3.139 \(\int (d+e x) \text {Li}_2(c (a+b x)) \, dx\)

Optimal. Leaf size=210 \[ -\frac {(-a c e+b c d+e)^2 \log (-a c-b c x+1)}{4 b^2 c^2 e}-\frac {(b d-a e)^2 \text {Li}_2(c (a+b x))}{2 b^2 e}-\frac {(-a c-b c x+1) (b d-a e) \log (-a c-b c x+1)}{2 b^2 c}+\frac {(d+e x)^2 \text {Li}_2(c (a+b x))}{2 e}-\frac {x (-a c e+b c d+e)}{4 b c}+\frac {(d+e x)^2 \log (-a c-b c x+1)}{4 e}-\frac {x (b d-a e)}{2 b}-\frac {(d+e x)^2}{8 e} \]

[Out]

-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*poly
log(2,c*(b*x+a))/b^2/e+1/2*(e*x+d)^2*polylog(2,c*(b*x+a))/e

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Rubi [A]  time = 0.20, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6598, 2418, 2389, 2295, 2393, 2391, 2395, 43} \[ -\frac {(b d-a e)^2 \text {PolyLog}(2,c (a+b x))}{2 b^2 e}+\frac {(d+e x)^2 \text {PolyLog}(2,c (a+b x))}{2 e}-\frac {(-a c e+b c d+e)^2 \log (-a c-b c x+1)}{4 b^2 c^2 e}-\frac {(-a c-b c x+1) (b d-a e) \log (-a c-b c x+1)}{2 b^2 c}-\frac {x (-a c e+b c d+e)}{4 b c}+\frac {(d+e x)^2 \log (-a c-b c x+1)}{4 e}-\frac {x (b d-a e)}{2 b}-\frac {(d+e x)^2}{8 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*PolyLog[2, c*(a + b*x)],x]

[Out]

-((b*d - a*e)*x)/(2*b) - ((b*c*d + e - a*c*e)*x)/(4*b*c) - (d + e*x)^2/(8*e) - ((b*c*d + e - a*c*e)^2*Log[1 -
a*c - b*c*x])/(4*b^2*c^2*e) - ((b*d - a*e)*(1 - a*c - b*c*x)*Log[1 - a*c - b*c*x])/(2*b^2*c) + ((d + e*x)^2*Lo
g[1 - a*c - b*c*x])/(4*e) - ((b*d - a*e)^2*PolyLog[2, c*(a + b*x)])/(2*b^2*e) + ((d + e*x)^2*PolyLog[2, c*(a +
b*x)])/(2*e)

Rule 43

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 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 2395

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 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> 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] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 6598

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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 161, normalized size = 0.77 \[ \frac {e \left (\left (-6 a^2 c^2-4 a c (b c x-2)+2 b^2 c^2 x^2-2\right ) \log (-a c-b c x+1)-4 a^2 c^2 \text {Li}_2(c (a+b x))-b c x (-6 a c+b c x+2)\right )}{8 b^2 c^2}+\frac {d (c (a+b x) \text {Li}_2(c (a+b x))-c (a+b x)+(c (a+b x)-1) \log (1-c (a+b x)))}{b c}+\frac {1}{2} e x^2 \text {Li}_2(a c+b x c) \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*PolyLog[2, c*(a + b*x)],x]

[Out]

(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

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fricas [A]  time = 0.63, size = 176, normalized size = 0.84 \[ -\frac {b^{2} c^{2} e x^{2} + 2 \, {\left (4 \, b^{2} c^{2} d - {\left (3 \, a b c^{2} - b c\right )} e\right )} x - 4 \, {\left (b^{2} c^{2} e x^{2} + 2 \, b^{2} c^{2} d x + 2 \, a b c^{2} d - a^{2} c^{2} e\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} e x^{2} + 4 \, {\left (a b c^{2} - b c\right )} d - {\left (3 \, a^{2} c^{2} - 4 \, a c + 1\right )} e + 2 \, {\left (2 \, b^{2} c^{2} d - a b c^{2} e\right )} x\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*polylog(2,c*(b*x+a)),x, algorithm="fricas")

[Out]

-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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )} {\rm Li}_2\left ({\left (b x + a\right )} c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*polylog(2,c*(b*x+a)),x, algorithm="giac")

[Out]

integrate((e*x + d)*dilog((b*x + a)*c), x)

