3.103 \(\int \frac{\log (\frac{-b c+a d}{d (a+b x)}) \log (\frac{e (c+d x)}{a+b x})}{(a+b x) (c+d x)} \, dx\)

Optimal. Leaf size=85 \[ \frac{\text{PolyLog}\left (2,\frac{b c-a d}{d (a+b x)}+1\right ) \log \left (\frac{e (c+d x)}{a+b x}\right )}{b c-a d}-\frac{\text{PolyLog}\left (3,\frac{b c-a d}{d (a+b x)}+1\right )}{b c-a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.134357, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 50, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2506, 6610} \[ \frac{\text{PolyLog}\left (2,\frac{b c-a d}{d (a+b x)}+1\right ) \log \left (\frac{e (c+d x)}{a+b x}\right )}{b c-a d}-\frac{\text{PolyLog}\left (3,\frac{b c-a d}{d (a+b x)}+1\right )}{b c-a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0237251, size = 68, normalized size = 0.8 \[ \frac{\text{PolyLog}\left (2,\frac{b (c+d x)}{d (a+b x)}\right ) \log \left (\frac{e (c+d x)}{a+b x}\right )-\text{PolyLog}\left (3,\frac{b (c+d x)}{d (a+b x)}\right )}{b c-a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[(e*(c + d*x))/(a + b*x)]*PolyLog[2, (b*(c + d*x))/(d*(a + b*x))] - PolyLog[3, (b*(c + d*x))/(d*(a + b*x))
])/(b*c - a*d)

________________________________________________________________________________________

Maple [F]  time = 1.318, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({\frac{ad-bc}{d \left ( bx+a \right ) }} \right ) \ln \left ({\frac{e \left ( dx+c \right ) }{bx+a}} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (\frac{{\left (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac{b c - a d}{{\left (b x + a\right )} d}\right )}{{\left (b x + a\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)/d/(b*x+a))*log(e*(d*x+c)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

integrate(log((d*x + c)*e/(b*x + a))*log(-(b*c - a*d)/((b*x + a)*d))/((b*x + a)*(d*x + c)), x)

________________________________________________________________________________________

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 + a d}\right ) \log \left (\frac{d e x + c e}{b x + a}\right )}{b d x^{2} + a c +{\left (b c + a d\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)/d/(b*x+a))*ln(e*(d*x+c)/(b*x+a))/(b*x+a)/(d*x+c),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (\frac{{\left (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac{b c - a d}{{\left (b x + a\right )} d}\right )}{{\left (b x + a\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)/d/(b*x+a))*log(e*(d*x+c)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="giac")

[Out]

integrate(log((d*x + c)*e/(b*x + a))*log(-(b*c - a*d)/((b*x + a)*d))/((b*x + a)*(d*x + c)), x)