3.5.8 \(\int \frac {\text {ArcTan}(a x)^3}{x (c+a^2 c x^2)^3} \, dx\) [408]

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

[Out]

3/128*a*x/c^3/(a^2*x^2+1)^2+141/256*a*x/c^3/(a^2*x^2+1)+141/256*arctan(a*x)/c^3-3/32*arctan(a*x)/c^3/(a^2*x^2+
1)^2-33/32*arctan(a*x)/c^3/(a^2*x^2+1)-3/16*a*x*arctan(a*x)^2/c^3/(a^2*x^2+1)^2-33/32*a*x*arctan(a*x)^2/c^3/(a
^2*x^2+1)-11/32*arctan(a*x)^3/c^3+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+arctan(a*x)^3*ln(2-2/(1-I*a*x))/c^3-3/2*I*arctan(a*x)^2*polylog(2,-1+2/(1-I*a*x))/c^3+3/2*ar
ctan(a*x)*polylog(3,-1+2/(1-I*a*x))/c^3+3/4*I*polylog(4,-1+2/(1-I*a*x))/c^3

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Rubi [A]
time = 0.50, antiderivative size = 332, normalized size of antiderivative = 1.00, 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 = {5086, 5044, 4988, 5004, 5112, 5116, 6745, 5050, 5012, 205, 211, 5020} \begin {gather*} \frac {\text {ArcTan}(a x)^3}{2 c^3 \left (a^2 x^2+1\right )}+\frac {\text {ArcTan}(a x)^3}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {33 a x \text {ArcTan}(a x)^2}{32 c^3 \left (a^2 x^2+1\right )}-\frac {3 a x \text {ArcTan}(a x)^2}{16 c^3 \left (a^2 x^2+1\right )^2}-\frac {33 \text {ArcTan}(a x)}{32 c^3 \left (a^2 x^2+1\right )}-\frac {3 \text {ArcTan}(a x)}{32 c^3 \left (a^2 x^2+1\right )^2}+\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 {3 i \text {ArcTan}(a x)^2 \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{2 c^3}+\frac {3 \text {ArcTan}(a x) \text {Li}_3\left (\frac {2}{1-i a x}-1\right )}{2 c^3}-\frac {i \text {ArcTan}(a x)^4}{4 c^3}-\frac {11 \text {ArcTan}(a x)^3}{32 c^3}+\frac {141 \text {ArcTan}(a x)}{256 c^3}+\frac {\text {ArcTan}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c^3}+\frac {3 i \text {Li}_4\left (\frac {2}{1-i a x}-1\right )}{4 c^3} \end {gather*}

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 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

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

Rule 4988

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 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 5012

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 5020

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]

Rule 5044

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*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 5050

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 5086

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 5112

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[I*(a + b*ArcTa
n[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 5116

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)*(PolyLo
g[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 6745

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]

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*}

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Mathematica [A]
time = 0.16, size = 208, normalized size = 0.63 \begin {gather*} \frac {-16 i \pi ^4+256 i \text {ArcTan}(a x)^4-576 \text {ArcTan}(a x) \cos (2 \text {ArcTan}(a x))+384 \text {ArcTan}(a x)^3 \cos (2 \text {ArcTan}(a x))-12 \text {ArcTan}(a x) \cos (4 \text {ArcTan}(a x))+32 \text {ArcTan}(a x)^3 \cos (4 \text {ArcTan}(a x))+1024 \text {ArcTan}(a x)^3 \log \left (1-e^{-2 i \text {ArcTan}(a x)}\right )+1536 i \text {ArcTan}(a x)^2 \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(a x)}\right )+1536 \text {ArcTan}(a x) \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(a x)}\right )-768 i \text {PolyLog}\left (4,e^{-2 i \text {ArcTan}(a x)}\right )+288 \sin (2 \text {ArcTan}(a x))-576 \text {ArcTan}(a x)^2 \sin (2 \text {ArcTan}(a x))+3 \sin (4 \text {ArcTan}(a x))-24 \text {ArcTan}(a x)^2 \sin (4 \text {ArcTan}(a x))}{1024 c^3} \end {gather*}

Antiderivative was successfully verified.

[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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 75.35, size = 1959, normalized size = 5.90

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1959\)
default	\(\text {Expression too large to display}\)	\(1959\)

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,method=_RETURNVERBOSE)

[Out]

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)-1/2/c^3*arctan(a*x)^3*ln(a^2*x^2+1)+1/c^
3*arctan(a*x)^3*ln(a*x)-3/4/c^3*(-2/3*I*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))+1/3*I*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)+1/3*I*arctan(a*x)^3*Pi*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^3+1/3*I*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+11/24*arctan(a*x)^3-2/3*I*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+2/3*I*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))^2-3*I*arctan(a*x)^2*(I+a*x)/(8*a*x-8*I)+3*I*arctan(a*x)^2*(a*x-I)/(8*a*x+8*I)-4/3*ar
ctan(a*x)^3*ln((1+I*a*x)/(a^2*x^2+1)^(1/2))+1/3*I*arctan(a*x)^4-8*I*polylog(4,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-8*
I*polylog(4,(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/64*arctan(a*x)*cos(4*arctan(a*x))-2/3*I*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))^3-1/3*I*arctan(a*x)^3*Pi*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1
)^2)^3+1/256*(8*arctan(a*x)^2-1)*sin(4*arctan(a*x))-4/3*arctan(a*x)^3*ln(2)-8*arctan(a*x)*polylog(3,-(1+I*a*x)
/(a^2*x^2+1)^(1/2))-8*arctan(a*x)*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(1/2))-4/3*arctan(a*x)^3*ln(1+(1+I*a*x)/(a^2
*x^2+1)^(1/2))-4/3*arctan(a*x)^3*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))+4/3*arctan(a*x)^3*ln((1+I*a*x)^2/(a^2*x^2+1
)-1)-1/3*I*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+2/3*I*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+2/3*I*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+2/3*I*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+2/3*I*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))^2-1/3*I*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)^2-2/3*I*arctan(a*x)^3*P
i*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))-2/3*I*arctan(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
+1/3*I*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))-1/3*I*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)-2/3*I*arctan(a*x)^3*Pi+4*I*
arctan(a*x)^2*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+4*I*arctan(a*x)^2*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))
+3*I*(I+a*x)/(16*a*x-16*I)-3*I*(a*x-I)/(16*a*x+16*I)-3/8*arctan(a*x)*(I+a*x)/(a*x-I)-3/8*arctan(a*x)*(a*x-I)/(
I+a*x))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, 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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, 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(arctan(a*x)^3/x/(a^2*c*x^2+c)^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 {{\mathrm {atan}\left (a\,x\right )}^3}{x\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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