3.2.32 \(\int x \text {PolyLog}(3,c (a+b x)) \, dx\) [132]

Optimal. Leaf size=198 \[ -\frac {3 a x}{4 b}+\frac {(1-a c) x}{8 b c}+\frac {x^2}{16}+\frac {(1-a c)^2 \log (1-a c-b c x)}{8 b^2 c^2}-\frac {1}{8} x^2 \log (1-a c-b c x)-\frac {3 a (1-a c-b c x) \log (1-a c-b c x)}{4 b^2 c}+\frac {3 a^2 \text {PolyLog}(2,c (a+b x))}{4 b^2}+\frac {a x \text {PolyLog}(2,c (a+b x))}{2 b}-\frac {1}{4} x^2 \text {PolyLog}(2,c (a+b x))-\frac {\left (a^2-b^2 x^2\right ) \text {PolyLog}(3,c (a+b x))}{2 b^2} \]

[Out]

-3/4*a*x/b+1/8*(-a*c+1)*x/b/c+1/16*x^2+1/8*(-a*c+1)^2*ln(-b*c*x-a*c+1)/b^2/c^2-1/8*x^2*ln(-b*c*x-a*c+1)-3/4*a*
(-b*c*x-a*c+1)*ln(-b*c*x-a*c+1)/b^2/c+3/4*a^2*polylog(2,c*(b*x+a))/b^2+1/2*a*x*polylog(2,c*(b*x+a))/b-1/4*x^2*
polylog(2,c*(b*x+a))-1/2*(-b^2*x^2+a^2)*polylog(3,c*(b*x+a))/b^2

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Rubi [A]
time = 0.19, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 12, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.091, Rules used = {6734, 6730, 2494, 2436, 2332, 2468, 2440, 2438, 6733, 45, 2463, 2442} \begin {gather*} -\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}+\frac {3 a^2 \text {Li}_2(c (a+b x))}{4 b^2}+\frac {(1-a c)^2 \log (-a c-b c x+1)}{8 b^2 c^2}-\frac {3 a (-a c-b c x+1) \log (-a c-b c x+1)}{4 b^2 c}-\frac {1}{4} x^2 \text {Li}_2(c (a+b x))+\frac {a x \text {Li}_2(c (a+b x))}{2 b}-\frac {1}{8} x^2 \log (-a c-b c x+1)+\frac {x (1-a c)}{8 b c}-\frac {3 a x}{4 b}+\frac {x^2}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[3, c*(a + b*x)],x]

[Out]

(-3*a*x)/(4*b) + ((1 - a*c)*x)/(8*b*c) + x^2/16 + ((1 - a*c)^2*Log[1 - a*c - b*c*x])/(8*b^2*c^2) - (x^2*Log[1
- a*c - b*c*x])/8 - (3*a*(1 - a*c - b*c*x)*Log[1 - a*c - b*c*x])/(4*b^2*c) + (3*a^2*PolyLog[2, c*(a + b*x)])/(
4*b^2) + (a*x*PolyLog[2, c*(a + b*x)])/(2*b) - (x^2*PolyLog[2, c*(a + b*x)])/4 - ((a^2 - b^2*x^2)*PolyLog[3, c
*(a + b*x)])/(2*b^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2332

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

Rule 2436

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 2438

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 2440

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 2442

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

Rule 2463

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 2468

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 2494

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 6730

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]

