3.235 \(\int \frac{A+B \log (\frac{e (a+b x)}{c+d x})}{f+g x} \, dx\)

Optimal. Leaf size=140 \[ -\frac{B \text{PolyLog}\left (2,\frac{b (f+g x)}{b f-a g}\right )}{g}+\frac{B \text{PolyLog}\left (2,\frac{d (f+g x)}{d f-c g}\right )}{g}+\frac{\log (f+g x) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{g}-\frac{B \log (f+g x) \log \left (-\frac{g (a+b x)}{b f-a g}\right )}{g}+\frac{B \log (f+g x) \log \left (-\frac{g (c+d x)}{d f-c g}\right )}{g} \]

[Out]

-((B*Log[-((g*(a + b*x))/(b*f - a*g))]*Log[f + g*x])/g) + ((A + B*Log[(e*(a + b*x))/(c + d*x)])*Log[f + g*x])/
g + (B*Log[-((g*(c + d*x))/(d*f - c*g))]*Log[f + g*x])/g - (B*PolyLog[2, (b*(f + g*x))/(b*f - a*g)])/g + (B*Po
lyLog[2, (d*(f + g*x))/(d*f - c*g)])/g

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Rubi [A]  time = 0.246955, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2524, 12, 2418, 2394, 2393, 2391} \[ -\frac{B \text{PolyLog}\left (2,\frac{b (f+g x)}{b f-a g}\right )}{g}+\frac{B \text{PolyLog}\left (2,\frac{d (f+g x)}{d f-c g}\right )}{g}+\frac{\log (f+g x) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{g}-\frac{B \log (f+g x) \log \left (-\frac{g (a+b x)}{b f-a g}\right )}{g}+\frac{B \log (f+g x) \log \left (-\frac{g (c+d x)}{d f-c g}\right )}{g} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[(e*(a + b*x))/(c + d*x)])/(f + g*x),x]

[Out]

-((B*Log[-((g*(a + b*x))/(b*f - a*g))]*Log[f + g*x])/g) + ((A + B*Log[(e*(a + b*x))/(c + d*x)])*Log[f + g*x])/
g + (B*Log[-((g*(c + d*x))/(d*f - c*g))]*Log[f + g*x])/g - (B*PolyLog[2, (b*(f + g*x))/(b*f - a*g)])/g + (B*Po
lyLog[2, (d*(f + g*x))/(d*f - c*g)])/g

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2394

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

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.0571215, size = 115, normalized size = 0.82 \[ \frac{-B \text{PolyLog}\left (2,\frac{b (f+g x)}{b f-a g}\right )+B \text{PolyLog}\left (2,\frac{d (f+g x)}{d f-c g}\right )+\log (f+g x) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )-B \log \left (\frac{g (a+b x)}{a g-b f}\right )+A+B \log \left (\frac{g (c+d x)}{c g-d f}\right )\right )}{g} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[(e*(a + b*x))/(c + d*x)])/(f + g*x),x]

[Out]

((A - B*Log[(g*(a + b*x))/(-(b*f) + a*g)] + B*Log[(e*(a + b*x))/(c + d*x)] + B*Log[(g*(c + d*x))/(-(d*f) + c*g
)])*Log[f + g*x] - B*PolyLog[2, (b*(f + g*x))/(b*f - a*g)] + B*PolyLog[2, (d*(f + g*x))/(d*f - c*g)])/g

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Maple [B]  time = 0.545, size = 1400, normalized size = 10. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*ln(e*(b*x+a)/(d*x+c)))/(g*x+f),x)

[Out]

d*A/g/(a*d-b*c)*ln((b*e/d+(a*d-b*c)*e/d/(d*x+c))*c*g-d*f*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)*a-A/g/(a*d
-b*c)*ln((b*e/d+(a*d-b*c)*e/d/(d*x+c))*c*g-d*f*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)*b*c-d*A/g/(a*d-b*c)*
ln(d*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-b*e)*a+A/g/(a*d-b*c)*ln(d*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-b*e)*b*c+d*B/(a*d-b
*c)*dilog(((c*g-d*f)*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)/(-a*e*g+b*e*f))/(c*g-d*f)*c*a-B/(a*d-b*c)*dilo
g(((c*g-d*f)*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)/(-a*e*g+b*e*f))/(c*g-d*f)*c^2*b-d^2*B/g/(a*d-b*c)*dilo
g(((c*g-d*f)*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)/(-a*e*g+b*e*f))/(c*g-d*f)*f*a+d*B/g/(a*d-b*c)*dilog(((
c*g-d*f)*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)/(-a*e*g+b*e*f))/(c*g-d*f)*f*b*c+d*B/(a*d-b*c)*ln(b*e/d+(a*
d-b*c)*e/d/(d*x+c))*ln(((c*g-d*f)*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)/(-a*e*g+b*e*f))/(c*g-d*f)*c*a-B/(
a*d-b*c)*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*ln(((c*g-d*f)*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)/(-a*e*g+b*e*
f))/(c*g-d*f)*c^2*b-d^2*B/g/(a*d-b*c)*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*ln(((c*g-d*f)*(b*e/d+(a*d-b*c)*e/d/(d*x+
c))-a*e*g+b*e*f)/(-a*e*g+b*e*f))/(c*g-d*f)*f*a+d*B/g/(a*d-b*c)*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*ln(((c*g-d*f)*(
b*e/d+(a*d-b*c)*e/d/(d*x+c))-a*e*g+b*e*f)/(-a*e*g+b*e*f))/(c*g-d*f)*f*b*c-d*B/g/(a*d-b*c)*dilog(-(d*(b*e/d+(a*
d-b*c)*e/d/(d*x+c))-b*e)/b/e)*a+B/g/(a*d-b*c)*dilog(-(d*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-b*e)/b/e)*b*c-d*B/g/(a*d
-b*c)*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*ln(-(d*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-b*e)/b/e)*a+B/g/(a*d-b*c)*ln(b*e/d+
(a*d-b*c)*e/d/(d*x+c))*ln(-(d*(b*e/d+(a*d-b*c)*e/d/(d*x+c))-b*e)/b/e)*b*c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -B \int -\frac{\log \left (b x + a\right ) - \log \left (d x + c\right ) + \log \left (e\right )}{g x + f}\,{d x} + \frac{A \log \left (g x + f\right )}{g} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(g*x+f),x, algorithm="maxima")

[Out]

-B*integrate(-(log(b*x + a) - log(d*x + c) + log(e))/(g*x + f), x) + A*log(g*x + f)/g

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \log \left (\frac{b e x + a e}{d x + c}\right ) + A}{g x + f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(g*x+f),x, algorithm="fricas")

[Out]

integral((B*log((b*e*x + a*e)/(d*x + c)) + A)/(g*x + f), x)

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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((A+B*ln(e*(b*x+a)/(d*x+c)))/(g*x+f),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \log \left (\frac{{\left (b x + a\right )} e}{d x + c}\right ) + A}{g x + f}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(g*x+f),x, algorithm="giac")

[Out]

integrate((B*log((b*x + a)*e/(d*x + c)) + A)/(g*x + f), x)