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

Optimal. Leaf size=496 \[ \frac{i b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{i b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{i b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2}{\left (c^2 d^2+e^2\right )^2}+\frac{c^2 (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e \left (c^2 d^2+e^2\right )^2}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}-\frac{2 b c^3 d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{b^2 c^2 e \log \left (c^2 x^2+1\right )}{2 \left (c^2 d^2+e^2\right )^2}+\frac{b^2 c^2 e \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac{b^2 c^3 d \tan ^{-1}(c x)}{\left (c^2 d^2+e^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.538568, antiderivative size = 496, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 15, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.833, Rules used = {4864, 4862, 706, 31, 635, 203, 260, 4856, 2402, 2315, 2447, 4984, 4884, 4920, 4854} \[ \frac{i b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{i b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{i b^2 c^3 d \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2}{\left (c^2 d^2+e^2\right )^2}+\frac{c^2 (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e \left (c^2 d^2+e^2\right )^2}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}-\frac{2 b c^3 d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{b^2 c^2 e \log \left (c^2 x^2+1\right )}{2 \left (c^2 d^2+e^2\right )^2}+\frac{b^2 c^2 e \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac{b^2 c^3 d \tan ^{-1}(c x)}{\left (c^2 d^2+e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4864

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a
+ b*ArcTan[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p -
1), (d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && N
eQ[q, -1]

Rule 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 31

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

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

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])

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 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[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 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 4984

