3.130 \(\int \frac{\text{PolyLog}(2,c (a+b x))}{x^4} \, dx\)

Optimal. Leaf size=276 \[ -\frac{b^3 \text{PolyLog}(2,c (a+b x))}{3 a^3}-\frac{b^3 \text{PolyLog}\left (2,1-\frac{b c x}{1-a c}\right )}{3 a^3}-\frac{\text{PolyLog}(2,c (a+b x))}{3 x^3}-\frac{b^3 c \log (x)}{3 a^2 (1-a c)}-\frac{b^3 \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)}{3 a^3}+\frac{b^3 c \log (-a c-b c x+1)}{3 a^2 (1-a c)}-\frac{b^2 \log (-a c-b c x+1)}{3 a^2 x}+\frac{b^3 c^2 \log (x)}{6 a (1-a c)^2}-\frac{b^3 c^2 \log (-a c-b c x+1)}{6 a (1-a c)^2}-\frac{b^2 c}{6 a x (1-a c)}+\frac{b \log (-a c-b c x+1)}{6 a x^2} \]

[Out]

-(b^2*c)/(6*a*(1 - a*c)*x) + (b^3*c^2*Log[x])/(6*a*(1 - a*c)^2) - (b^3*c*Log[x])/(3*a^2*(1 - a*c)) - (b^3*c^2*
Log[1 - a*c - b*c*x])/(6*a*(1 - a*c)^2) + (b^3*c*Log[1 - a*c - b*c*x])/(3*a^2*(1 - a*c)) + (b*Log[1 - a*c - b*
c*x])/(6*a*x^2) - (b^2*Log[1 - a*c - b*c*x])/(3*a^2*x) - (b^3*Log[(b*c*x)/(1 - a*c)]*Log[1 - a*c - b*c*x])/(3*
a^3) - (b^3*PolyLog[2, c*(a + b*x)])/(3*a^3) - PolyLog[2, c*(a + b*x)]/(3*x^3) - (b^3*PolyLog[2, 1 - (b*c*x)/(
1 - a*c)])/(3*a^3)

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Rubi [A]  time = 0.272902, antiderivative size = 276, normalized size of antiderivative = 1., number of steps used = 14, 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} \[ -\frac{b^3 \text{PolyLog}(2,c (a+b x))}{3 a^3}-\frac{b^3 \text{PolyLog}\left (2,1-\frac{b c x}{1-a c}\right )}{3 a^3}-\frac{\text{PolyLog}(2,c (a+b x))}{3 x^3}-\frac{b^3 c \log (x)}{3 a^2 (1-a c)}-\frac{b^3 \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)}{3 a^3}+\frac{b^3 c \log (-a c-b c x+1)}{3 a^2 (1-a c)}-\frac{b^2 \log (-a c-b c x+1)}{3 a^2 x}+\frac{b^3 c^2 \log (x)}{6 a (1-a c)^2}-\frac{b^3 c^2 \log (-a c-b c x+1)}{6 a (1-a c)^2}-\frac{b^2 c}{6 a x (1-a c)}+\frac{b \log (-a c-b c x+1)}{6 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)]/x^4,x]

[Out]

-(b^2*c)/(6*a*(1 - a*c)*x) + (b^3*c^2*Log[x])/(6*a*(1 - a*c)^2) - (b^3*c*Log[x])/(3*a^2*(1 - a*c)) - (b^3*c^2*
Log[1 - a*c - b*c*x])/(6*a*(1 - a*c)^2) + (b^3*c*Log[1 - a*c - b*c*x])/(3*a^2*(1 - a*c)) + (b*Log[1 - a*c - b*
c*x])/(6*a*x^2) - (b^2*Log[1 - a*c - b*c*x])/(3*a^2*x) - (b^3*Log[(b*c*x)/(1 - a*c)]*Log[1 - a*c - b*c*x])/(3*
a^3) - (b^3*PolyLog[2, c*(a + b*x)])/(3*a^3) - PolyLog[2, c*(a + b*x)]/(3*x^3) - (b^3*PolyLog[2, 1 - (b*c*x)/(
1 - a*c)])/(3*a^3)

