3.258 \(\int \frac {\log (\frac {a-c g+b x-d g x}{a+b x})}{(a+b x) (c+d x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {\text {Li}_2\left (\frac {g (c+d x)}{a+b x}\right )}{b c-a d} \]

[Out]

polylog(2,g*(d*x+c)/(b*x+a))/(-a*d+b*c)

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Rubi [A]  time = 0.12, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2516, 2502, 2315} \[ \frac {\text {PolyLog}\left (2,\frac {g (c+d x)}{a+b x}\right )}{b c-a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 2516

Int[Log[(e_.)*((f_.)*(v_)^(p_.)*(w_)^(q_.))^(r_.)]^(s_.)*(u_.), x_Symbol] :> Int[u*Log[e*(f*ExpandToSum[v, x]^
p*ExpandToSum[w, x]^q)^r]^s, x] /; FreeQ[{e, f, p, q, r, s}, x] && LinearQ[{v, w}, x] &&  !LinearMatchQ[{v, w}
, x] && AlgebraicFunctionQ[u, x]

Rubi steps

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

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Mathematica [B]  time = 0.18, size = 320, normalized size = 11.85 \[ \frac {-2 \text {Li}_2\left (\frac {(b-d g) (a+b x)}{(b c-a d) g}\right )+2 \text {Li}_2\left (\frac {(b-d g) (c+d x)}{b c-a d}\right )+\log ^2\left (\frac {g (b c-a d)}{(a+b x) (b-d g)}\right )+2 \log \left (-\frac {b (a+b x-c g-d g x)}{g (b c-a d)}\right ) \log \left (\frac {g (b c-a d)}{(a+b x) (b-d g)}\right )-2 \log \left (\frac {a+b x-c g-d g x}{a+b x}\right ) \log \left (\frac {g (b c-a d)}{(a+b x) (b-d g)}\right )+2 \log (c+d x) \log \left (-\frac {d (a+b x-c g-d g x)}{b c-a d}\right )-2 \log (c+d x) \log \left (\frac {a+b x-c g-d g x}{a+b x}\right )-2 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )-2 \log (c+d x) \log \left (\frac {d (a+b x)}{a d-b c}\right )}{2 b c-2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(a - c*g + b*x - d*g*x)/(a + b*x)]/((a + b*x)*(c + d*x)),x]

[Out]

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

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fricas [A]  time = 0.69, size = 38, normalized size = 1.41 \[ \frac {{\rm Li}_2\left (\frac {c g + {\left (d g - b\right )} x - a}{b x + a} + 1\right )}{b c - a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="fricas")

[Out]

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

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giac [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(log((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 45, normalized size = 1.67 \[ -\frac {\dilog \left (\frac {\left (a d -b c \right ) g}{\left (b x +a \right ) b}+\frac {-d g +b}{b}\right )}{a d -b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x)

[Out]

-dilog((a*d-b*c)/(b*x+a)/b*g+(-d*g+b)/b)/(a*d-b*c)

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maxima [B]  time = 0.82, size = 343, normalized size = 12.70 \[ {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} \log \left (-\frac {d g x + c g - b x - a}{b x + a}\right ) + \frac {\log \left (b x + a\right )^{2} - 2 \, \log \left (b x + a\right ) \log \left (d x + c\right )}{2 \, {\left (b c - a d\right )}} - \frac {\log \left (b x + a\right ) \log \left (\frac {{\left (d g - b\right )} a + {\left (b d g - b^{2}\right )} x}{b c g - a d g} + 1\right ) + {\rm Li}_2\left (-\frac {{\left (d g - b\right )} a + {\left (b d g - b^{2}\right )} x}{b c g - a d g}\right )}{b c - a d} + \frac {\log \left (d x + c\right ) \log \left (\frac {c d g - b c + {\left (d^{2} g - b d\right )} x}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {c d g - b c + {\left (d^{2} g - b d\right )} x}{b c - a d}\right )}{b c - a d} + \frac {\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )}{b c - a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\ln \left (\frac {a-c\,g+b\,x-d\,g\,x}{a+b\,x}\right )}{\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((a - c*g + b*x - d*g*x)/(a + b*x))/((a + b*x)*(c + d*x)),x)

[Out]

int(log((a - c*g + b*x - d*g*x)/(a + b*x))/((a + b*x)*(c + d*x)), 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((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x)

[Out]

Timed out

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