3.71 \(\int \frac {\log (-\frac {g (d+e x)}{e f-d g})}{f+g x} \, dx\)

Optimal. Leaf size=24 \[ -\frac {\text {Li}_2\left (\frac {e (f+g x)}{e f-d g}\right )}{g} \]

[Out]

-polylog(2,e*(g*x+f)/(-d*g+e*f))/g

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2393, 2391} \[ -\frac {\text {PolyLog}\left (2,\frac {e (f+g x)}{e f-d g}\right )}{g} \]

Antiderivative was successfully verified.

[In]

Int[Log[-((g*(d + e*x))/(e*f - d*g))]/(f + g*x),x]

[Out]

-(PolyLog[2, (e*(f + g*x))/(e*f - d*g)]/g)

Rule 2391

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

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
 b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
 + c*(e*f - d*g), 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \[ -\frac {\text {Li}_2\left (\frac {e (f+g x)}{e f-d g}\right )}{g} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[-((g*(d + e*x))/(e*f - d*g))]/(f + g*x),x]

[Out]

-(PolyLog[2, (e*(f + g*x))/(e*f - d*g)]/g)

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fricas [A]  time = 0.44, size = 27, normalized size = 1.12 \[ -\frac {{\rm Li}_2\left (\frac {e g x + d g}{e f - d g} + 1\right )}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-g*(e*x+d)/(-d*g+e*f))/(g*x+f),x, algorithm="fricas")

[Out]

-dilog((e*g*x + d*g)/(e*f - d*g) + 1)/g

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (-\frac {{\left (e x + d\right )} g}{e f - d g}\right )}{g x + f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-g*(e*x+d)/(-d*g+e*f))/(g*x+f),x, algorithm="giac")

[Out]

integrate(log(-(e*x + d)*g/(e*f - d*g))/(g*x + f), x)

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maple [A]  time = 0.04, size = 35, normalized size = 1.46 \[ -\frac {\dilog \left (\frac {e g x}{d g -e f}+\frac {d g}{d g -e f}\right )}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(-g*(e*x+d)/(-d*g+e*f))/(g*x+f),x)

[Out]

-1/g*dilog(e*g/(d*g-e*f)*x+d*g/(d*g-e*f))

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maxima [B]  time = 0.67, size = 102, normalized size = 4.25 \[ -\frac {\log \left (e x + d\right ) \log \left (g x + f\right )}{g} + \frac {\log \left (g x + f\right ) \log \left (-\frac {{\left (e x + d\right )} g}{e f - d g}\right )}{g} + \frac {\log \left (e x + d\right ) \log \left (\frac {e g x + d g}{e f - d g} + 1\right ) + {\rm Li}_2\left (-\frac {e g x + d g}{e f - d g}\right )}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-g*(e*x+d)/(-d*g+e*f))/(g*x+f),x, algorithm="maxima")

[Out]

-log(e*x + d)*log(g*x + f)/g + log(g*x + f)*log(-(e*x + d)*g/(e*f - d*g))/g + (log(e*x + d)*log((e*g*x + d*g)/
(e*f - d*g) + 1) + dilog(-(e*g*x + d*g)/(e*f - d*g)))/g

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mupad [B]  time = 0.39, size = 23, normalized size = 0.96 \[ -\frac {{\mathrm {Li}}_{\mathrm {2}}\left (\frac {g\,\left (d+e\,x\right )}{d\,g-e\,f}\right )}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((g*(d + e*x))/(d*g - e*f))/(f + g*x),x)

[Out]

-dilog((g*(d + e*x))/(d*g - e*f))/g

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (- \frac {d g}{- d g + e f} - \frac {e g x}{- d g + e f} \right )}}{f + g x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(-g*(e*x+d)/(-d*g+e*f))/(g*x+f),x)

[Out]

Integral(log(-d*g/(-d*g + e*f) - e*g*x/(-d*g + e*f))/(f + g*x), x)

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