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

Optimal. Leaf size=236 \[ \frac{\text{PolyLog}\left (3,-1+\frac{2}{1-i a x}\right )}{2 c^3}-\frac{i \tan ^{-1}(a x) \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{c^3}-\frac{11}{32 c^3 \left (a^2 x^2+1\right )}-\frac{1}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)^2}{2 c^3 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac{11 a x \tan ^{-1}(a x)}{16 c^3 \left (a^2 x^2+1\right )}-\frac{a x \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )^2}-\frac{i \tan ^{-1}(a x)^3}{3 c^3}-\frac{11 \tan ^{-1}(a x)^2}{32 c^3}+\frac{\log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)^2}{c^3} \]

[Out]

-1/(32*c^3*(1 + a^2*x^2)^2) - 11/(32*c^3*(1 + a^2*x^2)) - (a*x*ArcTan[a*x])/(8*c^3*(1 + a^2*x^2)^2) - (11*a*x*
ArcTan[a*x])/(16*c^3*(1 + a^2*x^2)) - (11*ArcTan[a*x]^2)/(32*c^3) + ArcTan[a*x]^2/(4*c^3*(1 + a^2*x^2)^2) + Ar
cTan[a*x]^2/(2*c^3*(1 + a^2*x^2)) - ((I/3)*ArcTan[a*x]^3)/c^3 + (ArcTan[a*x]^2*Log[2 - 2/(1 - I*a*x)])/c^3 - (
I*ArcTan[a*x]*PolyLog[2, -1 + 2/(1 - I*a*x)])/c^3 + PolyLog[3, -1 + 2/(1 - I*a*x)]/(2*c^3)

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Rubi [A]  time = 0.484914, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {4966, 4924, 4868, 4884, 4992, 6610, 4930, 4892, 261, 4896} \[ \frac{\text{PolyLog}\left (3,-1+\frac{2}{1-i a x}\right )}{2 c^3}-\frac{i \tan ^{-1}(a x) \text{PolyLog}\left (2,-1+\frac{2}{1-i a x}\right )}{c^3}-\frac{11}{32 c^3 \left (a^2 x^2+1\right )}-\frac{1}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac{\tan ^{-1}(a x)^2}{2 c^3 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac{11 a x \tan ^{-1}(a x)}{16 c^3 \left (a^2 x^2+1\right )}-\frac{a x \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )^2}-\frac{i \tan ^{-1}(a x)^3}{3 c^3}-\frac{11 \tan ^{-1}(a x)^2}{32 c^3}+\frac{\log \left (2-\frac{2}{1-i a x}\right ) \tan ^{-1}(a x)^2}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(32*c^3*(1 + a^2*x^2)^2) - 11/(32*c^3*(1 + a^2*x^2)) - (a*x*ArcTan[a*x])/(8*c^3*(1 + a^2*x^2)^2) - (11*a*x*
ArcTan[a*x])/(16*c^3*(1 + a^2*x^2)) - (11*ArcTan[a*x]^2)/(32*c^3) + ArcTan[a*x]^2/(4*c^3*(1 + a^2*x^2)^2) + Ar
cTan[a*x]^2/(2*c^3*(1 + a^2*x^2)) - ((I/3)*ArcTan[a*x]^3)/c^3 + (ArcTan[a*x]^2*Log[2 - 2/(1 - I*a*x)])/c^3 - (
I*ArcTan[a*x]*PolyLog[2, -1 + 2/(1 - I*a*x)])/c^3 + PolyLog[3, -1 + 2/(1 - I*a*x)]/(2*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 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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4896

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

Rubi steps

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

Mathematica [A]  time = 0.255242, size = 156, normalized size = 0.66 \[ \frac{768 i \tan ^{-1}(a x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(a x)}\right )+384 \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(a x)}\right )+256 i \tan ^{-1}(a x)^3+768 \tan ^{-1}(a x)^2 \log \left (1-e^{-2 i \tan ^{-1}(a x)}\right )-288 \tan ^{-1}(a x) \sin \left (2 \tan ^{-1}(a x)\right )-12 \tan ^{-1}(a x) \sin \left (4 \tan ^{-1}(a x)\right )+288 \tan ^{-1}(a x)^2 \cos \left (2 \tan ^{-1}(a x)\right )+24 \tan ^{-1}(a x)^2 \cos \left (4 \tan ^{-1}(a x)\right )-144 \cos \left (2 \tan ^{-1}(a x)\right )-3 \cos \left (4 \tan ^{-1}(a x)\right )-32 i \pi ^3}{768 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-32*I)*Pi^3 + (256*I)*ArcTan[a*x]^3 - 144*Cos[2*ArcTan[a*x]] + 288*ArcTan[a*x]^2*Cos[2*ArcTan[a*x]] - 3*Cos[
4*ArcTan[a*x]] + 24*ArcTan[a*x]^2*Cos[4*ArcTan[a*x]] + 768*ArcTan[a*x]^2*Log[1 - E^((-2*I)*ArcTan[a*x])] + (76
8*I)*ArcTan[a*x]*PolyLog[2, E^((-2*I)*ArcTan[a*x])] + 384*PolyLog[3, E^((-2*I)*ArcTan[a*x])] - 288*ArcTan[a*x]
*Sin[2*ArcTan[a*x]] - 12*ArcTan[a*x]*Sin[4*ArcTan[a*x]])/(768*c^3)

