3.400 \(\int \frac{\tan ^{-1}(a x)^3}{x (c+a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=240 \[ \frac{3 i \text{PolyLog}\left (4,-1+\frac{2}{1-i a x}\right )}{4 c^2}-\frac{3 i \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c^2}+\frac{3 \tan ^{-1}(a x) \text{PolyLog}\left (3,-1+\frac{2}{1-i a x}\right )}{2 c^2}+\frac{3 a x}{8 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^3}{2 c^2 \left (a^2 x^2+1\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{4 c^2 \left (a^2 x^2+1\right )}-\frac{3 \tan ^{-1}(a x)}{4 c^2 \left (a^2 x^2+1\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^2}-\frac{\tan ^{-1}(a x)^3}{4 c^2}+\frac{3 \tan ^{-1}(a x)}{8 c^2}+\frac{\log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c^2} \]

[Out]

(3*a*x)/(8*c^2*(1 + a^2*x^2)) + (3*ArcTan[a*x])/(8*c^2) - (3*ArcTan[a*x])/(4*c^2*(1 + a^2*x^2)) - (3*a*x*ArcTa
n[a*x]^2)/(4*c^2*(1 + a^2*x^2)) - ArcTan[a*x]^3/(4*c^2) + ArcTan[a*x]^3/(2*c^2*(1 + a^2*x^2)) - ((I/4)*ArcTan[
a*x]^4)/c^2 + (ArcTan[a*x]^3*Log[2 - 2/(1 - I*a*x)])/c^2 - (((3*I)/2)*ArcTan[a*x]^2*PolyLog[2, -1 + 2/(1 - I*a
*x)])/c^2 + (3*ArcTan[a*x]*PolyLog[3, -1 + 2/(1 - I*a*x)])/(2*c^2) + (((3*I)/4)*PolyLog[4, -1 + 2/(1 - I*a*x)]
)/c^2

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Rubi [A]  time = 0.428387, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4966, 4924, 4868, 4884, 4992, 4996, 6610, 4930, 4892, 199, 205} \[ \frac{3 i \text{PolyLog}\left (4,-1+\frac{2}{1-i a x}\right )}{4 c^2}-\frac{3 i \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c^2}+\frac{3 \tan ^{-1}(a x) \text{PolyLog}\left (3,-1+\frac{2}{1-i a x}\right )}{2 c^2}+\frac{3 a x}{8 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^3}{2 c^2 \left (a^2 x^2+1\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{4 c^2 \left (a^2 x^2+1\right )}-\frac{3 \tan ^{-1}(a x)}{4 c^2 \left (a^2 x^2+1\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^2}-\frac{\tan ^{-1}(a x)^3}{4 c^2}+\frac{3 \tan ^{-1}(a x)}{8 c^2}+\frac{\log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x)/(8*c^2*(1 + a^2*x^2)) + (3*ArcTan[a*x])/(8*c^2) - (3*ArcTan[a*x])/(4*c^2*(1 + a^2*x^2)) - (3*a*x*ArcTa
n[a*x]^2)/(4*c^2*(1 + a^2*x^2)) - ArcTan[a*x]^3/(4*c^2) + ArcTan[a*x]^3/(2*c^2*(1 + a^2*x^2)) - ((I/4)*ArcTan[
a*x]^4)/c^2 + (ArcTan[a*x]^3*Log[2 - 2/(1 - I*a*x)])/c^2 - (((3*I)/2)*ArcTan[a*x]^2*PolyLog[2, -1 + 2/(1 - I*a
*x)])/c^2 + (3*ArcTan[a*x]*PolyLog[3, -1 + 2/(1 - I*a*x)])/(2*c^2) + (((3*I)/4)*PolyLog[4, -1 + 2/(1 - I*a*x)]
)/c^2

Rule 4966

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/d, Int[
x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/d, Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTan[c*
x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && ILtQ[m, 0] &
& 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 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 4992

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcT
an[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 4996

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a
+ b*ArcTan[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[
k + 1, u])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[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 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rule 4892

