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

Optimal. Leaf size=332 \[ \frac{3 i \text{PolyLog}\left (4,-1+\frac{2}{1-i a x}\right )}{4 c^3}-\frac{3 i \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c^3}+\frac{3 \tan ^{-1}(a x) \text{PolyLog}\left (3,-1+\frac{2}{1-i a x}\right )}{2 c^3}+\frac{141 a x}{256 c^3 \left (a^2 x^2+1\right )}+\frac{3 a x}{128 c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)^3}{2 c^3 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^3}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac{33 a x \tan ^{-1}(a x)^2}{32 c^3 \left (a^2 x^2+1\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{16 c^3 \left (a^2 x^2+1\right )^2}-\frac{33 \tan ^{-1}(a x)}{32 c^3 \left (a^2 x^2+1\right )}-\frac{3 \tan ^{-1}(a x)}{32 c^3 \left (a^2 x^2+1\right )^2}-\frac{i \tan ^{-1}(a x)^4}{4 c^3}-\frac{11 \tan ^{-1}(a x)^3}{32 c^3}+\frac{141 \tan ^{-1}(a x)}{256 c^3}+\frac{\log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c^3} \]

[Out]

(3*a*x)/(128*c^3*(1 + a^2*x^2)^2) + (141*a*x)/(256*c^3*(1 + a^2*x^2)) + (141*ArcTan[a*x])/(256*c^3) - (3*ArcTa
n[a*x])/(32*c^3*(1 + a^2*x^2)^2) - (33*ArcTan[a*x])/(32*c^3*(1 + a^2*x^2)) - (3*a*x*ArcTan[a*x]^2)/(16*c^3*(1
+ a^2*x^2)^2) - (33*a*x*ArcTan[a*x]^2)/(32*c^3*(1 + a^2*x^2)) - (11*ArcTan[a*x]^3)/(32*c^3) + ArcTan[a*x]^3/(4
*c^3*(1 + a^2*x^2)^2) + ArcTan[a*x]^3/(2*c^3*(1 + a^2*x^2)) - ((I/4)*ArcTan[a*x]^4)/c^3 + (ArcTan[a*x]^3*Log[2
- 2/(1 - I*a*x)])/c^3 - (((3*I)/2)*ArcTan[a*x]^2*PolyLog[2, -1 + 2/(1 - I*a*x)])/c^3 + (3*ArcTan[a*x]*PolyLog
[3, -1 + 2/(1 - I*a*x)])/(2*c^3) + (((3*I)/4)*PolyLog[4, -1 + 2/(1 - I*a*x)])/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.709367, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 12, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {4966, 4924, 4868, 4884, 4992, 4996, 6610, 4930, 4892, 199, 205, 4900} \[ \frac{3 i \text{PolyLog}\left (4,-1+\frac{2}{1-i a x}\right )}{4 c^3}-\frac{3 i \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{2 c^3}+\frac{3 \tan ^{-1}(a x) \text{PolyLog}\left (3,-1+\frac{2}{1-i a x}\right )}{2 c^3}+\frac{141 a x}{256 c^3 \left (a^2 x^2+1\right )}+\frac{3 a x}{128 c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)^3}{2 c^3 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^3}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac{33 a x \tan ^{-1}(a x)^2}{32 c^3 \left (a^2 x^2+1\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{16 c^3 \left (a^2 x^2+1\right )^2}-\frac{33 \tan ^{-1}(a x)}{32 c^3 \left (a^2 x^2+1\right )}-\frac{3 \tan ^{-1}(a x)}{32 c^3 \left (a^2 x^2+1\right )^2}-\frac{i \tan ^{-1}(a x)^4}{4 c^3}-\frac{11 \tan ^{-1}(a x)^3}{32 c^3}+\frac{141 \tan ^{-1}(a x)}{256 c^3}+\frac{\log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x)/(128*c^3*(1 + a^2*x^2)^2) + (141*a*x)/(256*c^3*(1 + a^2*x^2)) + (141*ArcTan[a*x])/(256*c^3) - (3*ArcTa
n[a*x])/(32*c^3*(1 + a^2*x^2)^2) - (33*ArcTan[a*x])/(32*c^3*(1 + a^2*x^2)) - (3*a*x*ArcTan[a*x]^2)/(16*c^3*(1
+ a^2*x^2)^2) - (33*a*x*ArcTan[a*x]^2)/(32*c^3*(1 + a^2*x^2)) - (11*ArcTan[a*x]^3)/(32*c^3) + ArcTan[a*x]^3/(4
*c^3*(1 + a^2*x^2)^2) + ArcTan[a*x]^3/(2*c^3*(1 + a^2*x^2)) - ((I/4)*ArcTan[a*x]^4)/c^3 + (ArcTan[a*x]^3*Log[2
- 2/(1 - I*a*x)])/c^3 - (((3*I)/2)*ArcTan[a*x]^2*PolyLog[2, -1 + 2/(1 - I*a*x)])/c^3 + (3*ArcTan[a*x]*PolyLog
[3, -1 + 2/(1 - I*a*x)])/(2*c^3) + (((3*I)/4)*PolyLog[4, -1 + 2/(1 - I*a*x)])/c^3