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 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 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 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 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 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^4} \, dx &=-\frac{\text{Li}_2(c (a+b x))}{3 x^3}-\frac{1}{3} b \int \frac{\log (1-a c-b c x)}{x^3 (a+b x)} \, dx\\ &=-\frac{\text{Li}_2(c (a+b x))}{3 x^3}-\frac{1}{3} b \int \left (\frac{\log (1-a c-b c x)}{a x^3}-\frac{b \log (1-a c-b c x)}{a^2 x^2}+\frac{b^2 \log (1-a c-b c x)}{a^3 x}-\frac{b^3 \log (1-a c-b c x)}{a^3 (a+b x)}\right ) \, dx\\ &=-\frac{\text{Li}_2(c (a+b x))}{3 x^3}-\frac{b \int \frac{\log (1-a c-b c x)}{x^3} \, dx}{3 a}+\frac{b^2 \int \frac{\log (1-a c-b c x)}{x^2} \, dx}{3 a^2}-\frac{b^3 \int \frac{\log (1-a c-b c x)}{x} \, dx}{3 a^3}+\frac{b^4 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{3 a^3}\\ &=\frac{b \log (1-a c-b c x)}{6 a x^2}-\frac{b^2 \log (1-a c-b c x)}{3 a^2 x}-\frac{b^3 \log \left (\frac{b c x}{1-a c}\right ) \log (1-a c-b c x)}{3 a^3}-\frac{\text{Li}_2(c (a+b x))}{3 x^3}+\frac{b^3 \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{3 a^3}+\frac{\left (b^2 c\right ) \int \frac{1}{x^2 (1-a c-b c x)} \, dx}{6 a}-\frac{\left (b^3 c\right ) \int \frac{1}{x (1-a c-b c x)} \, dx}{3 a^2}-\frac{\left (b^4 c\right ) \int \frac{\log \left (-\frac{b c x}{-1+a c}\right )}{1-a c-b c x} \, dx}{3 a^3}\\ &=\frac{b \log (1-a c-b c x)}{6 a x^2}-\frac{b^2 \log (1-a c-b c x)}{3 a^2 x}-\frac{b^3 \log \left (\frac{b c x}{1-a c}\right ) \log (1-a c-b c x)}{3 a^3}-\frac{b^3 \text{Li}_2(c (a+b x))}{3 a^3}-\frac{\text{Li}_2(c (a+b x))}{3 x^3}-\frac{b^3 \text{Li}_2\left (1-\frac{b c x}{1-a c}\right )}{3 a^3}+\frac{\left (b^2 c\right ) \int \left (-\frac{1}{(-1+a c) x^2}+\frac{b c}{(-1+a c)^2 x}-\frac{b^2 c^2}{(-1+a c)^2 (-1+a c+b c x)}\right ) \, dx}{6 a}-\frac{\left (b^3 c\right ) \int \frac{1}{x} \, dx}{3 a^2 (1-a c)}-\frac{\left (b^4 c^2\right ) \int \frac{1}{1-a c-b c x} \, dx}{3 a^2 (1-a c)}\\ &=-\frac{b^2 c}{6 a (1-a c) x}+\frac{b^3 c^2 \log (x)}{6 a (1-a c)^2}-\frac{b^3 c \log (x)}{3 a^2 (1-a c)}-\frac{b^3 c^2 \log (1-a c-b c x)}{6 a (1-a c)^2}+\frac{b^3 c \log (1-a c-b c x)}{3 a^2 (1-a c)}+\frac{b \log (1-a c-b c x)}{6 a x^2}-\frac{b^2 \log (1-a c-b c x)}{3 a^2 x}-\frac{b^3 \log \left (\frac{b c x}{1-a c}\right ) \log (1-a c-b c x)}{3 a^3}-\frac{b^3 \text{Li}_2(c (a+b x))}{3 a^3}-\frac{\text{Li}_2(c (a+b x))}{3 x^3}-\frac{b^3 \text{Li}_2\left (1-\frac{b c x}{1-a c}\right )}{3 a^3}\\ \end{align*}

