Integrand size = 9, antiderivative size = 60 \[ \int \operatorname {PolyLog}(2,c (a+b x)) \, dx=-x-\frac {(1-a c-b c x) \log (1-a c-b c x)}{b c}+\frac {a \operatorname {PolyLog}(2,c (a+b x))}{b}+x \operatorname {PolyLog}(2,c (a+b x)) \]
Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int \operatorname {PolyLog}(2,c (a+b x)) \, dx=\frac {-c (a+b x)+(-1+c (a+b x)) \log (1-c (a+b x))+c (a+b x) \operatorname {PolyLog}(2,c (a+b x))}{b c} \]
(-(c*(a + b*x)) + (-1 + c*(a + b*x))*Log[1 - c*(a + b*x)] + c*(a + b*x)*Po lyLog[2, c*(a + b*x)])/(b*c)
Time = 0.38 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.889, Rules used = {7149, 25, 2868, 2840, 2838, 2894, 2836, 2732}
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.
\(\displaystyle \int \operatorname {PolyLog}(2,c (a+b x)) \, dx\) |
\(\Big \downarrow \) 7149 |
\(\displaystyle -\int -\log (1-c (a+b x))dx+a \int -\frac {\log (1-c (a+b x))}{a+b x}dx+x \operatorname {PolyLog}(2,c (a+b x))\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \log (1-c (a+b x))dx-a \int \frac {\log (1-c (a+b x))}{a+b x}dx+x \operatorname {PolyLog}(2,c (a+b x))\) |
\(\Big \downarrow \) 2868 |
\(\displaystyle -a \int \frac {\log (-a c-b x c+1)}{a+b x}dx+\int \log (1-c (a+b x))dx+x \operatorname {PolyLog}(2,c (a+b x))\) |
\(\Big \downarrow \) 2840 |
\(\displaystyle \int \log (1-c (a+b x))dx-\frac {a \int \frac {\log (1-c (a+b x))}{a+b x}d(a+b x)}{b}+x \operatorname {PolyLog}(2,c (a+b x))\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \int \log (1-c (a+b x))dx+x \operatorname {PolyLog}(2,c (a+b x))+\frac {a \operatorname {PolyLog}(2,c (a+b x))}{b}\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle \int \log (-a c-b x c+1)dx+x \operatorname {PolyLog}(2,c (a+b x))+\frac {a \operatorname {PolyLog}(2,c (a+b x))}{b}\) |
\(\Big \downarrow \) 2836 |
\(\displaystyle -\frac {\int \log (-a c-b x c+1)d(-a c-b x c+1)}{b c}+x \operatorname {PolyLog}(2,c (a+b x))+\frac {a \operatorname {PolyLog}(2,c (a+b x))}{b}\) |
\(\Big \downarrow \) 2732 |
\(\displaystyle x \operatorname {PolyLog}(2,c (a+b x))+\frac {a \operatorname {PolyLog}(2,c (a+b x))}{b}-\frac {(-a c-b c x+1) \log (-a c-b c x+1)+a c+b c x-1}{b c}\) |
-((-1 + a*c + b*c*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)]
3.2.26.3.1 Defintions of rubi rules used
Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x ] /; FreeQ[{c, n}, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : > Simp[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]
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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_ Symbol] :> Simp[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]
Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> In t[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] && !(BinomialMatch Q[u, x] && LinearMatchQ[v, x])
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] && Lin earQ[v, x] && !LinearMatchQ[v, x] && !(EqQ[n, 1] && MatchQ[c*v, (e_.)*((f _) + (g_.)*x) /; FreeQ[{e, f, g}, x]])
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)], x_Symbol] :> Simp[x*Poly Log[n, c*(a + b*x)^p], x] + (-Simp[p Int[PolyLog[n - 1, c*(a + b*x)^p], x ], x] + Simp[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]
