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

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

[Out]

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

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Rubi [A]  time = 0.0636605, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, 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 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 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]]

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*}

Mathematica [A]  time = 0.0244393, size = 88, normalized size = 1.8 \[ \frac{i \text{PolyLog}(2,-i a x)}{2 c}-\frac{i \text{PolyLog}(2,i a x)}{2 c}+\frac{i \text{PolyLog}\left (2,-\frac{a x+i}{-a x+i}\right )}{2 c}+\frac{\log \left (\frac{2 i}{-a x+i}\right ) \tan ^{-1}(a x)}{c} \]

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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Maple [B]  time = 0.049, size = 148, normalized size = 3. \begin{align*} -{\frac{\arctan \left ( ax \right ) \ln \left ( ax-i \right ) }{c}}+{\frac{\arctan \left ( ax \right ) \ln \left ( ax \right ) }{c}}+{\frac{{\frac{i}{2}}\ln \left ( ax \right ) \ln \left ( 1+iax \right ) }{c}}-{\frac{{\frac{i}{2}}\ln \left ( ax \right ) \ln \left ( 1-iax \right ) }{c}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+iax \right ) }{c}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( 1-iax \right ) }{c}}+{\frac{{\frac{i}{2}}\ln \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) \ln \left ( ax-i \right ) }{c}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( -{\frac{i}{2}} \left ( ax+i \right ) \right ) }{c}}-{\frac{{\frac{i}{4}} \left ( \ln \left ( ax-i \right ) \right ) ^{2}}{c}} \end{align*}

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

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Maxima [B]  time = 1.49263, size = 170, normalized size = 3.47 \begin{align*} \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 \left (x\right ) +{\rm Li}_2\left (-i \, a x\right )\right )}}{a c} - \frac{2 i \,{\left (\log \left (-i \, a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (i \, a x\right )\right )}}{a c}\right )} -{\left (\frac{\log \left (i \, a x + 1\right )}{c} - \frac{\log \left (x\right )}{c}\right )} \arctan \left (a x\right ) \end{align*}

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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Fricas [A]  time = 2.1571, size = 55, normalized size = 1.12 \begin{align*} -\frac{i \,{\rm Li}_2\left (\frac{a x + i}{a x - i} + 1\right )}{2 \, c} \end{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (a x\right )}{i \, a c x^{2} + c x}\,{d x} \end{align*}

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]

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