3.2.39 \(\int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx\) [139]

3.2.39.1 Optimal result
3.2.39.2 Mathematica [A] (verified)
3.2.39.3 Rubi [A] (verified)
3.2.39.4 Maple [A] (verified)
3.2.39.5 Fricas [A] (verification not implemented)
3.2.39.6 Sympy [A] (verification not implemented)
3.2.39.7 Maxima [A] (verification not implemented)
3.2.39.8 Giac [F]
3.2.39.9 Mupad [F(-1)]

3.2.39.1 Optimal result

Integrand size = 15, antiderivative size = 210 \[ \int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx=-\frac {(b d-a e) x}{2 b}-\frac {(b c d+e-a c e) x}{4 b c}-\frac {(d+e x)^2}{8 e}-\frac {(b c d+e-a c e)^2 \log (1-a c-b c x)}{4 b^2 c^2 e}-\frac {(b d-a e) (1-a c-b c x) \log (1-a c-b c x)}{2 b^2 c}+\frac {(d+e x)^2 \log (1-a c-b c x)}{4 e}-\frac {(b d-a e)^2 \operatorname {PolyLog}(2,c (a+b x))}{2 b^2 e}+\frac {(d+e x)^2 \operatorname {PolyLog}(2,c (a+b x))}{2 e} \]

output
-1/2*(-a*e+b*d)*x/b-1/4*(-a*c*e+b*c*d+e)*x/b/c-1/8*(e*x+d)^2/e-1/4*(-a*c*e 
+b*c*d+e)^2*ln(-b*c*x-a*c+1)/b^2/c^2/e-1/2*(-a*e+b*d)*(-b*c*x-a*c+1)*ln(-b 
*c*x-a*c+1)/b^2/c+1/4*(e*x+d)^2*ln(-b*c*x-a*c+1)/e-1/2*(-a*e+b*d)^2*polylo 
g(2,c*(b*x+a))/b^2/e+1/2*(e*x+d)^2*polylog(2,c*(b*x+a))/e
 
3.2.39.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.77 \[ \int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx=\frac {e \left (-b c x (2-6 a c+b c x)+\left (-2-6 a^2 c^2+2 b^2 c^2 x^2-4 a c (-2+b c x)\right ) \log (1-a c-b c x)-4 a^2 c^2 \operatorname {PolyLog}(2,c (a+b x))\right )}{8 b^2 c^2}+\frac {d (-c (a+b x)+(-1+c (a+b x)) \log (1-c (a+b x))+c (a+b x) \operatorname {PolyLog}(2,c (a+b x)))}{b c}+\frac {1}{2} e x^2 \operatorname {PolyLog}(2,a c+b c x) \]

input
Integrate[(d + e*x)*PolyLog[2, c*(a + b*x)],x]
 
output
(e*(-(b*c*x*(2 - 6*a*c + b*c*x)) + (-2 - 6*a^2*c^2 + 2*b^2*c^2*x^2 - 4*a*c 
*(-2 + b*c*x))*Log[1 - a*c - b*c*x] - 4*a^2*c^2*PolyLog[2, c*(a + b*x)]))/ 
(8*b^2*c^2) + (d*(-(c*(a + b*x)) + (-1 + c*(a + b*x))*Log[1 - c*(a + b*x)] 
 + c*(a + b*x)*PolyLog[2, c*(a + b*x)]))/(b*c) + (e*x^2*PolyLog[2, a*c + b 
*c*x])/2
 
3.2.39.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7152, 2865, 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.

\(\displaystyle \int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx\)

\(\Big \downarrow \) 7152

\(\displaystyle \frac {b \int \frac {(d+e x)^2 \log (-a c-b x c+1)}{a+b x}dx}{2 e}+\frac {(d+e x)^2 \operatorname {PolyLog}(2,c (a+b x))}{2 e}\)

