3.127 \(\int \frac {\text {Li}_2(c (a+b x))}{x} \, dx\)

Optimal. Leaf size=401 \[ \text {Li}_3\left (-\frac {b x}{a (1-c (a+b x))}\right )-\text {Li}_3\left (-\frac {b c x}{1-c (a+b x)}\right )-\text {Li}_3(1-c (a+b x))+\text {Li}_2\left (-\frac {b x}{a (1-c (a+b x))}\right ) \log \left (-\frac {a (1-c (a+b x))}{b x}\right )-\text {Li}_2\left (-\frac {b c x}{1-c (a+b x)}\right ) \log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\text {Li}_2\left (-\frac {b x}{a}\right ) \left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )+\log (x) \text {Li}_2(c (a+b x))+\text {Li}_2(1-c (a+b x)) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )+\frac {1}{2} \left (\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )+\frac {1}{2} \left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )^2+\log (x) \log \left (\frac {b x}{a}+1\right ) \log (1-c (a+b x))-\text {Li}_3\left (-\frac {b x}{a}\right ) \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6597, 2440, 2435} \[ \text {PolyLog}\left (3,-\frac {b x}{a (1-c (a+b x))}\right )-\text {PolyLog}\left (3,-\frac {b c x}{1-c (a+b x)}\right )-\text {PolyLog}(3,1-c (a+b x))+\log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {PolyLog}\left (2,-\frac {b x}{a (1-c (a+b x))}\right )-\log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {PolyLog}\left (2,-\frac {b c x}{1-c (a+b x)}\right )+\text {PolyLog}\left (2,-\frac {b x}{a}\right ) \left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )+\log (x) \text {PolyLog}(2,c (a+b x))+\left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right ) \text {PolyLog}(2,1-c (a+b x))-\text {PolyLog}\left (3,-\frac {b x}{a}\right )+\frac {1}{2} \left (\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )+\frac {1}{2} \left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )^2+\log (x) \log \left (\frac {b x}{a}+1\right ) \log (1-c (a+b x)) \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)]/x,x]

[Out]

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

Rule 2435

Int[(Log[(a_) + (b_.)*(x_)]*Log[(c_) + (d_.)*(x_)])/(x_), x_Symbol] :> Simp[Log[-((b*x)/a)]*Log[a + b*x]*Log[c
+ d*x], x] + (Simp[(1*(Log[-((b*x)/a)] - Log[-(((b*c - a*d)*x)/(a*(c + d*x)))] + Log[(b*c - a*d)/(b*(c + d*x)
)])*Log[(a*(c + d*x))/(c*(a + b*x))]^2)/2, x] - Simp[(1*(Log[-((b*x)/a)] - Log[-((d*x)/c)])*(Log[a + b*x] + Lo
g[(a*(c + d*x))/(c*(a + b*x))])^2)/2, x] + Simp[(Log[c + d*x] - Log[(a*(c + d*x))/(c*(a + b*x))])*PolyLog[2, 1
+ (b*x)/a], x] + Simp[(Log[a + b*x] + Log[(a*(c + d*x))/(c*(a + b*x))])*PolyLog[2, 1 + (d*x)/c], x] + Simp[Lo
g[(a*(c + d*x))/(c*(a + b*x))]*PolyLog[2, (c*(a + b*x))/(a*(c + d*x))], x] - Simp[Log[(a*(c + d*x))/(c*(a + b*
x))]*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))], x] - Simp[PolyLog[3, 1 + (b*x)/a], x] - Simp[PolyLog[3, 1 + (d*x
)/c], x] + Simp[PolyLog[3, (c*(a + b*x))/(a*(c + d*x))], x] - Simp[PolyLog[3, (d*(a + b*x))/(b*(c + d*x))], x]
) /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*(g_.))
*((k_) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/l, Subst[Int[x^r*(a + b*Log[c*(-((e*k - d*l)/l) + (e*x)/l)^n])
*(f + g*Log[h*(-((j*k - i*l)/l) + (j*x)/l)^m]), x], x, k + l*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k,
l, m, n}, x] && IntegerQ[r]

Rule 6597

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

Rubi steps

\begin {align*} \int \frac {\text {Li}_2(c (a+b x))}{x} \, dx &=\log (x) \text {Li}_2(c (a+b x))+b \int \frac {\log (x) \log (1-a c-b c x)}{a+b x} \, dx\\ &=\log (x) \text {Li}_2(c (a+b x))+\operatorname {Subst}\left (\int \frac {\log \left (-\frac {a}{b}+\frac {x}{b}\right ) \log \left (-\frac {-a b c-b (1-a c)}{b}-c x\right )}{x} \, dx,x,a+b x\right )\\ &=\log (x) \log \left (1+\frac {b x}{a}\right ) \log (1-c (a+b x))+\frac {1}{2} \left (\log \left (1+\frac {b x}{a}\right )+\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )-\frac {1}{2} \left (-\log (c (a+b x))+\log \left (1+\frac {b x}{a}\right )\right ) \left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )^2+\left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \text {Li}_2\left (-\frac {b x}{a}\right )+\log (x) \text {Li}_2(c (a+b x))+\log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {Li}_2\left (-\frac {b x}{a (1-c (a+b x))}\right )-\log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {Li}_2\left (-\frac {b c x}{1-c (a+b x)}\right )+\left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \text {Li}_2(1-c (a+b x))-\text {Li}_3\left (-\frac {b x}{a}\right )+\text {Li}_3\left (-\frac {b x}{a (1-c (a+b x))}\right )-\text {Li}_3\left (-\frac {b c x}{1-c (a+b x)}\right )-\text {Li}_3(1-c (a+b x))\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 422, normalized size = 1.05 \[ -\text {Li}_3(-a c-b x c+1)+\text {Li}_3\left (\frac {a (a c+b x c-1)}{b x}\right )-\text {Li}_3\left (\frac {a c+b x c-1}{b c x}\right )+\left (\text {Li}_2\left (\frac {a c+b x c-1}{b c x}\right )-\text {Li}_2\left (\frac {a (a c+b x c-1)}{b x}\right )\right ) \log \left (\frac {a (a c+b c x-1)}{b x}\right )+\text {Li}_2\left (-\frac {b x}{a}\right ) \left (\log (-a c-b c x+1)-\log \left (\frac {a (a c+b c x-1)}{b x}\right )\right )+\text {Li}_2(-a c-b x c+1) \left (\log \left (\frac {a (a c+b c x-1)}{b x}\right )+\log (x)\right )+\log (x) \text {Li}_2(a c+b x c)+\frac {1}{2} \left (\log \left (\frac {1-a c}{b c x}\right )-\log \left (\frac {(1-a c) (a+b x)}{b x}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (\frac {a (a c+b c x-1)}{b x}\right )+\left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \log (-a c-b c x+1) \log \left (\frac {a (a c+b c x-1)}{b x}\right )+\log (x) \log \left (\frac {b x}{a}+1\right ) \log (-a c-b c x+1)+\frac {1}{2} \left (\log \left (\frac {b x}{a}+1\right )-\log (c (a+b x))\right ) \log (-a c-b c x+1) (\log (-a c-b c x+1)-2 \log (x))-\text {Li}_3\left (-\frac {b x}{a}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm Li}_2\left (b c x + a c\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(dilog(b*c*x + a*c)/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog((b*x + a)*c)/x, x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {\polylog \left (2, c \left (b x +a \right )\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog((b*x + a)*c)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, c*(a + b*x))/x,x)

[Out]

int(polylog(2, c*(a + b*x))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{2}\left (a c + b c x\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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