\(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{c g+d g x} \, dx\) [33]

Optimal result

Integrand size = 33, antiderivative size = 80 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c g+d g x} \, dx=-\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{d g}-\frac {B n \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d g} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 80, 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 = {2543, 2458, 2378, 2370, 2352} \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c g+d g x} \, dx=-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d g}-\frac {B n \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d g} \]

[In]

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

[Out]

-(((A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[(b*c - a*d)/(b*(c + d*x))])/(d*g)) - (B*n*PolyLog[2, (d*(a + b*x
))/(b*(c + d*x))])/(d*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 2543

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))/((f_.) + (g_.)*(x_)), x_Symbo
l] :> Simp[(-Log[(b*c - a*d)/(b*(c + d*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)/(b*(c + d*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[d*f - c*g, 0]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.26 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c g+d g x} \, dx=\frac {\log (g (c+d x)) \left (2 A-2 B n \log \left (\frac {d (a+b x)}{-b c+a d}\right )+2 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B n \log (g (c+d x))\right )-2 B n \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{2 d g} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-1/2*B*((2*n*log(b*x + a)*log(d*x + c) - n*log(d*x + c)^2 - 2*log(d*x + c)*log((b*x + a)^n) + 2*log(d*x + c)*l
og((d*x + c)^n))/(d*g) - 2*integrate((n*log(b*x + a) + log(e))/(d*g*x + c*g), x)) + A*log(d*g*x + c*g)/(d*g)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 566 vs. \(2 (79) = 158\).

Time = 57.49 (sec) , antiderivative size = 566, normalized size of antiderivative = 7.08 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c g+d g x} \, dx=\frac {1}{2} \, {\left (\frac {{\left (B b^{3} c^{3} n - 3 \, B a b^{2} c^{2} d n + 3 \, B a^{2} b c d^{2} n - B a^{3} d^{3} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d g - \frac {2 \, {\left (b x + a\right )} b d^{2} g}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{3} g}{{\left (d x + c\right )}^{2}}} - \frac {B b^{4} c^{3} n - 3 \, B a b^{3} c^{2} d n - \frac {{\left (b x + a\right )} B b^{3} c^{3} d n}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} n + \frac {3 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} n}{d x + c} - B a^{3} b d^{3} n - \frac {3 \, {\left (b x + a\right )} B a^{2} b c d^{3} n}{d x + c} + \frac {{\left (b x + a\right )} B a^{3} d^{4} n}{d x + c} - B b^{4} c^{3} \log \left (e\right ) + 3 \, B a b^{3} c^{2} d \log \left (e\right ) - 3 \, B a^{2} b^{2} c d^{2} \log \left (e\right ) + B a^{3} b d^{3} \log \left (e\right ) - A b^{4} c^{3} + 3 \, A a b^{3} c^{2} d - 3 \, A a^{2} b^{2} c d^{2} + A a^{3} b d^{3}}{b^{3} d g - \frac {2 \, {\left (b x + a\right )} b^{2} d^{2} g}{d x + c} + \frac {{\left (b x + a\right )}^{2} b d^{3} g}{{\left (d x + c\right )}^{2}}} + \frac {{\left (B b^{3} c^{3} n - 3 \, B a b^{2} c^{2} d n + 3 \, B a^{2} b c d^{2} n - B a^{3} d^{3} n\right )} \log \left (-b + \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b^{2} d g} - \frac {{\left (B b^{3} c^{3} n - 3 \, B a b^{2} c^{2} d n + 3 \, B a^{2} b c d^{2} n - B a^{3} d^{3} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d g}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )}^{2} \]

[In]

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

[Out]

1/2*((B*b^3*c^3*n - 3*B*a*b^2*c^2*d*n + 3*B*a^2*b*c*d^2*n - B*a^3*d^3*n)*log((b*x + a)/(d*x + c))/(b^2*d*g - 2
*(b*x + a)*b*d^2*g/(d*x + c) + (b*x + a)^2*d^3*g/(d*x + c)^2) - (B*b^4*c^3*n - 3*B*a*b^3*c^2*d*n - (b*x + a)*B
*b^3*c^3*d*n/(d*x + c) + 3*B*a^2*b^2*c*d^2*n + 3*(b*x + a)*B*a*b^2*c^2*d^2*n/(d*x + c) - B*a^3*b*d^3*n - 3*(b*
x + a)*B*a^2*b*c*d^3*n/(d*x + c) + (b*x + a)*B*a^3*d^4*n/(d*x + c) - B*b^4*c^3*log(e) + 3*B*a*b^3*c^2*d*log(e)
- 3*B*a^2*b^2*c*d^2*log(e) + B*a^3*b*d^3*log(e) - A*b^4*c^3 + 3*A*a*b^3*c^2*d - 3*A*a^2*b^2*c*d^2 + A*a^3*b*d
^3)/(b^3*d*g - 2*(b*x + a)*b^2*d^2*g/(d*x + c) + (b*x + a)^2*b*d^3*g/(d*x + c)^2) + (B*b^3*c^3*n - 3*B*a*b^2*c
^2*d*n + 3*B*a^2*b*c*d^2*n - B*a^3*d^3*n)*log(-b + (b*x + a)*d/(d*x + c))/(b^2*d*g) - (B*b^3*c^3*n - 3*B*a*b^2
*c^2*d*n + 3*B*a^2*b*c*d^2*n - B*a^3*d^3*n)*log((b*x + a)/(d*x + c))/(b^2*d*g))*(b*c/(b*c - a*d)^2 - a*d/(b*c
- a*d)^2)^2

Mupad [F(-1)]

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

[In]

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

[Out]

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