3.1259 \(\int \frac{(d+e x^2)^2 (a+b \tan ^{-1}(c x))^2}{x^2} \, dx\)

Optimal. Leaf size=343 \[ -\frac{i b^2 e^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}-i b^2 c d^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )+\frac{2 i b^2 d e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c}-\frac{i e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}-\frac{2 b e^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}-i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c d^2 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d e x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac{4 b d e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{1}{3} e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{b^2 e^2 x}{3 c^2}-\frac{b^2 e^2 \tan ^{-1}(c x)}{3 c^3} \]

[Out]

(b^2*e^2*x)/(3*c^2) - (b^2*e^2*ArcTan[c*x])/(3*c^3) - (b*e^2*x^2*(a + b*ArcTan[c*x]))/(3*c) - I*c*d^2*(a + b*A
rcTan[c*x])^2 + ((2*I)*d*e*(a + b*ArcTan[c*x])^2)/c - ((I/3)*e^2*(a + b*ArcTan[c*x])^2)/c^3 - (d^2*(a + b*ArcT
an[c*x])^2)/x + 2*d*e*x*(a + b*ArcTan[c*x])^2 + (e^2*x^3*(a + b*ArcTan[c*x])^2)/3 + (4*b*d*e*(a + b*ArcTan[c*x
])*Log[2/(1 + I*c*x)])/c - (2*b*e^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3) + 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)] + ((2*I)*b^2*d*e*PolyLog[2, 1 - 2/(1
+ I*c*x)])/c - ((I/3)*b^2*e^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.575716, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {4980, 4846, 4920, 4854, 2402, 2315, 4852, 4924, 4868, 2447, 4916, 321, 203} \[ -\frac{i b^2 e^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}-i b^2 c d^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )+\frac{2 i b^2 d e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c}-\frac{i e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}-\frac{2 b e^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}-i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c d^2 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d e x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac{4 b d e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{1}{3} e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{b^2 e^2 x}{3 c^2}-\frac{b^2 e^2 \tan ^{-1}(c x)}{3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*e^2*x)/(3*c^2) - (b^2*e^2*ArcTan[c*x])/(3*c^3) - (b*e^2*x^2*(a + b*ArcTan[c*x]))/(3*c) - I*c*d^2*(a + b*A
rcTan[c*x])^2 + ((2*I)*d*e*(a + b*ArcTan[c*x])^2)/c - ((I/3)*e^2*(a + b*ArcTan[c*x])^2)/c^3 - (d^2*(a + b*ArcT
an[c*x])^2)/x + 2*d*e*x*(a + b*ArcTan[c*x])^2 + (e^2*x^3*(a + b*ArcTan[c*x])^2)/3 + (4*b*d*e*(a + b*ArcTan[c*x
])*Log[2/(1 + I*c*x)])/c - (2*b*e^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3) + 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)] + ((2*I)*b^2*d*e*PolyLog[2, 1 - 2/(1
+ I*c*x)])/c - ((I/3)*b^2*e^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^3

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 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 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 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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.766775, size = 349, normalized size = 1.02 \[ \frac{1}{3} \left (\frac{b^2 e^2 \left (i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+\left (c^3 x^3+i\right ) \tan ^{-1}(c x)^2-\tan ^{-1}(c x) \left (c^2 x^2+2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+1\right )+c x\right )}{c^3}+3 b^2 c d^2 \left (\tan ^{-1}(c x) \left (\left (-\frac{1}{c x}-i\right ) \tan ^{-1}(c x)+2 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )-i \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )\right )+\frac{6 b^2 d e \left (\tan ^{-1}(c x) \left ((c x-i) \tan ^{-1}(c x)+2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )\right )}{c}-\frac{3 a^2 d^2}{x}+6 a^2 d e x+a^2 e^2 x^3-\frac{3 a b d^2 \left (c x \left (\log \left (c^2 x^2+1\right )-2 \log (c x)\right )+2 \tan ^{-1}(c x)\right )}{x}+\frac{6 a b d e \left (2 c x \tan ^{-1}(c x)-\log \left (c^2 x^2+1\right )\right )}{c}+\frac{a b e^2 \left (-c^2 x^2+\log \left (c^2 x^2+1\right )+2 c^3 x^3 \tan ^{-1}(c x)\right )}{c^3}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.145, size = 997, normalized size = 2.9 \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^2,x)

