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

Optimal. Leaf size=250 \[ \frac {31 a^2 x}{64 c^3 \left (a^2 x^2+1\right )}+\frac {a^2 x}{32 c^3 \left (a^2 x^2+1\right )^2}-\frac {7 a^2 x \tan ^{-1}(a x)^2}{8 c^3 \left (a^2 x^2+1\right )}-\frac {a^2 x \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {7 a \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )}-\frac {a \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )^2}-\frac {i a \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{c^3}-\frac {5 a \tan ^{-1}(a x)^3}{8 c^3}-\frac {\tan ^{-1}(a x)^2}{c^3 x}-\frac {i a \tan ^{-1}(a x)^2}{c^3}+\frac {31 a \tan ^{-1}(a x)}{64 c^3}+\frac {2 a \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^3} \]

[Out]

1/32*a^2*x/c^3/(a^2*x^2+1)^2+31/64*a^2*x/c^3/(a^2*x^2+1)+31/64*a*arctan(a*x)/c^3-1/8*a*arctan(a*x)/c^3/(a^2*x^
2+1)^2-7/8*a*arctan(a*x)/c^3/(a^2*x^2+1)-I*a*arctan(a*x)^2/c^3-arctan(a*x)^2/c^3/x-1/4*a^2*x*arctan(a*x)^2/c^3
/(a^2*x^2+1)^2-7/8*a^2*x*arctan(a*x)^2/c^3/(a^2*x^2+1)-5/8*a*arctan(a*x)^3/c^3+2*a*arctan(a*x)*ln(2-2/(1-I*a*x
))/c^3-I*a*polylog(2,-1+2/(1-I*a*x))/c^3

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Rubi [A]  time = 0.55, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {4966, 4918, 4852, 4924, 4868, 2447, 4884, 4892, 4930, 199, 205, 4900} \[ -\frac {i a \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{c^3}+\frac {31 a^2 x}{64 c^3 \left (a^2 x^2+1\right )}+\frac {a^2 x}{32 c^3 \left (a^2 x^2+1\right )^2}-\frac {7 a^2 x \tan ^{-1}(a x)^2}{8 c^3 \left (a^2 x^2+1\right )}-\frac {a^2 x \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {7 a \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )}-\frac {a \tan ^{-1}(a x)}{8 c^3 \left (a^2 x^2+1\right )^2}-\frac {5 a \tan ^{-1}(a x)^3}{8 c^3}-\frac {\tan ^{-1}(a x)^2}{c^3 x}-\frac {i a \tan ^{-1}(a x)^2}{c^3}+\frac {31 a \tan ^{-1}(a x)}{64 c^3}+\frac {2 a \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/(32*c^3*(1 + a^2*x^2)^2) + (31*a^2*x)/(64*c^3*(1 + a^2*x^2)) + (31*a*ArcTan[a*x])/(64*c^3) - (a*ArcTan
[a*x])/(8*c^3*(1 + a^2*x^2)^2) - (7*a*ArcTan[a*x])/(8*c^3*(1 + a^2*x^2)) - (I*a*ArcTan[a*x]^2)/c^3 - ArcTan[a*
x]^2/(c^3*x) - (a^2*x*ArcTan[a*x]^2)/(4*c^3*(1 + a^2*x^2)^2) - (7*a^2*x*ArcTan[a*x]^2)/(8*c^3*(1 + a^2*x^2)) -
(5*a*ArcTan[a*x]^3)/(8*c^3) + (2*a*ArcTan[a*x]*Log[2 - 2/(1 - I*a*x)])/c^3 - (I*a*PolyLog[2, -1 + 2/(1 - I*a*
x)])/c^3

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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 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 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 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]

