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

Optimal. Leaf size=142 \[ -\frac{7 a}{16 c^3 \left (a^2 x^2+1\right )}-\frac{a}{16 c^3 \left (a^2 x^2+1\right )^2}-\frac{a \log \left (a^2 x^2+1\right )}{2 c^3}-\frac{7 a^2 x \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )}-\frac{a^2 x \tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac{a \log (x)}{c^3}-\frac{15 a \tan ^{-1}(a x)^2}{16 c^3}-\frac{\tan ^{-1}(a x)}{c^3 x} \]

[Out]

-a/(16*c^3*(1 + a^2*x^2)^2) - (7*a)/(16*c^3*(1 + a^2*x^2)) - ArcTan[a*x]/(c^3*x) - (a^2*x*ArcTan[a*x])/(4*c^3*
(1 + a^2*x^2)^2) - (7*a^2*x*ArcTan[a*x])/(8*c^3*(1 + a^2*x^2)) - (15*a*ArcTan[a*x]^2)/(16*c^3) + (a*Log[x])/c^
3 - (a*Log[1 + a^2*x^2])/(2*c^3)

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Rubi [A]  time = 0.262566, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.55, Rules used = {4966, 4918, 4852, 266, 36, 29, 31, 4884, 4892, 261, 4896} \[ -\frac{7 a}{16 c^3 \left (a^2 x^2+1\right )}-\frac{a}{16 c^3 \left (a^2 x^2+1\right )^2}-\frac{a \log \left (a^2 x^2+1\right )}{2 c^3}-\frac{7 a^2 x \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )}-\frac{a^2 x \tan ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac{a \log (x)}{c^3}-\frac{15 a \tan ^{-1}(a x)^2}{16 c^3}-\frac{\tan ^{-1}(a x)}{c^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(16*c^3*(1 + a^2*x^2)^2) - (7*a)/(16*c^3*(1 + a^2*x^2)) - ArcTan[a*x]/(c^3*x) - (a^2*x*ArcTan[a*x])/(4*c^3*
(1 + a^2*x^2)^2) - (7*a^2*x*ArcTan[a*x])/(8*c^3*(1 + a^2*x^2)) - (15*a*ArcTan[a*x]^2)/(16*c^3) + (a*Log[x])/c^
3 - (a*Log[1 + a^2*x^2])/(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 4918

