3.2.97 \(\int \frac {\text {ArcTan}(a x)}{x^2 (c+a^2 c x^2)^3} \, dx\) [197]

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

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {5086, 5038, 4946, 272, 36, 29, 31, 5004, 5012, 267, 5016} \begin {gather*} -\frac {7 a^2 x \text {ArcTan}(a x)}{8 c^3 \left (a^2 x^2+1\right )}-\frac {a^2 x \text {ArcTan}(a x)}{4 c^3 \left (a^2 x^2+1\right )^2}-\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 {15 a \text {ArcTan}(a x)^2}{16 c^3}-\frac {\text {ArcTan}(a x)}{c^3 x}+\frac {a \log (x)}{c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*a/(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 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 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 267

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 272

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 4946

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

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 5016

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]

Rule 5038

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 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]

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 \text {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 \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 c^3}-\frac {a^3 \text {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*}

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Mathematica [A]
time = 0.05, size = 118, normalized size = 0.83 \begin {gather*} \frac {-2 \left (8+25 a^2 x^2+15 a^4 x^4\right ) \text {ArcTan}(a x)-15 a x \left (1+a^2 x^2\right )^2 \text {ArcTan}(a x)^2+a x \left (-8-7 a^2 x^2+16 \left (1+a^2 x^2\right )^2 \log (x)-8 \left (1+a^2 x^2\right )^2 \log \left (1+a^2 x^2\right )\right )}{16 c^3 x \left (1+a^2 x^2\right )^2} \end {gather*}

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 = 0.08, size = 136, normalized size = 0.96

method	result	size
derivativedivides	\(a \left (-\frac {7 \arctan \left (a x \right ) a^{3} x^{3}}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {9 a x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {15 \arctan \left (a x \right )^{2}}{8 c^{3}}-\frac {\arctan \left (a x \right )}{c^{3} a x}-\frac {-\frac {15 \arctan \left (a x \right )^{2}}{2}+\frac {7}{2 \left (a^{2} x^{2}+1\right )}+4 \ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}-8 \ln \left (a x \right )}{8 c^{3}}\right )\)	\(136\)
default	\(a \left (-\frac {7 \arctan \left (a x \right ) a^{3} x^{3}}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {9 a x \arctan \left (a x \right )}{8 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {15 \arctan \left (a x \right )^{2}}{8 c^{3}}-\frac {\arctan \left (a x \right )}{c^{3} a x}-\frac {-\frac {15 \arctan \left (a x \right )^{2}}{2}+\frac {7}{2 \left (a^{2} x^{2}+1\right )}+4 \ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 \left (a^{2} x^{2}+1\right )^{2}}-8 \ln \left (a x \right )}{8 c^{3}}\right )\)	\(136\)
risch	\(\frac {15 a \dilog \left (\frac {1}{2}-\frac {i a x}{2}\right )}{32 c^{3}}+\frac {15 a \dilog \left (\frac {1}{2}+\frac {i a x}{2}\right )}{32 c^{3}}+\frac {a}{64 c^{3} \left (i a x -1\right )}-\frac {a}{64 c^{3} \left (i a x +1\right )^{2}}+\frac {a \ln \left (i a x \right )}{2 c^{3}}+\frac {15 a \ln \left (i a x +1\right )^{2}}{64 c^{3}}-\frac {7 a}{32 c^{3} \left (i a x +1\right )}-\frac {7 a}{32 c^{3} \left (-i a x +1\right )}+\frac {a}{64 c^{3} \left (-i a x -1\right )}-\frac {a}{64 c^{3} \left (-i a x +1\right )^{2}}+\frac {a \ln \left (-i a x \right )}{2 c^{3}}+\frac {15 a \ln \left (-i a x +1\right )^{2}}{64 c^{3}}-\frac {15 a \ln \left (\frac {1}{2}-\frac {i a x}{2}\right ) \ln \left (i a x +1\right )}{32 c^{3}}+\frac {3 a \ln \left (i a x +1\right )}{128 c^{3} \left (i a x -1\right )^{2}}-\frac {a \ln \left (i a x +1\right )}{32 c^{3} \left (i a x +1\right )^{2}}-\frac {7 a \ln \left (i a x +1\right )}{64 c^{3} \left (i a x -1\right )}+\frac {15 a \ln \left (\frac {1}{2}+\frac {i a x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i a x}{2}\right )}{16 c^{3}}-\frac {15 a \ln \left (\frac {1}{2}+\frac {i a x}{2}\right ) \ln \left (-i a x +1\right )}{32 c^{3}}+\frac {3 a \ln \left (-i a x +1\right )}{128 c^{3} \left (-i a x -1\right )^{2}}-\frac {a \ln \left (-i a x +1\right )}{32 c^{3} \left (-i a x +1\right )^{2}}-\frac {7 a \ln \left (-i a x +1\right )}{64 c^{3} \left (-i a x -1\right )}-\frac {7 a \ln \left (-i a x +1\right )}{32 c^{3} \left (-i a x +1\right )}-\frac {7 a \ln \left (i a x +1\right )}{32 c^{3} \left (i a x +1\right )}+\frac {i \ln \left (i a x +1\right )}{2 c^{3} x}-\frac {i \ln \left (-i a x +1\right )}{2 c^{3} x}-\frac {49 a \ln \left (a^{2} x^{2}+1\right )}{128 c^{3}}+\frac {a^{3} \ln \left (i a x +1\right ) x^{2}}{128 c^{3} \left (i a x -1\right )^{2}}+\frac {a^{3} \ln \left (-i a x +1\right ) x^{2}}{128 c^{3} \left (-i a x -1\right )^{2}}-\frac {7 i a^{2} \ln \left (i a x +1\right ) x}{64 c^{3} \left (i a x -1\right )}+\frac {i a^{2} \ln \left (i a x +1\right ) x}{64 c^{3} \left (i a x -1\right )^{2}}+\frac {7 i a^{2} \ln \left (-i a x +1\right ) x}{64 c^{3} \left (-i a x -1\right )}-\frac {i a^{2} \ln \left (-i a x +1\right ) x}{64 c^{3} \left (-i a x -1\right )^{2}}\)	\(640\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 181, normalized size = 1.27 \begin {gather*} -\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 {gather*}

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 = 1.12, size = 149, normalized size = 1.05 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 604 vs. \(2 (134) = 268\).
time = 1.50, size = 604, normalized size = 4.25 \begin {gather*} \begin {cases} \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} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((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)/(1
6*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), Ne(a, 0)), (0, True))

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

[Out]

sage0*x

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Mupad [B]
time = 0.57, size = 133, normalized size = 0.94 \begin {gather*} \frac {a\,\ln \left (x\right )}{c^3}-\frac {a\,\ln \left (a^2\,x^2+1\right )}{2\,c^3}-\frac {\frac {7\,a^3\,x^2}{2}+4\,a}{8\,a^4\,c^3\,x^4+16\,a^2\,c^3\,x^2+8\,c^3}-\frac {\mathrm {atan}\left (a\,x\right )\,\left (\frac {1}{a^2\,c^3}+\frac {25\,x^2}{8\,c^3}+\frac {15\,a^2\,x^4}{8\,c^3}\right )}{\frac {x}{a^2}+2\,x^3+a^2\,x^5}-\frac {15\,a\,{\mathrm {atan}\left (a\,x\right )}^2}{16\,c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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