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]

Rule 4900

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b*p*(d + e*x^2)^(q
+ 1)*(a + b*ArcTan[c*x])^(p - 1))/(4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q +
1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(b^2*p*(p - 1))/(4*(q + 1)^2), Int[(d + e*x^2)^q*(a + b*ArcTan[c*x])^(
p - 2), x], x] - Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e
}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && GtQ[p, 1] && NeQ[q, -3/2]

Rubi steps

\begin{align*} \int \frac{\tan ^{-1}(a x)^3}{x \left (c+a^2 c x^2\right )^3} \, dx &=-\left (a^2 \int \frac{x \tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^3} \, dx\right )+\frac{\int \frac{\tan ^{-1}(a x)^3}{x \left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=\frac{\tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{1}{4} (3 a) \int \frac{\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx+\frac{\int \frac{\tan ^{-1}(a x)^3}{x \left (c+a^2 c x^2\right )} \, dx}{c^2}-\frac{a^2 \int \frac{x \tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=-\frac{3 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac{3 a x \tan ^{-1}(a x)^2}{16 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)^3}{2 c^3 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^3}+\frac{1}{32} (3 a) \int \frac{1}{\left (c+a^2 c x^2\right )^3} \, dx+\frac{i \int \frac{\tan ^{-1}(a x)^3}{x (i+a x)} \, dx}{c^3}-\frac{(9 a) \int \frac{\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{16 c}-\frac{(3 a) \int \frac{\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{2 c}\\ &=\frac{3 a x}{128 c^3 \left (1+a^2 x^2\right )^2}-\frac{3 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac{3 a x \tan ^{-1}(a x)^2}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{33 a x \tan ^{-1}(a x)^2}{32 c^3 \left (1+a^2 x^2\right )}-\frac{11 \tan ^{-1}(a x)^3}{32 c^3}+\frac{\tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)^3}{2 c^3 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^3}+\frac{\tan ^{-1}(a x)^3 \log \left (2-\frac{2}{1-i a x}\right )}{c^3}-\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^3}+\frac{(9 a) \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx}{128 c}+\frac{\left (9 a^2\right ) \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{16 c}+\frac{\left (3 a^2\right ) \int \frac{x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{2 c}\\ &=\frac{3 a x}{128 c^3 \left (1+a^2 x^2\right )^2}+\frac{9 a x}{256 c^3 \left (1+a^2 x^2\right )}-\frac{3 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac{33 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{33 a x \tan ^{-1}(a x)^2}{32 c^3 \left (1+a^2 x^2\right )}-\frac{11 \tan ^{-1}(a x)^3}{32 c^3}+\frac{\tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)^3}{2 c^3 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^3}+\frac{\tan ^{-1}(a x)^3 \log \left (2-\frac{2}{1-i a x}\right )}{c^3}-\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{2 c^3}+\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^3}+\frac{(9 a) \int \frac{1}{c+a^2 c x^2} \, dx}{256 c^2}+\frac{(9 a) \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx}{32 c}+\frac{(3 a) \int \frac{1}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}\\ &=\frac{3 a x}{128 c^3 \left (1+a^2 x^2\right )^2}+\frac{141 a x}{256 c^3 \left (1+a^2 x^2\right )}+\frac{9 \tan ^{-1}(a x)}{256 c^3}-\frac{3 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac{33 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{33 a x \tan ^{-1}(a x)^2}{32 c^3 \left (1+a^2 x^2\right )}-\frac{11 \tan ^{-1}(a x)^3}{32 c^3}+\frac{\tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)^3}{2 c^3 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^3}+\frac{\tan ^{-1}(a x)^3 \log \left (2-\frac{2}{1-i a x}\right )}{c^3}-\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{2 c^3}+\frac{3 \tan ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1-i a x}\right )}{2 c^3}-\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^3}+\frac{(9 a) \int \frac{1}{c+a^2 c x^2} \, dx}{64 c^2}+\frac{(3 a) \int \frac{1}{c+a^2 c x^2} \, dx}{8 c^2}\\ &=\frac{3 a x}{128 c^3 \left (1+a^2 x^2\right )^2}+\frac{141 a x}{256 c^3 \left (1+a^2 x^2\right )}+\frac{141 \tan ^{-1}(a x)}{256 c^3}-\frac{3 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac{33 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )}-\frac{3 a x \tan ^{-1}(a x)^2}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{33 a x \tan ^{-1}(a x)^2}{32 c^3 \left (1+a^2 x^2\right )}-\frac{11 \tan ^{-1}(a x)^3}{32 c^3}+\frac{\tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\tan ^{-1}(a x)^3}{2 c^3 \left (1+a^2 x^2\right )}-\frac{i \tan ^{-1}(a x)^4}{4 c^3}+\frac{\tan ^{-1}(a x)^3 \log \left (2-\frac{2}{1-i a x}\right )}{c^3}-\frac{3 i \tan ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1-i a x}\right )}{2 c^3}+\frac{3 \tan ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1-i a x}\right )}{2 c^3}+\frac{3 i \text{Li}_4\left (-1+\frac{2}{1-i a x}\right )}{4 c^3}\\ \end{align*}