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

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 266

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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)}{x^2 \left (c+a^2 c x^2\right )^3} \, dx &=-\left (a^2 \int \frac{\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^3} \, dx\right )+\frac{\int \frac{\tan ^{-1}(a x)}{x^2 \left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=-\frac{a}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{a^2 x \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{\int \frac{\tan ^{-1}(a x)}{x^2 \left (c+a^2 c x^2\right )} \, dx}{c^2}-\frac{\left (3 a^2\right ) \int \frac{\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}-\frac{a^2 \int \frac{\tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=-\frac{a}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{a^2 x \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a^2 x \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac{7 a \tan ^{-1}(a x)^2}{16 c^3}+\frac{\int \frac{\tan ^{-1}(a x)}{x^2} \, dx}{c^3}-\frac{a^2 \int \frac{\tan ^{-1}(a x)}{c+a^2 c x^2} \, dx}{c^2}+\frac{\left (3 a^3\right ) \int \frac{x}{\left (c+a^2 c x^2\right )^2} \, dx}{8 c}+\frac{a^3 \int \frac{x}{\left (c+a^2 c x^2\right )^2} \, dx}{2 c}\\ &=-\frac{a}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a}{16 c^3 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)}{c^3 x}-\frac{a^2 x \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a^2 x \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac{15 a \tan ^{-1}(a x)^2}{16 c^3}+\frac{a \int \frac{1}{x \left (1+a^2 x^2\right )} \, dx}{c^3}\\ &=-\frac{a}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a}{16 c^3 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)}{c^3 x}-\frac{a^2 x \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a^2 x \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac{15 a \tan ^{-1}(a x)^2}{16 c^3}+\frac{a \operatorname{Subst}\left (\int \frac{1}{x \left (1+a^2 x\right )} \, dx,x,x^2\right )}{2 c^3}\\ &=-\frac{a}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a}{16 c^3 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)}{c^3 x}-\frac{a^2 x \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a^2 x \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac{15 a \tan ^{-1}(a x)^2}{16 c^3}+\frac{a \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{2 c^3}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{1+a^2 x} \, dx,x,x^2\right )}{2 c^3}\\ &=-\frac{a}{16 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a}{16 c^3 \left (1+a^2 x^2\right )}-\frac{\tan ^{-1}(a x)}{c^3 x}-\frac{a^2 x \tan ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{7 a^2 x \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac{15 a \tan ^{-1}(a x)^2}{16 c^3}+\frac{a \log (x)}{c^3}-\frac{a \log \left (1+a^2 x^2\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0939301, size = 118, normalized size = 0.83 \[ \frac{a x \left (-7 a^2 x^2+16 \left (a^2 x^2+1\right )^2 \log (x)-8 \left (a^2 x^2+1\right )^2 \log \left (a^2 x^2+1\right )-8\right )-15 a x \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^2-2 \left (15 a^4 x^4+25 a^2 x^2+8\right ) \tan ^{-1}(a x)}{16 c^3 x \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(8 + 25*a^2*x^2 + 15*a^4*x^4)*ArcTan[a*x] - 15*a*x*(1 + a^2*x^2)^2*ArcTan[a*x]^2 + a*x*(-8 - 7*a^2*x^2 + 1
6*(1 + a^2*x^2)^2*Log[x] - 8*(1 + a^2*x^2)^2*Log[1 + a^2*x^2]))/(16*c^3*x*(1 + a^2*x^2)^2)

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Maple [A]  time = 0.047, size = 135, normalized size = 1. \begin{align*} -{\frac{7\,\arctan \left ( ax \right ){a}^{4}{x}^{3}}{8\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{9\,{a}^{2}x\arctan \left ( ax \right ) }{8\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{15\,a \left ( \arctan \left ( ax \right ) \right ) ^{2}}{16\,{c}^{3}}}-{\frac{\arctan \left ( ax \right ) }{{c}^{3}x}}-{\frac{a}{16\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{2}}}-{\frac{a\ln \left ({a}^{2}{x}^{2}+1 \right ) }{2\,{c}^{3}}}-{\frac{7\,a}{16\,{c}^{3} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{a\ln \left ( ax \right ) }{{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8/c^3*arctan(a*x)/(a^2*x^2+1)^2*a^4*x^3-9/8*a^2*x*arctan(a*x)/c^3/(a^2*x^2+1)^2-15/16*a*arctan(a*x)^2/c^3-a
rctan(a*x)/c^3/x-1/16*a/c^3/(a^2*x^2+1)^2-1/2*a*ln(a^2*x^2+1)/c^3-7/16*a/c^3/(a^2*x^2+1)+a/c^3*ln(a*x)

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Maxima [A]  time = 1.62712, size = 244, normalized size = 1.72 \begin{align*} -\frac{1}{8} \,{\left (\frac{15 \, a^{4} x^{4} + 25 \, a^{2} x^{2} + 8}{a^{4} c^{3} x^{5} + 2 \, a^{2} c^{3} x^{3} + c^{3} x} + \frac{15 \, a \arctan \left (a x\right )}{c^{3}}\right )} \arctan \left (a x\right ) - \frac{{\left (7 \, a^{2} x^{2} - 15 \,{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 8 \,{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \log \left (a^{2} x^{2} + 1\right ) - 16 \,{\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \log \left (x\right ) + 8\right )} a}{16 \,{\left (a^{4} c^{3} x^{4} + 2 \, a^{2} c^{3} x^{2} + c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((15*a^4*x^4 + 25*a^2*x^2 + 8)/(a^4*c^3*x^5 + 2*a^2*c^3*x^3 + c^3*x) + 15*a*arctan(a*x)/c^3)*arctan(a*x)
- 1/16*(7*a^2*x^2 - 15*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 8*(a^4*x^4 + 2*a^2*x^2 + 1)*log(a^2*x^2 + 1)
- 16*(a^4*x^4 + 2*a^2*x^2 + 1)*log(x) + 8)*a/(a^4*c^3*x^4 + 2*a^2*c^3*x^2 + c^3)

