3.339 \(\int \frac{\log (\frac{a (1-c)+b (1+c) x}{a+b x})}{a^2-b^2 x^2} \, dx\)

Optimal. Leaf size=37 \[ \frac{\text{PolyLog}\left (2,1-\frac{a (1-c)+b (c+1) x}{a+b x}\right )}{2 a b} \]

[Out]

PolyLog[2, 1 - (a*(1 - c) + b*(1 + c)*x)/(a + b*x)]/(2*a*b)

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Rubi [A]  time = 0.0250285, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.026, Rules used = {2447} \[ \frac{\text{PolyLog}\left (2,1-\frac{a (1-c)+b (c+1) x}{a+b x}\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

PolyLog[2, 1 - (a*(1 - c) + b*(1 + c)*x)/(a + b*x)]/(2*a*b)

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

Mathematica [B]  time = 0.187185, size = 252, normalized size = 6.81 \[ \frac{2 \text{PolyLog}\left (2,\frac{(c+1) (a-b x)}{2 a}\right )-2 \text{PolyLog}\left (2,\frac{(c+1) (a+b x)}{2 a c}\right )-2 \text{PolyLog}\left (2,\frac{a-b x}{2 a}\right )+\log ^2\left (\frac{2 a c}{(c+1) (a+b x)}\right )+2 \log \left (-\frac{a (-c)+a+b (c+1) x}{2 a c}\right ) \log \left (\frac{2 a c}{(c+1) (a+b x)}\right )-2 \log \left (\frac{a (-c)+a+b (c+1) x}{a+b x}\right ) \log \left (\frac{2 a c}{(c+1) (a+b x)}\right )+2 \log (a-b x) \log \left (\frac{a (-c)+a+b (c+1) x}{2 a}\right )-2 \log (a-b x) \log \left (\frac{a (-c)+a+b (c+1) x}{a+b x}\right )-2 \log (a-b x) \log \left (\frac{a+b x}{2 a}\right )}{4 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[(2*a*c)/((1 + c)*(a + b*x))]^2 - 2*Log[a - b*x]*Log[(a + b*x)/(2*a)] + 2*Log[a - b*x]*Log[(a - a*c + b*(1
+ c)*x)/(2*a)] + 2*Log[(2*a*c)/((1 + c)*(a + b*x))]*Log[-(a - a*c + b*(1 + c)*x)/(2*a*c)] - 2*Log[a - b*x]*Lo
g[(a - a*c + b*(1 + c)*x)/(a + b*x)] - 2*Log[(2*a*c)/((1 + c)*(a + b*x))]*Log[(a - a*c + b*(1 + c)*x)/(a + b*x
)] - 2*PolyLog[2, (a - b*x)/(2*a)] + 2*PolyLog[2, ((1 + c)*(a - b*x))/(2*a)] - 2*PolyLog[2, ((1 + c)*(a + b*x)
)/(2*a*c)])/(4*a*b)

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Maple [A]  time = 0.064, size = 24, normalized size = 0.7 \begin{align*}{\frac{1}{2\,ab}{\it dilog} \left ( 1+c-2\,{\frac{ac}{bx+a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.26546, size = 332, normalized size = 8.97 \begin{align*} \frac{1}{2} \,{\left (\frac{\log \left (b x + a\right )}{a b} - \frac{\log \left (b x - a\right )}{a b}\right )} \log \left (\frac{b{\left (c + 1\right )} x - a{\left (c - 1\right )}}{b x + a}\right ) + \frac{\log \left (b x + a\right )^{2} - 2 \, \log \left (b x + a\right ) \log \left (b x - a\right )}{4 \, a b} + \frac{\log \left (b x - a\right ) \log \left (\frac{b{\left (c + 1\right )} x - a{\left (c + 1\right )}}{2 \, a} + 1\right ) +{\rm Li}_2\left (-\frac{b{\left (c + 1\right )} x - a{\left (c + 1\right )}}{2 \, a}\right )}{2 \, a b} + \frac{\log \left (b x + a\right ) \log \left (-\frac{b x + a}{2 \, a} + 1\right ) +{\rm Li}_2\left (\frac{b x + a}{2 \, a}\right )}{2 \, a b} - \frac{\log \left (b x + a\right ) \log \left (-\frac{b{\left (c + 1\right )} x + a{\left (c + 1\right )}}{2 \, a c} + 1\right ) +{\rm Li}_2\left (\frac{b{\left (c + 1\right )} x + a{\left (c + 1\right )}}{2 \, a c}\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.61004, size = 76, normalized size = 2.05 \begin{align*} \frac{{\rm Li}_2\left (\frac{a c -{\left (b c + b\right )} x - a}{b x + a} + 1\right )}{2 \, a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\log \left (\frac{b{\left (c + 1\right )} x - a{\left (c - 1\right )}}{b x + a}\right )}{b^{2} x^{2} - a^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-log((b*(c + 1)*x - a*(c - 1))/(b*x + a))/(b^2*x^2 - a^2), x)