Mathematica [A]  time = 0.303672, size = 210, normalized size = 0.76 \[ -\frac{b \left (2 b^2 \text{PolyLog}(2,c (a+b x))+2 b^2 \text{PolyLog}\left (2,\frac{a c+b c x-1}{a c-1}\right )-\frac{a^2 \log (-a c-b c x+1)}{x^2}-\frac{a^2 b c (-b c x \log (-a c-b c x+1)+a c+b c x \log (x)-1)}{x (a c-1)^2}-\frac{2 a b^2 c (\log (x)-\log (-a c-b c x+1))}{a c-1}+2 b^2 \log \left (\frac{b c x}{1-a c}\right ) \log (-a c-b c x+1)+\frac{2 a b \log (-a c-b c x+1)}{x}\right )}{6 a^3}-\frac{\text{PolyLog}(2,a c+b c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, c*(a + b*x)]/x^4,x]

[Out]

-(b*((-2*a*b^2*c*(Log[x] - Log[1 - a*c - b*c*x]))/(-1 + a*c) - (a^2*Log[1 - a*c - b*c*x])/x^2 + (2*a*b*Log[1 -
a*c - b*c*x])/x + 2*b^2*Log[(b*c*x)/(1 - a*c)]*Log[1 - a*c - b*c*x] - (a^2*b*c*(-1 + a*c + b*c*x*Log[x] - b*c
*x*Log[1 - a*c - b*c*x]))/((-1 + a*c)^2*x) + 2*b^2*PolyLog[2, c*(a + b*x)] + 2*b^2*PolyLog[2, (-1 + a*c + b*c*
x)/(-1 + a*c)]))/(6*a^3) - PolyLog[2, a*c + b*c*x]/(3*x^3)

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Maple [A]  time = 0.214, size = 376, normalized size = 1.4 \begin{align*} -{\frac{{\it polylog} \left ( 2,xbc+ac \right ) }{3\,{x}^{3}}}-{\frac{{b}^{3}\ln \left ( -xbc-ac+1 \right ) }{3\,{a}^{3}}\ln \left ( -{\frac{xbc}{ac-1}} \right ) }-{\frac{{b}^{3}}{3\,{a}^{3}}{\it dilog} \left ( -{\frac{xbc}{ac-1}} \right ) }+{\frac{{b}^{3}{c}^{2}\ln \left ( -xbc \right ) }{6\,a \left ( ac-1 \right ) ^{2}}}+{\frac{{b}^{2}{c}^{2}}{6\, \left ( ac-1 \right ) ^{2}x}}-{\frac{{b}^{2}c}{6\,a \left ( ac-1 \right ) ^{2}x}}-{\frac{{b}^{3}{c}^{2}\ln \left ( -xbc-ac+1 \right ) }{6\,a \left ( ac-1 \right ) ^{2}}}+{\frac{b{c}^{2}\ln \left ( -xbc-ac+1 \right ) a}{6\,{x}^{2} \left ( ac-1 \right ) ^{2}}}-{\frac{bc\ln \left ( -xbc-ac+1 \right ) }{3\,{x}^{2} \left ( ac-1 \right ) ^{2}}}+{\frac{b\ln \left ( -xbc-ac+1 \right ) }{6\,a{x}^{2} \left ( ac-1 \right ) ^{2}}}-{\frac{{b}^{3}{\it dilog} \left ( -xbc-ac+1 \right ) }{3\,{a}^{3}}}+{\frac{c{b}^{3}\ln \left ( -xbc \right ) }{3\,{a}^{2} \left ( ac-1 \right ) }}-{\frac{c{b}^{3}\ln \left ( -xbc-ac+1 \right ) }{3\,{a}^{2} \left ( ac-1 \right ) }}-{\frac{{b}^{2}c\ln \left ( -xbc-ac+1 \right ) }{3\,a \left ( ac-1 \right ) x}}+{\frac{{b}^{2}\ln \left ( -xbc-ac+1 \right ) }{3\,{a}^{2} \left ( ac-1 \right ) x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,c*(b*x+a))/x^4,x)

