3.2.30 \(\int \frac {\text {PolyLog}(2,c (a+b x))}{x^4} \, dx\) [130]

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

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 276, normalized size of antiderivative = 1.00, 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 = {6733, 46, 2463, 2442, 36, 29, 31, 2441, 2352, 2440, 2438} \begin {gather*} -\frac {b^3 \text {Li}_2(c (a+b x))}{3 a^3}-\frac {b^3 \text {Li}_2\left (1-\frac {b c x}{1-a c}\right )}{3 a^3}-\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 (x)}{3 a^2 (1-a c)}+\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 {\text {Li}_2(c (a+b x))}{3 x^3}+\frac {b \log (-a c-b c x+1)}{6 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)]/x^4,x]

[Out]

-1/6*(b^2*c)/(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 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 46

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] && Lt
Q[m + n + 2, 0])

Rule 2352

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

Rule 2438

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 2440

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 2441

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 2442

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 2463

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 6733

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^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 \text {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*}

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Mathematica [A]
time = 0.21, size = 210, normalized size = 0.76 \begin {gather*} -\frac {b \left (-\frac {2 a b^2 c (\log (x)-\log (1-a c-b c x))}{-1+a c}-\frac {a^2 \log (1-a c-b c x)}{x^2}+\frac {2 a b \log (1-a c-b c x)}{x}+2 b^2 \log \left (\frac {b c x}{1-a c}\right ) \log (1-a c-b c x)-\frac {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 \text {PolyLog}(2,c (a+b x))+2 b^2 \text {PolyLog}\left (2,\frac {-1+a c+b c x}{-1+a c}\right )\right )}{6 a^3}-\frac {\text {PolyLog}(2,a c+b c x)}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.34, size = 272, normalized size = 0.99

method	result	size
derivativedivides	\(c^{3} b^{3} \left (-\frac {\polylog \left (2, x b c +a c \right )}{3 x^{3} b^{3} c^{3}}-\frac {\dilog \left (-\frac {x b c}{a c -1}\right )+\ln \left (-x b c -a c +1\right ) \ln \left (-\frac {x b c}{a c -1}\right )}{3 a^{3} c^{3}}-\frac {\dilog \left (-x b c -a c +1\right )}{3 a^{3} c^{3}}-\frac {-\frac {\ln \left (-x b c \right )}{2 \left (a c -1\right )^{2}}-\frac {a}{2 \left (a c -1\right )^{2} x b}+\frac {1}{2 \left (a c -1\right )^{2} x b c}+\frac {\ln \left (-x b c -a c +1\right ) \left (-x b c +a c -1\right ) \left (-x b c -a c +1\right )}{2 x^{2} b^{2} c^{2} \left (a c -1\right )^{2}}}{3 a c}-\frac {-\frac {\ln \left (-x b c \right )}{a c -1}-\frac {\ln \left (-x b c -a c +1\right ) \left (-x b c -a c +1\right )}{\left (a c -1\right ) x b c}}{3 a^{2} c^{2}}\right )\)	\(272\)
default	\(c^{3} b^{3} \left (-\frac {\polylog \left (2, x b c +a c \right )}{3 x^{3} b^{3} c^{3}}-\frac {\dilog \left (-\frac {x b c}{a c -1}\right )+\ln \left (-x b c -a c +1\right ) \ln \left (-\frac {x b c}{a c -1}\right )}{3 a^{3} c^{3}}-\frac {\dilog \left (-x b c -a c +1\right )}{3 a^{3} c^{3}}-\frac {-\frac {\ln \left (-x b c \right )}{2 \left (a c -1\right )^{2}}-\frac {a}{2 \left (a c -1\right )^{2} x b}+\frac {1}{2 \left (a c -1\right )^{2} x b c}+\frac {\ln \left (-x b c -a c +1\right ) \left (-x b c +a c -1\right ) \left (-x b c -a c +1\right )}{2 x^{2} b^{2} c^{2} \left (a c -1\right )^{2}}}{3 a c}-\frac {-\frac {\ln \left (-x b c \right )}{a c -1}-\frac {\ln \left (-x b c -a c +1\right ) \left (-x b c -a c +1\right )}{\left (a c -1\right ) x b c}}{3 a^{2} c^{2}}\right )\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 302, normalized size = 1.09 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {Li}_{2}\left (a c + b c x\right )}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, c*(a + b*x))/x^4,x)

[Out]

int(polylog(2, c*(a + b*x))/x^4, x)

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