3.3.10 \(\int (a+b \log (c x^n)) \text {Li}_2(e x) \, dx\) [210]

Optimal. Leaf size=106 \[ 3 b n x-x \left (a+b \log \left (c x^n\right )\right )+\frac {2 b n (1-e x) \log (1-e x)}{e}-\frac {(1-e x) \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{e}-\frac {b n \text {Li}_2(e x)}{e}-b n x \text {Li}_2(e x)+x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x) \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2428, 2436, 2332, 2417, 2458, 45, 2393, 2352} \begin {gather*} x \text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-b n x \text {PolyLog}(2,e x)-\frac {b n \text {PolyLog}(2,e x)}{e}-\frac {(1-e x) \log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )+\frac {2 b n (1-e x) \log (1-e x)}{e}+3 b n x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])*PolyLog[2, e*x],x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 2393

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2417

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> With[
{u = IntHide[Log[d*(e + f*x^m)^r], x]}, Dist[(a + b*Log[c*x^n])^p, u, x] - Dist[b*n*p, Int[Dist[(a + b*Log[c*x
^n])^(p - 1)/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] && RationalQ[m] && (EqQ[
p, 1] || (FractionQ[m] && IntegerQ[1/m]) || (EqQ[r, 1] && EqQ[m, 1] && EqQ[d*e, 1]))

Rule 2428

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*PolyLog[k_, (e_.)*(x_)^(q_.)], x_Symbol] :> Simp[(-b)*n*x*PolyLog[k,
e*x^q], x] + (-Dist[q, Int[PolyLog[k - 1, e*x^q]*(a + b*Log[c*x^n]), x], x] + Dist[b*n*q, Int[PolyLog[k - 1,
e*x^q], x], x] + Simp[x*PolyLog[k, e*x^q]*(a + b*Log[c*x^n]), x]) /; FreeQ[{a, b, c, e, n, q}, x] && IGtQ[k, 0
]

Rule 2436

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

Rule 2458

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]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 113, normalized size = 1.07 \begin {gather*} \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right ) \left (-x+\left (-\frac {1}{e}+x\right ) \log (1-e x)+x \text {Li}_2(e x)\right )+\frac {b n (3 e x+2 \log (1-e x)-2 e x \log (1-e x)+\log (x) (-e x+(-1+e x) \log (1-e x))+(-1-e x+e x \log (x)) \text {Li}_2(e x))}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])*PolyLog[2, e*x],x]

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (a +b \ln \left (c \,x^{n}\right )\right ) \polylog \left (2, e x \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))*polylog(2,e*x),x)

[Out]

int((a+b*ln(c*x^n))*polylog(2,e*x),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*dilog(x*e)*e - x*e + (x*e - 1)*log(-x*e + 1))*a*e^(-1) + (((x*e*log(x^n) - (n*e - e*log(c))*x)*dilog(x*e) -
((2*n*e - e*log(c))*x - n*log(x))*log(-x*e + 1) - (x*e - (x*e - 1)*log(-x*e + 1))*log(x^n))*e^(-1) - integrat
e(-((3*n*e - e*log(c))*x - n*log(x) - n)/(x*e - 1), x))*b

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Fricas [A]
time = 0.36, size = 146, normalized size = 1.38 \begin {gather*} {\left ({\left (3 \, b n - a\right )} x e - {\left ({\left (b n - a\right )} x e + b n\right )} {\rm Li}_2\left (x e\right ) - {\left ({\left (2 \, b n - a\right )} x e - 2 \, b n + a\right )} \log \left (-x e + 1\right ) + {\left (b x {\rm Li}_2\left (x e\right ) e - b x e + {\left (b x e - b\right )} \log \left (-x e + 1\right )\right )} \log \left (c\right ) + {\left (b n x {\rm Li}_2\left (x e\right ) e - b n x e + {\left (b n x e - b n\right )} \log \left (-x e + 1\right )\right )} \log \left (x\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*b*n - a)*x*e - ((b*n - a)*x*e + b*n)*dilog(x*e) - ((2*b*n - a)*x*e - 2*b*n + a)*log(-x*e + 1) + (b*x*dilog
(x*e)*e - b*x*e + (b*x*e - b)*log(-x*e + 1))*log(c) + (b*n*x*dilog(x*e)*e - b*n*x*e + (b*n*x*e - b*n)*log(-x*e
+ 1))*log(x))*e^(-1)

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Sympy [A]
time = 6.96, size = 136, normalized size = 1.28 \begin {gather*} \begin {cases} - a x \operatorname {Li}_{1}\left (e x\right ) + a x \operatorname {Li}_{2}\left (e x\right ) - a x + \frac {a \operatorname {Li}_{1}\left (e x\right )}{e} + 2 b n x \operatorname {Li}_{1}\left (e x\right ) - b n x \operatorname {Li}_{2}\left (e x\right ) + 3 b n x - b x \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right ) + b x \log {\left (c x^{n} \right )} \operatorname {Li}_{2}\left (e x\right ) - b x \log {\left (c x^{n} \right )} - \frac {2 b n \operatorname {Li}_{1}\left (e x\right )}{e} - \frac {b n \operatorname {Li}_{2}\left (e x\right )}{e} + \frac {b \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))*polylog(2,e*x),x)

[Out]

Piecewise((-a*x*polylog(1, e*x) + a*x*polylog(2, e*x) - a*x + a*polylog(1, e*x)/e + 2*b*n*x*polylog(1, e*x) -
b*n*x*polylog(2, e*x) + 3*b*n*x - b*x*log(c*x**n)*polylog(1, e*x) + b*x*log(c*x**n)*polylog(2, e*x) - b*x*log(
c*x**n) - 2*b*n*polylog(1, e*x)/e - b*n*polylog(2, e*x)/e + b*log(c*x**n)*polylog(1, e*x)/e, Ne(e, 0)), (0, Tr
ue))

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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((a+b*log(c*x^n))*polylog(2,e*x),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*dilog(x*e), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, e*x)*(a + b*log(c*x^n)),x)

[Out]

int(polylog(2, e*x)*(a + b*log(c*x^n)), x)

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