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Maple [C]  time = 0.559, size = 2176, normalized size = 9.2 \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)^2/x/(a^2*c*x^2+c)^3,x)

[Out]

1/2*I/c^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))*arctan(a*x)^2-3/32*I/c^3/(a*x+I)+3/32*I/c^3/(a*x-I)-1/512/c^3/(a*x+I)^2
*a^2*x^2-1/512/c^3/(a*x-I)^2*a^2*x^2+3/32/c^3/(a*x+I)*a*x+3/32/c^3/(a*x-I)*a*x-1/128*I/c^3*arctan(a*x)/(a*x+I)
^2+1/128*I/c^3*arctan(a*x)/(a*x-I)^2-2*I/c^3*arctan(a*x)*polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/2*I/c^3*Pi*
arctan(a*x)^2-2*I/c^3*arctan(a*x)*polylog(2,(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/4*I/c^3*Pi*arctan(a*x)^2*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)^2*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*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))*arctan(a*x)^2+1/128*I/c^3*arctan(a*x)/(a*x+I)^2*a^2*x^2-3/2*I/c^3*arctan(a*x)/(8*a*x+8*I)*a*x+3/2*I/c^3*ar
ctan(a*x)/(8*a*x-8*I)*a*x-1/128*I/c^3*arctan(a*x)/(a*x-I)^2*a^2*x^2-1/3*I*arctan(a*x)^3/c^3+1/4*arctan(a*x)^2/
c^3/(a^2*x^2+1)^2+1/2*arctan(a*x)^2/c^3/(a^2*x^2+1)-1/2*I/c^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*arctan(a*x)^2-1/2*I/c^3*Pi*arctan(a*x)^2*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)^2-1/2*I/c^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*arctan(a*x)^2+1/2*I/c^3*Pi*arctan(a*x)^2*
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^3*Pi*arctan(a*x)^2*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))+1/4*I/c^3*Pi*arctan(a*x)^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)^2-1/2*I/c^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*arctan(a*
x)^2-1/4*I/c^3*Pi*arctan(a*x)^2*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/c^3*arctan(a*x)^2*ln(a*x)-1/2/c^3*arctan(a*x)^2*ln(a^2
*x^2+1)+2/c^3*polylog(3,-(1+I*a*x)/(a^2*x^2+1)^(1/2))+2/c^3*polylog(3,(1+I*a*x)/(a^2*x^2+1)^(1/2))+1/512/c^3/(
a*x+I)^2+1/512/c^3/(a*x-I)^2+1/c^3*arctan(a*x)^2*ln(2)-1/c^3*arctan(a*x)^2*ln((1+I*a*x)^2/(a^2*x^2+1)-1)-11/32
*arctan(a*x)^2/c^3-3/2/c^3*arctan(a*x)/(8*a*x+8*I)-3/2/c^3*arctan(a*x)/(8*a*x-8*I)+1/c^3*arctan(a*x)^2*ln(1+(1
+I*a*x)/(a^2*x^2+1)^(1/2))+1/c^3*arctan(a*x)^2*ln((1+I*a*x)/(a^2*x^2+1)^(1/2))+1/2*I/c^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*arctan(a*x)^2+1/2*I/c^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*arctan(a*x)^2-1/4*I/c^3*Pi*arctan(a*x)^2*csgn(I*(1+I*a*x)^2/(a^2*x^2+1))^3-1/4*I/
c^3*Pi*arctan(a*x)^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)^3+1/4*I/c^3*Pi*arctan(a*x)^
2*csgn(I*((1+I*a*x)^2/(a^2*x^2+1)+1)^2)^3+1/256*I/c^3/(a*x+I)^2*a*x-1/256*I/c^3/(a*x-I)^2*a*x-1/2*I/c^3*Pi*csg
n(((1+I*a*x)^2/(a^2*x^2+1)-1)/((1+I*a*x)^2/(a^2*x^2+1)+1))^2*arctan(a*x)^2+1/64/c^3*arctan(a*x)/(a*x+I)^2*a*x+
1/64/c^3*arctan(a*x)/(a*x-I)^2*a*x+1/c^3*arctan(a*x)^2*ln(1-(1+I*a*x)/(a^2*x^2+1)^(1/2))

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

[Out]

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

[Out]

integral(arctan(a*x)^2/(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}^{2}{\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)**2/x/(a**2*c*x**2+c)**3,x)

[Out]

Integral(atan(a*x)**2/(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 )^{2}}{{\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)^2/x/(a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

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