3.129 \(\int \frac {\text {Li}_2(c (a+b x))}{x^3} \, dx\)

Optimal. Leaf size=173 \[ \frac {b^2 \text {Li}_2(c (a+b x))}{2 a^2}+\frac {b^2 \text {Li}_2\left (1-\frac {b c x}{1-a c}\right )}{2 a^2}+\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (-a c-b c x+1)}{2 a^2}+\frac {b^2 c \log (x)}{2 a (1-a c)}-\frac {b^2 c \log (-a c-b c x+1)}{2 a (1-a c)}-\frac {\text {Li}_2(c (a+b x))}{2 x^2}+\frac {b \log (-a c-b c x+1)}{2 a x} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {6598, 44, 2416, 2395, 36, 29, 31, 2394, 2315, 2393, 2391} \[ \frac {b^2 \text {PolyLog}(2,c (a+b x))}{2 a^2}+\frac {b^2 \text {PolyLog}\left (2,1-\frac {b c x}{1-a c}\right )}{2 a^2}-\frac {\text {PolyLog}(2,c (a+b x))}{2 x^2}+\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (-a c-b c x+1)}{2 a^2}+\frac {b^2 c \log (x)}{2 a (1-a c)}-\frac {b^2 c \log (-a c-b c x+1)}{2 a (1-a c)}+\frac {b \log (-a c-b c x+1)}{2 a x} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)]/x^3,x]

[Out]

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

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 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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 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 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 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 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 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 \frac {\text {Li}_2(c (a+b x))}{x^3} \, dx &=-\frac {\text {Li}_2(c (a+b x))}{2 x^2}-\frac {1}{2} b \int \frac {\log (1-a c-b c x)}{x^2 (a+b x)} \, dx\\ &=-\frac {\text {Li}_2(c (a+b x))}{2 x^2}-\frac {1}{2} b \int \left (\frac {\log (1-a c-b c x)}{a x^2}-\frac {b \log (1-a c-b c x)}{a^2 x}+\frac {b^2 \log (1-a c-b c x)}{a^2 (a+b x)}\right ) \, dx\\ &=-\frac {\text {Li}_2(c (a+b x))}{2 x^2}-\frac {b \int \frac {\log (1-a c-b c x)}{x^2} \, dx}{2 a}+\frac {b^2 \int \frac {\log (1-a c-b c x)}{x} \, dx}{2 a^2}-\frac {b^3 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{2 a^2}\\ &=\frac {b \log (1-a c-b c x)}{2 a x}+\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}-\frac {\text {Li}_2(c (a+b x))}{2 x^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 a^2}+\frac {\left (b^2 c\right ) \int \frac {1}{x (1-a c-b c x)} \, dx}{2 a}+\frac {\left (b^3 c\right ) \int \frac {\log \left (-\frac {b c x}{-1+a c}\right )}{1-a c-b c x} \, dx}{2 a^2}\\ &=\frac {b \log (1-a c-b c x)}{2 a x}+\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}+\frac {b^2 \text {Li}_2(c (a+b x))}{2 a^2}-\frac {\text {Li}_2(c (a+b x))}{2 x^2}+\frac {b^2 \text {Li}_2\left (1-\frac {b c x}{1-a c}\right )}{2 a^2}+\frac {\left (b^2 c\right ) \int \frac {1}{x} \, dx}{2 a (1-a c)}+\frac {\left (b^3 c^2\right ) \int \frac {1}{1-a c-b c x} \, dx}{2 a (1-a c)}\\ &=\frac {b^2 c \log (x)}{2 a (1-a c)}-\frac {b^2 c \log (1-a c-b c x)}{2 a (1-a c)}+\frac {b \log (1-a c-b c x)}{2 a x}+\frac {b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)}{2 a^2}+\frac {b^2 \text {Li}_2(c (a+b x))}{2 a^2}-\frac {\text {Li}_2(c (a+b x))}{2 x^2}+\frac {b^2 \text {Li}_2\left (1-\frac {b c x}{1-a c}\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 131, normalized size = 0.76 \[ \frac {b x \left (b x (a c-1) \text {Li}_2\left (\frac {a c+b x c-1}{a c-1}\right )-a b c x \log (x)+\left (a (a c+b c x-1)+b x (a c-1) \log \left (\frac {b c x}{1-a c}\right )\right ) \log (-a c-b c x+1)\right )-(a c-1) \left (a^2-b^2 x^2\right ) \text {Li}_2(c (a+b x))}{2 a^2 x^2 (a c-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm Li}_2\left (b c x + a c\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(dilog(b*c*x + a*c)/x^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog((b*x + a)*c)/x^3, x)

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maple [A]  time = 0.02, size = 195, normalized size = 1.13 \[ -\frac {\polylog \left (2, b c x +a c \right )}{2 x^{2}}+\frac {b^{2} \dilog \left (-b c x -a c +1\right )}{2 a^{2}}-\frac {b^{2} c \ln \left (-b c x \right )}{2 a \left (a c -1\right )}+\frac {b^{2} c \ln \left (-b c x -a c +1\right )}{2 a \left (a c -1\right )}+\frac {b c \ln \left (-b c x -a c +1\right )}{2 \left (a c -1\right ) x}-\frac {b \ln \left (-b c x -a c +1\right )}{2 a \left (a c -1\right ) x}+\frac {b^{2} \ln \left (-b c x -a c +1\right ) \ln \left (-\frac {x b c}{a c -1}\right )}{2 a^{2}}+\frac {b^{2} \dilog \left (-\frac {x b c}{a c -1}\right )}{2 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 193, normalized size = 1.12 \[ -\frac {b^{2} c \log \relax (x)}{2 \, {\left (a^{2} c - a\right )}} - \frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} b^{2}}{2 \, a^{2}} + \frac {{\left (\log \left (-b c x - a c + 1\right ) \log \left (-\frac {b c x + a c - 1}{a c - 1} + 1\right ) + {\rm Li}_2\left (\frac {b c x + a c - 1}{a c - 1}\right )\right )} b^{2}}{2 \, a^{2}} - \frac {{\left (a^{2} c - a\right )} {\rm Li}_2\left (b c x + a c\right ) - {\left (b^{2} c x^{2} + {\left (a b c - b\right )} x\right )} \log \left (-b c x - a c + 1\right )}{2 \, {\left (a^{2} c - a\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^2*c*log(x)/(a^2*c - a) - 1/2*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*b^2/a^2
+ 1/2*(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^2/
a^2 - 1/2*((a^2*c - a)*dilog(b*c*x + a*c) - (b^2*c*x^2 + (a*b*c - b)*x)*log(-b*c*x - a*c + 1))/((a^2*c - a)*x^
2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, c*(a + b*x))/x^3,x)

[Out]

int(polylog(2, c*(a + b*x))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{2}\left (a c + b c x\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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