\(\Big \downarrow \) 2865

\(\displaystyle \frac {b \int \left (\frac {\log (-a c-b x c+1) (b d-a e)^2}{b^2 (a+b x)}+\frac {e \log (-a c-b x c+1) (b d-a e)}{b^2}+\frac {e (d+e x) \log (-a c-b x c+1)}{b}\right )dx}{2 e}+\frac {(d+e x)^2 \operatorname {PolyLog}(2,c (a+b x))}{2 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (-\frac {(-a c e+b c d+e)^2 \log (-a c-b c x+1)}{2 b^3 c^2}-\frac {(b d-a e)^2 \operatorname {PolyLog}(2,c (a+b x))}{b^3}-\frac {e (-a c-b c x+1) (b d-a e) \log (-a c-b c x+1)}{b^3 c}-\frac {e x (-a c e+b c d+e)}{2 b^2 c}-\frac {e x (b d-a e)}{b^2}+\frac {(d+e x)^2 \log (-a c-b c x+1)}{2 b}-\frac {(d+e x)^2}{4 b}\right )}{2 e}+\frac {(d+e x)^2 \operatorname {PolyLog}(2,c (a+b x))}{2 e}\)

input
Int[(d + e*x)*PolyLog[2, c*(a + b*x)],x]
 
output
((d + e*x)^2*PolyLog[2, c*(a + b*x)])/(2*e) + (b*(-((e*(b*d - a*e)*x)/b^2) 
 - (e*(b*c*d + e - a*c*e)*x)/(2*b^2*c) - (d + e*x)^2/(4*b) - ((b*c*d + e - 
 a*c*e)^2*Log[1 - a*c - b*c*x])/(2*b^3*c^2) - (e*(b*d - a*e)*(1 - a*c - b* 
c*x)*Log[1 - a*c - b*c*x])/(b^3*c) + ((d + e*x)^2*Log[1 - a*c - b*c*x])/(2 
*b) - ((b*d - a*e)^2*PolyLog[2, c*(a + b*x)])/b^3))/(2*e)
 

3.2.39.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2865
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Sy 
mbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, 
Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunctionQ[ 
RFx, x] && IntegerQ[p]
 

rule 7152
Int[((d_.) + (e_.)*(x_))^(m_.)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Sy 
mbol] :> Simp[(d + e*x)^(m + 1)*(PolyLog[2, c*(a + b*x)]/(e*(m + 1))), x] + 
 Simp[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]
 
3.2.39.4 Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.08

method result size
parts \(\frac {\operatorname {polylog}\left (2, c \left (b x +a \right )\right ) e \,x^{2}}{2}+\operatorname {polylog}\left (2, c \left (b x +a \right )\right ) d x -\frac {-\frac {2 a c e \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )-2 d b c \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )+e \left (\frac {\left (-b c x -a c +1\right )^{2} \ln \left (-b c x -a c +1\right )}{2}-\frac {\left (-b c x -a c +1\right )^{2}}{4}\right )-e \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )}{b^{2} c}+\frac {a c \left (a e -2 d b \right ) \operatorname {dilog}\left (-b c x -a c +1\right )}{b^{2}}}{2 c}\) \(227\)
derivativedivides \(\frac {-\frac {\operatorname {polylog}\left (2, b c x +a c \right ) a e \left (b c x +a c \right )}{b}+\operatorname {polylog}\left (2, b c x +a c \right ) d \left (b c x +a c \right )+\frac {\operatorname {polylog}\left (2, b c x +a c \right ) e \left (b c x +a c \right )^{2}}{2 b c}-\frac {-2 a c e \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )+2 d b c \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )-e \left (\frac {\left (-b c x -a c +1\right )^{2} \ln \left (-b c x -a c +1\right )}{2}-\frac {\left (-b c x -a c +1\right )^{2}}{4}\right )+e \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )}{2 b c}}{b c}\) \(249\)
default \(\frac {-\frac {\operatorname {polylog}\left (2, b c x +a c \right ) a e \left (b c x +a c \right )}{b}+\operatorname {polylog}\left (2, b c x +a c \right ) d \left (b c x +a c \right )+\frac {\operatorname {polylog}\left (2, b c x +a c \right ) e \left (b c x +a c \right )^{2}}{2 b c}-\frac {-2 a c e \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )+2 d b c \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )-e \left (\frac {\left (-b c x -a c +1\right )^{2} \ln \left (-b c x -a c +1\right )}{2}-\frac {\left (-b c x -a c +1\right )^{2}}{4}\right )+e \left (\left (-b c x -a c +1\right ) \ln \left (-b c x -a c +1\right )-1+b c x +a c \right )}{2 b c}}{b c}\) \(249\)
parallelrisch \(\frac {4 x^{2} \operatorname {polylog}\left (2, c \left (b x +a \right )\right ) b^{2} c^{2} e +2 x^{2} \ln \left (1-c \left (b x +a \right )\right ) b^{2} c^{2} e -x^{2} b^{2} c^{2} e +8 x \operatorname {polylog}\left (2, c \left (b x +a \right )\right ) b^{2} c^{2} d -4 x \ln \left (1-c \left (b x +a \right )\right ) a b \,c^{2} e +8 x \ln \left (1-c \left (b x +a \right )\right ) b^{2} c^{2} d +6 x a b \,c^{2} e -8 x \,b^{2} c^{2} d -4 \operatorname {polylog}\left (2, c \left (b x +a \right )\right ) a^{2} c^{2} e +8 \operatorname {polylog}\left (2, c \left (b x +a \right )\right ) a b \,c^{2} d -6 \ln \left (1-c \left (b x +a \right )\right ) a^{2} c^{2} e +8 \ln \left (1-c \left (b x +a \right )\right ) a b \,c^{2} d -11 a^{2} e \,c^{2}+16 a b \,c^{2} d -2 b x e c +8 \ln \left (1-c \left (b x +a \right )\right ) a c e -8 \ln \left (1-c \left (b x +a \right )\right ) b c d +9 a e c -8 b c d -2 \ln \left (1-c \left (b x +a \right )\right ) e -2 e}{8 b^{2} c^{2}}\) \(301\)

