Integrand size = 11, antiderivative size = 198 \[ \int x \operatorname {PolyLog}(3,c (a+b x)) \, dx=-\frac {3 a x}{4 b}+\frac {(1-a c) x}{8 b c}+\frac {x^2}{16}+\frac {(1-a c)^2 \log (1-a c-b c x)}{8 b^2 c^2}-\frac {1}{8} x^2 \log (1-a c-b c x)-\frac {3 a (1-a c-b c x) \log (1-a c-b c x)}{4 b^2 c}+\frac {3 a^2 \operatorname {PolyLog}(2,c (a+b x))}{4 b^2}+\frac {a x \operatorname {PolyLog}(2,c (a+b x))}{2 b}-\frac {1}{4} x^2 \operatorname {PolyLog}(2,c (a+b x))-\frac {\left (a^2-b^2 x^2\right ) \operatorname {PolyLog}(3,c (a+b x))}{2 b^2} \]
-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*po lylog(2,c*(b*x+a))-1/2*(-b^2*x^2+a^2)*polylog(3,c*(b*x+a))/b^2
Time = 0.04 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00 \[ \int x \operatorname {PolyLog}(3,c (a+b x)) \, dx=\frac {2 a c-15 a^2 c^2+2 b c x-14 a b c^2 x+b^2 c^2 x^2+2 \log (1-a c-b c x)-16 a c \log (1-a c-b c x)+14 a^2 c^2 \log (1-a c-b c x)+12 a b c^2 x \log (1-a c-b c x)-2 b^2 c^2 x^2 \log (1-a c-b c x)+4 c^2 \left (3 a^2+2 a b x-b^2 x^2\right ) \operatorname {PolyLog}(2,c (a+b x))-8 c^2 \left (a^2-b^2 x^2\right ) \operatorname {PolyLog}(3,c (a+b x))}{16 b^2 c^2} \]
(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)
Time = 0.49 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {7153, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \operatorname {PolyLog}(3,c (a+b x)) \, dx\) |
\(\Big \downarrow \) 7153 |
\(\displaystyle \frac {\int (a \operatorname {PolyLog}(2,c (a+b x))-b x \operatorname {PolyLog}(2,c (a+b x)))dx}{2 b}-\frac {\left (a^2-b^2 x^2\right ) \operatorname {PolyLog}(3,c (a+b x))}{2 b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {3 a^2 \operatorname {PolyLog}(2,c (a+b x))}{2 b}+\frac {(1-a c)^2 \log (-a c-b c x+1)}{4 b c^2}-\frac {1}{2} b x^2 \operatorname {PolyLog}(2,c (a+b x))+a x \operatorname {PolyLog}(2,c (a+b x))-\frac {1}{4} b x^2 \log (-a c-b c x+1)-\frac {3 a (-a c-b c x+1) \log (-a c-b c x+1)}{2 b c}+\frac {x (1-a c)}{4 c}-\frac {3 a x}{2}+\frac {b x^2}{8}}{2 b}-\frac {\left (a^2-b^2 x^2\right ) \operatorname {PolyLog}(3,c (a+b x))}{2 b^2}\) |
((-3*a*x)/2 + ((1 - a*c)*x)/(4*c) + (b*x^2)/8 + ((1 - a*c)^2*Log[1 - a*c - b*c*x])/(4*b*c^2) - (b*x^2*Log[1 - a*c - b*c*x])/4 - (3*a*(1 - a*c - b*c* x)*Log[1 - a*c - b*c*x])/(2*b*c) + (3*a^2*PolyLog[2, c*(a + b*x)])/(2*b) + a*x*PolyLog[2, c*(a + b*x)] - (b*x^2*PolyLog[2, c*(a + b*x)])/2)/(2*b) - ((a^2 - b^2*x^2)*PolyLog[3, c*(a + b*x)])/(2*b^2)
3.2.32.3.1 Defintions of rubi rules used
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] + Simp[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]
\[\int x \operatorname {polylog}\left (3, c \left (b x +a \right )\right )d x\]
Time = 0.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.75 \[ \int x \operatorname {PolyLog}(3,c (a+b x)) \, dx=\frac {b^{2} c^{2} x^{2} - 2 \, {\left (7 \, a b c^{2} - b c\right )} x - 4 \, {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2}\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} x^{2} - 6 \, a b c^{2} x - 7 \, a^{2} c^{2} + 8 \, a c - 1\right )} \log \left (-b c x - a c + 1\right ) + 8 \, {\left (b^{2} c^{2} x^{2} - a^{2} c^{2}\right )} {\rm polylog}\left (3, b c x + a c\right )}{16 \, b^{2} c^{2}} \]
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)
\[ \int x \operatorname {PolyLog}(3,c (a+b x)) \, dx=\int x \operatorname {Li}_{3}\left (a c + b c x\right )\, dx \]
Time = 0.20 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.97 \[ \int x \operatorname {PolyLog}(3,c (a+b x)) \, dx=-\frac {3 \, {\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} a^{2}}{4 \, b^{2}} - \frac {a^{2} {\rm Li}_{3}(b c x + a c)}{2 \, b^{2}} + \frac {8 \, b^{2} c^{2} x^{2} {\rm Li}_{3}(b c x + a c) + b^{2} c^{2} x^{2} - 2 \, {\left (7 \, a b c^{2} - b c\right )} x - 4 \, {\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} x^{2} - 6 \, a b c^{2} x - 7 \, a^{2} c^{2} + 8 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{16 \, b^{2} c^{2}} \]
-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)
\[ \int x \operatorname {PolyLog}(3,c (a+b x)) \, dx=\int { x {\rm Li}_{3}({\left (b x + a\right )} c) \,d x } \]
Timed out. \[ \int x \operatorname {PolyLog}(3,c (a+b x)) \, dx=\int x\,\mathrm {polylog}\left (3,c\,\left (a+b\,x\right )\right ) \,d x \]