Rule 6733

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 6734

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] && G
tQ[n, 0] && IntegerQ[m] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \text {Li}_3(c (a+b x)) \, dx &=-\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}+\frac {\int (a \text {Li}_2(c (a+b x))-b x \text {Li}_2(c (a+b x))) \, dx}{2 b}\\ &=-\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}-\frac {1}{2} \int x \text {Li}_2(c (a+b x)) \, dx+\frac {a \int \text {Li}_2(c (a+b x)) \, dx}{2 b}\\ &=\frac {a x \text {Li}_2(c (a+b x))}{2 b}-\frac {1}{4} x^2 \text {Li}_2(c (a+b x))-\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}+\frac {a \int \log (1-c (a+b x)) \, dx}{2 b}-\frac {a^2 \int \frac {\log (1-c (a+b x))}{a+b x} \, dx}{2 b}-\frac {1}{4} b \int \frac {x^2 \log (1-a c-b c x)}{a+b x} \, dx\\ &=\frac {a x \text {Li}_2(c (a+b x))}{2 b}-\frac {1}{4} x^2 \text {Li}_2(c (a+b x))-\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}+\frac {a \int \log (1-a c-b c x) \, dx}{2 b}-\frac {a^2 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{2 b}-\frac {1}{4} b \int \left (-\frac {a \log (1-a c-b c x)}{b^2}+\frac {x \log (1-a c-b c x)}{b}+\frac {a^2 \log (1-a c-b c x)}{b^2 (a+b x)}\right ) \, dx\\ &=\frac {a x \text {Li}_2(c (a+b x))}{2 b}-\frac {1}{4} x^2 \text {Li}_2(c (a+b x))-\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}-\frac {1}{4} \int x \log (1-a c-b c x) \, dx-\frac {a^2 \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 b^2}+\frac {a \int \log (1-a c-b c x) \, dx}{4 b}-\frac {a^2 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{4 b}-\frac {a \text {Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{2 b^2 c}\\ &=-\frac {a x}{2 b}-\frac {1}{8} x^2 \log (1-a c-b c x)-\frac {a (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}+\frac {a^2 \text {Li}_2(c (a+b x))}{2 b^2}+\frac {a x \text {Li}_2(c (a+b x))}{2 b}-\frac {1}{4} x^2 \text {Li}_2(c (a+b x))-\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}-\frac {a^2 \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{4 b^2}-\frac {a \text {Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{4 b^2 c}-\frac {1}{8} (b c) \int \frac {x^2}{1-a c-b c x} \, dx\\ &=-\frac {3 a x}{4 b}-\frac {1}{8} x^2 \log (1-a c-b c x)-\frac {3 a (1-a c-b c x) \log (1-a c-b c x)}{4 b^2 c}+\frac {3 a^2 \text {Li}_2(c (a+b x))}{4 b^2}+\frac {a x \text {Li}_2(c (a+b x))}{2 b}-\frac {1}{4} x^2 \text {Li}_2(c (a+b x))-\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}-\frac {1}{8} (b c) \int \left (\frac {-1+a c}{b^2 c^2}-\frac {x}{b c}-\frac {(-1+a c)^2}{b^2 c^2 (-1+a c+b c x)}\right ) \, dx\\ &=-\frac {3 a x}{4 b}+\frac {(1-a c) x}{8 b c}+\frac {x^2}{16}+\frac {(1-a c)^2 \log (1-a c-b c x)}{8 b^2 c^2}-\frac {1}{8} x^2 \log (1-a c-b c x)-\frac {3 a (1-a c-b c x) \log (1-a c-b c x)}{4 b^2 c}+\frac {3 a^2 \text {Li}_2(c (a+b x))}{4 b^2}+\frac {a x \text {Li}_2(c (a+b x))}{2 b}-\frac {1}{4} x^2 \text {Li}_2(c (a+b x))-\frac {\left (a^2-b^2 x^2\right ) \text {Li}_3(c (a+b x))}{2 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 198, normalized size = 1.00 \begin {gather*} \frac {2 a c-15 a^2 c^2+2 b c x-14 a b c^2 x+b^2 c^2 x^2+2 \log (1-a c-b c x)-16 a c \log (1-a c-b c x)+14 a^2 c^2 \log (1-a c-b c x)+12 a b c^2 x \log (1-a c-b c x)-2 b^2 c^2 x^2 \log (1-a c-b c x)+4 c^2 \left (3 a^2+2 a b x-b^2 x^2\right ) \text {PolyLog}(2,c (a+b x))-8 c^2 \left (a^2-b^2 x^2\right ) \text {PolyLog}(3,c (a+b x))}{16 b^2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*c - 15*a^2*c^2 + 2*b*c*x - 14*a*b*c^2*x + b^2*c^2*x^2 + 2*Log[1 - a*c - b*c*x] - 16*a*c*Log[1 - a*c - b*c
*x] + 14*a^2*c^2*Log[1 - a*c - b*c*x] + 12*a*b*c^2*x*Log[1 - a*c - b*c*x] - 2*b^2*c^2*x^2*Log[1 - a*c - b*c*x]
+ 4*c^2*(3*a^2 + 2*a*b*x - b^2*x^2)*PolyLog[2, c*(a + b*x)] - 8*c^2*(a^2 - b^2*x^2)*PolyLog[3, c*(a + b*x)])/
(16*b^2*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 193, normalized size = 0.97 \begin {gather*} -\frac {3 \, {\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^{2}}{4 \, b^{2}} - \frac {a^{2} {\rm Li}_{3}(b c x + a c)}{2 \, b^{2}} + \frac {8 \, b^{2} c^{2} x^{2} {\rm Li}_{3}(b c x + a c) + b^{2} c^{2} x^{2} - 2 \, {\left (7 \, a b c^{2} - b c\right )} x - 4 \, {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} x^{2} - 6 \, a b c^{2} x - 7 \, a^{2} c^{2} + 8 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{16 \, b^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a^2/b^2 - 1/2*a^2*polylog(3, b*c*x + a
*c)/b^2 + 1/16*(8*b^2*c^2*x^2*polylog(3, b*c*x + a*c) + b^2*c^2*x^2 - 2*(7*a*b*c^2 - b*c)*x - 4*(b^2*c^2*x^2 -
2*a*b*c^2*x)*dilog(b*c*x + a*c) - 2*(b^2*c^2*x^2 - 6*a*b*c^2*x - 7*a^2*c^2 + 8*a*c - 1)*log(-b*c*x - a*c + 1)
)/(b^2*c^2)

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Fricas [A]
time = 0.39, size = 149, normalized size = 0.75 \begin {gather*} \frac {b^{2} c^{2} x^{2} - 2 \, {\left (7 \, a b c^{2} - b c\right )} x - 4 \, {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2}\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} x^{2} - 6 \, a b c^{2} x - 7 \, a^{2} c^{2} + 8 \, a c - 1\right )} \log \left (-b c x - a c + 1\right ) + 8 \, {\left (b^{2} c^{2} x^{2} - a^{2} c^{2}\right )} {\rm polylog}\left (3, b c x + a c\right )}{16 \, b^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(b^2*c^2*x^2 - 2*(7*a*b*c^2 - b*c)*x - 4*(b^2*c^2*x^2 - 2*a*b*c^2*x - 3*a^2*c^2)*dilog(b*c*x + a*c) - 2*(
b^2*c^2*x^2 - 6*a*b*c^2*x - 7*a^2*c^2 + 8*a*c - 1)*log(-b*c*x - a*c + 1) + 8*(b^2*c^2*x^2 - a^2*c^2)*polylog(3
, b*c*x + a*c))/(b^2*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\mathrm {polylog}\left (3,c\,\left (a+b\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(3, c*(a + b*x)),x)

[Out]

int(x*polylog(3, c*(a + b*x)), x)

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