\(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{f+\frac {g}{x}} \, dx\) [4]

Optimal result

Integrand size = 32, antiderivative size = 217 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f+\frac {g}{x}} \, dx=\frac {A x}{f}+\frac {B (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b f}-\frac {B (b c-a d) n \log (c+d x)}{b d f}+\frac {B g n \log \left (\frac {f (a+b x)}{a f-b g}\right ) \log (g+f x)}{f^2}-\frac {g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (g+f x)}{f^2}-\frac {B g n \log \left (\frac {f (c+d x)}{c f-d g}\right ) \log (g+f x)}{f^2}+\frac {B g n \operatorname {PolyLog}\left (2,-\frac {b (g+f x)}{a f-b g}\right )}{f^2}-\frac {B g n \operatorname {PolyLog}\left (2,-\frac {d (g+f x)}{c f-d g}\right )}{f^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2608, 2535, 31, 2545, 2441, 2440, 2438} \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f+\frac {g}{x}} \, dx=-\frac {g \log (f x+g) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{f^2}+\frac {B (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b f}-\frac {B n (b c-a d) \log (c+d x)}{b d f}+\frac {B g n \operatorname {PolyLog}\left (2,-\frac {b (g+f x)}{a f-b g}\right )}{f^2}+\frac {B g n \log (f x+g) \log \left (\frac {f (a+b x)}{a f-b g}\right )}{f^2}+\frac {A x}{f}-\frac {B g n \operatorname {PolyLog}\left (2,-\frac {d (g+f x)}{c f-d g}\right )}{f^2}-\frac {B g n \log (f x+g) \log \left (\frac {f (c+d x)}{c f-d g}\right )}{f^2} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2438

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 2440

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 2441

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 2535

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

Rule 2545

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

Rule 2608

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.85 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f+\frac {g}{x}} \, dx=\frac {A f x+\frac {B f (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b}-\frac {B (b c-a d) f n \log (c+d x)}{b d}-g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (g+f x)+B g n \left (\left (\log \left (\frac {f (a+b x)}{a f-b g}\right )-\log \left (\frac {f (c+d x)}{c f-d g}\right )\right ) \log (g+f x)+\operatorname {PolyLog}\left (2,\frac {b (g+f x)}{-a f+b g}\right )-\operatorname {PolyLog}\left (2,\frac {d (g+f x)}{-c f+d g}\right )\right )}{f^2} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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