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

Optimal. Leaf size=109 \[ \frac{\log \left (\frac{e (c+d x)}{a+b x}\right ) \text{PolyLog}\left (2,\frac{(e+f x) (b c-a d)}{(a+b x) (d e-c f)}+1\right )}{b c-a d}-\frac{\text{PolyLog}\left (3,\frac{(e+f x) (b c-a d)}{(a+b x) (d e-c f)}+1\right )}{b c-a d} \]

[Out]

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

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Rubi [A]  time = 0.163821, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 62, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.032, Rules used = {2506, 6610} \[ \frac{\log \left (\frac{e (c+d x)}{a+b x}\right ) \text{PolyLog}\left (2,\frac{(e+f x) (b c-a d)}{(a+b x) (d e-c f)}+1\right )}{b c-a d}-\frac{\text{PolyLog}\left (3,\frac{(e+f x) (b c-a d)}{(a+b x) (d e-c f)}+1\right )}{b c-a d} \]

Antiderivative was successfully verified.

[In]

Int[(Log[(e*(c + d*x))/(a + b*x)]*Log[((-(b*c) + a*d)*(e + f*x))/((d*e - c*f)*(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)*(e + f*x))/((d*e - c*f)*(a + b*x))])/(b*c - a*d) - P
olyLog[3, 1 + ((b*c - a*d)*(e + f*x))/((d*e - c*f)*(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{e (c+d x)}{a+b x}\right ) \log \left (\frac{(-b c+a d) (e+f x)}{(d e-c f) (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) (e+f x)}{(d e-c f) (a+b x)}\right )}{b c-a d}+\int \frac{\text{Li}_2\left (1-\frac{(-b c+a d) (e+f x)}{(d e-c f) (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) (e+f x)}{(d e-c f) (a+b x)}\right )}{b c-a d}-\frac{\text{Li}_3\left (1+\frac{(b c-a d) (e+f x)}{(d e-c f) (a+b x)}\right )}{b c-a d}\\ \end{align*}

Mathematica [A]  time = 0.0266982, size = 96, normalized size = 0.88 \[ \frac{\log \left (\frac{e (c+d x)}{a+b x}\right ) \text{PolyLog}\left (2,\frac{(c+d x) (b e-a f)}{(a+b x) (d e-c f)}\right )-\text{PolyLog}\left (3,\frac{(c+d x) (b e-a f)}{(a+b x) (d e-c f)}\right )}{b c-a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 2.268, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({\frac{e \left ( dx+c \right ) }{bx+a}} \right ) \ln \left ({\frac{ \left ( ad-bc \right ) \left ( fx+e \right ) }{ \left ( -cf+de \right ) \left ( bx+a \right ) }} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left (-\frac{{\left (b c - a d\right )} f x +{\left (b c - a d\right )} e}{a d e - a c f +{\left (b d e - b c f\right )} x}\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(e*(d*x+c)/(b*x+a))*log((a*d-b*c)*(f*x+e)/(-c*f+d*e)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="frica
s")

[Out]

integral(log(-((b*c - a*d)*f*x + (b*c - a*d)*e)/(a*d*e - a*c*f + (b*d*e - b*c*f)*x))*log((d*e*x + c*e)/(b*x +
a))/(b*d*x^2 + a*c + (b*c + a*d)*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*(d*x+c)/(b*x+a))*ln((a*d-b*c)*(f*x+e)/(-c*f+d*e)/(b*x+a))/(b*x+a)/(d*x+c),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 (d x + c\right )} e}{b x + a}\right ) \log \left (-\frac{{\left (b c - a d\right )}{\left (f x + e\right )}}{{\left (d e - c f\right )}{\left (b x + a\right )}}\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(e*(d*x+c)/(b*x+a))*log((a*d-b*c)*(f*x+e)/(-c*f+d*e)/(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)*(f*x + e)/((d*e - c*f)*(b*x + a)))/((b*x + a)*(d*x + c))
, x)