3.174 \(\int \frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x^2} \, dx\)

Optimal. Leaf size=156 \[ -\frac {\text {Li}_2(c x) (h \log (1-c x)+g)}{x}+\frac {\log (1-c x) (h \log (1-c x)+g)}{x}+c \log \left (1-\frac {1}{1-c x}\right ) (2 h \log (1-c x)+g)-2 c h \text {Li}_2\left (\frac {1}{1-c x}\right )-c h \text {Li}_3(c x)-2 c h \text {Li}_3(1-c x)+c h \text {Li}_2(c x) \log (1-c x)+2 c h \text {Li}_2(1-c x) \log (1-c x)+c h \log (c x) \log ^2(1-c x) \]

[Out]

c*h*ln(c*x)*ln(-c*x+1)^2+ln(-c*x+1)*(g+h*ln(-c*x+1))/x+c*(g+2*h*ln(-c*x+1))*ln(1-1/(-c*x+1))+c*h*ln(-c*x+1)*po
lylog(2,c*x)-(g+h*ln(-c*x+1))*polylog(2,c*x)/x-2*c*h*polylog(2,1/(-c*x+1))+2*c*h*ln(-c*x+1)*polylog(2,-c*x+1)-
c*h*polylog(3,c*x)-2*c*h*polylog(3,-c*x+1)

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Rubi [A]  time = 0.36, antiderivative size = 165, normalized size of antiderivative = 1.06, number of steps used = 19, number of rules used = 15, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6603, 2439, 2410, 2391, 2390, 2301, 2411, 2344, 2316, 2315, 6589, 6596, 2396, 2433, 2374} \[ -\frac {\text {PolyLog}(2,c x) (h \log (1-c x)+g)}{x}-2 c h \text {PolyLog}(2,c x)-c h \text {PolyLog}(3,c x)-2 c h \text {PolyLog}(3,1-c x)+c h \log (1-c x) \text {PolyLog}(2,c x)+2 c h \log (1-c x) \text {PolyLog}(2,1-c x)+\frac {\log (1-c x) (h \log (1-c x)+g)}{x}-\frac {c (h \log (1-c x)+g)^2}{2 h}+c g \log (x)-\frac {1}{2} c h \log ^2(1-c x)+c h \log (c x) \log ^2(1-c x) \]

Antiderivative was successfully verified.

[In]

Int[((g + h*Log[1 - c*x])*PolyLog[2, c*x])/x^2,x]

[Out]

c*g*Log[x] - (c*h*Log[1 - c*x]^2)/2 + c*h*Log[c*x]*Log[1 - c*x]^2 + (Log[1 - c*x]*(g + h*Log[1 - c*x]))/x - (c
*(g + h*Log[1 - c*x])^2)/(2*h) - 2*c*h*PolyLog[2, c*x] + c*h*Log[1 - c*x]*PolyLog[2, c*x] - ((g + h*Log[1 - c*
x])*PolyLog[2, c*x])/x + 2*c*h*Log[1 - c*x]*PolyLog[2, 1 - c*x] - c*h*PolyLog[3, c*x] - 2*c*h*PolyLog[3, 1 - c
*x]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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 2316

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

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 0]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[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 2396

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*
(f + g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n])^p)/g, x] - Dist[(b*e*n*p)/g, Int[(Log[(e*(f + g*x))/(e*f -
d*g)]*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*
f - d*g, 0] && IGtQ[p, 1]

Rule 2410

Int[(Log[(c_.)*((d_) + (e_.)*(x_))]*(x_)^(m_.))/((f_) + (g_.)*(x_)), x_Symbol] :> Int[ExpandIntegrand[Log[c*(d
+ e*x)], x^m/(f + g*x), x], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[e*f - d*g, 0] && EqQ[c*d, 1] && IntegerQ[m
]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*(x_)^(r_.), x_Symbol] :> Simp[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]))/(r +
1), x] + (-Dist[(g*j*m)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[(b*e*n*
p)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f + g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /
; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0] && IntegerQ[r] && (EqQ[p, 1] || GtQ[r, 0]) && N
eQ[r, -1]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6596

