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

Optimal. Leaf size=49 \[ \frac {i \text {Li}_2\left (\frac {2}{i a x+1}-1\right )}{2 c}+\frac {\log \left (2-\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{c} \]

[Out]

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 = 0.06, antiderivative size = 49, 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 = {1593, 4868, 2447} \[ \frac {i \text {PolyLog}\left (2,-1+\frac {2}{1+i a x}\right )}{2 c}+\frac {\log \left (2-\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]/(c*x + I*a*c*x^2),x]

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)}{c x+i a c x^2} \, dx &=\int \frac {\tan ^{-1}(a x)}{x (c+i a c x)} \, dx\\ &=\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 {\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 = 88, normalized size = 1.80 \[ \frac {i \text {Li}_2(-i a x)}{2 c}-\frac {i \text {Li}_2(i a x)}{2 c}+\frac {i \text {Li}_2\left (-\frac {a x+i}{i-a x}\right )}{2 c}+\frac {\log \left (\frac {2 i}{-a x+i}\right ) \tan ^{-1}(a x)}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 21, normalized size = 0.43 \[ -\frac {i \, {\rm Li}_2\left (\frac {a x + i}{a x - i} + 1\right )}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*dilog((a*x + I)/(a*x - I) + 1)/c

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

[Out]

sage0*x

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maple [B]  time = 0.06, size = 148, normalized size = 3.02 \[ \frac {\arctan \left (a x \right ) \ln \left (a x \right )}{c}-\frac {\arctan \left (a x \right ) \ln \left (a x -i\right )}{c}+\frac {i \ln \left (a x \right ) \ln \left (i a x +1\right )}{2 c}-\frac {i \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2 c}+\frac {i \dilog \left (i a x +1\right )}{2 c}-\frac {i \dilog \left (-i a x +1\right )}{2 c}+\frac {i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{2 c}+\frac {i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{2 c}-\frac {i \ln \left (a x -i\right )^{2}}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 126, normalized size = 2.57 \[ \frac {1}{4} \, a {\left (-\frac {i \, \log \left (i \, a x + 1\right )^{2}}{a c} + \frac {2 i \, {\left (\log \left (i \, a x + 1\right ) \log \left (-\frac {1}{2} i \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} i \, a x + \frac {1}{2}\right )\right )}}{a c} + \frac {2 i \, {\left (\log \left (i \, a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-i \, a x\right )\right )}}{a c} - \frac {2 i \, {\left (\log \left (-i \, a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (i \, a x\right )\right )}}{a c}\right )} - {\left (\frac {\log \left (i \, a x + 1\right )}{c} - \frac {\log \relax (x)}{c}\right )} \arctan \left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)/(c*x + a*c*x^2*1i),x)

[Out]

int(atan(a*x)/(c*x + a*c*x^2*1i), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*Integral(atan(a*x)/(a*x**2 - I*x), x)/c

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