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

Optimal. Leaf size=473 \[ \frac{i b e \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}+\frac{i b e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac{i b e \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^2}-\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 d^2}+\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 d^2}-\frac{b^2 e \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d^2}+\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 d^2}-\frac{i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d}-\frac{e \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}+\frac{e \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^2}-\frac{2 e \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.603513, antiderivative size = 473, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.524, Rules used = {4876, 4852, 4924, 4868, 2447, 4850, 4988, 4884, 4994, 6610, 4858} \[ \frac{i b e \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}+\frac{i b e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac{i b e \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^2}-\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 d^2}+\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 d^2}-\frac{b^2 e \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d^2}+\frac{b^2 e \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 d^2}-\frac{i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d}-\frac{e \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}+\frac{e \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^2}-\frac{2 e \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{2 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*c*(a + b*ArcTan[c*x])^2)/d - (a + b*ArcTan[c*x])^2/(d*x) - (2*e*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 +
I*c*x)])/d^2 - (e*(a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/d^2 + (e*(a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x)
)/((c*d + I*e)*(1 - I*c*x))])/d^2 + (2*b*c*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/d + (I*b*e*(a + b*ArcTa
n[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/d^2 - (I*b^2*c*PolyLog[2, -1 + 2/(1 - I*c*x)])/d + (I*b*e*(a + b*ArcTan
[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/d^2 - (I*b*e*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d^2 - (
I*b*e*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/d^2 - (b^2*e*PolyLog[3, 1
- 2/(1 - I*c*x)])/(2*d^2) + (b^2*e*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*d^2) - (b^2*e*PolyLog[3, -1 + 2/(1 + I*c
*x)])/(2*d^2) + (b^2*e*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(2*d^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 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 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 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{\left (a+b \tan ^{-1}(c x)\right )^2}{x^2 (d+e x)} \, dx &=\int \left (\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x^2}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}+\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (d+e x)}\right ) \, dx\\ &=\frac{\int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx}{d}-\frac{e \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx}{d^2}+\frac{e^2 \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d+e x} \, dx}{d^2}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )}{d^2}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{d^2}+\frac{e \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 )}{d^2}+\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{d^2}-\frac{i b e \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 )}{d^2}-\frac{b^2 e \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 d^2}+\frac{b^2 e \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^2}+\frac{(2 b c) \int \frac{a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac{(4 b c e) \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}{d^2}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )}{d^2}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{d^2}+\frac{e \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 )}{d^2}+\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{d^2}-\frac{i b e \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 )}{d^2}-\frac{b^2 e \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 d^2}+\frac{b^2 e \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^2}+\frac{(2 i b c) \int \frac{a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx}{d}-\frac{(2 b c e) \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}{d^2}+\frac{(2 b c e) \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}{d^2}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )}{d^2}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{d^2}+\frac{e \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 )}{d^2}+\frac{2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )}{d}+\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{d^2}+\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{d^2}-\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d^2}-\frac{i b e \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 )}{d^2}-\frac{b^2 e \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 d^2}+\frac{b^2 e \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^2}-\frac{\left (2 b^2 c^2\right ) \int \frac{\log \left (2-\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d}-\frac{\left (i b^2 c e\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^2}+\frac{\left (i b^2 c e\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^2}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{2 e \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )}{d^2}-\frac{e \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{d^2}+\frac{e \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 )}{d^2}+\frac{2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )}{d}+\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{d^2}-\frac{i b^2 c \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )}{d}+\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{d^2}-\frac{i b e \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d^2}-\frac{i b e \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 )}{d^2}-\frac{b^2 e \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 d^2}+\frac{b^2 e \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 d^2}-\frac{b^2 e \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{2 d^2}+\frac{b^2 e \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d^2}\\ \end{align*}

Mathematica [F]  time = 120.59, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^2 (d+e x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 6.674, size = 40579, normalized size = 85.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2}{\left (\frac{e \log \left (e x + d\right )}{d^{2}} - \frac{e \log \left (x\right )}{d^{2}} - \frac{1}{d x}\right )} - \frac{4 \, b^{2} \arctan \left (c x\right )^{2} - b^{2} \log \left (c^{2} x^{2} + 1\right )^{2} - d x \int \frac{12 \,{\left (b^{2} c^{2} d x^{2} + b^{2} d\right )} \arctan \left (c x\right )^{2} +{\left (b^{2} c^{2} d x^{2} + b^{2} d\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + 8 \,{\left (b^{2} c d x + 4 \, a b d +{\left (4 \, a b c^{2} d + b^{2} c e\right )} x^{2}\right )} \arctan \left (c x\right ) - 4 \,{\left (b^{2} c^{2} e x^{3} + b^{2} c^{2} d x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{c^{2} d e x^{5} + c^{2} d^{2} x^{4} + d e x^{3} + d^{2} x^{2}}\,{d x}}{16 \, d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(e*log(e*x + d)/d^2 - e*log(x)/d^2 - 1/(d*x)) - 1/16*(4*b^2*arctan(c*x)^2 - b^2*log(c^2*x^2 + 1)^2 - 16*d*
x*integrate(1/16*(12*(b^2*c^2*d*x^2 + b^2*d)*arctan(c*x)^2 + (b^2*c^2*d*x^2 + b^2*d)*log(c^2*x^2 + 1)^2 + 8*(b
^2*c*d*x + 4*a*b*d + (4*a*b*c^2*d + b^2*c*e)*x^2)*arctan(c*x) - 4*(b^2*c^2*e*x^3 + b^2*c^2*d*x^2)*log(c^2*x^2
+ 1))/(c^2*d*e*x^5 + c^2*d^2*x^4 + d*e*x^3 + d^2*x^2), x))/(d*x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (c x\right )^{2} + 2 \, a b \arctan \left (c x\right ) + a^{2}}{e x^{3} + d x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctan(c*x)^2 + 2*a*b*arctan(c*x) + a^2)/(e*x^3 + d*x^2), x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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