3.1.88 \(\int \frac {\log (\frac {c x}{a+b x})}{a+b x} \, dx\) [88]

Optimal. Leaf size=46 \[ -\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}-\frac {\text {Li}_2\left (1-\frac {a}{a+b x}\right )}{b} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2544, 2458, 2378, 2370, 2352} \begin {gather*} -\frac {\text {PolyLog}\left (2,1-\frac {a}{a+b x}\right )}{b}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[(c*x)/(a + b*x)]/(a + b*x),x]

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2370

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

Rule 2378

Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Dist[1/n, Subst[Int[(a
+ b*Log[c*x])/(x*(d + e*x^(r/n))), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IntegerQ[r/n]

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2544

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))/((f_.) + (g_.)*(x_)), x_S
ymbol] :> Simp[(-Log[(b*c - a*d)/(b*(c + d*x))])*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])/g), x] + Dist[B*n*(
(b*c - a*d)/g), Int[Log[(b*c - a*d)/(b*(c + d*x))]/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g
, A, B, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && EqQ[d*f - c*g, 0]

Rubi steps

\begin {align*} \int \frac {\log \left (\frac {c x}{a+b x}\right )}{a+b x} \, dx &=-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}+\frac {a \int \frac {\log \left (\frac {a}{a+b x}\right )}{x (a+b x)} \, dx}{b}\\ &=-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}+\frac {a \text {Subst}\left (\int \frac {\log \left (\frac {a}{x}\right )}{x \left (-\frac {a}{b}+\frac {x}{b}\right )} \, dx,x,a+b x\right )}{b^2}\\ &=-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}-\frac {a \text {Subst}\left (\int \frac {\log (a x)}{\left (-\frac {a}{b}+\frac {1}{b x}\right ) x} \, dx,x,\frac {1}{a+b x}\right )}{b^2}\\ &=-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}-\frac {a \text {Subst}\left (\int \frac {\log (a x)}{\frac {1}{b}-\frac {a x}{b}} \, dx,x,\frac {1}{a+b x}\right )}{b^2}\\ &=-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}-\frac {\text {Li}_2\left (\frac {b x}{a+b x}\right )}{b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 84, normalized size = 1.83 \begin {gather*} \frac {\log \left (-\frac {b x}{a}\right ) \log \left (\frac {a}{a+b x}\right )}{b}+\frac {\log ^2\left (\frac {a}{a+b x}\right )}{2 b}-\frac {\log \left (\frac {a}{a+b x}\right ) \log \left (\frac {c x}{a+b x}\right )}{b}-\frac {\text {Li}_2\left (\frac {a+b x}{a}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(96\) vs. \(2(46)=92\).
time = 1.66, size = 97, normalized size = 2.11

method	result	size
derivativedivides	\(-\frac {\dilog \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}-\frac {\ln \left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) \ln \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}\)	\(97\)
default	\(-\frac {\dilog \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}-\frac {\ln \left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) \ln \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}\)	\(97\)
risch	\(-\frac {\dilog \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}-\frac {\ln \left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) \ln \left (-\frac {\left (\frac {c}{b}-\frac {a c}{b \left (b x +a \right )}\right ) b -c}{c}\right )}{b}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*x/(b*x+a))/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-dilog(-((c/b-a*c/b/(b*x+a))*b-c)/c)/b-ln(c/b-a*c/b/(b*x+a))*ln(-((c/b-a*c/b/(b*x+a))*b-c)/c)/b

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (45) = 90\).
time = 0.27, size = 95, normalized size = 2.07 \begin {gather*} \frac {\log \left (b x + a\right ) \log \left (\frac {c x}{b x + a}\right )}{b} - \frac {\frac {c \log \left (b x + a\right )^{2}}{b} - \frac {2 \, {\left (\log \left (\frac {b x}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a}\right )\right )} c}{b}}{2 \, c} + \frac {{\left (c \log \left (b x + a\right ) - c \log \left (x\right )\right )} \log \left (b x + a\right )}{b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*x/(b*x+a))/(b*x+a),x, algorithm="maxima")

[Out]

log(b*x + a)*log(c*x/(b*x + a))/b - 1/2*(c*log(b*x + a)^2/b - 2*(log(b*x/a + 1)*log(x) + dilog(-b*x/a))*c/b)/c
+ (c*log(b*x + a) - c*log(x))*log(b*x + a)/(b*c)

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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(c*x/(b*x+a))/(b*x+a),x, algorithm="fricas")

[Out]

integral(log(c*x/(b*x + a))/(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*x/(b*x+a))/(b*x+a),x)

[Out]

Integral(log(c*x/(a + b*x))/(a + b*x), x)

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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(c*x/(b*x+a))/(b*x+a),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\ln \left (\frac {c\,x}{a+b\,x}\right )}{a+b\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((c*x)/(a + b*x))/(a + b*x),x)

[Out]

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

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