\(\int \frac {(A+B \log (\frac {e (a+b x)^2}{(c+d x)^2}))^2}{a g+b g x} \, dx\) [132]

Optimal result

Integrand size = 34, antiderivative size = 132 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{a g+b g x} \, dx=-\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b g}+\frac {4 B \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b g}+\frac {8 B^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b g} \]

[Out]

-(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))^2*ln(1-b*(d*x+c)/d/(b*x+a))/b/g+4*B*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2))*polylog(2
,b*(d*x+c)/d/(b*x+a))/b/g+8*B^2*polylog(3,b*(d*x+c)/d/(b*x+a))/b/g

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2550, 2379, 2421, 6724} \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{a g+b g x} \, dx=\frac {4 B \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{b g}-\frac {\log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )^2}{b g}+\frac {8 B^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b g} \]

[In]

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

[Out]

-(((A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])^2*Log[1 - (b*(c + d*x))/(d*(a + b*x))])/(b*g)) + (4*B*(A + B*Log[(
e*(a + b*x)^2)/(c + d*x)^2])*PolyLog[2, (b*(c + d*x))/(d*(a + b*x))])/(b*g) + (8*B^2*PolyLog[3, (b*(c + d*x))/
(d*(a + b*x))])/(b*g)

Rule 2379

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

Rule 2421

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

Rule 2550

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_)
)^(m_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/b)^m, Subst[Int[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] && EqQ[n + mn, 0] && IGtQ[n, 0]
&& NeQ[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[b*f - a*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rule 6724

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (A+B \log \left (e x^2\right )\right )^2}{x (b-d x)} \, dx,x,\frac {a+b x}{c+d x}\right )}{g} \\ & = -\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b g}+\frac {(4 B) \text {Subst}\left (\int \frac {\log \left (1-\frac {b}{d x}\right ) \left (A+B \log \left (e x^2\right )\right )}{x} \, dx,x,\frac {a+b x}{c+d x}\right )}{b g} \\ & = -\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b g}+\frac {4 B \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b g}-\frac {\left (8 B^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {b}{d x}\right )}{x} \, dx,x,\frac {a+b x}{c+d x}\right )}{b g} \\ & = -\frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b g}+\frac {4 B \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right ) \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b g}+\frac {8 B^2 \text {Li}_3\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.96 \[ \int \frac {\left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2}{a g+b g x} \, dx=\frac {-2 A B \log ^2\left (\frac {-b c+a d}{d (a+b x)}\right )+A^2 \log (a+b x)-2 A B \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )-B^2 \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log ^2\left (\frac {e (a+b x)^2}{(c+d x)^2}\right )-4 A B \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {b (c+d x)}{b c-a d}\right )+4 A B \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )+4 B^2 \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right ) \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )+8 B^2 \operatorname {PolyLog}\left (3,\frac {b (c+d x)}{d (a+b x)}\right )}{b g} \]

[In]

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

[Out]

(-2*A*B*Log[(-(b*c) + a*d)/(d*(a + b*x))]^2 + A^2*Log[a + b*x] - 2*A*B*Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(
e*(a + b*x)^2)/(c + d*x)^2] - B^2*Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(e*(a + b*x)^2)/(c + d*x)^2]^2 - 4*A*B
*Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(b*(c + d*x))/(b*c - a*d)] + 4*A*B*PolyLog[2, (d*(a + b*x))/(-(b*c) + a
*d)] + 4*B^2*Log[(e*(a + b*x)^2)/(c + d*x)^2]*PolyLog[2, (b*(c + d*x))/(d*(a + b*x))] + 8*B^2*PolyLog[3, (b*(c
+ d*x))/(d*(a + b*x))])/(b*g)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

(Integral(A**2/(a + b*x), x) + Integral(B**2*log(a**2*e/(c**2 + 2*c*d*x + d**2*x**2) + 2*a*b*e*x/(c**2 + 2*c*d
*x + d**2*x**2) + b**2*e*x**2/(c**2 + 2*c*d*x + d**2*x**2))**2/(a + b*x), x) + Integral(2*A*B*log(a**2*e/(c**2
+ 2*c*d*x + d**2*x**2) + 2*a*b*e*x/(c**2 + 2*c*d*x + d**2*x**2) + b**2*e*x**2/(c**2 + 2*c*d*x + d**2*x**2))/(
a + b*x), x))/g

Maxima [F]

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

[In]

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

[Out]

4*B^2*log(b*x + a)*log(d*x + c)^2/(b*g) + A^2*log(b*g*x + a*g)/(b*g) - integrate(-(B^2*b*c*log(e)^2 + 2*A*B*b*
c*log(e) + 4*(B^2*b*d*x + B^2*b*c)*log(b*x + a)^2 + (B^2*b*d*log(e)^2 + 2*A*B*b*d*log(e))*x + 4*(B^2*b*c*log(e
) + A*B*b*c + (B^2*b*d*log(e) + A*B*b*d)*x)*log(b*x + a) - 4*(B^2*b*c*log(e) + A*B*b*c + (B^2*b*d*log(e) + A*B
*b*d)*x + 2*(2*B^2*b*d*x + (b*c + a*d)*B^2)*log(b*x + a))*log(d*x + c))/(b^2*d*g*x^2 + a*b*c*g + (b^2*c*g + a*
b*d*g)*x), x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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