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

Optimal. Leaf size=64 \[ -\frac {i \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{2 c}-\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4924, 4868, 2447} \[ -\frac {i \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 103, normalized size = 1.61 \[ \frac {i \text {Li}_2(-i a x)}{2 c}-\frac {i \text {Li}_2(i a x)}{2 c}+\frac {i \text {Li}_2\left (-\frac {a x+i}{i-a x}\right )}{2 c}+\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {\log \left (\frac {2 i}{-a x+i}\right ) \tan ^{-1}(a x)}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )}{a^{2} c x^{3} + c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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maple [B]  time = 0.11, size = 251, normalized size = 3.92 \[ \frac {\arctan \left (a x \right ) \ln \left (a x \right )}{c}-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 c}+\frac {i \ln \left (a x \right ) \ln \left (i a x +1\right )}{2 c}-\frac {i \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2 c}+\frac {i \dilog \left (i a x +1\right )}{2 c}-\frac {i \dilog \left (-i a x +1\right )}{2 c}-\frac {i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c}+\frac {i \ln \left (a x -i\right )^{2}}{8 c}+\frac {i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c}+\frac {i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c}+\frac {i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c}-\frac {i \ln \left (a x +i\right )^{2}}{8 c}-\frac {i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c}-\frac {i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {atan}\left (a\,x\right )}{x\,\left (c\,a^2\,x^2+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)/(x*(c + a^2*c*x^2)),x)

[Out]

int(atan(a*x)/(x*(c + a^2*c*x^2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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