3.90 \(\int \frac{\log (\frac{a}{a+b x}) \log ^2(\frac{c x}{a+b x})}{x (a+b x)} \, dx\)

Optimal. Leaf size=82 \[ -\frac{\text{PolyLog}\left (2,1-\frac{a}{a+b x}\right ) \log ^2\left (\frac{c x}{a+b x}\right )}{a}+\frac{2 \text{PolyLog}\left (3,1-\frac{a}{a+b x}\right ) \log \left (\frac{c x}{a+b x}\right )}{a}-\frac{2 \text{PolyLog}\left (4,1-\frac{a}{a+b x}\right )}{a} \]

[Out]

-((Log[(c*x)/(a + b*x)]^2*PolyLog[2, 1 - a/(a + b*x)])/a) + (2*Log[(c*x)/(a + b*x)]*PolyLog[3, 1 - a/(a + b*x)
])/a - (2*PolyLog[4, 1 - a/(a + b*x)])/a

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Rubi [A]  time = 0.168258, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {2506, 2508, 6610} \[ -\frac{\text{PolyLog}\left (2,1-\frac{a}{a+b x}\right ) \log ^2\left (\frac{c x}{a+b x}\right )}{a}+\frac{2 \text{PolyLog}\left (3,1-\frac{a}{a+b x}\right ) \log \left (\frac{c x}{a+b x}\right )}{a}-\frac{2 \text{PolyLog}\left (4,1-\frac{a}{a+b x}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Log[(c*x)/(a + b*x)]^2*PolyLog[2, 1 - a/(a + b*x)])/a) + (2*Log[(c*x)/(a + b*x)]*PolyLog[3, 1 - a/(a + b*x)
])/a - (2*PolyLog[4, 1 - a/(a + b*x)])/a

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 2508

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(u_)*PolyLog[n_, v_],
x_Symbol] :> With[{g = Simplify[(v*(c + d*x))/(a + b*x)], h = Simplify[u*(a + b*x)*(c + d*x)]}, Simp[(h*PolyL
og[n + 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[n + 1, v]*Lo
g[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, n, 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{a}{a+b x}\right ) \log ^2\left (\frac{c x}{a+b x}\right )}{x (a+b x)} \, dx &=-\frac{\log ^2\left (\frac{c x}{a+b x}\right ) \text{Li}_2\left (1-\frac{a}{a+b x}\right )}{a}+2 \int \frac{\log \left (\frac{c x}{a+b x}\right ) \text{Li}_2\left (1-\frac{a}{a+b x}\right )}{x (a+b x)} \, dx\\ &=-\frac{\log ^2\left (\frac{c x}{a+b x}\right ) \text{Li}_2\left (1-\frac{a}{a+b x}\right )}{a}+\frac{2 \log \left (\frac{c x}{a+b x}\right ) \text{Li}_3\left (1-\frac{a}{a+b x}\right )}{a}-2 \int \frac{\text{Li}_3\left (1-\frac{a}{a+b x}\right )}{x (a+b x)} \, dx\\ &=-\frac{\log ^2\left (\frac{c x}{a+b x}\right ) \text{Li}_2\left (1-\frac{a}{a+b x}\right )}{a}+\frac{2 \log \left (\frac{c x}{a+b x}\right ) \text{Li}_3\left (1-\frac{a}{a+b x}\right )}{a}-\frac{2 \text{Li}_4\left (1-\frac{a}{a+b x}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.013578, size = 76, normalized size = 0.93 \[ -\frac{\text{PolyLog}\left (2,\frac{b x}{a+b x}\right ) \log ^2\left (\frac{c x}{a+b x}\right )}{a}+\frac{2 \text{PolyLog}\left (3,\frac{b x}{a+b x}\right ) \log \left (\frac{c x}{a+b x}\right )}{a}-\frac{2 \text{PolyLog}\left (4,\frac{b x}{a+b x}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Log[(c*x)/(a + b*x)]^2*PolyLog[2, (b*x)/(a + b*x)])/a) + (2*Log[(c*x)/(a + b*x)]*PolyLog[3, (b*x)/(a + b*x)
])/a - (2*PolyLog[4, (b*x)/(a + b*x)])/a

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Maple [F]  time = 0.531, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x \left ( bx+a \right ) }\ln \left ({\frac{a}{bx+a}} \right ) \left ( \ln \left ({\frac{cx}{bx+a}} \right ) \right ) ^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a/(b*x+a))*ln(c*x/(b*x+a))^2/x/(b*x+a),x)

[Out]

int(ln(a/(b*x+a))*ln(c*x/(b*x+a))^2/x/(b*x+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\log \left (b x + a\right )^{4} - 4 \, \log \left (b x + a\right )^{3} \log \left (x\right )}{4 \, a} + \int \frac{a \log \left (a\right ) \log \left (c\right )^{2} + 2 \, a \log \left (a\right ) \log \left (c\right ) \log \left (x\right ) + a \log \left (a\right ) \log \left (x\right )^{2} +{\left (a{\left (\log \left (a\right ) + 2 \, \log \left (c\right )\right )} +{\left (3 \, b x + 2 \, a\right )} \log \left (x\right )\right )} \log \left (b x + a\right )^{2} -{\left (2 \, a{\left (\log \left (a\right ) + \log \left (c\right )\right )} \log \left (x\right ) + a \log \left (x\right )^{2} +{\left (2 \, \log \left (a\right ) \log \left (c\right ) + \log \left (c\right )^{2}\right )} a\right )} \log \left (b x + a\right )}{a b x^{2} + a^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a/(b*x+a))*log(c*x/(b*x+a))^2/x/(b*x+a),x, algorithm="maxima")

[Out]

1/4*(log(b*x + a)^4 - 4*log(b*x + a)^3*log(x))/a + integrate((a*log(a)*log(c)^2 + 2*a*log(a)*log(c)*log(x) + a
*log(a)*log(x)^2 + (a*(log(a) + 2*log(c)) + (3*b*x + 2*a)*log(x))*log(b*x + a)^2 - (2*a*(log(a) + log(c))*log(
x) + a*log(x)^2 + (2*log(a)*log(c) + log(c)^2)*a)*log(b*x + a))/(a*b*x^2 + 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{c x}{b x + a}\right )^{2} \log \left (\frac{a}{b x + a}\right )}{b x^{2} + a x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a/(b*x+a))*log(c*x/(b*x+a))^2/x/(b*x+a),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(a/(b*x+a))*ln(c*x/(b*x+a))**2/x/(b*x+a),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (\frac{c x}{b x + a}\right )^{2} \log \left (\frac{a}{b x + a}\right )}{{\left (b x + a\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a/(b*x+a))*log(c*x/(b*x+a))^2/x/(b*x+a),x, algorithm="giac")

[Out]

integrate(log(c*x/(b*x + a))^2*log(a/(b*x + a))/((b*x + a)*x), x)