3.136 \(\int \frac {\text {Li}_3(c (a+b x))}{x^3} \, dx\)

Optimal. Leaf size=629 \[ \frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {Li}_3(c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_2(c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_2\left (1-\frac {b c x}{1-a c}\right )}{2 a^2}-\frac {b^2 \text {Li}_3\left (-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \text {Li}_3\left (-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac {b^2 \text {Li}_3(1-c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_2\left (-\frac {b x}{a}\right ) \left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )}{2 a^2}-\frac {b^2 \log (x) \text {Li}_2(c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_2\left (-\frac {b x}{a (1-c (a+b x))}\right ) \log \left (-\frac {a (1-c (a+b x))}{b x}\right )}{2 a^2}+\frac {b^2 \text {Li}_2\left (-\frac {b c x}{1-c (a+b x)}\right ) \log \left (-\frac {a (1-c (a+b x))}{b x}\right )}{2 a^2}-\frac {b^2 \text {Li}_2(1-c (a+b x)) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )}{2 a^2}-\frac {b^2 \left (\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac {b^2 \left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )^2}{4 a^2}-\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (-a c-b c x+1)}{2 a^2}-\frac {b^2 \log (x) \log \left (\frac {b x}{a}+1\right ) \log (1-c (a+b x))}{2 a^2}+\frac {b^2 \text {Li}_3\left (-\frac {b x}{a}\right )}{2 a^2}-\frac {b \text {Li}_2(c (a+b x))}{2 a x} \]

[Out]

-1/2*b^2*ln(b*c*x/(-a*c+1))*ln(-b*c*x-a*c+1)/a^2-1/2*b^2*ln(x)*ln(1+b*x/a)*ln(1-c*(b*x+a))/a^2-1/4*b^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/a^2-1/4*b^2*
(ln(c*(b*x+a))-ln(1+b*x/a))*(ln(x)+ln(-a*(1-c*(b*x+a))/b/x))^2/a^2-1/2*b^2*(ln(1-c*(b*x+a))-ln(-a*(1-c*(b*x+a)
)/b/x))*polylog(2,-b*x/a)/a^2-1/2*b^2*polylog(2,c*(b*x+a))/a^2-1/2*b*polylog(2,c*(b*x+a))/a/x-1/2*b^2*ln(x)*po
lylog(2,c*(b*x+a))/a^2-1/2*b^2*polylog(2,1-b*c*x/(-a*c+1))/a^2-1/2*b^2*ln(-a*(1-c*(b*x+a))/b/x)*polylog(2,-b*x
/a/(1-c*(b*x+a)))/a^2+1/2*b^2*ln(-a*(1-c*(b*x+a))/b/x)*polylog(2,-b*c*x/(1-c*(b*x+a)))/a^2-1/2*b^2*(ln(x)+ln(-
a*(1-c*(b*x+a))/b/x))*polylog(2,1-c*(b*x+a))/a^2+1/2*b^2*polylog(3,-b*x/a)/a^2+1/2*(b^2-a^2/x^2)*polylog(3,c*(
b*x+a))/a^2-1/2*b^2*polylog(3,-b*x/a/(1-c*(b*x+a)))/a^2+1/2*b^2*polylog(3,-b*c*x/(1-c*(b*x+a)))/a^2+1/2*b^2*po
lylog(3,1-c*(b*x+a))/a^2

________________________________________________________________________________________

Rubi [A]  time = 0.66, antiderivative size = 629, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 13, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6599, 6598, 36, 29, 31, 2416, 2394, 2315, 2393, 2391, 6597, 2440, 2435} \[ \frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {PolyLog}(3,c (a+b x))}{2 a^2}-\frac {b^2 \text {PolyLog}(2,c (a+b x))}{2 a^2}-\frac {b^2 \text {PolyLog}\left (2,1-\frac {b c x}{1-a c}\right )}{2 a^2}-\frac {b^2 \text {PolyLog}\left (3,-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \text {PolyLog}\left (3,-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac {b^2 \text {PolyLog}(3,1-c (a+b x))}{2 a^2}-\frac {b^2 \text {PolyLog}\left (2,-\frac {b x}{a}\right ) \left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )}{2 a^2}-\frac {b^2 \log (x) \text {PolyLog}(2,c (a+b x))}{2 a^2}-\frac {b^2 \log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {PolyLog}\left (2,-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {PolyLog}\left (2,-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}-\frac {b^2 \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right ) \text {PolyLog}(2,1-c (a+b x))}{2 a^2}+\frac {b^2 \text {PolyLog}\left (3,-\frac {b x}{a}\right )}{2 a^2}-\frac {b \text {PolyLog}(2,c (a+b x))}{2 a x}-\frac {b^2 \left (\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )}{4 a^2}-\frac {b^2 \left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )^2}{4 a^2}-\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (-a c-b c x+1)}{2 a^2}-\frac {b^2 \log (x) \log \left (\frac {b x}{a}+1\right ) \log (1-c (a+b x))}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, c*(a + b*x)]/x^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

