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

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

[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)

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Rubi [A]  time = 0.196508, antiderivative size = 210, normalized size of antiderivative = 1., 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 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]

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 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 2295

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

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 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 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 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])

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*}

Mathematica [A]  time = 0.0900849, size = 161, normalized size = 0.77 \[ \frac{e \left (-4 a^2 c^2 \text{PolyLog}(2,c (a+b x))+\left (-6 a^2 c^2-4 a c (b c x-2)+2 b^2 c^2 x^2-2\right ) \log (-a c-b c x+1)-b c x (-6 a c+b c x+2)\right )}{8 b^2 c^2}+\frac{d (c (a+b x) \text{PolyLog}(2,c (a+b x))-c (a+b x)+(c (a+b x)-1) \log (1-c (a+b x)))}{b c}+\frac{1}{2} e x^2 \text{PolyLog}(2,a c+b c x) \]

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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Maple [A]  time = 0.049, size = 292, normalized size = 1.4 \begin{align*} -{\frac{{\it polylog} \left ( 2,xbc+ac \right ){a}^{2}e}{2\,{b}^{2}}}+{\it polylog} \left ( 2,xbc+ac \right ) xd+{\frac{{\it polylog} \left ( 2,xbc+ac \right ) ad}{b}}+{\frac{{\it polylog} \left ( 2,xbc+ac \right ) e{x}^{2}}{2}}+{\frac{3\,e}{8\,{b}^{2}{c}^{2}}}-{\frac{5\,ae}{4\,{b}^{2}c}}+{\frac{d}{bc}}+{\frac{e\ln \left ( -xbc-ac+1 \right ){x}^{2}}{4}}+{\frac{\ln \left ( -xbc-ac+1 \right ) ad}{b}}+\ln \left ( -xbc-ac+1 \right ) xd+{\frac{3\,axe}{4\,b}}+{\frac{7\,{a}^{2}e}{8\,{b}^{2}}}-{\frac{\ln \left ( -xbc-ac+1 \right ) xae}{2\,b}}-{\frac{e\ln \left ( -xbc-ac+1 \right ) }{4\,{b}^{2}{c}^{2}}}-{\frac{e{x}^{2}}{8}}-{\frac{ex}{4\,bc}}-dx-{\frac{ad}{b}}-{\frac{\ln \left ( -xbc-ac+1 \right ) d}{bc}}-{\frac{3\,\ln \left ( -xbc-ac+1 \right ){a}^{2}e}{4\,{b}^{2}}}+{\frac{\ln \left ( -xbc-ac+1 \right ) ae}{{b}^{2}c}} \end{align*}

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+3/8/b^2/c^2*e-5/4/b^2/c*a*e+1/b/c*d+1/4*e*ln(-b*c*x-a*c+1)*x^2+1/b*ln(-b*c*x-a*c+1)*a*d+ln(-b*c*x-a
*c+1)*x*d+3/4/b*a*e*x+7/8/b^2*a^2*e-1/2/b*ln(-b*c*x-a*c+1)*x*a*e-1/4/b^2/c^2*e*ln(-b*c*x-a*c+1)-1/8*e*x^2-1/4/
b/c*x*e-d*x-a*d/b-1/b/c*ln(-b*c*x-a*c+1)*d-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 = 1.01942, size = 286, normalized size = 1.36 \begin{align*} -\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}} \end{align*}

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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Fricas [A]  time = 2.49073, size = 377, normalized size = 1.8 \begin{align*} -\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}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}{\rm Li}_2\left ({\left (b x + a\right )} c\right )\,{d x} \end{align*}

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)