3.143 \(\int \frac{\text{PolyLog}(2,c (a+b x))}{(d+e x)^3} \, dx\)

Optimal. Leaf size=278 \[ \frac{b^2 \text{PolyLog}(2,c (a+b x))}{2 e (b d-a e)^2}+\frac{b^2 \text{PolyLog}\left (2,\frac{e (-a c-b c x+1)}{-a c e+b c d+e}\right )}{2 e (b d-a e)^2}-\frac{\text{PolyLog}(2,c (a+b x))}{2 e (d+e x)^2}+\frac{b^2 c \log (-a c-b c x+1)}{2 e (b d-a e) (-a c e+b c d+e)}-\frac{b^2 c \log (d+e x)}{2 e (b d-a e) (-a c e+b c d+e)}+\frac{b^2 \log (-a c-b c x+1) \log \left (\frac{b c (d+e x)}{-a c e+b c d+e}\right )}{2 e (b d-a e)^2}-\frac{b \log (-a c-b c x+1)}{2 e (d+e x) (b d-a e)} \]

[Out]

(b^2*c*Log[1 - a*c - b*c*x])/(2*e*(b*d - a*e)*(b*c*d + e - a*c*e)) - (b*Log[1 - a*c - b*c*x])/(2*e*(b*d - a*e)
*(d + e*x)) - (b^2*c*Log[d + e*x])/(2*e*(b*d - a*e)*(b*c*d + e - a*c*e)) + (b^2*Log[1 - a*c - b*c*x]*Log[(b*c*
(d + e*x))/(b*c*d + e - a*c*e)])/(2*e*(b*d - a*e)^2) + (b^2*PolyLog[2, c*(a + b*x)])/(2*e*(b*d - a*e)^2) - Pol
yLog[2, c*(a + b*x)]/(2*e*(d + e*x)^2) + (b^2*PolyLog[2, (e*(1 - a*c - b*c*x))/(b*c*d + e - a*c*e)])/(2*e*(b*d
- a*e)^2)

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Rubi [A]  time = 0.270028, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.471, Rules used = {6598, 2418, 2393, 2391, 2395, 36, 31, 2394} \[ \frac{b^2 \text{PolyLog}(2,c (a+b x))}{2 e (b d-a e)^2}+\frac{b^2 \text{PolyLog}\left (2,\frac{e (-a c-b c x+1)}{-a c e+b c d+e}\right )}{2 e (b d-a e)^2}-\frac{\text{PolyLog}(2,c (a+b x))}{2 e (d+e x)^2}+\frac{b^2 c \log (-a c-b c x+1)}{2 e (b d-a e) (-a c e+b c d+e)}-\frac{b^2 c \log (d+e x)}{2 e (b d-a e) (-a c e+b c d+e)}+\frac{b^2 \log (-a c-b c x+1) \log \left (\frac{b c (d+e x)}{-a c e+b c d+e}\right )}{2 e (b d-a e)^2}-\frac{b \log (-a c-b c x+1)}{2 e (d+e x) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)]/(d + e*x)^3,x]

[Out]

(b^2*c*Log[1 - a*c - b*c*x])/(2*e*(b*d - a*e)*(b*c*d + e - a*c*e)) - (b*Log[1 - a*c - b*c*x])/(2*e*(b*d - a*e)
*(d + e*x)) - (b^2*c*Log[d + e*x])/(2*e*(b*d - a*e)*(b*c*d + e - a*c*e)) + (b^2*Log[1 - a*c - b*c*x]*Log[(b*c*
(d + e*x))/(b*c*d + e - a*c*e)])/(2*e*(b*d - a*e)^2) + (b^2*PolyLog[2, c*(a + b*x)])/(2*e*(b*d - a*e)^2) - Pol
yLog[2, c*(a + b*x)]/(2*e*(d + e*x)^2) + (b^2*PolyLog[2, (e*(1 - a*c - b*c*x))/(b*c*d + e - a*c*e)])/(2*e*(b*d
- a*e)^2)

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 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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rubi steps

