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

Optimal. Leaf size=150 \[ -\frac {\text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}+\frac {2 \text {Li}_3\left (1-\frac {b c-a d}{b (c+d x)}\right ) \log \left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}-\frac {2 \text {Li}_4\left (1-\frac {b c-a d}{b (c+d x)}\right )}{g (b c-a d)} \]

[Out]

-ln(e*(b*x+a)/(d*x+c))^2*polylog(2,1+(a*d-b*c)/b/(d*x+c))/(-a*d+b*c)/g+2*ln(e*(b*x+a)/(d*x+c))*polylog(3,1+(a*
d-b*c)/b/(d*x+c))/(-a*d+b*c)/g-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 = 150, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {2506, 2508, 6610} \[ -\frac {\text {PolyLog}\left (2,1-\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}+\frac {2 \text {PolyLog}\left (3,1-\frac {b c-a d}{b (c+d x)}\right ) \log \left (\frac {e (a+b x)}{c+d x}\right )}{g (b c-a d)}-\frac {2 \text {PolyLog}\left (4,1-\frac {b c-a d}{b (c+d x)}\right )}{g (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Log[(e*(a + b*x))/(c + d*x)]^2*PolyLog[2, 1 - (b*c - a*d)/(b*(c + d*x))])/((b*c - a*d)*g)) + (2*Log[(e*(a +
b*x))/(c + d*x)]*PolyLog[3, 1 - (b*c - a*d)/(b*(c + d*x))])/((b*c - a*d)*g) - (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 \left (\frac {b c-a d}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x) (a g+b g x)} \, dx &=-\frac {\log ^2\left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 \int \frac {\log \left (\frac {e (a+b x)}{c+d x}\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 (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 \log \left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_3\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac {2 \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 (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_2\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}+\frac {2 \log \left (\frac {e (a+b x)}{c+d x}\right ) \text {Li}_3\left (1-\frac {b c-a d}{b (c+d x)}\right )}{(b c-a d) g}-\frac {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 [A]  time = 0.04, size = 110, normalized size = 0.73 \[ \frac {-\text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right ) \log ^2\left (\frac {e (a+b x)}{c+d x}\right )+2 \text {Li}_3\left (\frac {d (a+b x)}{b (c+d x)}\right ) \log \left (\frac {e (a+b x)}{c+d x}\right )-2 \text {Li}_4\left (\frac {d (a+b x)}{b (c+d x)}\right )}{g (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (\frac {b c - a d}{b d x + b c}\right ) \log \left (\frac {b e x + a e}{d x + c}\right )^{2}}{b d g x^{2} + a c g + {\left (b c + a d\right )} g x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((b*c - a*d)/(b*d*x + b*c))*log((b*e*x + a*e)/(d*x + c))^2/(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 \[ \int \frac {\log \left (\frac {{\left (b x + a\right )} e}{d x + c}\right )^{2} \log \left (\frac {b c - a d}{{\left (d x + c\right )} b}\right )}{{\left (b g x + a g\right )} {\left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((b*x + a)*e/(d*x + c))^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 = 1.09, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (\frac {\left (b x +a \right ) e}{d x +c}\right )^{2} \ln \left (\frac {-a d +b c}{\left (d x +c \right ) b}\right )}{\left (d x +c \right ) \left (b g x +a g \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {4 \, \log \left (b x + a\right ) \log \left (d x + c\right )^{3} - \log \left (d x + c\right )^{4}}{4 \, {\left (b c g - a d g\right )}} - \int \frac {{\left ({\left (d \log \left (b c - a d\right ) - d \log \relax (b)\right )} a - {\left (c \log \left (b c - a d\right ) - c \log \relax (b)\right )} b\right )} \log \left (b x + a\right )^{2} + {\left ({\left (d \log \left (b c - a d\right ) - d \log \relax (b) + 2 \, d \log \relax (e)\right )} a - {\left (c {\left (\log \left (b c - a d\right ) + 2 \, \log \relax (e)\right )} - c \log \relax (b)\right )} b - {\left (3 \, b d x + 2 \, b c + a d\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )^{2} + {\left (d \log \left (b c - a d\right ) \log \relax (e)^{2} - d \log \relax (b) \log \relax (e)^{2}\right )} a - {\left (c \log \left (b c - a d\right ) \log \relax (e)^{2} - c \log \relax (b) \log \relax (e)^{2}\right )} b + 2 \, {\left ({\left (d \log \left (b c - a d\right ) \log \relax (e) - d \log \relax (b) \log \relax (e)\right )} a - {\left (c \log \left (b c - a d\right ) \log \relax (e) - c \log \relax (b) \log \relax (e)\right )} b\right )} \log \left (b x + a\right ) + {\left ({\left (b c - a d\right )} \log \left (b x + a\right )^{2} - {\left (2 \, d \log \left (b c - a d\right ) \log \relax (e) - 2 \, d \log \relax (b) \log \relax (e) + d \log \relax (e)^{2}\right )} a - {\left (2 \, c \log \relax (b) \log \relax (e) - {\left (2 \, \log \left (b c - a d\right ) \log \relax (e) + \log \relax (e)^{2}\right )} c\right )} b - 2 \, {\left ({\left (d \log \left (b c - a d\right ) - d \log \relax (b) + d \log \relax (e)\right )} a - {\left (c {\left (\log \left (b c - a d\right ) + \log \relax (e)\right )} - c \log \relax (b)\right )} b\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{a b c^{2} g - a^{2} c d g + {\left (b^{2} c d g - a b d^{2} g\right )} x^{2} + {\left (b^{2} c^{2} g - a^{2} d^{2} g\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*log(b*x + a)*log(d*x + c)^3 - log(d*x + c)^4)/(b*c*g - a*d*g) - integrate((((d*log(b*c - a*d) - d*log(
b))*a - (c*log(b*c - a*d) - c*log(b))*b)*log(b*x + a)^2 + ((d*log(b*c - a*d) - d*log(b) + 2*d*log(e))*a - (c*(
log(b*c - a*d) + 2*log(e)) - c*log(b))*b - (3*b*d*x + 2*b*c + a*d)*log(b*x + a))*log(d*x + c)^2 + (d*log(b*c -
a*d)*log(e)^2 - d*log(b)*log(e)^2)*a - (c*log(b*c - a*d)*log(e)^2 - c*log(b)*log(e)^2)*b + 2*((d*log(b*c - a*
d)*log(e) - d*log(b)*log(e))*a - (c*log(b*c - a*d)*log(e) - c*log(b)*log(e))*b)*log(b*x + a) + ((b*c - a*d)*lo
g(b*x + a)^2 - (2*d*log(b*c - a*d)*log(e) - 2*d*log(b)*log(e) + d*log(e)^2)*a - (2*c*log(b)*log(e) - (2*log(b*
c - a*d)*log(e) + log(e)^2)*c)*b - 2*((d*log(b*c - a*d) - d*log(b) + d*log(e))*a - (c*(log(b*c - a*d) + log(e)
) - c*log(b))*b)*log(b*x + a))*log(d*x + c))/(a*b*c^2*g - a^2*c*d*g + (b^2*c*d*g - a*b*d^2*g)*x^2 + (b^2*c^2*g
- a^2*d^2*g)*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {d \int \frac {\log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}^{3}}{c + d x}\, dx}{3 g \left (a d - b c\right )} - \frac {\log {\left (\frac {- a d + b c}{b \left (c + d x\right )} \right )} \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}^{3}}{3 a d g - 3 b c g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*Integral(log(a*e/(c + d*x) + b*e*x/(c + d*x))**3/(c + d*x), x)/(3*g*(a*d - b*c)) - log((-a*d + b*c)/(b*(c +
d*x)))*log(e*(a + b*x)/(c + d*x))**3/(3*a*d*g - 3*b*c*g)

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