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

Optimal. Leaf size=341 \[ \frac{i b^2 c \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac{i b^2 c \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^2 d^2+e^2}-\frac{i b^2 c \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{c^2 d^2+e^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^2+e^2}+\frac{c^2 d \left (a+b \tan ^{-1}(c x)\right )^2}{e \left (c^2 d^2+e^2\right )}-\frac{2 b c \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2+e^2}+\frac{2 b c \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2+e^2}+\frac{2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{c^2 d^2+e^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{e (d+e x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.3701, antiderivative size = 341, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4864, 4856, 2402, 2315, 2447, 4984, 4884, 4920, 4854} \[ \frac{i b^2 c \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{c^2 d^2+e^2}+\frac{i b^2 c \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^2 d^2+e^2}-\frac{i b^2 c \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{c^2 d^2+e^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^2+e^2}+\frac{c^2 d \left (a+b \tan ^{-1}(c x)\right )^2}{e \left (c^2 d^2+e^2\right )}-\frac{2 b c \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2+e^2}+\frac{2 b c \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2+e^2}+\frac{2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{c^2 d^2+e^2}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.88078, size = 300, normalized size = 0.88 \[ \frac{b^2 \left (-\frac{c d \left (i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )-\frac{1}{2} \pi \log \left (c^2 x^2+1\right )-i \tan ^{-1}(c x) \left (\pi -2 \tan ^{-1}\left (\frac{c d}{e}\right )\right )-2 \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right ) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )+2 \tan ^{-1}\left (\frac{c d}{e}\right ) \log \left (\sin \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )\right )-\pi \log \left (1+e^{-2 i \tan ^{-1}(c x)}\right )\right )}{c^2 d^2+e^2}-\frac{\tan ^{-1}(c x)^2 e^{i \tan ^{-1}\left (\frac{c d}{e}\right )}}{e \sqrt{\frac{c^2 d^2}{e^2}+1}}+\frac{x \tan ^{-1}(c x)^2}{d+e x}\right )}{d}-\frac{a^2}{e (d+e x)}+\frac{a b \left (c (d+e x) \left (2 \log (c (d+e x))-\log \left (c^2 x^2+1\right )\right )-2 \tan ^{-1}(c x) \left (e-c^2 d x\right )\right )}{\left (c^2 d^2+e^2\right ) (d+e x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.102, size = 698, normalized size = 2.1 \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)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\left ({\left (\frac{2 \, c d \arctan \left (c x\right )}{c^{2} d^{2} e + e^{3}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{2} + e^{2}} + \frac{2 \, \log \left (e x + d\right )}{c^{2} d^{2} + e^{2}}\right )} c - \frac{2 \, \arctan \left (c x\right )}{e^{2} x + d e}\right )} a b - \frac{\frac{1}{4} \,{\left (28 \, \arctan \left (c x\right )^{2} - 4 \,{\left (e^{2} x + d e\right )} \int \frac{36 \,{\left (c^{2} e x^{2} + e\right )} \arctan \left (c x\right )^{2} + 3 \,{\left (c^{2} e x^{2} + e\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + 56 \,{\left (c e x + c d\right )} \arctan \left (c x\right ) - 12 \,{\left (c^{2} e x^{2} + c^{2} d x\right )} \log \left (c^{2} x^{2} + 1\right )}{4 \,{\left (c^{2} e^{3} x^{4} + 2 \, c^{2} d e^{2} x^{3} + 2 \, d e^{2} x + d^{2} e +{\left (c^{2} d^{2} e + e^{3}\right )} x^{2}\right )}}\,{d x} - 3 \, \log \left (c^{2} x^{2} + 1\right )^{2}\right )} b^{2}}{16 \,{\left (e^{2} x + d e\right )}} - \frac{a^{2}}{e^{2} x + d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2} x^{2} + 2 \, d e x + d^{2}}, 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)^2,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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