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> I
nt[ExpandIntegrand[(a + b*ArcTan[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& IGtQ[p, 0] && EqQ[e, c^2*d] && IGtQ[m, 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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(d+e x)^3} \, dx &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac{(b c) \int \left (\frac{e^2 \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)^2}+\frac{2 c^2 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right )^2 (d+e x)}+\frac{\left (c^4 d^2-c^2 e^2-2 c^4 d e x\right ) \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right )^2 \left (1+c^2 x^2\right )}\right ) \, dx}{e}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac{(b c) \int \frac{\left (c^4 d^2-c^2 e^2-2 c^4 d e x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{e \left (c^2 d^2+e^2\right )^2}+\frac{\left (2 b c^3 d e\right ) \int \frac{a+b \tan ^{-1}(c x)}{d+e x} \, dx}{\left (c^2 d^2+e^2\right )^2}+\frac{(b c e) \int \frac{a+b \tan ^{-1}(c x)}{(d+e x)^2} \, dx}{c^2 d^2+e^2}\\ &=-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{\left (2 b^2 c^4 d\right ) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{\left (c^2 d^2+e^2\right )^2}-\frac{\left (2 b^2 c^4 d\right ) \int \frac{\log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{1+c^2 x^2} \, dx}{\left (c^2 d^2+e^2\right )^2}+\frac{(b c) \int \left (\frac{c^4 d^2 \left (1-\frac{e^2}{c^2 d^2}\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}-\frac{2 c^4 d e x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}\right ) \, dx}{e \left (c^2 d^2+e^2\right )^2}+\frac{\left (b^2 c^2\right ) \int \frac{1}{(d+e x) \left (1+c^2 x^2\right )} \, dx}{c^2 d^2+e^2}\\ &=-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{\left (b^2 c^2\right ) \int \frac{c^2 d-c^2 e x}{1+c^2 x^2} \, dx}{\left (c^2 d^2+e^2\right )^2}+\frac{\left (2 i b^2 c^3 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{\left (2 b c^5 d\right ) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{\left (c^2 d^2+e^2\right )^2}+\frac{\left (b^2 c^2 e^2\right ) \int \frac{1}{d+e x} \, dx}{\left (c^2 d^2+e^2\right )^2}+\frac{\left (b c^3 (c d-e) (c d+e)\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{e \left (c^2 d^2+e^2\right )^2}\\ &=-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac{i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2}{\left (c^2 d^2+e^2\right )^2}+\frac{c^2 (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e \left (c^2 d^2+e^2\right )^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{b^2 c^2 e \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{\left (2 b c^4 d\right ) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{\left (c^2 d^2+e^2\right )^2}+\frac{\left (b^2 c^4 d\right ) \int \frac{1}{1+c^2 x^2} \, dx}{\left (c^2 d^2+e^2\right )^2}-\frac{\left (b^2 c^4 e\right ) \int \frac{x}{1+c^2 x^2} \, dx}{\left (c^2 d^2+e^2\right )^2}\\ &=\frac{b^2 c^3 d \tan ^{-1}(c x)}{\left (c^2 d^2+e^2\right )^2}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac{i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2}{\left (c^2 d^2+e^2\right )^2}+\frac{c^2 (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e \left (c^2 d^2+e^2\right )^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{b^2 c^2 e \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{b^2 c^2 e \log \left (1+c^2 x^2\right )}{2 \left (c^2 d^2+e^2\right )^2}+\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{\left (2 b^2 c^4 d\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{\left (c^2 d^2+e^2\right )^2}\\ &=\frac{b^2 c^3 d \tan ^{-1}(c x)}{\left (c^2 d^2+e^2\right )^2}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac{i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2}{\left (c^2 d^2+e^2\right )^2}+\frac{c^2 (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e \left (c^2 d^2+e^2\right )^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{b^2 c^2 e \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{b^2 c^2 e \log \left (1+c^2 x^2\right )}{2 \left (c^2 d^2+e^2\right )^2}+\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{\left (2 i b^2 c^3 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{\left (c^2 d^2+e^2\right )^2}\\ &=\frac{b^2 c^3 d \tan ^{-1}(c x)}{\left (c^2 d^2+e^2\right )^2}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac{i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2}{\left (c^2 d^2+e^2\right )^2}+\frac{c^2 (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e \left (c^2 d^2+e^2\right )^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{b^2 c^2 e \log (d+e x)}{\left (c^2 d^2+e^2\right )^2}+\frac{2 b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{b^2 c^2 e \log \left (1+c^2 x^2\right )}{2 \left (c^2 d^2+e^2\right )^2}+\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{\left (c^2 d^2+e^2\right )^2}+\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{\left (c^2 d^2+e^2\right )^2}-\frac{i b^2 c^3 d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{\left (c^2 d^2+e^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 6.03814, size = 479, normalized size = 0.97 \[ \frac{b^2 c^2 \left (-\frac{2 c d \left (i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )-\frac{1}{2} \pi \log \left (c^2 x^2+1\right )-i \tan ^{-1}(c x) \left (\pi -2 \tan ^{-1}\left (\frac{c d}{e}\right )\right )-2 \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right ) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )+2 \tan ^{-1}\left (\frac{c d}{e}\right ) \log \left (\sin \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )\right )-\pi \log \left (1+e^{-2 i \tan ^{-1}(c x)}\right )\right )}{c^2 d^2+e^2}+\frac{2 c d e \log \left (\frac{c (d+e x)}{\sqrt{c^2 x^2+1}}\right )-2 e^2 \tan ^{-1}(c x)}{c^3 d^3+c d e^2}-\frac{2 \tan ^{-1}(c x)^2 e^{i \tan ^{-1}\left (\frac{c d}{e}\right )}}{e \sqrt{\frac{c^2 d^2}{e^2}+1}}-\frac{e \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2}{c^2 (d+e x)^2}+\frac{2 x \tan ^{-1}(c x) \left (c d \tan ^{-1}(c x)+e\right )}{c d (d+e x)}\right )}{2 \left (c^2 d^2+e^2\right )}-\frac{a^2}{2 e (d+e x)^2}+\frac{a b \left (c (d+e x) \left (c^2 \left (-d^2\right )-c^2 d \log \left (c^2 x^2+1\right ) (d+e x)+2 c^2 d (d+e x) \log (c (d+e x))-e^2\right )+\tan ^{-1}(c x) \left (-c^2 e \left (3 d^2+2 d e x+e^2 x^2\right )+c^4 d^2 x (2 d+e x)-e^3\right )\right )}{\left (c^2 d^2+e^2\right )^2 (d+e x)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-a^2/(2*e*(d + e*x)^2) + (a*b*((-e^3 + c^4*d^2*x*(2*d + e*x) - c^2*e*(3*d^2 + 2*d*e*x + e^2*x^2))*ArcTan[c*x]
+ c*(d + e*x)*(-(c^2*d^2) - e^2 + 2*c^2*d*(d + e*x)*Log[c*(d + e*x)] - c^2*d*(d + e*x)*Log[1 + c^2*x^2])))/((c
^2*d^2 + e^2)^2*(d + e*x)^2) + (b^2*c^2*((-2*E^(I*ArcTan[(c*d)/e])*ArcTan[c*x]^2)/(Sqrt[1 + (c^2*d^2)/e^2]*e)
- (e*(1 + c^2*x^2)*ArcTan[c*x]^2)/(c^2*(d + e*x)^2) + (2*x*ArcTan[c*x]*(e + c*d*ArcTan[c*x]))/(c*d*(d + e*x))
+ (-2*e^2*ArcTan[c*x] + 2*c*d*e*Log[(c*(d + e*x))/Sqrt[1 + c^2*x^2]])/(c^3*d^3 + c*d*e^2) - (2*c*d*((-I)*(Pi -
2*ArcTan[(c*d)/e])*ArcTan[c*x] - Pi*Log[1 + E^((-2*I)*ArcTan[c*x])] - 2*(ArcTan[(c*d)/e] + ArcTan[c*x])*Log[1
- E^((2*I)*(ArcTan[(c*d)/e] + ArcTan[c*x]))] - (Pi*Log[1 + c^2*x^2])/2 + 2*ArcTan[(c*d)/e]*Log[Sin[ArcTan[(c*
d)/e] + ArcTan[c*x]]] + I*PolyLog[2, E^((2*I)*(ArcTan[(c*d)/e] + ArcTan[c*x]))]))/(c^2*d^2 + e^2)))/(2*(c^2*d^
2 + e^2))

