3.147 \(\int \frac{(a+b \tan ^{-1}(c x))^2}{x^3 (d+e x)} \, dx\)

Optimal. Leaf size=591 \[ -\frac{i b e^2 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac{i b e^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac{i b e^2 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac{i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac{b^2 e^2 \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 d^3}-\frac{b^2 e^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 d^3}+\frac{b^2 e^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d^3}-\frac{b^2 e^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 d^3}+\frac{i b^2 c e \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d^2}-\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac{e^2 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac{2 e^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}+\frac{i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}+\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac{2 b c e \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac{b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d}+\frac{b^2 c^2 \log (x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.84186, antiderivative size = 591, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 16, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.762, Rules used = {4876, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 2447, 4850, 4988, 4994, 6610, 4858} \[ -\frac{i b e^2 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac{i b e^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac{i b e^2 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac{i b e^2 \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac{b^2 e^2 \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 d^3}-\frac{b^2 e^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 d^3}+\frac{b^2 e^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d^3}-\frac{b^2 e^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 d^3}+\frac{i b^2 c e \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d^2}-\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac{e^2 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d^3}+\frac{2 e^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}+\frac{i c e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}+\frac{e \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac{2 b c e \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac{b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d}+\frac{b^2 c^2 \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^2/(x^3*(d + e*x)),x]

[Out]

-((b*c*(a + b*ArcTan[c*x]))/(d*x)) - (c^2*(a + b*ArcTan[c*x])^2)/(2*d) + (I*c*e*(a + b*ArcTan[c*x])^2)/d^2 - (
a + b*ArcTan[c*x])^2/(2*d*x^2) + (e*(a + b*ArcTan[c*x])^2)/(d^2*x) + (2*e^2*(a + b*ArcTan[c*x])^2*ArcTanh[1 -
2/(1 + I*c*x)])/d^3 + (b^2*c^2*Log[x])/d + (e^2*(a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/d^3 - (e^2*(a + b*Ar
cTan[c*x])^2*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/d^3 - (b^2*c^2*Log[1 + c^2*x^2])/(2*d) - (2*b*c*e
*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/d^2 - (I*b*e^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])
/d^3 + (I*b^2*c*e*PolyLog[2, -1 + 2/(1 - I*c*x)])/d^2 - (I*b*e^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c
*x)])/d^3 + (I*b*e^2*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d^3 + (I*b*e^2*(a + b*ArcTan[c*x])*Po
lyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/d^3 + (b^2*e^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*d^3)
- (b^2*e^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*d^3) + (b^2*e^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/(2*d^3) - (b^2*e
^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(2*d^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 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 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), 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] && LtQ[m, -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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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

Mathematica [F]  time = 180.003, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*ArcTan[c*x])^2/(x^3*(d + e*x)),x]

[Out]

$Aborted

________________________________________________________________________________________

Maple [C]  time = 14.738, size = 2861, normalized size = 4.8 \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^3/(e*x+d),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a^{2}{\left (\frac{2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac{2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac{2 \, e x - d}{d^{2} x^{2}}\right )} + \frac{2 \, d^{2} x^{2} \int \frac{12 \,{\left (b^{2} c^{2} d^{2} x^{2} + b^{2} d^{2}\right )} \arctan \left (c x\right )^{2} +{\left (b^{2} c^{2} d^{2} x^{2} + b^{2} d^{2}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} - 4 \,{\left (2 \, b^{2} c e^{2} x^{3} - b^{2} c d^{2} x - 8 \, a b d^{2} -{\left (8 \, a b c^{2} d^{2} - b^{2} c d e\right )} x^{2}\right )} \arctan \left (c x\right ) + 2 \,{\left (2 \, b^{2} c^{2} e^{2} x^{4} + b^{2} c^{2} d e x^{3} - b^{2} c^{2} d^{2} x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{2} e x^{6} + c^{2} d^{3} x^{5} + d^{2} e x^{4} + d^{3} x^{3}}\,{d x} + 4 \,{\left (2 \, b^{2} e x - b^{2} d\right )} \arctan \left (c x\right )^{2} -{\left (2 \, b^{2} e x - b^{2} d\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{32 \, d^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/x^3/(e*x+d),x, algorithm="maxima")

[Out]

-1/2*a^2*(2*e^2*log(e*x + d)/d^3 - 2*e^2*log(x)/d^3 - (2*e*x - d)/(d^2*x^2)) + 1/32*(32*d^2*x^2*integrate(1/16
*(12*(b^2*c^2*d^2*x^2 + b^2*d^2)*arctan(c*x)^2 + (b^2*c^2*d^2*x^2 + b^2*d^2)*log(c^2*x^2 + 1)^2 - 4*(2*b^2*c*e
^2*x^3 - b^2*c*d^2*x - 8*a*b*d^2 - (8*a*b*c^2*d^2 - b^2*c*d*e)*x^2)*arctan(c*x) + 2*(2*b^2*c^2*e^2*x^4 + b^2*c
^2*d*e*x^3 - b^2*c^2*d^2*x^2)*log(c^2*x^2 + 1))/(c^2*d^2*e*x^6 + c^2*d^3*x^5 + d^2*e*x^4 + d^3*x^3), x) + 4*(2
*b^2*e*x - b^2*d)*arctan(c*x)^2 - (2*b^2*e*x - b^2*d)*log(c^2*x^2 + 1)^2)/(d^2*x^2)

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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^{4} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/x^3/(e*x+d),x, algorithm="fricas")

[Out]

integral((b^2*arctan(c*x)^2 + 2*a*b*arctan(c*x) + a^2)/(e*x^4 + d*x^3), 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((a+b*atan(c*x))**2/x**3/(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}}{{\left (e x + d\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/x^3/(e*x+d),x, algorithm="giac")

[Out]

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