3.280 \(\int \frac {\log (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}})}{1-a^2 x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {\text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2518} \[ \frac {\text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[Log[1 + (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/(1 - a^2*x^2),x]

[Out]

PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/a

Rule 2518

Int[Log[v_]*(u_), x_Symbol] :> With[{w = DerivativeDivides[v, u*(1 - v), x]}, Simp[w*PolyLog[2, 1 - v], x] /;
!FalseQ[w]]

Rubi steps

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

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Mathematica [B]  time = 0.65, size = 134, normalized size = 4.62 \[ \frac {\text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )-2 \left (-\text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )+\text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x) \left (\log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )\right )+4 \log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{4 a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Log[1 + (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]]/(1 - a^2*x^2),x]

[Out]

(4*ArcTanh[a*x]*Log[1 + (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]] + PolyLog[2, -E^(-2*ArcTanh[a*x])] - 2*(ArcTanh[a*x]*
(Log[1 + E^(-2*ArcTanh[a*x])] - Log[1 - I/E^ArcTanh[a*x]] + Log[1 + I/E^ArcTanh[a*x]]) - PolyLog[2, (-I)/E^Arc
Tanh[a*x]] + PolyLog[2, I/E^ArcTanh[a*x]]))/(4*a)

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fricas [A]  time = 0.44, size = 37, normalized size = 1.28 \[ \frac {{\rm Li}_2\left (-\frac {a x - \sqrt {a x + 1} \sqrt {a x - 1} + 1}{a x + 1} + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1+I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

dilog(-(a*x - sqrt(a*x + 1)*sqrt(a*x - 1) + 1)/(a*x + 1) + 1)/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\log \left (\frac {i \, \sqrt {-a x + 1}}{\sqrt {a x + 1}} + 1\right )}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1+I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-log(I*sqrt(-a*x + 1)/sqrt(a*x + 1) + 1)/(a^2*x^2 - 1), x)

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maple [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (1+\frac {i \sqrt {-a x +1}}{\sqrt {a x +1}}\right )}{-a^{2} x^{2}+1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(1+I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x)

[Out]

int(ln(1+I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, {\left (\log \left (a x + 1\right ) - \log \left (-a x + 1\right )\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} + 2 \, \log \left (a x + 1\right ) \log \left (-a x + 1\right ) - \log \left (-a x + 1\right )^{2} - 4 \, {\left (\log \left (a x + 1\right ) - \log \left (-a x + 1\right )\right )} \log \left (\sqrt {a x + 1} + i \, \sqrt {-a x + 1}\right )}{8 \, a} - \int -\frac {\sqrt {a x + 1} {\left (\log \left (a x + 1\right ) - \log \left (-a x + 1\right )\right )}}{2 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {a x + 1} + {\left (2 i \, a^{2} x^{2} - 2 i\right )} \sqrt {-a x + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1+I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-1/8*(2*(log(a*x + 1) - log(-a*x + 1))*log(a*x + 1) - log(a*x + 1)^2 + 2*log(a*x + 1)*log(-a*x + 1) - log(-a*x
+ 1)^2 - 4*(log(a*x + 1) - log(-a*x + 1))*log(sqrt(a*x + 1) + I*sqrt(-a*x + 1)))/a - integrate(-sqrt(a*x + 1)
*(log(a*x + 1) - log(-a*x + 1))/(2*(a^2*x^2 - 1)*sqrt(a*x + 1) + (2*I*a^2*x^2 - 2*I)*sqrt(-a*x + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int -\frac {\ln \left (1+\frac {\sqrt {1-a\,x}\,1{}\mathrm {i}}{\sqrt {a\,x+1}}\right )}{a^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-log(((1 - a*x)^(1/2)*1i)/(a*x + 1)^(1/2) + 1)/(a^2*x^2 - 1),x)

[Out]

int(-log(((1 - a*x)^(1/2)*1i)/(a*x + 1)^(1/2) + 1)/(a^2*x^2 - 1), x)

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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(ln(1+I*(-a*x+1)**(1/2)/(a*x+1)**(1/2))/(-a**2*x**2+1),x)

[Out]

Timed out

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