\(\int \frac {(A+B \log (e (\frac {a+b x}{c+d x})^n))^2}{(f+g x)^2} \, dx\) [72]

Optimal result

Integrand size = 32, antiderivative size = 206 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(f+g x)^2} \, dx=\frac {(a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b f-a g) (f+g x)}+\frac {2 B (b c-a d) n \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (1-\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g) (d f-c g)}+\frac {2 B^2 (b c-a d) n^2 \operatorname {PolyLog}\left (2,\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g) (d f-c g)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2553, 2355, 2354, 2438} \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(f+g x)^2} \, dx=\frac {2 B n (b c-a d) \log \left (1-\frac {(a+b x) (d f-c g)}{(c+d x) (b f-a g)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{(b f-a g) (d f-c g)}+\frac {(a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{(f+g x) (b f-a g)}+\frac {2 B^2 n^2 (b c-a d) \operatorname {PolyLog}\left (2,\frac {(d f-c g) (a+b x)}{(b f-a g) (c+d x)}\right )}{(b f-a g) (d f-c g)} \]

[In]

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

[Out]

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

Rule 2354

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

Rule 2355

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_))^2, x_Symbol] :> Simp[x*((a + b*Log[c*x^n])
^p/(d*(d + e*x))), x] - Dist[b*n*(p/d), Int[(a + b*Log[c*x^n])^(p - 1)/(d + e*x), x], x] /; FreeQ[{a, b, c, d,
e, n, p}, x] && GtQ[p, 0]

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 2553

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.), x_Symbol] :> Dist[b*c - a*d, Subst[Int[(b*f - a*g - (d*f - c*g)*x)^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m +
2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && Inte
gerQ[m] && IGtQ[p, 0]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(418\) vs. \(2(206)=412\).

Time = 0.28 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.03 \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(f+g x)^2} \, dx=\frac {-\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{f+g x}+\frac {B n \left (2 b (d f-c g) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-2 d (b f-a g) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)+2 (b c-a d) g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (f+g x)-b B (d f-c g) n \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )\right )+B d (b f-a g) n \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )\right )-2 B (b c-a d) g n \left (\left (\log \left (\frac {g (a+b x)}{-b f+a g}\right )-\log \left (\frac {g (c+d x)}{-d f+c g}\right )\right ) \log (f+g x)+\operatorname {PolyLog}\left (2,\frac {b (f+g x)}{b f-a g}\right )-\operatorname {PolyLog}\left (2,\frac {d (f+g x)}{d f-c g}\right )\right )\right )}{(b f-a g) (d f-c g)}}{g} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {{\left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}^{2}}{\left (g x +f \right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(f+g x)^2} \, dx=\int { \frac {{\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(f+g x)^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(f+g x)^2} \, dx=\int { \frac {{\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]

[In]

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

[Out]

2*A*B*n*(b*log(b*x + a)/(b*f*g - a*g^2) - d*log(d*x + c)/(d*f*g - c*g^2) + (b*c - a*d)*log(g*x + f)/(b*d*f^2 +
a*c*g^2 - (b*c + a*d)*f*g)) - B^2*(log((d*x + c)^n)^2/(g^2*x + f*g) + integrate(-(d*g*x*log(e)^2 + c*g*log(e)
^2 + (d*g*x + c*g)*log((b*x + a)^n)^2 + 2*(d*g*x*log(e) + c*g*log(e))*log((b*x + a)^n) + 2*(d*f*n + (g*n - g*l
og(e))*d*x - c*g*log(e) - (d*g*x + c*g)*log((b*x + a)^n))*log((d*x + c)^n))/(d*g^3*x^3 + c*f^2*g + (2*d*f*g^2
+ c*g^3)*x^2 + (d*f^2*g + 2*c*f*g^2)*x), x)) - 2*A*B*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)/(g^2*x + f*g) - A^
2/(g^2*x + f*g)

Giac [F]

\[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(f+g x)^2} \, dx=\int { \frac {{\left (B \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right ) + A\right )}^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(f+g x)^2} \, dx=\int \frac {{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^2}{{\left (f+g\,x\right )}^2} \,d x \]

[In]

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

[Out]

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