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

Optimal. Leaf size=27 \[ \frac{\text{PolyLog}\left (2,\frac{c (a-b x)}{a+b x}\right )}{2 a b} \]

[Out]

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

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Rubi [A]  time = 0.0739229, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.051, Rules used = {2502, 2315} \[ \frac{\text{PolyLog}\left (2,\frac{c (a-b x)}{a+b x}\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2502

Int[Log[((e_.)*((c_.) + (d_.)*(x_)))/((a_.) + (b_.)*(x_))]*(u_), x_Symbol] :> With[{g = Coeff[Simplify[1/(u*(a
+ b*x))], x, 0], h = Coeff[Simplify[1/(u*(a + b*x))], x, 1]}, -Dist[(b - d*e)/(h*(b*c - a*d)), Subst[Int[Log[
e*x]/(1 - e*x), x], x, (c + d*x)/(a + b*x)], x] /; EqQ[g*(b - d*e) - h*(a - c*e), 0]] /; FreeQ[{a, b, c, d, e}
, x] && NeQ[b*c - a*d, 0] && LinearQ[Simplify[1/(u*(a + b*x))], x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [B]  time = 0.147699, size = 252, normalized size = 9.33 \[ \frac{2 \text{PolyLog}\left (2,\frac{(c+1) (a-b x)}{2 a}\right )-2 \text{PolyLog}\left (2,\frac{(c+1) (a+b x)}{2 a c}\right )-2 \text{PolyLog}\left (2,\frac{a-b x}{2 a}\right )+\log ^2\left (\frac{2 a c}{(c+1) (a+b x)}\right )+2 \log \left (-\frac{a (-c)+a+b (c+1) x}{2 a c}\right ) \log \left (\frac{2 a c}{(c+1) (a+b x)}\right )-2 \log \left (\frac{a (-c)+a+b (c+1) x}{a+b x}\right ) \log \left (\frac{2 a c}{(c+1) (a+b x)}\right )+2 \log (a-b x) \log \left (\frac{a (-c)+a+b (c+1) x}{2 a}\right )-2 \log (a-b x) \log \left (\frac{a (-c)+a+b (c+1) x}{a+b x}\right )-2 \log (a-b x) \log \left (\frac{a+b x}{2 a}\right )}{4 a b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(a*(1 - c) + b*(1 + c)*x)/(a + b*x)]/((a - b*x)*(a + b*x)),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 = 0.06, size = 24, normalized size = 0.9 \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*x+a)/(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.19238, size = 332, normalized size = 12.3 \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*x+a)/(b*x+a),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 = 1.66643, size = 76, normalized size = 2.81 \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*x+a)/(b*x+a),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*x+a)/(b*x+a),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 )}{{\left (b x + a\right )}{\left (b x - a\right )}}\,{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*x+a)/(b*x+a),x, algorithm="giac")

[Out]

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