Time = 0.68 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(\frac {\left (b c x +a c \right ) \operatorname {polylog}\left (2, b c x +a c \right )-\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )+1-b c x -a c}{b c}\) | \(63\) |
default | \(\frac {\left (b c x +a c \right ) \operatorname {polylog}\left (2, b c x +a c \right )-\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )+1-b c x -a c}{b c}\) | \(63\) |
parts | \(x \operatorname {polylog}\left (2, c \left (b x +a \right )\right )+\frac {-c \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )+a \,c^{2} \operatorname {dilog}\left (-b c x -a c +1\right )}{b \,c^{2}}\) | \(74\) |
parallelrisch | \(\frac {x \operatorname {polylog}\left (2, c \left (b x +a \right )\right ) a^{2} b \,c^{2}+x \ln \left (1-c \left (b x +a \right )\right ) a^{2} b \,c^{2}-x \,a^{2} b \,c^{2}+\operatorname {polylog}\left (2, c \left (b x +a \right )\right ) a^{3} c^{2}+\ln \left (1-c \left (b x +a \right )\right ) a^{3} c^{2}-x \operatorname {polylog}\left (2, c \left (b x +a \right )\right ) a b c -x \ln \left (1-c \left (b x +a \right )\right ) a b c +x a b c -\operatorname {polylog}\left (2, c \left (b x +a \right )\right ) a^{2} c -2 \ln \left (1-c \left (b x +a \right )\right ) a^{2} c +a \ln \left (1-c \left (b x +a \right )\right )}{a b c \left (a c -1\right )}\) | \(183\) |
Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \operatorname {PolyLog}(2,c (a+b x)) \, dx=-\frac {b c x - {\left (b c x + a c\right )} {\rm Li}_2\left (b c x + a c\right ) - {\left (b c x + a c - 1\right )} \log \left (-b c x - a c + 1\right )}{b c} \]
Time = 0.73 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \operatorname {PolyLog}(2,c (a+b x)) \, dx=\begin {cases} 0 & \text {for}\: b = 0 \wedge c = 0 \\x \operatorname {Li}_{2}\left (a c\right ) & \text {for}\: b = 0 \\0 & \text {for}\: c = 0 \\- \frac {a \operatorname {Li}_{1}\left (a c + b c x\right )}{b} + \frac {a \operatorname {Li}_{2}\left (a c + b c x\right )}{b} - x \operatorname {Li}_{1}\left (a c + b c x\right ) + x \operatorname {Li}_{2}\left (a c + b c x\right ) - x + \frac {\operatorname {Li}_{1}\left (a c + b c x\right )}{b c} & \text {otherwise} \end {cases} \]
Piecewise((0, Eq(b, 0) & Eq(c, 0)), (x*polylog(2, a*c), Eq(b, 0)), (0, Eq( c, 0)), (-a*polylog(1, a*c + b*c*x)/b + a*polylog(2, a*c + b*c*x)/b - x*po lylog(1, a*c + b*c*x) + x*polylog(2, a*c + b*c*x) - x + polylog(1, a*c + b *c*x)/(b*c), True))
Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.50 \[ \int \operatorname {PolyLog}(2,c (a+b x)) \, dx=-\frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} a}{b} + \frac {b c x {\rm Li}_2\left (b c x + a c\right ) - b c x + {\left (b c x + a c - 1\right )} \log \left (-b c x - a c + 1\right )}{b c} \]
-(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a/b + (b*c*x*dilog(b*c*x + a*c) - b*c*x + (b*c*x + a*c - 1)*log(-b*c*x - a*c + 1 ))/(b*c)
\[ \int \operatorname {PolyLog}(2,c (a+b x)) \, dx=\int { {\rm Li}_2\left ({\left (b x + a\right )} c\right ) \,d x } \]
Time = 5.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \operatorname {PolyLog}(2,c (a+b x)) \, dx=\frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b}-x-\frac {\ln \left (1-c\,\left (a+b\,x\right )\right )}{b\,c}+\frac {\ln \left (1-c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b} \]