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

Rule 6603

Int[((g_.) + Log[(f_.)*((d_.) + (e_.)*(x_))^(n_.)]*(h_.))*(x_)^(m_.)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x
_Symbol] :> Simp[(x^(m + 1)*(g + h*Log[f*(d + e*x)^n])*PolyLog[2, c*(a + b*x)])/(m + 1), x] + (Dist[b/(m + 1),
Int[ExpandIntegrand[(g + h*Log[f*(d + e*x)^n])*Log[1 - a*c - b*c*x], x^(m + 1)/(a + b*x), x], x], x] - Dist[(
e*h*n)/(m + 1), Int[ExpandIntegrand[PolyLog[2, c*(a + b*x)], x^(m + 1)/(d + e*x), x], x], x]) /; FreeQ[{a, b,
c, d, e, f, g, h, n}, x] && IntegerQ[m] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x^2} \, dx &=-\frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x}-(c h) \int \left (\frac {\text {Li}_2(c x)}{x}-\frac {c \text {Li}_2(c x)}{-1+c x}\right ) \, dx-\int \frac {\log (1-c x) (g+h \log (1-c x))}{x^2} \, dx\\ &=\frac {\log (1-c x) (g+h \log (1-c x))}{x}-\frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x}+c \int \frac {g+h \log (1-c x)}{x (1-c x)} \, dx+(c h) \int \frac {\log (1-c x)}{x (1-c x)} \, dx-(c h) \int \frac {\text {Li}_2(c x)}{x} \, dx+\left (c^2 h\right ) \int \frac {\text {Li}_2(c x)}{-1+c x} \, dx\\ &=\frac {\log (1-c x) (g+h \log (1-c x))}{x}+c h \log (1-c x) \text {Li}_2(c x)-\frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x}-c h \text {Li}_3(c x)+(c h) \int \frac {\log ^2(1-c x)}{x} \, dx+(c h) \int \left (\frac {\log (1-c x)}{x}-\frac {c \log (1-c x)}{-1+c x}\right ) \, dx-\operatorname {Subst}\left (\int \frac {g+h \log (x)}{x \left (\frac {1}{c}-\frac {x}{c}\right )} \, dx,x,1-c x\right )\\ &=c h \log (c x) \log ^2(1-c x)+\frac {\log (1-c x) (g+h \log (1-c x))}{x}+c h \log (1-c x) \text {Li}_2(c x)-\frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x}-c h \text {Li}_3(c x)-c \operatorname {Subst}\left (\int \frac {g+h \log (x)}{x} \, dx,x,1-c x\right )+(c h) \int \frac {\log (1-c x)}{x} \, dx-\left (c^2 h\right ) \int \frac {\log (1-c x)}{-1+c x} \, dx+\left (2 c^2 h\right ) \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx-\operatorname {Subst}\left (\int \frac {g+h \log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x\right )\\ &=c g \log (x)+c h \log (c x) \log ^2(1-c x)+\frac {\log (1-c x) (g+h \log (1-c x))}{x}-\frac {c (g+h \log (1-c x))^2}{2 h}-c h \text {Li}_2(c x)+c h \log (1-c x) \text {Li}_2(c x)-\frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x}-c h \text {Li}_3(c x)-h \operatorname {Subst}\left (\int \frac {\log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x\right )-(c h) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x\right )-(2 c h) \operatorname {Subst}\left (\int \frac {\log (x) \log \left (c \left (\frac {1}{c}-\frac {x}{c}\right )\right )}{x} \, dx,x,1-c x\right )\\ &=c g \log (x)-\frac {1}{2} c h \log ^2(1-c x)+c h \log (c x) \log ^2(1-c x)+\frac {\log (1-c x) (g+h \log (1-c x))}{x}-\frac {c (g+h \log (1-c x))^2}{2 h}-2 c h \text {Li}_2(c x)+c h \log (1-c x) \text {Li}_2(c x)-\frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x}+2 c h \log (1-c x) \text {Li}_2(1-c x)-c h \text {Li}_3(c x)-(2 c h) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )\\ &=c g \log (x)-\frac {1}{2} c h \log ^2(1-c x)+c h \log (c x) \log ^2(1-c x)+\frac {\log (1-c x) (g+h \log (1-c x))}{x}-\frac {c (g+h \log (1-c x))^2}{2 h}-2 c h \text {Li}_2(c x)+c h \log (1-c x) \text {Li}_2(c x)-\frac {(g+h \log (1-c x)) \text {Li}_2(c x)}{x}+2 c h \log (1-c x) \text {Li}_2(1-c x)-c h \text {Li}_3(c x)-2 c h \text {Li}_3(1-c x)\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 150, normalized size = 0.96 \[ \frac {g (-\text {Li}_2(c x)+c x \log (x)+(1-c x) \log (1-c x))}{x}+h \left (-c \text {Li}_3(c x)-2 c \text {Li}_3(1-c x)+\frac {(c x-1) \text {Li}_2(c x) \log (1-c x)}{x}+2 c \text {Li}_2(1-c x) (\log (1-c x)+1)-c \log ^2(1-c x)+c \log (c x) \log ^2(1-c x)+\frac {\log ^2(1-c x)}{x}+2 c \log (c x) \log (1-c x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((g + h*Log[1 - c*x])*PolyLog[2, c*x])/x^2,x]

