3.1.14 \(\int \frac {(a+b \text {ArcTan}(c x))^2}{(d+e x)^3} \, dx\) [14]

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

[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)/(e*x+d)+I*c^3*d*(a+b*arctan(c*x))^2/
(c^2*d^2+e^2)^2+1/2*c^2*(c*d-e)*(c*d+e)*(a+b*arctan(c*x))^2/e/(c^2*d^2+e^2)^2-1/2*(a+b*arctan(c*x))^2/e/(e*x+d
)^2-2*b*c^3*d*(a+b*arctan(c*x))*ln(2/(1-I*c*x))/(c^2*d^2+e^2)^2+2*b*c^3*d*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/(c
^2*d^2+e^2)^2+b^2*c^2*e*ln(e*x+d)/(c^2*d^2+e^2)^2+2*b*c^3*d*(a+b*arctan(c*x))*ln(2*c*(e*x+d)/(c*d+I*e)/(1-I*c*
x))/(c^2*d^2+e^2)^2-1/2*b^2*c^2*e*ln(c^2*x^2+1)/(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*(e*x+d)/(c*d+I*e)/(1-I
*c*x))/(c^2*d^2+e^2)^2

________________________________________________________________________________________

Rubi [A]
time = 0.39, antiderivative size = 496, normalized size of antiderivative = 1.00, 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 = {4974, 4972, 720, 31, 649, 209, 266, 4966, 2449, 2352, 2497, 5104, 5004, 5040, 4964} \begin {gather*} \frac {c^2 (c d-e) (c d+e) (a+b \text {ArcTan}(c x))^2}{2 e \left (c^2 d^2+e^2\right )^2}-\frac {b c (a+b \text {ArcTan}(c x))}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac {i c^3 d (a+b \text {ArcTan}(c x))^2}{\left (c^2 d^2+e^2\right )^2}-\frac {2 b c^3 d \log \left (\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))}{\left (c^2 d^2+e^2\right )^2}+\frac {2 b c^3 d \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{\left (c^2 d^2+e^2\right )^2}+\frac {2 b c^3 d (a+b \text {ArcTan}(c x)) \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 {(a+b \text {ArcTan}(c x))^2}{2 e (d+e x)^2}+\frac {b^2 c^3 d \text {ArcTan}(c x)}{\left (c^2 d^2+e^2\right )^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 {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}{i c x+1}\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 {gather*}

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 31

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

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 649

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 720

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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

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 2497

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 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 4966

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 4972

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 4974

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 5004

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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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 5104