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*(a + b*ArcTan[c*x])
^p)/(2*d*(d + e*x^2)), x] + (-Dist[(b*c*p)/2, Int[(x*(a + b*ArcTan[c*x])^(p - 1))/(d + e*x^2)^2, x], x] + Simp
[(a + b*ArcTan[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p,
0]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(a x)^3}{x \left (c+a^2 c x^2\right )^2} \, dx &=-\left (a^2 \int \frac{x \tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx\right )+\frac{\int \frac{\tan ^{-1}(a x)^3}{x \left (c+a^2 c x^2\right )} \, dx}{c}\\ &=\frac{\tan ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^2}-\frac{1}{2} (3 a) \int \frac{\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx+\frac{i \int \frac{\tan ^{-1}(a x)^3}{x (i+a x)} \, dx}{c^2}\\ &=-\frac{3 a x \tan ^{-1}(a x)^2}{4 c^2 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)^3}{4 c^2}+\frac{\tan ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^2}+\frac{\tan ^{-1}(a x)^3 \log \left (2-\frac{2}{1-i a x}\right )}{c^2}+\frac{1}{2} \left (3 a^2\right ) \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx-\frac{(3 a) \int \frac{\tan ^{-1}(a x)^2 \log \left (2-\frac{2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c^2}\\ &=-\frac{3 \tan ^{-1}(a x)}{4 c^2 \left (1+a^2 x^2\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{4 c^2 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)^3}{4 c^2}+\frac{\tan ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^2}+\frac{\tan ^{-1}(a x)^3 \log \left (2-\frac{2}{1-i a x}\right )}{c^2}-\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{2 c^2}+\frac{1}{4} (3 a) \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx+\frac{(3 i a) \int \frac{\tan ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c^2}\\ &=\frac{3 a x}{8 c^2 \left (1+a^2 x^2\right )}-\frac{3 \tan ^{-1}(a x)}{4 c^2 \left (1+a^2 x^2\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{4 c^2 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)^3}{4 c^2}+\frac{\tan ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^2}+\frac{\tan ^{-1}(a x)^3 \log \left (2-\frac{2}{1-i a x}\right )}{c^2}-\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{2 c^2}+\frac{3 \tan ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1-i a x}\right )}{2 c^2}-\frac{(3 a) \int \frac{\text{Li}_3\left (-1+\frac{2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{2 c^2}+\frac{(3 a) \int \frac{1}{c+a^2 c x^2} \, dx}{8 c}\\ &=\frac{3 a x}{8 c^2 \left (1+a^2 x^2\right )}+\frac{3 \tan ^{-1}(a x)}{8 c^2}-\frac{3 \tan ^{-1}(a x)}{4 c^2 \left (1+a^2 x^2\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{4 c^2 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)^3}{4 c^2}+\frac{\tan ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^2}+\frac{\tan ^{-1}(a x)^3 \log \left (2-\frac{2}{1-i a x}\right )}{c^2}-\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{2 c^2}+\frac{3 \tan ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1-i a x}\right )}{2 c^2}+\frac{3 i \text{Li}_4\left (-1+\frac{2}{1-i a x}\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.22227, size = 156, normalized size = 0.65 \[ \frac{96 i \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(a x)}\right )+96 \tan ^{-1}(a x) \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(a x)}\right )-48 i \text{PolyLog}\left (4,e^{-2 i \tan ^{-1}(a x)}\right )+16 i \tan ^{-1}(a x)^4+64 \tan ^{-1}(a x)^3 \log \left (1-e^{-2 i \tan ^{-1}(a x)}\right )-24 \tan ^{-1}(a x)^2 \sin \left (2 \tan ^{-1}(a x)\right )+12 \sin \left (2 \tan ^{-1}(a x)\right )+16 \tan ^{-1}(a x)^3 \cos \left (2 \tan ^{-1}(a x)\right )-24 \tan ^{-1}(a x) \cos \left (2 \tan ^{-1}(a x)\right )-i \pi ^4}{64 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*Pi^4 + (16*I)*ArcTan[a*x]^4 - 24*ArcTan[a*x]*Cos[2*ArcTan[a*x]] + 16*ArcTan[a*x]^3*Cos[2*ArcTan[a*x]] +
64*ArcTan[a*x]^3*Log[1 - E^((-2*I)*ArcTan[a*x])] + (96*I)*ArcTan[a*x]^2*PolyLog[2, E^((-2*I)*ArcTan[a*x])] + 9
6*ArcTan[a*x]*PolyLog[3, E^((-2*I)*ArcTan[a*x])] - (48*I)*PolyLog[4, E^((-2*I)*ArcTan[a*x])] + 12*Sin[2*ArcTan
[a*x]] - 24*ArcTan[a*x]^2*Sin[2*ArcTan[a*x]])/(64*c^2)