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Fricas [A]  time = 1.73396, size = 333, normalized size = 2.35 \begin{align*} -\frac{7 \, a^{3} x^{3} + 15 \,{\left (a^{5} x^{5} + 2 \, a^{3} x^{3} + a x\right )} \arctan \left (a x\right )^{2} + 8 \, a x + 2 \,{\left (15 \, a^{4} x^{4} + 25 \, a^{2} x^{2} + 8\right )} \arctan \left (a x\right ) + 8 \,{\left (a^{5} x^{5} + 2 \, a^{3} x^{3} + a x\right )} \log \left (a^{2} x^{2} + 1\right ) - 16 \,{\left (a^{5} x^{5} + 2 \, a^{3} x^{3} + a x\right )} \log \left (x\right )}{16 \,{\left (a^{4} c^{3} x^{5} + 2 \, a^{2} c^{3} x^{3} + c^{3} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(7*a^3*x^3 + 15*(a^5*x^5 + 2*a^3*x^3 + a*x)*arctan(a*x)^2 + 8*a*x + 2*(15*a^4*x^4 + 25*a^2*x^2 + 8)*arct
an(a*x) + 8*(a^5*x^5 + 2*a^3*x^3 + a*x)*log(a^2*x^2 + 1) - 16*(a^5*x^5 + 2*a^3*x^3 + a*x)*log(x))/(a^4*c^3*x^5
+ 2*a^2*c^3*x^3 + c^3*x)

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Sympy [B]  time = 4.04583, size = 602, normalized size = 4.24 \begin{align*} \frac{16 a^{5} x^{5} \log{\left (x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{8 a^{5} x^{5} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{15 a^{5} x^{5} \operatorname{atan}^{2}{\left (a x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{30 a^{4} x^{4} \operatorname{atan}{\left (a x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} + \frac{32 a^{3} x^{3} \log{\left (x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{16 a^{3} x^{3} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{30 a^{3} x^{3} \operatorname{atan}^{2}{\left (a x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{7 a^{3} x^{3}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{50 a^{2} x^{2} \operatorname{atan}{\left (a x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} + \frac{16 a x \log{\left (x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{8 a x \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{15 a x \operatorname{atan}^{2}{\left (a x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{8 a x}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} - \frac{16 \operatorname{atan}{\left (a x \right )}}{16 a^{4} c^{3} x^{5} + 32 a^{2} c^{3} x^{3} + 16 c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*a**5*x**5*log(x)/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 8*a**5*x**5*log(x**2 + a**(-2))/(16*
a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 15*a**5*x**5*atan(a*x)**2/(16*a**4*c**3*x**5 + 32*a**2*c**3*
x**3 + 16*c**3*x) - 30*a**4*x**4*atan(a*x)/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) + 32*a**3*x**3*
log(x)/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 16*a**3*x**3*log(x**2 + a**(-2))/(16*a**4*c**3*x*
*5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 30*a**3*x**3*atan(a*x)**2/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c*
*3*x) - 7*a**3*x**3/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 50*a**2*x**2*atan(a*x)/(16*a**4*c**3
*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) + 16*a*x*log(x)/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 8
*a*x*log(x**2 + a**(-2))/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 15*a*x*atan(a*x)**2/(16*a**4*c*
*3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 8*a*x/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x) - 16*atan
(a*x)/(16*a**4*c**3*x**5 + 32*a**2*c**3*x**3 + 16*c**3*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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