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 \[ -\frac {\text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}+\frac {2 n \text {Li}_3\left (1-\frac {b c-a d}{b (c+d x)}\right ) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g (b c-a d)}-\frac {2 n^2 \text {Li}_4\left (1-\frac {b c-a d}{b (c+d x)}\right )}{g (b c-a d)} \]

[Out]

-ln(e*((b*x+a)/(d*x+c))^n)^2*polylog(2,1+(a*d-b*c)/b/(d*x+c))/(-a*d+b*c)/g+2*n*ln(e*((b*x+a)/(d*x+c))^n)*polyl
og(3,1+(a*d-b*c)/b/(d*x+c))/(-a*d+b*c)/g-2*n^2*polylog(4,1+(a*d-b*c)/b/(d*x+c))/(-a*d+b*c)/g

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Rubi [A]  time = 0.25, antiderivative size = 160, normalized size of antiderivative = 1.00, 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*}

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Mathematica [B]  time = 0.45, size = 559, normalized size = 3.49 \[ \frac {\log \left (\frac {a+b x}{c+d x}\right ) \log \left (\frac {b c-a d}{b c+b d x}\right ) \left (3 \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )-3 n \log \left (\frac {a+b x}{c+d x}\right ) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+n^2 \log ^2\left (\frac {a+b x}{c+d x}\right )\right )+3 n \left (-2 \text {Li}_3\left (\frac {d (a+b x)}{b (c+d x)}\right )+2 \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right ) \log \left (\frac {a+b x}{c+d x}\right )+\log \left (\frac {b c-a d}{b c+b d x}\right ) \log ^2\left (\frac {a+b x}{c+d x}\right )\right ) \left (n \log \left (\frac {a+b x}{c+d x}\right )-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+\frac {3}{2} \left (2 \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )-2 \log \left (\frac {a}{b}+x\right ) \log (c+d x)+2 \log (c+d x) \log \left (\frac {a+b x}{c+d x}\right )+2 \log \left (\frac {a}{b}+x\right ) \log \left (\frac {b (c+d x)}{b c-a d}\right )-\log ^2\left (\frac {c}{d}+x\right )+2 \log (c+d x) \log \left (\frac {c}{d}+x\right )\right ) \left (\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )-n \log \left (\frac {a+b x}{c+d x}\right )\right )^2-\left (n^2 \left (6 \text {Li}_4\left (\frac {d (a+b x)}{b (c+d x)}\right )+3 \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right ) \log ^2\left (\frac {a+b x}{c+d x}\right )-6 \text {Li}_3\left (\frac {d (a+b x)}{b (c+d x)}\right ) \log \left (\frac {a+b x}{c+d x}\right )+\log \left (\frac {b c-a d}{b c+b d x}\right ) \log ^3\left (\frac {a+b x}{c+d x}\right )\right )\right )}{3 g (b c-a d)} \]

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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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac {b c - a d}{b d x + b c}\right )}{b d g x^{2} + a c g + {\left (b c + a d\right )} g x}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}}\,{d x} \]

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)

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maple [F]  time = 10.22, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )^{2} \ln \left (\frac {-a d +b c}{\left (d x +c \right ) b}\right )}{\left (d x +c \right ) \left (b g x +a g \right )}\, dx \]

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)/(d*x+c)/b)/(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)/(d*x+c)/b)/(d*x+c)/(b*g*x+a*g),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (-\frac {a\,d-b\,c}{b\,\left (c+d\,x\right )}\right )\,{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2}{\left (a\,g+b\,g\,x\right )\,\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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