3.135 \(\int \frac {\text {Li}_3(c (a+b x))}{x^2} \, dx\)

Optimal. Leaf size=486 \[ -\frac {2 b \text {Li}_3(c (a+b x))}{a}+\frac {\left (b-\frac {a}{x}\right ) \text {Li}_3(c (a+b x))}{a}+\frac {b \text {Li}_3\left (-\frac {b x}{a (1-c (a+b x))}\right )}{a}-\frac {b \text {Li}_3\left (-\frac {b c x}{1-c (a+b x)}\right )}{a}-\frac {b \text {Li}_3(1-c (a+b x))}{a}+\frac {b \text {Li}_2\left (-\frac {b x}{a (1-c (a+b x))}\right ) \log \left (-\frac {a (1-c (a+b x))}{b x}\right )}{a}-\frac {b \text {Li}_2\left (-\frac {b c x}{1-c (a+b x)}\right ) \log \left (-\frac {a (1-c (a+b x))}{b x}\right )}{a}+\frac {b \text {Li}_2\left (-\frac {b x}{a}\right ) \left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )}{a}+\frac {b \log (x) \text {Li}_2(c (a+b x))}{a}+\frac {b \text {Li}_2(1-c (a+b x)) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )}{a}+\frac {b \left (\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )}{2 a}+\frac {b \left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )^2}{2 a}+\frac {b \log (x) \log \left (\frac {b x}{a}+1\right ) \log (1-c (a+b x))}{a}-\frac {b \text {Li}_3\left (-\frac {b x}{a}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.56, antiderivative size = 486, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6599, 6597, 2440, 2435, 6589} \[ -\frac {2 b \text {PolyLog}(3,c (a+b x))}{a}+\frac {\left (b-\frac {a}{x}\right ) \text {PolyLog}(3,c (a+b x))}{a}+\frac {b \text {PolyLog}\left (3,-\frac {b x}{a (1-c (a+b x))}\right )}{a}-\frac {b \text {PolyLog}\left (3,-\frac {b c x}{1-c (a+b x)}\right )}{a}-\frac {b \text {PolyLog}(3,1-c (a+b x))}{a}+\frac {b \log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {PolyLog}\left (2,-\frac {b x}{a (1-c (a+b x))}\right )}{a}-\frac {b \log \left (-\frac {a (1-c (a+b x))}{b x}\right ) \text {PolyLog}\left (2,-\frac {b c x}{1-c (a+b x)}\right )}{a}+\frac {b \text {PolyLog}\left (2,-\frac {b x}{a}\right ) \left (\log (1-c (a+b x))-\log \left (-\frac {a (1-c (a+b x))}{b x}\right )\right )}{a}+\frac {b \log (x) \text {PolyLog}(2,c (a+b x))}{a}+\frac {b \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right ) \text {PolyLog}(2,1-c (a+b x))}{a}-\frac {b \text {PolyLog}\left (3,-\frac {b x}{a}\right )}{a}+\frac {b \left (\log \left (\frac {1-a c}{1-c (a+b x)}\right )-\log \left (\frac {(1-a c) (a+b x)}{a (1-c (a+b x))}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (-\frac {a (1-c (a+b x))}{b x}\right )}{2 a}+\frac {b \left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \left (\log \left (-\frac {a (1-c (a+b x))}{b x}\right )+\log (x)\right )^2}{2 a}+\frac {b \log (x) \log \left (\frac {b x}{a}+1\right ) \log (1-c (a+b x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, c*(a + b*x)]/x^2,x]

[Out]

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

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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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

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Mathematica [A]  time = 0.70, size = 477, normalized size = 0.98 \[ \frac {b \left (-\text {Li}_3(c (a+b x))-\text {Li}_3(-a c-b x c+1)+\text {Li}_3\left (\frac {a (a c+b x c-1)}{b x}\right )-\text {Li}_3\left (\frac {a c+b x c-1}{b c x}\right )+\left (\text {Li}_2\left (\frac {a c+b x c-1}{b c x}\right )-\text {Li}_2\left (\frac {a (a c+b x c-1)}{b x}\right )\right ) \log \left (\frac {a (a c+b c x-1)}{b x}\right )+\text {Li}_2\left (-\frac {b x}{a}\right ) \left (\log (-a c-b c x+1)-\log \left (\frac {a (a c+b c x-1)}{b x}\right )\right )+(\log (x)-\log (a+b x)) \text {Li}_2(c (a+b x))+\log (a+b x) \text {Li}_2(c (a+b x))+\text {Li}_2(-a c-b x c+1) \left (\log \left (\frac {a (a c+b c x-1)}{b x}\right )+\log (x)\right )+\frac {1}{2} \left (\log \left (\frac {1-a c}{b c x}\right )-\log \left (\frac {(1-a c) (a+b x)}{b x}\right )+\log \left (\frac {b x}{a}+1\right )\right ) \log ^2\left (\frac {a (a c+b c x-1)}{b x}\right )+\left (\log (c (a+b x))-\log \left (\frac {b x}{a}+1\right )\right ) \log (-a c-b c x+1) \log \left (\frac {a (a c+b c x-1)}{b x}\right )+\log (x) \log \left (\frac {b x}{a}+1\right ) \log (-a c-b c x+1)+\frac {1}{2} \left (\log \left (\frac {b x}{a}+1\right )-\log (c (a+b x))\right ) \log (-a c-b c x+1) (\log (-a c-b c x+1)-2 \log (x))-\text {Li}_3\left (-\frac {b x}{a}\right )\right )}{a}-\frac {\text {Li}_3(c (a+b x))}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(PolyLog[3, c*(a + b*x)]/x) + (b*(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 + b*x])*PolyLog[2, c*(a + b*x)] + Log[a + b*x]*Poly
Log[2, c*(a + b*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)]) - Po
lyLog[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

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm polylog}\left (3, b c x + a c\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,c*(b*x+a))/x^2,x, algorithm="fricas")

[Out]

integral(polylog(3, b*c*x + a*c)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,c*(b*x+a))/x^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \frac {\polylog \left (3, c \left (b x +a \right )\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,c*(b*x+a))/x^2,x)

[Out]

int(polylog(3,c*(b*x+a))/x^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,c*(b*x+a))/x^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {polylog}\left (3,c\,\left (a+b\,x\right )\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, c*(a + b*x))/x^2,x)

[Out]

int(polylog(3, c*(a + b*x))/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{3}\left (a c + b c x\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,c*(b*x+a))/x**2,x)

[Out]

Integral(polylog(3, a*c + b*c*x)/x**2, x)

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