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)^2}{x^2 \left (c+a^2 c x^2\right )^3} \, dx &=-\left (a^2 \int \frac {\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^2 \left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=-\frac {a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )^2}-\frac {a^2 x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {1}{8} a^2 \int \frac {1}{\left (c+a^2 c x^2\right )^3} \, dx+\frac {\int \frac {\tan ^{-1}(a x)^2}{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)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}-\frac {a^2 \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=\frac {a^2 x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac {a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )^2}-\frac {a^2 x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a^2 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}-\frac {7 a \tan ^{-1}(a x)^3}{24 c^3}+\frac {\int \frac {\tan ^{-1}(a x)^2}{x^2} \, dx}{c^3}-\frac {a^2 \int \frac {\tan ^{-1}(a x)^2}{c+a^2 c x^2} \, dx}{c^2}+\frac {\left (3 a^2\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{32 c}+\frac {\left (3 a^3\right ) \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}+\frac {a^3 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{c}\\ &=\frac {a^2 x}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 a^2 x}{64 c^3 \left (1+a^2 x^2\right )}-\frac {a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac {\tan ^{-1}(a x)^2}{c^3 x}-\frac {a^2 x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a^2 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}-\frac {5 a \tan ^{-1}(a x)^3}{8 c^3}+\frac {(2 a) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx}{c^3}+\frac {\left (3 a^2\right ) \int \frac {1}{c+a^2 c x^2} \, dx}{64 c^2}+\frac {\left (3 a^2\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{8 c}+\frac {a^2 \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{2 c}\\ &=\frac {a^2 x}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {31 a^2 x}{64 c^3 \left (1+a^2 x^2\right )}+\frac {3 a \tan ^{-1}(a x)}{64 c^3}-\frac {a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac {i a \tan ^{-1}(a x)^2}{c^3}-\frac {\tan ^{-1}(a x)^2}{c^3 x}-\frac {a^2 x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a^2 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}-\frac {5 a \tan ^{-1}(a x)^3}{8 c^3}+\frac {(2 i a) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c^3}+\frac {\left (3 a^2\right ) \int \frac {1}{c+a^2 c x^2} \, dx}{16 c^2}+\frac {a^2 \int \frac {1}{c+a^2 c x^2} \, dx}{4 c^2}\\ &=\frac {a^2 x}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {31 a^2 x}{64 c^3 \left (1+a^2 x^2\right )}+\frac {31 a \tan ^{-1}(a x)}{64 c^3}-\frac {a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac {i a \tan ^{-1}(a x)^2}{c^3}-\frac {\tan ^{-1}(a x)^2}{c^3 x}-\frac {a^2 x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a^2 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}-\frac {5 a \tan ^{-1}(a x)^3}{8 c^3}+\frac {2 a \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c^3}-\frac {\left (2 a^2\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c^3}\\ &=\frac {a^2 x}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {31 a^2 x}{64 c^3 \left (1+a^2 x^2\right )}+\frac {31 a \tan ^{-1}(a x)}{64 c^3}-\frac {a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a \tan ^{-1}(a x)}{8 c^3 \left (1+a^2 x^2\right )}-\frac {i a \tan ^{-1}(a x)^2}{c^3}-\frac {\tan ^{-1}(a x)^2}{c^3 x}-\frac {a^2 x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {7 a^2 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}-\frac {5 a \tan ^{-1}(a x)^3}{8 c^3}+\frac {2 a \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c^3}-\frac {i a \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{c^3}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 139, normalized size = 0.56 \[ -\frac {256 i a x \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )+160 a x \tan ^{-1}(a x)^3+8 \tan ^{-1}(a x)^2 \left (32 i a x+16 a x \sin \left (2 \tan ^{-1}(a x)\right )+a x \sin \left (4 \tan ^{-1}(a x)\right )+32\right )-a x \left (64 \sin \left (2 \tan ^{-1}(a x)\right )+\sin \left (4 \tan ^{-1}(a x)\right )\right )+4 a x \tan ^{-1}(a x) \left (-128 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )+32 \cos \left (2 \tan ^{-1}(a x)\right )+\cos \left (4 \tan ^{-1}(a x)\right )\right )}{256 c^3 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/256*(160*a*x*ArcTan[a*x]^3 + 4*a*x*ArcTan[a*x]*(32*Cos[2*ArcTan[a*x]] + Cos[4*ArcTan[a*x]] - 128*Log[1 - E^
((2*I)*ArcTan[a*x])]) + (256*I)*a*x*PolyLog[2, E^((2*I)*ArcTan[a*x])] - a*x*(64*Sin[2*ArcTan[a*x]] + Sin[4*Arc
Tan[a*x]]) + 8*ArcTan[a*x]^2*(32 + (32*I)*a*x + 16*a*x*Sin[2*ArcTan[a*x]] + a*x*Sin[4*ArcTan[a*x]]))/(c^3*x)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )^{2}}{a^{6} c^{3} x^{8} + 3 \, a^{4} c^{3} x^{6} + 3 \, a^{2} c^{3} x^{4} + c^{3} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(a*x)^2/(a^6*c^3*x^8 + 3*a^4*c^3*x^6 + 3*a^2*c^3*x^4 + c^3*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.18, size = 440, normalized size = 1.76 \[ -\frac {\arctan \left (a x \right )^{2}}{c^{3} x}-\frac {7 \arctan \left (a x \right )^{2} a^{4} x^{3}}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {9 a^{2} x \arctan \left (a x \right )^{2}}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {5 a \arctan \left (a x \right )^{3}}{8 c^{3}}+\frac {2 a \arctan \left (a x \right ) \ln \left (a x \right )}{c^{3}}-\frac {a \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{c^{3}}-\frac {a \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {7 a \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {i a \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{2 c^{3}}+\frac {i a \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{2 c^{3}}+\frac {i a \ln \left (a x -i\right )^{2}}{4 c^{3}}+\frac {i a \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{2 c^{3}}+\frac {i a \dilog \left (i a x +1\right )}{c^{3}}-\frac {i a \ln \left (a x \right ) \ln \left (-i a x +1\right )}{c^{3}}-\frac {i a \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{2 c^{3}}-\frac {i a \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{2 c^{3}}+\frac {i a \ln \left (a x \right ) \ln \left (i a x +1\right )}{c^{3}}-\frac {i a \ln \left (a x +i\right )^{2}}{4 c^{3}}-\frac {i a \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{2 c^{3}}-\frac {i a \dilog \left (-i a x +1\right )}{c^{3}}+\frac {31 x^{3} a^{4}}{64 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {33 a^{2} x}{64 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {31 a \arctan \left (a x \right )}{64 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(a*x)^2/c^3/x-7/8/c^3*arctan(a*x)^2/(a^2*x^2+1)^2*a^4*x^3-9/8*a^2*x*arctan(a*x)^2/c^3/(a^2*x^2+1)^2-5/8
*a*arctan(a*x)^3/c^3+2*a/c^3*arctan(a*x)*ln(a*x)-a/c^3*arctan(a*x)*ln(a^2*x^2+1)-1/8*a*arctan(a*x)/c^3/(a^2*x^
2+1)^2-7/8*a*arctan(a*x)/c^3/(a^2*x^2+1)+I*a/c^3*dilog(1+I*a*x)-I*a/c^3*dilog(1-I*a*x)+1/2*I*a/c^3*ln(I+a*x)*l
n(a^2*x^2+1)+1/4*I*a/c^3*ln(a*x-I)^2+1/2*I*a/c^3*dilog(-1/2*I*(I+a*x))+1/2*I*a/c^3*ln(a*x-I)*ln(-1/2*I*(I+a*x)
)-1/2*I*a/c^3*dilog(1/2*I*(a*x-I))-I*a/c^3*ln(a*x)*ln(1-I*a*x)-1/2*I*a/c^3*ln(I+a*x)*ln(1/2*I*(a*x-I))+I*a/c^3
*ln(a*x)*ln(1+I*a*x)-1/4*I*a/c^3*ln(I+a*x)^2-1/2*I*a/c^3*ln(a*x-I)*ln(a^2*x^2+1)+31/64/c^3/(a^2*x^2+1)^2*x^3*a
^4+33/64*a^2*x/c^3/(a^2*x^2+1)^2+31/64*a*arctan(a*x)/c^3

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x^2\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{a^{6} x^{8} + 3 a^{4} x^{6} + 3 a^{2} x^{4} + x^{2}}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a*x)**2/(a**6*x**8 + 3*a**4*x**6 + 3*a**2*x**4 + x**2), x)/c**3

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