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maple [A]  time = 0.01, size = 292, normalized size = 1.39 \[ -\frac {\polylog \left (2, b c x +a c \right ) a^{2} e}{2 b^{2}}+\polylog \left (2, b c x +a c \right ) x d +\frac {\polylog \left (2, b c x +a c \right ) a d}{b}+\frac {\polylog \left (2, b c x +a c \right ) e \,x^{2}}{2}-\frac {\ln \left (-b c x -a c +1\right ) x a e}{2 b}+\frac {3 e}{8 b^{2} c^{2}}+\frac {d}{b c}-\frac {5 a e}{4 b^{2} c}-\frac {e \ln \left (-b c x -a c +1\right )}{4 b^{2} c^{2}}+\frac {3 a e x}{4 b}+\ln \left (-b c x -a c +1\right ) x d +\frac {\ln \left (-b c x -a c +1\right ) a d}{b}+\frac {e \ln \left (-b c x -a c +1\right ) x^{2}}{4}+\frac {7 a^{2} e}{8 b^{2}}-d x -\frac {a d}{b}-\frac {\ln \left (-b c x -a c +1\right ) d}{b c}-\frac {e \,x^{2}}{8}-\frac {e x}{4 b c}-\frac {3 \ln \left (-b c x -a c +1\right ) a^{2} e}{4 b^{2}}+\frac {\ln \left (-b c x -a c +1\right ) a e}{b^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b^2*polylog(2,b*c*x+a*c)*a^2*e+polylog(2,b*c*x+a*c)*x*d+1/b*polylog(2,b*c*x+a*c)*a*d+1/2*polylog(2,b*c*x+
a*c)*e*x^2-1/2/b*ln(-b*c*x-a*c+1)*x*a*e+3/8/b^2/c^2*e+1/b/c*d-5/4/b^2/c*a*e-1/4/b^2/c^2*e*ln(-b*c*x-a*c+1)+3/4
/b*a*e*x+ln(-b*c*x-a*c+1)*x*d+1/b*ln(-b*c*x-a*c+1)*a*d+1/4*e*ln(-b*c*x-a*c+1)*x^2+7/8/b^2*a^2*e-d*x-1/b*a*d-1/
b/c*ln(-b*c*x-a*c+1)*d-1/8*e*x^2-1/4/b/c*e*x-3/4/b^2*ln(-b*c*x-a*c+1)*a^2*e+1/b^2/c*ln(-b*c*x-a*c+1)*a*e

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maxima [A]  time = 0.33, size = 212, normalized size = 1.01 \[ -\frac {{\left (2 \, a b d - a^{2} e\right )} {\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )}}{2 \, b^{2}} - \frac {b^{2} c^{2} e x^{2} + 2 \, {\left (4 \, b^{2} c^{2} d - {\left (3 \, a b c^{2} - b c\right )} e\right )} x - 4 \, {\left (b^{2} c^{2} e x^{2} + 2 \, b^{2} c^{2} d x\right )} {\rm Li}_2\left (b c x + a c\right ) - 2 \, {\left (b^{2} c^{2} e x^{2} + 4 \, {\left (a b c^{2} - b c\right )} d - {\left (3 \, a^{2} c^{2} - 4 \, a c + 1\right )} e + 2 \, {\left (2 \, b^{2} c^{2} d - a b c^{2} e\right )} x\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*polylog(2,c*(b*x+a)),x, algorithm="maxima")

[Out]

-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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )\,\left (d+e\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.46, size = 252, normalized size = 1.20 \[ \begin {cases} 0 & \text {for}\: b = 0 \wedge c = 0 \\\left (d x + \frac {e x^{2}}{2}\right ) \operatorname {Li}_{2}\left (a c\right ) & \text {for}\: b = 0 \\0 & \text {for}\: c = 0 \\\frac {3 a^{2} e \operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2}} - \frac {a^{2} e \operatorname {Li}_{2}\left (a c + b c x\right )}{2 b^{2}} - \frac {a d \operatorname {Li}_{1}\left (a c + b c x\right )}{b} + \frac {a d \operatorname {Li}_{2}\left (a c + b c x\right )}{b} + \frac {a e x \operatorname {Li}_{1}\left (a c + b c x\right )}{2 b} + \frac {3 a e x}{4 b} - \frac {a e \operatorname {Li}_{1}\left (a c + b c x\right )}{b^{2} c} - d x \operatorname {Li}_{1}\left (a c + b c x\right ) + d x \operatorname {Li}_{2}\left (a c + b c x\right ) - d x - \frac {e x^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{4} + \frac {e x^{2} \operatorname {Li}_{2}\left (a c + b c x\right )}{2} - \frac {e x^{2}}{8} + \frac {d \operatorname {Li}_{1}\left (a c + b c x\right )}{b c} - \frac {e x}{4 b c} + \frac {e \operatorname {Li}_{1}\left (a c + b c x\right )}{4 b^{2} c^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*polylog(2,c*(b*x+a)),x)

[Out]

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*pol
ylog(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*
polylog(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))

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