3.133 \(\int \text {Li}_3(c (a+b x)) \, dx\)

Optimal. Leaf size=84 \[ x (-\text {Li}_2(c (a+b x)))+x \text {Li}_3(c (a+b x))-\frac {a \text {Li}_2(c (a+b x))}{b}+\frac {a \text {Li}_3(c (a+b x))}{b}+\frac {(-a c-b c x+1) \log (-a c-b c x+1)}{b c}+x \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.889, Rules used = {6595, 2444, 2389, 2295, 2421, 2393, 2391, 6589} \[ x (-\text {PolyLog}(2,c (a+b x)))+x \text {PolyLog}(3,c (a+b x))-\frac {a \text {PolyLog}(2,c (a+b x))}{b}+\frac {a \text {PolyLog}(3,c (a+b x))}{b}+\frac {(-a c-b c x+1) \log (-a c-b c x+1)}{b c}+x \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, c*(a + b*x)],x]

[Out]

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

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

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 2421

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(Binomial
MatchQ[u, x] && LinearMatchQ[v, x])

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

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 6595

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 66, normalized size = 0.79 \[ \frac {(a+b x) \left (-\text {Li}_2(c (a+b x))+\text {Li}_3(c (a+b x))+\frac {\log (1-c (a+b x))}{c (a+b x)}-\log (1-c (a+b x))+1\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, c*(a + b*x)],x]

[Out]

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

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fricas [C]  time = 0.98, size = 73, normalized size = 0.87 \[ \frac {b c x - {\left (b c x + a c\right )} {\rm Li}_2\left (b c x + a c\right ) - {\left (b c x + a c - 1\right )} \log \left (-b c x - a c + 1\right ) + {\left (b c x + a c\right )} {\rm polylog}\left (3, b c x + a c\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c*x - (b*c*x + a*c)*dilog(b*c*x + a*c) - (b*c*x + a*c - 1)*log(-b*c*x - a*c + 1) + (b*c*x + a*c)*polylog(3,
b*c*x + a*c))/(b*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(polylog(3, (b*x + a)*c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 120, normalized size = 1.43 \[ \frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} a}{b} + \frac {a {\rm Li}_{3}(b c x + a c)}{b} - \frac {b c x {\rm Li}_2\left (b c x + a c\right ) - b c x {\rm Li}_{3}(b c x + a c) - b c x + {\left (b c x + a c - 1\right )} \log \left (-b c x - a c + 1\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a/b + a*polylog(3, b*c*x + a*c)/b - (b*c*x*
dilog(b*c*x + a*c) - b*c*x*polylog(3, b*c*x + a*c) - b*c*x + (b*c*x + a*c - 1)*log(-b*c*x - a*c + 1))/(b*c)

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mupad [B]  time = 2.18, size = 77, normalized size = 0.92 \[ x-\frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b}+\frac {\mathrm {polylog}\left (3,c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b}+\frac {\ln \left (c\,\left (a+b\,x\right )-1\right )}{b\,c}-\frac {\ln \left (1-c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, c*(a + b*x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(polylog(3, c*(a + b*x)), x)

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