\begin{align*} \int \frac{\text{Li}_2(c (a+b x))}{(d+e x)^3} \, dx &=-\frac{\text{Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac{b \int \frac{\log (1-a c-b c x)}{(a+b x) (d+e x)^2} \, dx}{2 e}\\ &=-\frac{\text{Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac{b \int \left (\frac{b^2 \log (1-a c-b c x)}{(b d-a e)^2 (a+b x)}-\frac{e \log (1-a c-b c x)}{(b d-a e) (d+e x)^2}-\frac{b e \log (1-a c-b c x)}{(b d-a e)^2 (d+e x)}\right ) \, dx}{2 e}\\ &=-\frac{\text{Li}_2(c (a+b x))}{2 e (d+e x)^2}+\frac{b^2 \int \frac{\log (1-a c-b c x)}{d+e x} \, dx}{2 (b d-a e)^2}-\frac{b^3 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{2 e (b d-a e)^2}+\frac{b \int \frac{\log (1-a c-b c x)}{(d+e x)^2} \, dx}{2 (b d-a e)}\\ &=-\frac{b \log (1-a c-b c x)}{2 e (b d-a e) (d+e x)}+\frac{b^2 \log (1-a c-b c x) \log \left (\frac{b c (d+e x)}{b c d+e-a c e}\right )}{2 e (b d-a e)^2}-\frac{\text{Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 e (b d-a e)^2}+\frac{\left (b^3 c\right ) \int \frac{\log \left (-\frac{b c (d+e x)}{-b c d-(1-a c) e}\right )}{1-a c-b c x} \, dx}{2 e (b d-a e)^2}-\frac{\left (b^2 c\right ) \int \frac{1}{(1-a c-b c x) (d+e x)} \, dx}{2 e (b d-a e)}\\ &=-\frac{b \log (1-a c-b c x)}{2 e (b d-a e) (d+e x)}+\frac{b^2 \log (1-a c-b c x) \log \left (\frac{b c (d+e x)}{b c d+e-a c e}\right )}{2 e (b d-a e)^2}+\frac{b^2 \text{Li}_2(c (a+b x))}{2 e (b d-a e)^2}-\frac{\text{Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{e x}{-b c d-(1-a c) e}\right )}{x} \, dx,x,1-a c-b c x\right )}{2 e (b d-a e)^2}-\frac{\left (b^2 c\right ) \int \frac{1}{d+e x} \, dx}{2 (b d-a e) (b c d+e-a c e)}-\frac{\left (b^3 c^2\right ) \int \frac{1}{1-a c-b c x} \, dx}{2 e (b d-a e) (b c d+e-a c e)}\\ &=\frac{b^2 c \log (1-a c-b c x)}{2 e (b d-a e) (b c d+e-a c e)}-\frac{b \log (1-a c-b c x)}{2 e (b d-a e) (d+e x)}-\frac{b^2 c \log (d+e x)}{2 e (b d-a e) (b c d+e-a c e)}+\frac{b^2 \log (1-a c-b c x) \log \left (\frac{b c (d+e x)}{b c d+e-a c e}\right )}{2 e (b d-a e)^2}+\frac{b^2 \text{Li}_2(c (a+b x))}{2 e (b d-a e)^2}-\frac{\text{Li}_2(c (a+b x))}{2 e (d+e x)^2}+\frac{b^2 \text{Li}_2\left (\frac{e (1-a c-b c x)}{b c d+e-a c e}\right )}{2 e (b d-a e)^2}\\ \end{align*}

Mathematica [A]  time = 0.420508, size = 190, normalized size = 0.68 \[ \frac{\frac{b \left (b \text{PolyLog}\left (2,\frac{e (a c+b c x-1)}{e (a c-1)-b c d}\right )+b \text{PolyLog}(2,c (a+b x))+b \log (-a c-b c x+1) \log \left (\frac{b c (d+e x)}{-a c e+b c d+e}\right )-\frac{(b d-a e) \log (-a c-b c x+1)}{d+e x}+\frac{b c (b d-a e) (\log (-a c-b c x+1)-\log (d+e x))}{-a c e+b c d+e}\right )}{(b d-a e)^2}-\frac{\text{PolyLog}(2,c (a+b x))}{(d+e x)^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, c*(a + b*x)]/(d + e*x)^3,x]

[Out]

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

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Maple [A]  time = 0.462, size = 437, normalized size = 1.6 \begin{align*} -{\frac{{b}^{2}{c}^{2}{\it polylog} \left ( 2,xbc+ac \right ) }{2\, \left ( bcex+bcd \right ) ^{2}e}}+{\frac{{b}^{2}}{2\,e \left ( ae-bd \right ) ^{2}}{\it dilog} \left ({\frac{ace-bcd+ \left ( -xbc-ac+1 \right ) e-e}{ace-bcd-e}} \right ) }+{\frac{{b}^{2}\ln \left ( -xbc-ac+1 \right ) }{2\,e \left ( ae-bd \right ) ^{2}}\ln \left ({\frac{ace-bcd+ \left ( -xbc-ac+1 \right ) e-e}{ace-bcd-e}} \right ) }-{\frac{{b}^{2}c\ln \left ( ace-bcd+ \left ( -xbc-ac+1 \right ) e-e \right ) }{2\,e \left ( ae-bd \right ) \left ( ace-bcd-e \right ) }}-{\frac{{b}^{3}{c}^{2}\ln \left ( -xbc-ac+1 \right ) x}{ \left ( 2\,ae-2\,bd \right ) \left ( ace-bcd-e \right ) \left ( -bcex-bcd \right ) }}-{\frac{{b}^{2}{c}^{2}\ln \left ( -xbc-ac+1 \right ) a}{ \left ( 2\,ae-2\,bd \right ) \left ( ace-bcd-e \right ) \left ( -bcex-bcd \right ) }}+{\frac{{b}^{2}c\ln \left ( -xbc-ac+1 \right ) }{ \left ( 2\,ae-2\,bd \right ) \left ( ace-bcd-e \right ) \left ( -bcex-bcd \right ) }}+{\frac{{b}^{2}{\it dilog} \left ( -xbc-ac+1 \right ) }{2\,e \left ( ae-bd \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,c*(b*x+a))/(e*x+d)^3,x)

