Pi^3 + ((2*I)/3)*ArcTan[c*x]^3 + ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - ArcTan[c*x]^2*L
og[1 + E^((2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, -E
^((2*I)*ArcTan[c*x])] + PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2 - PolyLog[3, -E^((2*I)*ArcTan[c*x])]/2)

________________________________________________________________________________________

Maple [C]  time = 4.632, size = 1549, 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((e*x^2+d)^2*(a+b*arctan(c*x))^2/x,x)

[Out]

a^2*x^2*d*e-1/4*b^2/c^4*arctan(c*x)^2*e^2+b^2*d^2*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/12*b^2*e^2
*x^2/c^2-2*a*b*d*e*x/c-2*b^2*d*e*x*arctan(c*x)/c+1/2*a*b*e^2*x/c^3+1/2*b^2*e^2*x*arctan(c*x)/c^3+b^2*arctan(c*
x)^2*x^2*d*e+2*a*b*arctan(c*x)*d^2*ln(c*x)+I*a*b*d^2*dilog(1+I*c*x)-I*a*b*d^2*dilog(1-I*c*x)-1/6*a*b/c*x^3*e^2
-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)^2-1/2*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))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arct
an(c*x)^2-1/2*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)^2-1/6*b^2/c*arctan(c*x)*x^3*e^2+1/2*a*b*arctan(c*x)*x^4*e^2-1/2*a*b/c^4*arctan(
c*x)*e^2+b^2/c^2*arctan(c*x)^2*d*e-2*b^2/c^2*e*d*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+I*b^2*d^2*arctan(c*x)*polylog(2
,-(1+I*c*x)^2/(c^2*x^2+1))-2*I*b^2*d^2*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*I*b^2*d^2*Pi*ar
ctan(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))-2/3*I*b^2/c^4*arctan(c*x)*e^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))-1/2*b^2*d^2*poly
log(3,-(1+I*c*x)^2/(c^2*x^2+1))+1/4*a^2*x^4*e^2+b^2*d^2*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^2*ar
ctan(c*x)^2*d^2*ln(c*x)-b^2*d^2*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+2/3*b^2/c^4*e^2*ln((1+I*c*x)^2/(c^
2*x^2+1)+1)+1/4*b^2*arctan(c*x)^2*x^4*e^2+2*a*b/c^2*arctan(c*x)*d*e+I*a*b*d^2*ln(c*x)*ln(1+I*c*x)-I*a*b*d^2*ln
(c*x)*ln(1-I*c*x)+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))^2*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+2*I*b^2/c^2*arctan(c*x)*e
*d+2*a*b*arctan(c*x)*x^2*d*e+a^2*d^2*ln(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))*csgn(I*((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+1/12*b^2
/c^4*e^2

________________________________________________________________________________________

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

[Out]

1/4*a^2*e^2*x^4 + 12*b^2*c^2*e^2*integrate(1/16*x^6*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*c^2*e^2*integrate(1/
16*x^6*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32*a*b*c^2*e^2*integrate(1/16*x^6*arctan(c*x)/(c^2*x^3 + x), x)
+ b^2*c^2*e^2*integrate(1/16*x^6*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 24*b^2*c^2*d*e*integrate(1/16*x^4*arctan
(c*x)^2/(c^2*x^3 + x), x) + 2*b^2*c^2*d*e*integrate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 64*a*b*c^2
*d*e*integrate(1/16*x^4*arctan(c*x)/(c^2*x^3 + x), x) + 4*b^2*c^2*d*e*integrate(1/16*x^4*log(c^2*x^2 + 1)/(c^2
*x^3 + x), x) + 12*b^2*c^2*d^2*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^3 + x), x) + 32*a*b*c^2*d^2*integrate(1
/16*x^2*arctan(c*x)/(c^2*x^3 + x), x) + 1/96*b^2*d^2*log(c^2*x^2 + 1)^3 + a^2*d*e*x^2 - 2*b^2*c*e^2*integrate(
1/16*x^5*arctan(c*x)/(c^2*x^3 + x), x) - 8*b^2*c*d*e*integrate(1/16*x^3*arctan(c*x)/(c^2*x^3 + x), x) + 12*b^2
*e^2*integrate(1/16*x^4*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*e^2*integrate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x
^3 + x), x) + 32*a*b*e^2*integrate(1/16*x^4*arctan(c*x)/(c^2*x^3 + x), x) + 24*b^2*d*e*integrate(1/16*x^2*arct
an(c*x)^2/(c^2*x^3 + x), x) + 64*a*b*d*e*integrate(1/16*x^2*arctan(c*x)/(c^2*x^3 + x), x) + 12*b^2*d^2*integra
te(1/16*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*d^2*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32*a*b
*d^2*integrate(1/16*arctan(c*x)/(c^2*x^3 + x), x) + 1/48*b^2*d*e*log(c^2*x^2 + 1)^3/c^2 + a^2*d^2*log(x) + 1/1
6*(b^2*e^2*x^4 + 4*b^2*d*e*x^2)*arctan(c*x)^2 - 1/64*(b^2*e^2*x^4 + 4*b^2*d*e*x^2)*log(c^2*x^2 + 1)^2

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a^{2} e^{2} x^{4} + 2 \, a^{2} d e x^{2} + a^{2} d^{2} +{\left (b^{2} e^{2} x^{4} + 2 \, b^{2} d e x^{2} + b^{2} d^{2}\right )} \arctan \left (c x\right )^{2} + 2 \,{\left (a b e^{2} x^{4} + 2 \, a b d e x^{2} + a b d^{2}\right )} \arctan \left (c x\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*e^2*x^4 + 2*a^2*d*e*x^2 + a^2*d^2 + (b^2*e^2*x^4 + 2*b^2*d*e*x^2 + b^2*d^2)*arctan(c*x)^2 + 2*(a
*b*e^2*x^4 + 2*a*b*d*e*x^2 + a*b*d^2)*arctan(c*x))/x, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right )^{2} \left (d + e x^{2}\right )^{2}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c*x))**2*(d + e*x**2)**2/x, x)

________________________________________________________________________________________

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

[Out]

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