Mathematica [A]  time = 0.313079, size = 208, normalized size = 0.63 \[ \frac{1536 i \tan ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(a x)}\right )+1536 \tan ^{-1}(a x) \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(a x)}\right )-768 i \text{PolyLog}\left (4,e^{-2 i \tan ^{-1}(a x)}\right )+256 i \tan ^{-1}(a x)^4+1024 \tan ^{-1}(a x)^3 \log \left (1-e^{-2 i \tan ^{-1}(a x)}\right )-576 \tan ^{-1}(a x)^2 \sin \left (2 \tan ^{-1}(a x)\right )-24 \tan ^{-1}(a x)^2 \sin \left (4 \tan ^{-1}(a x)\right )+288 \sin \left (2 \tan ^{-1}(a x)\right )+3 \sin \left (4 \tan ^{-1}(a x)\right )+384 \tan ^{-1}(a x)^3 \cos \left (2 \tan ^{-1}(a x)\right )+32 \tan ^{-1}(a x)^3 \cos \left (4 \tan ^{-1}(a x)\right )-576 \tan ^{-1}(a x) \cos \left (2 \tan ^{-1}(a x)\right )-12 \tan ^{-1}(a x) \cos \left (4 \tan ^{-1}(a x)\right )-16 i \pi ^4}{1024 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-16*I)*Pi^4 + (256*I)*ArcTan[a*x]^4 - 576*ArcTan[a*x]*Cos[2*ArcTan[a*x]] + 384*ArcTan[a*x]^3*Cos[2*ArcTan[a*
x]] - 12*ArcTan[a*x]*Cos[4*ArcTan[a*x]] + 32*ArcTan[a*x]^3*Cos[4*ArcTan[a*x]] + 1024*ArcTan[a*x]^3*Log[1 - E^(
(-2*I)*ArcTan[a*x])] + (1536*I)*ArcTan[a*x]^2*PolyLog[2, E^((-2*I)*ArcTan[a*x])] + 1536*ArcTan[a*x]*PolyLog[3,
E^((-2*I)*ArcTan[a*x])] - (768*I)*PolyLog[4, E^((-2*I)*ArcTan[a*x])] + 288*Sin[2*ArcTan[a*x]] - 576*ArcTan[a*
x]^2*Sin[2*ArcTan[a*x]] + 3*Sin[4*ArcTan[a*x]] - 24*ArcTan[a*x]^2*Sin[4*ArcTan[a*x]])/(1024*c^3)

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Maple [C]  time = 3.697, size = 2463, normalized size = 7.4 \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)^3,x)

[Out]