input
int((e*x+d)*polylog(2,c*(b*x+a)),x,method=_RETURNVERBOSE)
 
output
1/2*polylog(2,c*(b*x+a))*e*x^2+polylog(2,c*(b*x+a))*d*x-1/2/c*(-1/b^2/c*(2 
*a*c*e*((-b*c*x-a*c+1)*ln(-b*c*x-a*c+1)-1+b*c*x+a*c)-2*d*b*c*((-b*c*x-a*c+ 
1)*ln(-b*c*x-a*c+1)-1+b*c*x+a*c)+e*(1/2*(-b*c*x-a*c+1)^2*ln(-b*c*x-a*c+1)- 
1/4*(-b*c*x-a*c+1)^2)-e*((-b*c*x-a*c+1)*ln(-b*c*x-a*c+1)-1+b*c*x+a*c))+a*c 
*(a*e-2*b*d)/b^2*dilog(-b*c*x-a*c+1))
 
3.2.39.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.84 \[ \int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx=-\frac {b^{2} c^{2} e x^{2} + 2 \, {\left (4 \, b^{2} c^{2} d - {\left (3 \, a b c^{2} - b c\right )} e\right )} x - 4 \, {\left (b^{2} c^{2} e x^{2} + 2 \, b^{2} c^{2} d x + 2 \, a b c^{2} d - a^{2} c^{2} e\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} e x^{2} + 4 \, {\left (a b c^{2} - b c\right )} d - {\left (3 \, a^{2} c^{2} - 4 \, a c + 1\right )} e + 2 \, {\left (2 \, b^{2} c^{2} d - a b c^{2} e\right )} x\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \]

input
integrate((e*x+d)*polylog(2,c*(b*x+a)),x, algorithm="fricas")
 
output
-1/8*(b^2*c^2*e*x^2 + 2*(4*b^2*c^2*d - (3*a*b*c^2 - b*c)*e)*x - 4*(b^2*c^2 
*e*x^2 + 2*b^2*c^2*d*x + 2*a*b*c^2*d - a^2*c^2*e)*dilog(b*c*x + a*c) - 2*( 
b^2*c^2*e*x^2 + 4*(a*b*c^2 - b*c)*d - (3*a^2*c^2 - 4*a*c + 1)*e + 2*(2*b^2 
*c^2*d - a*b*c^2*e)*x)*log(-b*c*x - a*c + 1))/(b^2*c^2)
 