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Maple [C]  time = 1.752, size = 2089, normalized size = 8.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*arctan(a*x)^3/c^2+3/2/c^2/(16*a*x+16*I)+1/2*I/c^2*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1
+I*a*x)^2/(a^2*x^2+1)+1))^3+3/16/c^2*arctan(a*x)/(a*x-I)*a*x-1/4*I/c^2*arctan(a*x)^3*Pi*csgn(I*(1+I*a*x)^2/(a^
2*x^2+1))^3+3/2*I/c^2/(16*a*x+16*I)*a*x-1/4*I/c^2*arctan(a*x)^3*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2
/(a^2*x^2+1)+1)^2)^3+1/c^2*arctan(a*x)^3*ln(2)+6/c^2*arctan(a*x)*polylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/c^2
*arctan(a*x)^3*ln((1+I*a*x)/(a^2*x^2+1)^(1/2))-1/c^2*arctan(a*x)^3*ln((1+I*a*x)^2/(a^2*x^2+1)-1)+1/c^2*arctan(
a*x)^3*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))+6/c^2*arctan(a*x)*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*I/c^2*poly
log(4,(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*I/c^2*polylog(4,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-3/2/c^2*arctan(a*x)^2/(8*a*
x+8*I)-3/2/c^2*arctan(a*x)^2/(8*a*x-8*I)+1/c^2*arctan(a*x)^3*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))+3/16/c^2*arctan
(a*x)/(a*x+I)*a*x+3/2/c^2/(16*a*x-16*I)+1/2*arctan(a*x)^3/c^2/(a^2*x^2+1)-1/4*I*arctan(a*x)^4/c^2-1/2/c^2*arct
an(a*x)^3*ln(a^2*x^2+1)+1/c^2*arctan(a*x)^3*ln(a*x)+1/4*I/c^2*arctan(a*x)^3*Pi*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)
+1)^2)*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2-1/4*I/c^2*arctan(a*x)^3*Pi*csgn(I*(1+I*
a*x)/(a^2*x^2+1)^(1/2))^2*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))-1/2*I/c^2*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)^2/(a^2*
x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2+1/4*I/c
^2*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1))^2*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)+1/2*I/c^2*arct
an(a*x)^3*Pi*csgn(I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^2+1/4*I/c^2*arctan(a*x)^3*Pi*
csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2-1/2*I/c^2*arct
an(a*x)^3*Pi*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1
))^2-1/2*I/c^2*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a
*x)^2/(a^2*x^2+1)+1))^2+1/2*I/c^2*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)
+1))*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))-1/2*I/c^2*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x
)^2/(a^2*x^2+1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^2-3/2*I/c^2*arctan(a*x)^2/(8*a*x+8*I)*a*x+3/2*I/c^2*
arctan(a*x)^2/(8*a*x-8*I)*a*x-1/2*I/c^2*arctan(a*x)^3*Pi*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^
2+1)+1))^2+1/2*I/c^2*arctan(a*x)^3*Pi*csgn(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^3+1/4*I/c^
2*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-3/2*I/c^2/(16*a*x-16*I)*a*x-3*I/c^2*arctan(a*x)^2*p
olylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+3/16*I/c^2*arctan(a*x)/(a*x-I)-3/16*I/c^2*arctan(a*x)/(a*x+I)+1/2*I/c^2
*arctan(a*x)^3*Pi-3*I/c^2*arctan(a*x)^2*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))-1/4*I/c^2*arctan(a*x)^3*Pi*csgn
(I/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/
(a^2*x^2+1)+1)^2)+1/2*I/c^2*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1))*csgn(I/((1+I*a*x)^2/(a^2*x^2+
1)+1))*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (a x\right )^{3}}{a^{4} c^{2} x^{5} + 2 \, a^{2} c^{2} x^{3} + c^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(a*x)^3/(a^4*c^2*x^5 + 2*a^2*c^2*x^3 + c^2*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{atan}^{3}{\left (a x \right )}}{a^{4} x^{5} + 2 a^{2} x^{3} + x}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)**3/x/(a**2*c*x**2+c)**2,x)

[Out]

Integral(atan(a*x)**3/(a**4*x**5 + 2*a**2*x**3 + x), x)/c**2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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