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

Rule 2315

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

Rule 2391

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

Rule 2393

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

Rule 2394

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

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]

Rule 6598

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

Rule 6599

Int[(x_)^(m_.)*PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)], x_Symbol] :> -Simp[((a^(m + 1) - b^(m + 1)*x^(m
+ 1))*PolyLog[n, c*(a + b*x)^p])/((m + 1)*b^(m + 1)), x] + Dist[p/((m + 1)*b^m), Int[ExpandIntegrand[PolyLog[n
- 1, c*(a + b*x)^p], (a^(m + 1) - b^(m + 1)*x^(m + 1))/(a + b*x), x], x], x] /; FreeQ[{a, b, c, p}, x] && GtQ
[n, 0] && IntegerQ[m] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\text {Li}_3(c (a+b x))}{x^3} \, dx &=\frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {Li}_3(c (a+b x))}{2 a^2}-\frac {1}{2} b^3 \int \left (-\frac {\text {Li}_2(c (a+b x))}{a b^2 x^2}+\frac {\text {Li}_2(c (a+b x))}{a^2 b x}\right ) \, dx\\ &=\frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {Li}_3(c (a+b x))}{2 a^2}+\frac {b \int \frac {\text {Li}_2(c (a+b x))}{x^2} \, dx}{2 a}-\frac {b^2 \int \frac {\text {Li}_2(c (a+b x))}{x} \, dx}{2 a^2}\\ &=-\frac {b \text {Li}_2(c (a+b x))}{2 a x}-\frac {b^2 \log (x) \text {Li}_2(c (a+b x))}{2 a^2}+\frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {Li}_3(c (a+b x))}{2 a^2}-\frac {b^2 \int \frac {\log (1-a c-b c x)}{x (a+b x)} \, dx}{2 a}-\frac {b^3 \int \frac {\log (x) \log (1-a c-b c x)}{a+b x} \, dx}{2 a^2}\\ &=-\frac {b \text {Li}_2(c (a+b x))}{2 a x}-\frac {b^2 \log (x) \text {Li}_2(c (a+b x))}{2 a^2}+\frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {Li}_3(c (a+b x))}{2 a^2}-\frac {b^2 \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 )}{2 a^2}-\frac {b^2 \int \left (\frac {\log (1-a c-b c x)}{a x}-\frac {b \log (1-a c-b c x)}{a (a+b x)}\right ) \, dx}{2 a}\\ &=-\frac {b^2 \log (x) \log \left (1+\frac {b x}{a}\right ) \log (1-c (a+b x))}{2 a^2}-\frac {b^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 )}{4 a^2}-\frac {b^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}{4 a^2}-\frac {b^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 )}{2 a^2}-\frac {b \text {Li}_2(c (a+b x))}{2 a x}-\frac {b^2 \log (x) \text {Li}_2(c (a+b x))}{2 a^2}-\frac {b^2 \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 )}{2 a^2}+\frac {b^2 \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 )}{2 a^2}-\frac {b^2 \left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \text {Li}_2(1-c (a+b x))}{2 a^2}+\frac {b^2 \text {Li}_3\left (-\frac {b x}{a}\right )}{2 a^2}+\frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {Li}_3(c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_3\left (-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \text {Li}_3\left (-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac {b^2 \text {Li}_3(1-c (a+b x))}{2 a^2}-\frac {b^2 \int \frac {\log (1-a c-b c x)}{x} \, dx}{2 a^2}+\frac {b^3 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{2 a^2}\\ &=-\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}-\frac {b^2 \log (x) \log \left (1+\frac {b x}{a}\right ) \log (1-c (a+b x))}{2 a^2}-\frac {b^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 )}{4 a^2}-\frac {b^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}{4 a^2}-\frac {b^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 )}{2 a^2}-\frac {b \text {Li}_2(c (a+b x))}{2 a x}-\frac {b^2 \log (x) \text {Li}_2(c (a+b x))}{2 a^2}-\frac {b^2 \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 )}{2 a^2}+\frac {b^2 \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 )}{2 a^2}-\frac {b^2 \left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \text {Li}_2(1-c (a+b x))}{2 a^2}+\frac {b^2 \text {Li}_3\left (-\frac {b x}{a}\right )}{2 a^2}+\frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {Li}_3(c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_3\left (-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \text {Li}_3\left (-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac {b^2 \text {Li}_3(1-c (a+b x))}{2 a^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 a^2}-\frac {\left (b^3 c\right ) \int \frac {\log \left (-\frac {b c x}{-1+a c}\right )}{1-a c-b c x} \, dx}{2 a^2}\\ &=-\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}-\frac {b^2 \log (x) \log \left (1+\frac {b x}{a}\right ) \log (1-c (a+b x))}{2 a^2}-\frac {b^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 )}{4 a^2}-\frac {b^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}{4 a^2}-\frac {b^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 )}{2 a^2}-\frac {b^2 \text {Li}_2(c (a+b x))}{2 a^2}-\frac {b \text {Li}_2(c (a+b x))}{2 a x}-\frac {b^2 \log (x) \text {Li}_2(c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_2\left (1-\frac {b c x}{1-a c}\right )}{2 a^2}-\frac {b^2 \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 )}{2 a^2}+\frac {b^2 \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 )}{2 a^2}-\frac {b^2 \left (\log (x)+\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right ) \text {Li}_2(1-c (a+b x))}{2 a^2}+\frac {b^2 \text {Li}_3\left (-\frac {b x}{a}\right )}{2 a^2}+\frac {\left (b^2-\frac {a^2}{x^2}\right ) \text {Li}_3(c (a+b x))}{2 a^2}-\frac {b^2 \text {Li}_3\left (-\frac {b x}{a (1-c (a+b x))}\right )}{2 a^2}+\frac {b^2 \text {Li}_3\left (-\frac {b c x}{1-c (a+b x)}\right )}{2 a^2}+\frac {b^2 \text {Li}_3(1-c (a+b x))}{2 a^2}\\ \end {align*}