[Out]

-1/3*polylog(2,b*c*x+a*c)/x^3-1/3*b^3/a^3*ln(-b*c*x-a*c+1)*ln(-b*c*x/(a*c-1))-1/3*b^3/a^3*dilog(-b*c*x/(a*c-1)
)+1/6*b^3*c^2/a/(a*c-1)^2*ln(-x*b*c)+1/6*b^2*c^2/(a*c-1)^2/x-1/6*b^2*c/a/(a*c-1)^2/x-1/6*b^3*c^2/a*ln(-b*c*x-a
*c+1)/(a*c-1)^2+1/6*b*c^2*ln(-b*c*x-a*c+1)/x^2/(a*c-1)^2*a-1/3*b*c*ln(-b*c*x-a*c+1)/x^2/(a*c-1)^2+1/6*b/a*ln(-
b*c*x-a*c+1)/x^2/(a*c-1)^2-1/3*b^3/a^3*dilog(-b*c*x-a*c+1)+1/3*b^3*c/a^2/(a*c-1)*ln(-x*b*c)-1/3*b^3*c/a^2*ln(-
b*c*x-a*c+1)/(a*c-1)-1/3*b^2*c/a*ln(-b*c*x-a*c+1)/(a*c-1)/x+1/3*b^2/a^2*ln(-b*c*x-a*c+1)/(a*c-1)/x

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Maxima [A]  time = 1.02696, size = 408, normalized size = 1.48 \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^{3}}{3 \, a^{3}} - \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^{3}}{3 \, a^{3}} + \frac{{\left (3 \, a b^{3} c^{2} - 2 \, b^{3} c\right )} \log \left (x\right )}{6 \,{\left (a^{4} c^{2} - 2 \, a^{3} c + a^{2}\right )}} + \frac{{\left (a^{2} b^{2} c^{2} - a b^{2} c\right )} x^{2} - 2 \,{\left (a^{4} c^{2} - 2 \, a^{3} c + a^{2}\right )}{\rm Li}_2\left (b c x + a c\right ) -{\left ({\left (3 \, a b^{3} c^{2} - 2 \, b^{3} c\right )} x^{3} + 2 \,{\left (a^{2} b^{2} c^{2} - 2 \, a b^{2} c + b^{2}\right )} x^{2} -{\left (a^{3} b c^{2} - 2 \, a^{2} b c + a b\right )} x\right )} \log \left (-b c x - a c + 1\right )}{6 \,{\left (a^{4} c^{2} - 2 \, a^{3} c + a^{2}\right )} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/x^4,x, algorithm="maxima")

[Out]

1/3*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*b^3/a^3 - 1/3*(log(-b*c*x - a*c + 1)*lo
g(-(b*c*x + a*c - 1)/(a*c - 1) + 1) + dilog((b*c*x + a*c - 1)/(a*c - 1)))*b^3/a^3 + 1/6*(3*a*b^3*c^2 - 2*b^3*c
)*log(x)/(a^4*c^2 - 2*a^3*c + a^2) + 1/6*((a^2*b^2*c^2 - a*b^2*c)*x^2 - 2*(a^4*c^2 - 2*a^3*c + a^2)*dilog(b*c*
x + a*c) - ((3*a*b^3*c^2 - 2*b^3*c)*x^3 + 2*(a^2*b^2*c^2 - 2*a*b^2*c + b^2)*x^2 - (a^3*b*c^2 - 2*a^2*b*c + a*b
)*x)*log(-b*c*x - a*c + 1))/((a^4*c^2 - 2*a^3*c + a^2)*x^3)

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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^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/x^4,x, algorithm="fricas")

[Out]

integral(dilog(b*c*x + a*c)/x^4, 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**4,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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/x^4,x, algorithm="giac")

[Out]

integrate(dilog((b*x + a)*c)/x^4, x)