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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 4.05, size = 479, normalized size = 0.97 \begin {gather*} -\frac {a^2}{2 e (d+e x)^2}+\frac {a b \left (\left (-e^3+c^4 d^2 x (2 d+e x)-c^2 e \left (3 d^2+2 d e x+e^2 x^2\right )\right ) \text {ArcTan}(c x)+c (d+e x) \left (-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 \left (1+c^2 x^2\right )\right )\right )}{\left (c^2 d^2+e^2\right )^2 (d+e x)^2}+\frac {b^2 c^2 \left (-\frac {2 e^{i \text {ArcTan}\left (\frac {c d}{e}\right )} \text {ArcTan}(c x)^2}{\sqrt {1+\frac {c^2 d^2}{e^2}} e}-\frac {e \left (1+c^2 x^2\right ) \text {ArcTan}(c x)^2}{c^2 (d+e x)^2}+\frac {2 x \text {ArcTan}(c x) (e+c d \text {ArcTan}(c x))}{c d (d+e x)}+\frac {-2 e^2 \text {ArcTan}(c x)+2 c d e \log \left (\frac {c (d+e x)}{\sqrt {1+c^2 x^2}}\right )}{c^3 d^3+c d e^2}-\frac {2 c d \left (-i \left (\pi -2 \text {ArcTan}\left (\frac {c d}{e}\right )\right ) \text {ArcTan}(c x)-\pi \log \left (1+e^{-2 i \text {ArcTan}(c x)}\right )-2 \left (\text {ArcTan}\left (\frac {c d}{e}\right )+\text {ArcTan}(c x)\right ) \log \left (1-e^{2 i \left (\text {ArcTan}\left (\frac {c d}{e}\right )+\text {ArcTan}(c x)\right )}\right )-\frac {1}{2} \pi \log \left (1+c^2 x^2\right )+2 \text {ArcTan}\left (\frac {c d}{e}\right ) \log \left (\sin \left (\text {ArcTan}\left (\frac {c d}{e}\right )+\text {ArcTan}(c x)\right )\right )+i \text {PolyLog}\left (2,e^{2 i \left (\text {ArcTan}\left (\frac {c d}{e}\right )+\text {ArcTan}(c x)\right )}\right )\right )}{c^2 d^2+e^2}\right )}{2 \left (c^2 d^2+e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a^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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 964 vs. \(2 (478 ) = 956\).
time = 5.14, size = 965, normalized size = 1.95 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-a*b*c^3/(c*e*x+c*d)^2/e*arctan(c*x)+I*b^2*c^4/(c^2*d^2+e^2)^2*d*ln(c*e*x+c*d)*ln((I*e-c*e*x)/(c*d+I*e))-
1/2*I*b^2*c^4/(c^2*d^2+e^2)^2*d*ln(c*x+I)*ln(1/2*I*(c*x-I))+1/2*I*b^2*c^4/(c^2*d^2+e^2)^2*d*ln(c*x+I)*ln(c^2*x
^2+1)-I*b^2*c^4/(c^2*d^2+e^2)^2*d*ln(c*e*x+c*d)*ln((I*e+c*e*x)/(I*e-c*d))-1/2*I*b^2*c^4/(c^2*d^2+e^2)^2*d*ln(c
*x-I)*ln(c^2*x^2+1)+1/2*I*b^2*c^4/(c^2*d^2+e^2)^2*d*ln(c*x-I)*ln(-1/2*I*(c*x+I))+a*b*c^5/e/(c^2*d^2+e^2)^2*arc
tan(c*x)*d^2-a*b*c^4/(c^2*d^2+e^2)^2*d*ln(c^2*x^2+1)-a*b*c^3*e/(c^2*d^2+e^2)^2*arctan(c*x)+2*a*b*c^4*d/(c^2*d^
2+e^2)^2*ln(c*e*x+c*d)+1/2*b^2*c^5/e/(c^2*d^2+e^2)^2*arctan(c*x)^2*d^2-b^2*c^4*arctan(c*x)/(c^2*d^2+e^2)^2*d*l
n(c^2*x^2+1)+2*b^2*c^4*arctan(c*x)*d/(c^2*d^2+e^2)^2*ln(c*e*x+c*d)+I*b^2*c^4/(c^2*d^2+e^2)^2*d*dilog((I*e-c*e*
x)/(c*d+I*e))-1/4*I*b^2*c^4/(c^2*d^2+e^2)^2*d*ln(c*x+I)^2-1/2*I*b^2*c^4/(c^2*d^2+e^2)^2*d*dilog(1/2*I*(c*x-I))
-I*b^2*c^4/(c^2*d^2+e^2)^2*d*dilog((I*e+c*e*x)/(I*e-c*d))+1/4*I*b^2*c^4/(c^2*d^2+e^2)^2*d*ln(c*x-I)^2+1/2*I*b^
2*c^4/(c^2*d^2+e^2)^2*d*dilog(-1/2*I*(c*x+I))-a*b*c^3/(c^2*d^2+e^2)/(c*e*x+c*d)-1/2*b^2*c^3/(c*e*x+c*d)^2/e*ar
ctan(c*x)^2-1/2*b^2*c^3*e/(c^2*d^2+e^2)^2*arctan(c*x)^2-b^2*c^3*arctan(c*x)/(c^2*d^2+e^2)/(c*e*x+c*d)-1/2*b^2*
c^3*e/(c^2*d^2+e^2)^2*ln(c^2*x^2+1)+b^2*c^4/(c^2*d^2+e^2)^2*d*arctan(c*x)+b^2*c^3*e/(c^2*d^2+e^2)^2*ln(c*e*x+c
*d)-1/2*a^2*c^3/(c*e*x+c*d)^2/e)

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)/(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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