-3/1024/c^3/(a*x+I)^2*a*x-3/1024/c^3/(a*x-I)^2*a*x-9/32*I/c^3*arctan(a*x)/(a*x+I)+9/32*I/c^3*arctan(a*x)/(a*x-
I)-11/32*arctan(a*x)^3/c^3+9/4/c^3/(16*a*x+16*I)-1/2/c^3*arctan(a*x)^3*ln(a^2*x^2+1)+1/c^3*arctan(a*x)^3*ln(a*
x)+1/c^3*arctan(a*x)^3*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))+6/c^3*arctan(a*x)*polylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1
/2))+3/512/c^3*arctan(a*x)/(a*x-I)^2+3/512/c^3*arctan(a*x)/(a*x+I)^2-9/4/c^3*arctan(a*x)^2/(8*a*x+8*I)-9/4/c^3
*arctan(a*x)^2/(8*a*x-8*I)+6*I/c^3*polylog(4,(1+I*a*x)/(a^2*x^2+1)^(1/2))+6*I/c^3*polylog(4,-(1+I*a*x)/(a^2*x^
2+1)^(1/2))+3/2048*I/c^3/(a*x+I)^2-3/2048*I/c^3/(a*x-I)^2+1/c^3*arctan(a*x)^3*ln(2)+1/c^3*arctan(a*x)^3*ln((1+
I*a*x)/(a^2*x^2+1)^(1/2))-1/c^3*arctan(a*x)^3*ln((1+I*a*x)^2/(a^2*x^2+1)-1)+1/c^3*arctan(a*x)^3*ln(1-(1+I*a*x)
/(a^2*x^2+1)^(1/2))+6/c^3*arctan(a*x)*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/2*I/c^3*Pi*arctan(a*x)^3*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))+1/4*I/c^3*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+1/4*I/c^3*Pi*arctan(a*x)^3*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)-1/4*I/c^3*Pi*arctan(a*x)^3*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^3*Pi*arctan(a*x)^3*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)^2-1/4*I/c^3*Pi*arctan(a*x)^3*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)+1/4*arctan(a*x)^3/c^3/(a^2*x^2+1)^2+1/2*arctan(a*x)^3/c
^3/(a^2*x^2+1)-1/4*I*arctan(a*x)^4/c^3-3/256*I/c^3*arctan(a*x)^2/(a*x+I)^2+3/256*I/c^3*arctan(a*x)^2/(a*x-I)^2
-3*I/c^3*arctan(a*x)^2*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))-3*I/c^3*arctan(a*x)^2*polylog(2,-(1+I*a*x)/(a^2*
x^2+1)^(1/2))+1/2*I/c^3*arctan(a*x)^3*Pi+1/2*I/c^3*Pi*arctan(a*x)^3*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^3*Pi*arctan(a*x)^3
*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
^3*Pi*arctan(a*x)^3*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^3*Pi*arctan(a*x)^3*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+1/2*I/c^3*Pi*arctan(a*x)^3*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))^2-1/2*I/c^3*Pi*arctan(a*x)^3*csgn(I/((1+I*a*x)^2/(a^2*x^2+1)+1))*csg
n(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2-3/256*I/c^3*arctan(a*x)/(a*x-I)^2*a*x+3/256*I/c
^3*arctan(a*x)/(a*x+I)^2*a*x-9/4*I/c^3*arctan(a*x)^2/(8*a*x+8*I)*a*x+3/256*I/c^3*arctan(a*x)^2/(a*x+I)^2*a^2*x
^2-3/256*I/c^3*arctan(a*x)^2/(a*x-I)^2*a^2*x^2+9/4*I/c^3*arctan(a*x)^2/(8*a*x-8*I)*a*x+9/4/c^3/(16*a*x-16*I)+3
/128/c^3*arctan(a*x)^2/(a*x+I)^2*a*x+3/128/c^3*arctan(a*x)^2/(a*x-I)^2*a*x-3/512/c^3*arctan(a*x)/(a*x-I)^2*a^2
*x^2+9/32/c^3*arctan(a*x)/(a*x+I)*a*x-3/512/c^3*arctan(a*x)/(a*x+I)^2*a^2*x^2+9/32/c^3*arctan(a*x)/(a*x-I)*a*x
+1/2*I/c^3*Pi*arctan(a*x)^3*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^3-1/2*I/c^3*Pi*arc
tan(a*x)^3*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^3*Pi*arctan(a*x)^3*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^3*Pi*arctan(a*x)^3*csgn(I*((1+I*a*x)^2/(a^2*x^
2+1)+1)^2)^3-1/4*I/c^3*Pi*arctan(a*x)^3*csgn(I*(1+I*a*x)^2/(a^2*x^2+1)/((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3-1/4*I/
c^3*arctan(a*x)^3*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^3+9/4*I/c^3/(16*a*x+16*I)*a*x-3/2048*I/c^3/(a*x+I)^2*a^2*
x^2+3/2048*I/c^3/(a*x-I)^2*a^2*x^2-9/4*I/c^3/(16*a*x-16*I)*a*x

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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 )}^{3} 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)^3,x, algorithm="maxima")

[Out]

integrate(arctan(a*x)^3/((a^2*c*x^2 + c)^3*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^{6} c^{3} x^{7} + 3 \, a^{4} c^{3} x^{5} + 3 \, a^{2} c^{3} x^{3} + c^{3} 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)^3,x, algorithm="fricas")

[Out]

integral(arctan(a*x)^3/(a^6*c^3*x^7 + 3*a^4*c^3*x^5 + 3*a^2*c^3*x^3 + c^3*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^{6} x^{7} + 3 a^{4} x^{5} + 3 a^{2} x^{3} + x}\, dx}{c^{3}} \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)**3,x)

[Out]

Integral(atan(a*x)**3/(a**6*x**7 + 3*a**4*x**5 + 3*a**2*x**3 + x), x)/c**3

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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 )}^{3} 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)^3,x, algorithm="giac")

[Out]

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