3.92 \(\int \frac{\log ^2(e (\frac{a+b x}{c+d x})^n) \log (\frac{b c-a d}{b (c+d x)})}{(c+d x) (a g+b g x)} \, dx\)

Optimal. Leaf size=160 \[ -\frac{\text{PolyLog}\left (2,1-\frac{b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}+\frac{2 n \text{PolyLog}\left (3,1-\frac{b c-a d}{b (c+d x)}\right ) \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}-\frac{2 n^2 \text{PolyLog}\left (4,1-\frac{b c-a d}{b (c+d x)}\right )}{g (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.250151, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 58, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.052, Rules used = {2506, 2508, 6610} \[ -\frac{\text{PolyLog}\left (2,1-\frac{b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}+\frac{2 n \text{PolyLog}\left (3,1-\frac{b c-a d}{b (c+d x)}\right ) \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}-\frac{2 n^2 \text{PolyLog}\left (4,1-\frac{b c-a d}{b (c+d x)}\right )}{g (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2506

Int[Log[v_]*Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(u_), x_Symbo
l] :> With[{g = Simplify[((v - 1)*(c + d*x))/(a + b*x)], h = Simplify[u*(a + b*x)*(c + d*x)]}, -Simp[(h*PolyLo
g[2, 1 - v]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(b*c - a*d), x] + Dist[h*p*r*s, Int[(PolyLog[2, 1 - v]*Log
[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{g, h}, x]] /; FreeQ[{a, b,
c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && EqQ[p + q, 0]

Rule 2508

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(u_)*PolyLog[n_, v_],
x_Symbol] :> With[{g = Simplify[(v*(c + d*x))/(a + b*x)], h = Simplify[u*(a + b*x)*(c + d*x)]}, Simp[(h*PolyL
og[n + 1, v]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(b*c - a*d), x] - Dist[h*p*r*s, Int[(PolyLog[n + 1, v]*Lo
g[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{g, h}, x]] /; FreeQ[{a, b,
c, d, e, f, n, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && EqQ[p + q, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{\log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \log \left (\frac{b c-a d}{b (c+d x)}\right )}{(c+d x) (a g+b g x)} \, dx &=-\frac{\log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_2\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac{(2 n) \int \frac{\log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_2\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)} \, dx}{g}\\ &=-\frac{\log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_2\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac{2 n \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_3\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac{\left (2 n^2\right ) \int \frac{\text{Li}_3\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(a+b x) (c+d x)} \, dx}{g}\\ &=-\frac{\log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_2\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac{2 n \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right ) \text{Li}_3\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac{2 n^2 \text{Li}_4\left (1-\frac{b c-a d}{b (c+d x)}\right )}{(b c-a d) g}\\ \end{align*}

Mathematica [B]  time = 0.455601, size = 559, normalized size = 3.49 \[ \frac{3 n \left (-2 \text{PolyLog}\left (3,\frac{d (a+b x)}{b (c+d x)}\right )+2 \log \left (\frac{a+b x}{c+d x}\right ) \text{PolyLog}\left (2,\frac{d (a+b x)}{b (c+d x)}\right )+\log \left (\frac{b c-a d}{b c+b d x}\right ) \log ^2\left (\frac{a+b x}{c+d x}\right )\right ) \left (n \log \left (\frac{a+b x}{c+d x}\right )-\log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )\right )+\frac{3}{2} \left (2 \text{PolyLog}\left (2,\frac{d (a+b x)}{a d-b c}\right )-2 \log \left (\frac{a}{b}+x\right ) \log (c+d x)+2 \log (c+d x) \log \left (\frac{a+b x}{c+d x}\right )+2 \log \left (\frac{a}{b}+x\right ) \log \left (\frac{b (c+d x)}{b c-a d}\right )-\log ^2\left (\frac{c}{d}+x\right )+2 \log (c+d x) \log \left (\frac{c}{d}+x\right )\right ) \left (\log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )-n \log \left (\frac{a+b x}{c+d x}\right )\right )^2+n^2 \left (-\left (6 \text{PolyLog}\left (4,\frac{d (a+b x)}{b (c+d x)}\right )+3 \log ^2\left (\frac{a+b x}{c+d x}\right ) \text{PolyLog}\left (2,\frac{d (a+b x)}{b (c+d x)}\right )-6 \log \left (\frac{a+b x}{c+d x}\right ) \text{PolyLog}\left (3,\frac{d (a+b x)}{b (c+d x)}\right )+\log \left (\frac{b c-a d}{b c+b d x}\right ) \log ^3\left (\frac{a+b x}{c+d x}\right )\right )\right )+\log \left (\frac{a+b x}{c+d x}\right ) \log \left (\frac{b c-a d}{b c+b d x}\right ) \left (3 \log ^2\left (e \left (\frac{a+b x}{c+d x}\right )^n\right )-3 n \log \left (\frac{a+b x}{c+d x}\right ) \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+n^2 \log ^2\left (\frac{a+b x}{c+d x}\right )\right )}{3 g (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[(a + b*x)/(c + d*x)]*(3*Log[e*((a + b*x)/(c + d*x))^n]^2 - 3*n*Log[e*((a + b*x)/(c + d*x))^n]*Log[(a + b*
x)/(c + d*x)] + n^2*Log[(a + b*x)/(c + d*x)]^2)*Log[(b*c - a*d)/(b*c + b*d*x)] + (3*(Log[e*((a + b*x)/(c + d*x
))^n] - n*Log[(a + b*x)/(c + d*x)])^2*(-Log[c/d + x]^2 - 2*Log[a/b + x]*Log[c + d*x] + 2*Log[c/d + x]*Log[c +
d*x] + 2*Log[(a + b*x)/(c + d*x)]*Log[c + d*x] + 2*Log[a/b + x]*Log[(b*(c + d*x))/(b*c - a*d)] + 2*PolyLog[2,
(d*(a + b*x))/(-(b*c) + a*d)]))/2 + 3*n*(-Log[e*((a + b*x)/(c + d*x))^n] + n*Log[(a + b*x)/(c + d*x)])*(Log[(a
+ b*x)/(c + d*x)]^2*Log[(b*c - a*d)/(b*c + b*d*x)] + 2*Log[(a + b*x)/(c + d*x)]*PolyLog[2, (d*(a + b*x))/(b*(
c + d*x))] - 2*PolyLog[3, (d*(a + b*x))/(b*(c + d*x))]) - n^2*(Log[(a + b*x)/(c + d*x)]^3*Log[(b*c - a*d)/(b*c
+ b*d*x)] + 3*Log[(a + b*x)/(c + d*x)]^2*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))] - 6*Log[(a + b*x)/(c + d*x)]
*PolyLog[3, (d*(a + b*x))/(b*(c + d*x))] + 6*PolyLog[4, (d*(a + b*x))/(b*(c + d*x))]))/(3*(b*c - a*d)*g)

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Maple [F]  time = 5.443, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( dx+c \right ) \left ( bgx+ag \right ) } \left ( \ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) ^{2}\ln \left ({\frac{-ad+bc}{b \left ( dx+c \right ) }} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac{b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )}{\left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac{b c - a d}{b d x + b c}\right )}{b d g x^{2} + a c g +{\left (b c + a d\right )} g x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (e \left (\frac{b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac{b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )}{\left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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