\(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{a g+b g x} \, dx\) [5]

Optimal result

Integrand size = 33, antiderivative size = 84 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx=-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b g}+\frac {B n \operatorname {PolyLog}\left (2,1+\frac {b c-a d}{d (a+b x)}\right )}{b g} \]

[Out]

-ln((a*d-b*c)/d/(b*x+a))*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b/g+B*n*polylog(2,1+(-a*d+b*c)/d/(b*x+a))/b/g

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2541, 2458, 2378, 2370, 2352} \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx=\frac {B n \operatorname {PolyLog}\left (2,\frac {b c-a d}{d (a+b x)}+1\right )}{b g}-\frac {\log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b g} \]

[In]

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

[Out]

-((Log[-((b*c - a*d)/(d*(a + b*x)))]*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(b*g)) + (B*n*PolyLog[2, 1 + (b*c
- a*d)/(d*(a + b*x))])/(b*g)

Rule 2352

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

Rule 2370

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

Rule 2378

Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Dist[1/n, Subst[Int[(a
+ b*Log[c*x])/(x*(d + e*x^(r/n))), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IntegerQ[r/n]

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2541

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx=\frac {\log (g (a+b x)) \left (-B n \log (g (a+b x))+2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B n \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )\right )+2 B n \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )}{2 b g} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}{b g x +a g}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx=\int { \frac {B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A}{b g x + a g} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx=\frac {\int \frac {A}{a + b x}\, dx + \int \frac {B \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{a + b x}\, dx}{g} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx=\int { \frac {B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A}{b g x + a g} \,d x } \]

[In]

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

[Out]

B*((log(b*x + a)*log((b*x + a)^n) - log(b*x + a)*log((d*x + c)^n))/(b*g) + integrate((b*d*x*log(e) + b*c*log(e
) - (b*c*n - a*d*n)*log(b*x + a))/(b^2*d*g*x^2 + a*b*c*g + (b^2*c*g + a*b*d*g)*x), x)) + A*log(b*g*x + a*g)/(b
*g)

Giac [F]

\[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx=\int { \frac {B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A}{b g x + a g} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a g+b g x} \, dx=\int \frac {A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{a\,g+b\,g\,x} \,d x \]

[In]

int((A + B*log(e*((a + b*x)/(c + d*x))^n))/(a*g + b*g*x),x)

[Out]

int((A + B*log(e*((a + b*x)/(c + d*x))^n))/(a*g + b*g*x), x)