3.2.3 \(\int \frac {\log (\frac {-b c+a d}{d (a+b x)}) \log (\frac {e (c+d x)}{a+b x})}{(a+b x) (c+d x)} \, dx\) [103]

Optimal. Leaf size=85 \[ \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \text {Li}_2\left (1+\frac {b c-a d}{d (a+b x)}\right )}{b c-a d}-\frac {\text {Li}_3\left (1+\frac {b c-a d}{d (a+b x)}\right )}{b c-a d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2588, 6745} \begin {gather*} \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} \end {gather*}

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 2588

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)*PolyL
og[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]*(L
og[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 6745

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*}

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Mathematica [A]
time = 0.02, size = 68, normalized size = 0.80 \begin {gather*} \frac {\log \left (\frac {e (c+d x)}{a+b x}\right ) \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )-\text {Li}_3\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b c-a d} \end {gather*}

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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs. \(2(85)=170\).
time = 3.71, size = 186, normalized size = 2.19

method	result	size
default	\(\frac {\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )^{2} \ln \left (-\frac {\frac {e \left (d x +c \right ) b}{b x +a}-e d}{e d}\right )}{2 a d -2 c b}-\frac {\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )^{2} \ln \left (1-\frac {b \left (d x +c \right )}{d \left (b x +a \right )}\right )}{2 \left (a d -c b \right )}-\frac {\ln \left (\frac {e \left (d x +c \right )}{b x +a}\right ) \polylog \left (2, \frac {b \left (d x +c \right )}{d \left (b x +a \right )}\right )}{a d -c b}+\frac {\polylog \left (3, \frac {b \left (d x +c \right )}{d \left (b x +a \right )}\right )}{a d -c b}\)	\(186\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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