[Out]

-1/2*I*c*b^2*dilog(1/2*I*(c*x-I))*d^2+1/3*b^2*e^2*x/c^2-1/3*b^2*e^2*arctan(c*x)/c^3+1/3*a^2*x^3*e^2-a^2*d^2/x-
1/6*I*b^2/c^3*ln(c*x+I)*ln(c^2*x^2+1)*e^2-2*b^2/c*arctan(c*x)*ln(c^2*x^2+1)*d*e+I*b^2/c*ln(c*x+I)*ln(c^2*x^2+1
)*d*e+I*b^2/c*ln(c*x-I)*ln(-1/2*I*(c*x+I))*d*e-I*b^2/c*ln(c*x-I)*ln(c^2*x^2+1)*d*e-I*b^2/c*ln(c*x+I)*ln(1/2*I*
(c*x-I))*d*e-b^2*arctan(c*x)^2*d^2/x+2*a^2*e*d*x+1/3*b^2*arctan(c*x)^2*x^3*e^2-1/2*I*b^2/c*ln(c*x+I)^2*d*e-1/6
*I*b^2/c^3*ln(c*x-I)*ln(-1/2*I*(c*x+I))*e^2+1/6*I*b^2/c^3*ln(c*x-I)*ln(c^2*x^2+1)*e^2+1/6*I*b^2/c^3*ln(c*x+I)*
ln(1/2*I*(c*x-I))*e^2+I*b^2/c*dilog(-1/2*I*(c*x+I))*d*e+4*a*b*arctan(c*x)*e*d*x-2*a*b/c*ln(c^2*x^2+1)*d*e-1/2*
I*c*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))*d^2+1/2*I*c*b^2*ln(c*x+I)*ln(c^2*x^2+1)*d^2+I*c*b^2*d^2*ln(c*x)*ln(1+I*c*x
)+1/2*I*b^2/c*ln(c*x-I)^2*d*e-I*b^2/c*dilog(1/2*I*(c*x-I))*d*e-I*c*b^2*d^2*ln(c*x)*ln(1-I*c*x)+1/2*I*c*b^2*ln(
c*x-I)*ln(-1/2*I*(c*x+I))*d^2-1/2*I*c*b^2*ln(c*x-I)*ln(c^2*x^2+1)*d^2-c*a*b*ln(c^2*x^2+1)*d^2+I*c*b^2*d^2*dilo
g(1+I*c*x)+2*b^2*arctan(c*x)^2*e*d*x+2/3*a*b*arctan(c*x)*x^3*e^2-1/3*a*b/c*x^2*e^2-1/3*b^2/c*arctan(c*x)*x^2*e
^2+1/3*b^2/c^3*arctan(c*x)*ln(c^2*x^2+1)*e^2+1/3*a*b/c^3*ln(c^2*x^2+1)*e^2+1/4*I*c*b^2*ln(c*x-I)^2*d^2+1/2*I*c
*b^2*dilog(-1/2*I*(c*x+I))*d^2-1/4*I*c*b^2*ln(c*x+I)^2*d^2-I*c*b^2*d^2*dilog(1-I*c*x)-1/12*I*b^2/c^3*ln(c*x-I)
^2*e^2+1/6*I*b^2/c^3*dilog(1/2*I*(c*x-I))*e^2-1/6*I*b^2/c^3*dilog(-1/2*I*(c*x+I))*e^2+1/12*I*b^2/c^3*ln(c*x+I)
^2*e^2-2*a*b*arctan(c*x)*d^2/x+2*c*b^2*arctan(c*x)*d^2*ln(c*x)-c*b^2*arctan(c*x)*ln(c^2*x^2+1)*d^2+2*c*a*b*d^2
*ln(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((e*x^2+d)^2*(a+b*arctan(c*x))^2/x^2,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{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^{2}}, 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^2,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^2, x)

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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^{2}}\, 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**2,x)

[Out]

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

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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^{2}}\,{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^2,x, algorithm="giac")

[Out]

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