[Out]

(g*(c*x*Log[x] + (1 - c*x)*Log[1 - c*x] - PolyLog[2, c*x]))/x + h*(2*c*Log[c*x]*Log[1 - c*x] - c*Log[1 - c*x]^
2 + Log[1 - c*x]^2/x + c*Log[c*x]*Log[1 - c*x]^2 + ((-1 + c*x)*Log[1 - c*x]*PolyLog[2, c*x])/x + 2*c*(1 + Log[
1 - c*x])*PolyLog[2, 1 - c*x] - c*PolyLog[3, c*x] - 2*c*PolyLog[3, 1 - c*x])

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {h {\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right ) + g {\rm Li}_2\left (c x\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g+h*log(-c*x+1))*polylog(2,c*x)/x^2,x, algorithm="fricas")

[Out]

integral((h*dilog(c*x)*log(-c*x + 1) + g*dilog(c*x))/x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (h \log \left (-c x + 1\right ) + g\right )} {\rm Li}_2\left (c x\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g+h*log(-c*x+1))*polylog(2,c*x)/x^2,x, algorithm="giac")

[Out]

integrate((h*log(-c*x + 1) + g)*dilog(c*x)/x^2, x)

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \frac {\left (g +h \ln \left (-c x +1\right )\right ) \polylog \left (2, c x \right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g+h*ln(-c*x+1))*polylog(2,c*x)/x^2,x)

[Out]

int((g+h*ln(-c*x+1))*polylog(2,c*x)/x^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (c \log \relax (x) - \frac {{\left (c x - 1\right )} \log \left (-c x + 1\right ) + {\rm Li}_2\left (c x\right )}{x}\right )} g + h \int \frac {{\rm Li}_2\left (c x\right ) \log \left (-c x + 1\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g+h*log(-c*x+1))*polylog(2,c*x)/x^2,x, algorithm="maxima")

[Out]

(c*log(x) - ((c*x - 1)*log(-c*x + 1) + dilog(c*x))/x)*g + h*integrate(dilog(c*x)*log(-c*x + 1)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((g + h*log(1 - c*x))*polylog(2, c*x))/x^2,x)

[Out]

int(((g + h*log(1 - c*x))*polylog(2, c*x))/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g + h \log {\left (- c x + 1 \right )}\right ) \operatorname {Li}_{2}\left (c x\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g+h*ln(-c*x+1))*polylog(2,c*x)/x**2,x)

[Out]

Integral((g + h*log(-c*x + 1))*polylog(2, c*x)/x**2, x)

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