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

[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)

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Rubi [A]  time = 0.247639, antiderivative size = 150, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0376798, size = 110, normalized size = 0.73 \[ \frac{-\text{PolyLog}\left (2,\frac{d (a+b x)}{b (c+d x)}\right ) \log ^2\left (\frac{e (a+b x)}{c+d x}\right )+2 \text{PolyLog}\left (3,\frac{d (a+b x)}{b (c+d x)}\right ) \log \left (\frac{e (a+b x)}{c+d x}\right )-2 \text{PolyLog}\left (4,\frac{d (a+b x)}{b (c+d x)}\right )}{g (b c-a d)} \]

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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Maple [F]  time = 1.102, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( dx+c \right ) \left ( bgx+ag \right ) }\ln \left ({\frac{-ad+bc}{b \left ( dx+c \right ) }} \right ) \left ( \ln \left ({\frac{e \left ( bx+a \right ) }{dx+c}} \right ) \right ) ^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(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]

int(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)

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

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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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((-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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)