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

[Out]

ln(e*(d*x+c)/(b*x+a))*polylog(2,1+(-a*d+b*c)*(f*x+e)/(-c*f+d*e)/(b*x+a))/(-a*d+b*c)-polylog(3,1+(-a*d+b*c)*(f*
x+e)/(-c*f+d*e)/(b*x+a))/(-a*d+b*c)

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Rubi [A]  time = 0.16, antiderivative size = 109, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.03, size = 96, normalized size = 0.88 \[ \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \text {Li}_2\left (\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )-\text {Li}_3\left (\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{b c-a d} \]

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

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

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)

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maple [F]  time = 2.77, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (\frac {\left (d x +c \right ) e}{b x +a}\right ) \ln \left (\frac {\left (a d -b c \right ) \left (f x +e \right )}{\left (-c f +d e \right ) \left (b x +a \right )}\right )}{\left (b x +a \right ) \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln((d*x+c)/(b*x+a)*e)*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((d*x+c)/(b*x+a)*e)*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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (\log \left (b x + a\right )^{2} - 2 \, {\left (\log \left (b x + a\right ) - \log \relax (e)\right )} \log \left (d x + c\right ) + \log \left (d x + c\right )^{2} - 2 \, \log \left (b x + a\right ) \log \relax (e)\right )} \log \left (f x + e\right )}{2 \, {\left (b c - a d\right )}} + \int \frac {2 \, {\left (e \log \left (-b c + a d\right ) \log \relax (e) - e \log \left (d e - c f\right ) \log \relax (e)\right )} b c + {\left (b d f x^{2} + 2 \, b c e - {\left (2 \, d e - c f\right )} a + {\left (3 \, b c f - a d f\right )} x\right )} \log \left (b x + a\right )^{2} - 2 \, {\left (d e \log \left (-b c + a d\right ) \log \relax (e) - d e \log \left (d e - c f\right ) \log \relax (e)\right )} a + 2 \, {\left ({\left (f \log \left (-b c + a d\right ) \log \relax (e) - f \log \left (d e - c f\right ) \log \relax (e)\right )} b c - {\left (d f \log \left (-b c + a d\right ) \log \relax (e) - d f \log \left (d e - c f\right ) \log \relax (e)\right )} a\right )} x - 2 \, {\left (b d f x^{2} \log \relax (e) - {\left (e {\left (\log \left (d e - c f\right ) - \log \relax (e)\right )} - e \log \left (-b c + a d\right )\right )} b c + {\left (d e {\left (\log \left (d e - c f\right ) - \log \relax (e)\right )} - d e \log \left (-b c + a d\right ) + c f \log \relax (e)\right )} a + {\left ({\left (f \log \left (-b c + a d\right ) - f \log \left (d e - c f\right ) + 2 \, f \log \relax (e)\right )} b c - {\left (d f \log \left (-b c + a d\right ) - d f \log \left (d e - c f\right )\right )} a\right )} x\right )} \log \left (b x + a\right ) + 2 \, {\left (b d f x^{2} \log \relax (e) + {\left (e \log \left (-b c + a d\right ) - e \log \left (d e - c f\right )\right )} b c - {\left (d e \log \left (-b c + a d\right ) - d e \log \left (d e - c f\right ) - c f \log \relax (e)\right )} a + {\left ({\left (f \log \left (-b c + a d\right ) - f \log \left (d e - c f\right ) + f \log \relax (e)\right )} b c - {\left (d f \log \left (-b c + a d\right ) - {\left (f \log \left (d e - c f\right ) + f \log \relax (e)\right )} d\right )} a\right )} x - {\left (b d f x^{2} + 2 \, b c f x + b c e - {\left (d e - c f\right )} a\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{2 \, {\left (a b c^{2} e - a^{2} c d e + {\left (b^{2} c d f - a b d^{2} f\right )} x^{3} - {\left (a b d^{2} e + a^{2} d^{2} f - {\left (c d e + c^{2} f\right )} b^{2}\right )} x^{2} + {\left (b^{2} c^{2} e + a b c^{2} f - {\left (d^{2} e + c d f\right )} a^{2}\right )} x\right )}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

log(e*(c + d*x)/(a + b*x))**2*log((e + f*x)*(a*d - b*c)/((a + b*x)*(-c*f + d*e)))/(2*a*d - 2*b*c) - (a*f - b*e
)*Integral(log(c*e/(a + b*x) + d*e*x/(a + b*x))**2/(a*e + a*f*x + b*e*x + b*f*x**2), x)/(2*(a*d - b*c))

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