3.128 \(\int \frac{\text{PolyLog}(2,c (a+b x))}{x^2} \, dx\)

Optimal. Leaf size=84 \[ -\frac{b \text{PolyLog}(2,c (a+b x))}{a}-\frac{\text{PolyLog}(2,c (a+b x))}{x}-\frac{b \text{PolyLog}\left (2,1-\frac{b c x}{1-a c}\right )}{a}-\frac{b \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)}{a} \]

[Out]

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

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Rubi [A]  time = 0.116564, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {6598, 36, 29, 31, 2416, 2394, 2315, 2393, 2391} \[ -\frac{b \text{PolyLog}(2,c (a+b x))}{a}-\frac{\text{PolyLog}(2,c (a+b x))}{x}-\frac{b \text{PolyLog}\left (2,1-\frac{b c x}{1-a c}\right )}{a}-\frac{b \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)}{a} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)]/x^2,x]

[Out]

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

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 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

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]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

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]

Rubi steps

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

Mathematica [A]  time = 0.0519712, size = 73, normalized size = 0.87 \[ -\frac{(a+b x) \text{PolyLog}(2,c (a+b x))+b x \left (\text{PolyLog}\left (2,\frac{a c+b c x-1}{a c-1}\right )+\log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)\right )}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.13, size = 85, normalized size = 1. \begin{align*} -{\frac{{\it polylog} \left ( 2,xbc+ac \right ) }{x}}-{\frac{b\ln \left ( -xbc-ac+1 \right ) }{a}\ln \left ( -{\frac{xbc}{ac-1}} \right ) }-{\frac{b}{a}{\it dilog} \left ( -{\frac{xbc}{ac-1}} \right ) }-{\frac{b{\it dilog} \left ( -xbc-ac+1 \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.981607, size = 154, normalized size = 1.83 \begin{align*} \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}{a} - \frac{{\left (\log \left (-b c x - a c + 1\right ) \log \left (-\frac{b c x + a c - 1}{a c - 1} + 1\right ) +{\rm Li}_2\left (\frac{b c x + a c - 1}{a c - 1}\right )\right )} b}{a} - \frac{{\rm Li}_2\left (b c x + a c\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*b/a - (log(-b*c*x - a*c + 1)*log(-(b*c*x +
a*c - 1)/(a*c - 1) + 1) + dilog((b*c*x + a*c - 1)/(a*c - 1)))*b/a - dilog(b*c*x + a*c)/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 )}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(dilog(b*c*x + a*c)/x^2, 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))/x**2,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 )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog((b*x + a)*c)/x^2, x)