3.2.39.6 Sympy [A] (verification not implemented)

Time = 1.86 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.20 \[ \int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx=\begin {cases} 0 & \text {for}\: b = 0 \wedge c = 0 \\\left (d x + \frac {e x^{2}}{2}\right ) \operatorname {Li}_{2}\left (a c\right ) & \text {for}\: b = 0 \\0 & \text {for}\: c = 0 \\\frac {3 a^{2} e \operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2}} - \frac {a^{2} e \operatorname {Li}_{2}\left (a c + b c x\right )}{2 b^{2}} - \frac {a d \operatorname {Li}_{1}\left (a c + b c x\right )}{b} + \frac {a d \operatorname {Li}_{2}\left (a c + b c x\right )}{b} + \frac {a e x \operatorname {Li}_{1}\left (a c + b c x\right )}{2 b} + \frac {3 a e x}{4 b} - \frac {a e \operatorname {Li}_{1}\left (a c + b c x\right )}{b^{2} c} - d x \operatorname {Li}_{1}\left (a c + b c x\right ) + d x \operatorname {Li}_{2}\left (a c + b c x\right ) - d x - \frac {e x^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{4} + \frac {e x^{2} \operatorname {Li}_{2}\left (a c + b c x\right )}{2} - \frac {e x^{2}}{8} + \frac {d \operatorname {Li}_{1}\left (a c + b c x\right )}{b c} - \frac {e x}{4 b c} + \frac {e \operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2} c^{2}} & \text {otherwise} \end {cases} \]

input
integrate((e*x+d)*polylog(2,c*(b*x+a)),x)
 
output
Piecewise((0, Eq(b, 0) & Eq(c, 0)), ((d*x + e*x**2/2)*polylog(2, a*c), Eq( 
b, 0)), (0, Eq(c, 0)), (3*a**2*e*polylog(1, a*c + b*c*x)/(4*b**2) - a**2*e 
*polylog(2, a*c + b*c*x)/(2*b**2) - a*d*polylog(1, a*c + b*c*x)/b + a*d*po 
lylog(2, a*c + b*c*x)/b + a*e*x*polylog(1, a*c + b*c*x)/(2*b) + 3*a*e*x/(4 
*b) - a*e*polylog(1, a*c + b*c*x)/(b**2*c) - d*x*polylog(1, a*c + b*c*x) + 
 d*x*polylog(2, a*c + b*c*x) - d*x - e*x**2*polylog(1, a*c + b*c*x)/4 + e* 
x**2*polylog(2, a*c + b*c*x)/2 - e*x**2/8 + d*polylog(1, a*c + b*c*x)/(b*c 
) - e*x/(4*b*c) + e*polylog(1, a*c + b*c*x)/(4*b**2*c**2), True))
 
3.2.39.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.01 \[ \int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx=-\frac {{\left (2 \, a b d - a^{2} e\right )} {\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )}}{2 \, b^{2}} - \frac {b^{2} c^{2} e x^{2} + 2 \, {\left (4 \, b^{2} c^{2} d - {\left (3 \, a b c^{2} - b c\right )} e\right )} x - 4 \, {\left (b^{2} c^{2} e x^{2} + 2 \, b^{2} c^{2} d x\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} e x^{2} + 4 \, {\left (a b c^{2} - b c\right )} d - {\left (3 \, a^{2} c^{2} - 4 \, a c + 1\right )} e + 2 \, {\left (2 \, b^{2} c^{2} d - a b c^{2} e\right )} x\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \]

input
integrate((e*x+d)*polylog(2,c*(b*x+a)),x, algorithm="maxima")
 
output
-1/2*(2*a*b*d - a^2*e)*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b* 
c*x - a*c + 1))/b^2 - 1/8*(b^2*c^2*e*x^2 + 2*(4*b^2*c^2*d - (3*a*b*c^2 - b 
*c)*e)*x - 4*(b^2*c^2*e*x^2 + 2*b^2*c^2*d*x)*dilog(b*c*x + a*c) - 2*(b^2*c 
^2*e*x^2 + 4*(a*b*c^2 - b*c)*d - (3*a^2*c^2 - 4*a*c + 1)*e + 2*(2*b^2*c^2* 
d - a*b*c^2*e)*x)*log(-b*c*x - a*c + 1))/(b^2*c^2)
 
3.2.39.8 Giac [F]

\[ \int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx=\int { {\left (e x + d\right )} {\rm Li}_2\left ({\left (b x + a\right )} c\right ) \,d x } \]

input
integrate((e*x+d)*polylog(2,c*(b*x+a)),x, algorithm="giac")
 
output
integrate((e*x + d)*dilog((b*x + a)*c), x)
 
3.2.39.9 Mupad [F(-1)]

Timed out. \[ \int (d+e x) \operatorname {PolyLog}(2,c (a+b x)) \, dx=\int \mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )\,\left (d+e\,x\right ) \,d x \]

input
int(polylog(2, c*(a + b*x))*(d + e*x),x)
 
output
int(polylog(2, c*(a + b*x))*(d + e*x), x)