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Mathematica [A]  time = 1.73, size = 573, normalized size = 0.91 \[ \frac {\frac {b x \left (b x \left (\text {Li}_2\left (\frac {b c x}{1-a c}\right )+\text {Li}_2(-a c-b x c+1)+\text {Li}_3(c (a+b x))+\text {Li}_3(-a c-b x c+1)-\text {Li}_3\left (\frac {a (a c+b x c-1)}{b x}\right )+\text {Li}_3\left (\frac {a c+b x c-1}{b c x}\right )+\left (\text {Li}_2\left (\frac {a (a c+b x c-1)}{b x}\right )-\text {Li}_2\left (\frac {a c+b x c-1}{b c x}\right )\right ) \log \left (\frac {a (a c+b c x-1)}{b x}\right )-\text {Li}_2\left (-\frac {b x}{a}\right ) \left (\log (-a c-b c x+1)-\log \left (\frac {a (a c+b c x-1)}{b x}\right )\right )-\log (a+b x) \text {Li}_2(c (a+b x))-\text {Li}_2(-a c-b x c+1) \left (\log \left (\frac {a (a c+b c x-1)}{b x}\right )+\log (x)\right )-\frac {1}{2} \left (\log \left (\frac {1-a c}{b c x}\right )-\log \left (\frac {(1-a c) (a+b x)}{b x}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (\frac {a (a c+b c x-1)}{b x}\right )-\left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \log (-a c-b c x+1) \log \left (\frac {a (a c+b c x-1)}{b x}\right )+\log (c (a+b x)) \log (-a c-b c x+1)-\log (x) \log \left (\frac {b x}{a}+1\right ) \log (-a c-b c x+1)+\frac {1}{2} \left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \log (-a c-b c x+1) (\log (-a c-b c x+1)-2 \log (x))-\log (x) \left (\log (-a c-b c x+1)-\log \left (\frac {b c x}{a c-1}+1\right )\right )+\text {Li}_3\left (-\frac {b x}{a}\right )\right )-(-b x \log (a+b x)+a+b x \log (x)) \text {Li}_2(c (a+b x))\right )}{a^2}-\text {Li}_3(c (a+b x))}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, c*(a + b*x)]/x^3,x]

[Out]

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

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fricas [F]  time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm polylog}\left (3, b c x + a c\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,c*(b*x+a))/x^3,x, algorithm="fricas")

[Out]

integral(polylog(3, b*c*x + a*c)/x^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_{3}({\left (b x + a\right )} c)}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,c*(b*x+a))/x^3,x, algorithm="giac")

[Out]

integrate(polylog(3, (b*x + a)*c)/x^3, x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \frac {\polylog \left (3, c \left (b x +a \right )\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,c*(b*x+a))/x^3,x)

[Out]

int(polylog(3,c*(b*x+a))/x^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {{\rm Li}_2\left (b c x + a c\right )}{2 \, {\left (b x^{3} + a x^{2}\right )}}\,{d x} - \frac {{\rm Li}_{3}(b c x + a c)}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,c*(b*x+a))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {polylog}\left (3,c\,\left (a+b\,x\right )\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, c*(a + b*x))/x^3,x)

[Out]

int(polylog(3, c*(a + b*x))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{3}\left (a c + b c x\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,c*(b*x+a))/x**3,x)

[Out]

Integral(polylog(3, a*c + b*c*x)/x**3, x)

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