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Maple [B]  time = 0.102, size = 961, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2*a*b/(c^2*d^2+e^2)/(c*e*x+c*d)-1/2*b^2*c^2*e*ln(c^2*x^2+1)/(c^2*d^2+e^2)^2-1/2*c^2*b^2/(c*e*x+c*d)^2/e*arc
tan(c*x)^2-c^2*b^2*arctan(c*x)/(c^2*d^2+e^2)/(c*e*x+c*d)+c^2*b^2*e/(c^2*d^2+e^2)^2*ln(c*e*x+c*d)-1/2*c^2*b^2*e
/(c^2*d^2+e^2)^2*arctan(c*x)^2+b^2*c^3*d*arctan(c*x)/(c^2*d^2+e^2)^2-1/2*c^2*a^2/(c*e*x+c*d)^2/e-I*c^3*b^2*d/(
c^2*d^2+e^2)^2*ln(c*e*x+c*d)*ln((I*e+e*c*x)/(I*e-d*c))+1/2*I*c^3*b^2*d/(c^2*d^2+e^2)^2*ln(c*x+I)*ln(c^2*x^2+1)
+1/2*I*c^3*b^2*d/(c^2*d^2+e^2)^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-1/2*I*c^3*b^2*d/(c^2*d^2+e^2)^2*ln(c*x-I)*ln(c^2
*x^2+1)-1/2*I*c^3*b^2*d/(c^2*d^2+e^2)^2*ln(c*x+I)*ln(1/2*I*(c*x-I))+I*c^3*b^2*d/(c^2*d^2+e^2)^2*ln(c*e*x+c*d)*
ln((I*e-e*c*x)/(d*c+I*e))+c^4*a*b/e/(c^2*d^2+e^2)^2*arctan(c*x)*d^2-c^2*a*b*e/(c^2*d^2+e^2)^2*arctan(c*x)+2*c^
3*a*b*d/(c^2*d^2+e^2)^2*ln(c*e*x+c*d)+I*c^3*b^2*d/(c^2*d^2+e^2)^2*dilog((I*e-e*c*x)/(d*c+I*e))+1/2*c^4*b^2/e/(
c^2*d^2+e^2)^2*arctan(c*x)^2*d^2-I*c^3*b^2*d/(c^2*d^2+e^2)^2*dilog((I*e+e*c*x)/(I*e-d*c))-1/4*I*c^3*b^2*d/(c^2
*d^2+e^2)^2*ln(c*x+I)^2+1/2*I*c^3*b^2*d/(c^2*d^2+e^2)^2*dilog(-1/2*I*(c*x+I))+1/4*I*c^3*b^2*d/(c^2*d^2+e^2)^2*
ln(c*x-I)^2-1/2*I*c^3*b^2*d/(c^2*d^2+e^2)^2*dilog(1/2*I*(c*x-I))+2*c^3*b^2*arctan(c*x)*d/(c^2*d^2+e^2)^2*ln(c*
e*x+c*d)-c^3*b^2*arctan(c*x)/(c^2*d^2+e^2)^2*d*ln(c^2*x^2+1)-c^2*a*b/(c*e*x+c*d)^2/e*arctan(c*x)-c^3*a*b/(c^2*
d^2+e^2)^2*d*ln(c^2*x^2+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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