[Out]

-1/2*b^2*c^2/(b*c*e*x+b*c*d)^2/e*polylog(2,b*c*x+a*c)+1/2*b^2/e/(a*e-b*d)^2*dilog((a*c*e-b*c*d+(-b*c*x-a*c+1)*
e-e)/(a*c*e-b*c*d-e))+1/2*b^2/e/(a*e-b*d)^2*ln(-b*c*x-a*c+1)*ln((a*c*e-b*c*d+(-b*c*x-a*c+1)*e-e)/(a*c*e-b*c*d-
e))-1/2*b^2*c/e/(a*e-b*d)/(a*c*e-b*c*d-e)*ln(a*c*e-b*c*d+(-b*c*x-a*c+1)*e-e)-1/2*b^3*c^2/(a*e-b*d)*ln(-b*c*x-a
*c+1)/(a*c*e-b*c*d-e)/(-b*c*e*x-b*c*d)*x-1/2*b^2*c^2/(a*e-b*d)*ln(-b*c*x-a*c+1)/(a*c*e-b*c*d-e)/(-b*c*e*x-b*c*
d)*a+1/2*b^2*c/(a*e-b*d)*ln(-b*c*x-a*c+1)/(a*c*e-b*c*d-e)/(-b*c*e*x-b*c*d)+1/2*b^2/e/(a*e-b*d)^2*dilog(-b*c*x-
a*c+1)

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Maxima [A]  time = 1.00782, size = 512, normalized size = 1.84 \begin{align*} \frac{b^{2} c \log \left (b c x + a c - 1\right )}{2 \,{\left (b^{2} c d^{2} e -{\left (2 \, a b c - b\right )} d e^{2} +{\left (a^{2} c - a\right )} e^{3}\right )}} - \frac{b^{2} c \log \left (e x + d\right )}{2 \,{\left (b^{2} c d^{2} e -{\left (2 \, a b c - b\right )} d e^{2} +{\left (a^{2} c - a\right )} e^{3}\right )}} - \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 )} b^{2}}{2 \,{\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )}} + \frac{{\left (\log \left (-b c x - a c + 1\right ) \log \left (\frac{b c e x +{\left (a c - 1\right )} e}{b c d -{\left (a c - 1\right )} e} + 1\right ) +{\rm Li}_2\left (-\frac{b c e x +{\left (a c - 1\right )} e}{b c d -{\left (a c - 1\right )} e}\right )\right )} b^{2}}{2 \,{\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )}} - \frac{{\left (b d - a e\right )}{\rm Li}_2\left (b c x + a c\right ) +{\left (b e x + b d\right )} \log \left (-b c x - a c + 1\right )}{2 \,{\left (b d^{3} e - a d^{2} e^{2} +{\left (b d e^{3} - a e^{4}\right )} x^{2} + 2 \,{\left (b d^{2} e^{2} - a d e^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/(e*x+d)^3,x, algorithm="maxima")

[Out]

1/2*b^2*c*log(b*c*x + a*c - 1)/(b^2*c*d^2*e - (2*a*b*c - b)*d*e^2 + (a^2*c - a)*e^3) - 1/2*b^2*c*log(e*x + d)/
(b^2*c*d^2*e - (2*a*b*c - b)*d*e^2 + (a^2*c - a)*e^3) - 1/2*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b
*c*x - a*c + 1))*b^2/(b^2*d^2*e - 2*a*b*d*e^2 + a^2*e^3) + 1/2*(log(-b*c*x - a*c + 1)*log((b*c*e*x + (a*c - 1)
*e)/(b*c*d - (a*c - 1)*e) + 1) + dilog(-(b*c*e*x + (a*c - 1)*e)/(b*c*d - (a*c - 1)*e)))*b^2/(b^2*d^2*e - 2*a*b
*d*e^2 + a^2*e^3) - 1/2*((b*d - a*e)*dilog(b*c*x + a*c) + (b*e*x + b*d)*log(-b*c*x - a*c + 1))/(b*d^3*e - a*d^
2*e^2 + (b*d*e^3 - a*e^4)*x^2 + 2*(b*d^2*e^2 - a*d*e^3)*x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\rm Li}_2\left (b c x + a c\right )}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/(e*x+d)^3,x, algorithm="fricas")

[Out]

integral(dilog(b*c*x + a*c)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)

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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(polylog(2,c*(b*x+a))/(e*x+d)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/(e*x+d)^3,x, algorithm="giac")

[Out]

integrate(dilog((b*x + a)*c)/(e*x + d)^3, x)