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

Optimal. Leaf size=117 \[ -\frac {a x}{4 c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{2 c^2 \left (a^2 x^2+1\right )}-\frac {i \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{2 c^2}-\frac {i \tan ^{-1}(a x)^2}{2 c^2}-\frac {\tan ^{-1}(a x)}{4 c^2}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^2} \]

[Out]

-1/4*a*x/c^2/(a^2*x^2+1)-1/4*arctan(a*x)/c^2+1/2*arctan(a*x)/c^2/(a^2*x^2+1)-1/2*I*arctan(a*x)^2/c^2+arctan(a*
x)*ln(2-2/(1-I*a*x))/c^2-1/2*I*polylog(2,-1+2/(1-I*a*x))/c^2

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4966, 4924, 4868, 2447, 4930, 199, 205} \[ -\frac {i \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c^2}-\frac {a x}{4 c^2 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{2 c^2 \left (a^2 x^2+1\right )}-\frac {i \tan ^{-1}(a x)^2}{2 c^2}-\frac {\tan ^{-1}(a x)}{4 c^2}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x)/(4*c^2*(1 + a^2*x^2)) - ArcTan[a*x]/(4*c^2) + ArcTan[a*x]/(2*c^2*(1 + a^2*x^2)) - ((I/2)*ArcTan[a*x]^2)
/c^2 + (ArcTan[a*x]*Log[2 - 2/(1 - I*a*x)])/c^2 - ((I/2)*PolyLog[2, -1 + 2/(1 - I*a*x)])/c^2

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

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 72, normalized size = 0.62 \[ -\frac {4 i \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )+4 i \tan ^{-1}(a x)^2+\sin \left (2 \tan ^{-1}(a x)\right )-2 \tan ^{-1}(a x) \left (\cos \left (2 \tan ^{-1}(a x)\right )+4 \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )\right )}{8 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/8*((4*I)*ArcTan[a*x]^2 - 2*ArcTan[a*x]*(Cos[2*ArcTan[a*x]] + 4*Log[1 - E^((2*I)*ArcTan[a*x])]) + (4*I)*Poly
Log[2, E^((2*I)*ArcTan[a*x])] + Sin[2*ArcTan[a*x]])/c^2

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )}{a^{4} c^{2} x^{5} + 2 \, a^{2} c^{2} x^{3} + c^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(a*x)/(a^4*c^2*x^5 + 2*a^2*c^2*x^3 + c^2*x), x)

________________________________________________________________________________________

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

[Out]

sage0*x

________________________________________________________________________________________

maple [B]  time = 0.12, size = 298, normalized size = 2.55 \[ \frac {\arctan \left (a x \right ) \ln \left (a x \right )}{c^{2}}-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 c^{2}}+\frac {\arctan \left (a x \right )}{2 c^{2} \left (a^{2} x^{2}+1\right )}-\frac {a x}{4 c^{2} \left (a^{2} x^{2}+1\right )}-\frac {\arctan \left (a x \right )}{4 c^{2}}-\frac {i \dilog \left (-i a x +1\right )}{2 c^{2}}-\frac {i \ln \left (a x +i\right )^{2}}{8 c^{2}}+\frac {i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c^{2}}+\frac {i \ln \left (a x \right ) \ln \left (i a x +1\right )}{2 c^{2}}+\frac {i \dilog \left (i a x +1\right )}{2 c^{2}}+\frac {i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c^{2}}-\frac {i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c^{2}}-\frac {i \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2 c^{2}}+\frac {i \ln \left (a x -i\right )^{2}}{8 c^{2}}-\frac {i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c^{2}}-\frac {i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c^{2}}+\frac {i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*arctan(a*x)*ln(a*x)-1/2/c^2*arctan(a*x)*ln(a^2*x^2+1)+1/2*arctan(a*x)/c^2/(a^2*x^2+1)-1/4*a*x/c^2/(a^2*x
^2+1)-1/4*arctan(a*x)/c^2-1/4*I/c^2*dilog(1/2*I*(a*x-I))-1/8*I/c^2*ln(I+a*x)^2+1/4*I/c^2*ln(a*x-I)*ln(-1/2*I*(
I+a*x))+1/2*I/c^2*ln(a*x)*ln(1+I*a*x)+1/4*I/c^2*ln(I+a*x)*ln(a^2*x^2+1)-1/2*I/c^2*dilog(1-I*a*x)+1/4*I/c^2*dil
og(-1/2*I*(I+a*x))-1/2*I/c^2*ln(a*x)*ln(1-I*a*x)+1/8*I/c^2*ln(a*x-I)^2-1/4*I/c^2*ln(I+a*x)*ln(1/2*I*(a*x-I))-1
/4*I/c^2*ln(a*x-I)*ln(a^2*x^2+1)+1/2*I/c^2*dilog(1+I*a*x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atan}\left (a\,x\right )}{x\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [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(atan(a*x)/x/(a**2*c*x**2+c)**2,x)

[Out]

Timed out

________________________________________________________________________________________