\(\int \frac {\log (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}})}{1-a^2 x^2} \, dx\) [280]

Optimal result

Integrand size = 39, antiderivative size = 29 \[ \int \frac {\log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2598} \[ \int \frac {\log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a} \]

[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 2598

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*} \text {integral}& = \frac {\text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(134\) vs. \(2(29)=58\).

Time = 0.53 (sec) , antiderivative size = 134, normalized size of antiderivative = 4.62 \[ \int \frac {\log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\frac {4 \text {arctanh}(a x) \log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(a x)}\right )-2 \left (\text {arctanh}(a x) \left (\log \left (1+e^{-2 \text {arctanh}(a x)}\right )-\log \left (1-i e^{-\text {arctanh}(a x)}\right )+\log \left (1+i e^{-\text {arctanh}(a x)}\right )\right )-\operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )+\operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )\right )}{4 a} \]

[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)

Maple [F]

\[\int \frac {\ln \left (1+\frac {i \sqrt {-a x +1}}{\sqrt {a x +1}}\right )}{-x^{2} a^{2}+1}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {\log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\frac {{\rm Li}_2\left (-\frac {a x - \sqrt {a x + 1} \sqrt {a x - 1} + 1}{a x + 1} + 1\right )}{a} \]

[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

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int { -\frac {\log \left (\frac {i \, \sqrt {-a x + 1}}{\sqrt {a x + 1}} + 1\right )}{a^{2} x^{2} - 1} \,d x } \]

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

Giac [F]

\[ \int \frac {\log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int { -\frac {\log \left (\frac {i \, \sqrt {-a x + 1}}{\sqrt {a x + 1}} + 1\right )}{a^{2} x^{2} - 1} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (1+\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{1-a^2 x^2} \, dx=\int -\frac {\ln \left (1+\frac {\sqrt {1-a\,x}\,1{}\mathrm {i}}{\sqrt {a\,x+1}}\right )}{a^2\,x^2-1} \,d x \]

[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)