3.2.3 \(\int (d x)^m \text {PolyLog}(3,a x) \, dx\) [103]

Optimal. Leaf size=102 \[ -\frac {a (d x)^{2+m} \, _2F_1(1,2+m;3+m;a x)}{d^2 (1+m)^3 (2+m)}-\frac {(d x)^{1+m} \log (1-a x)}{d (1+m)^3}-\frac {(d x)^{1+m} \text {PolyLog}(2,a x)}{d (1+m)^2}+\frac {(d x)^{1+m} \text {PolyLog}(3,a x)}{d (1+m)} \]

[Out]

-a*(d*x)^(2+m)*hypergeom([1, 2+m],[3+m],a*x)/d^2/(1+m)^3/(2+m)-(d*x)^(1+m)*ln(-a*x+1)/d/(1+m)^3-(d*x)^(1+m)*po
lylog(2,a*x)/d/(1+m)^2+(d*x)^(1+m)*polylog(3,a*x)/d/(1+m)

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Rubi [A]
time = 0.05, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6726, 2442, 66} \begin {gather*} -\frac {a (d x)^{m+2} \, _2F_1(1,m+2;m+3;a x)}{d^2 (m+1)^3 (m+2)}-\frac {\text {Li}_2(a x) (d x)^{m+1}}{d (m+1)^2}+\frac {\text {Li}_3(a x) (d x)^{m+1}}{d (m+1)}-\frac {\log (1-a x) (d x)^{m+1}}{d (m+1)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*PolyLog[3, a*x],x]

[Out]

-((a*(d*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, a*x])/(d^2*(1 + m)^3*(2 + m))) - ((d*x)^(1 + m)*Log[1 -
a*x])/(d*(1 + m)^3) - ((d*x)^(1 + m)*PolyLog[2, a*x])/(d*(1 + m)^2) + ((d*x)^(1 + m)*PolyLog[3, a*x])/(d*(1 +
m))

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 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 6726

Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[(d*x)^(m + 1)*(PolyLog[n
, a*(b*x^p)^q]/(d*(m + 1))), x] - Dist[p*(q/(m + 1)), Int[(d*x)^m*PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ
[{a, b, d, m, p, q}, x] && NeQ[m, -1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int (d x)^m \text {Li}_3(a x) \, dx &=\frac {(d x)^{1+m} \text {Li}_3(a x)}{d (1+m)}-\frac {\int (d x)^m \text {Li}_2(a x) \, dx}{1+m}\\ &=-\frac {(d x)^{1+m} \text {Li}_2(a x)}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_3(a x)}{d (1+m)}-\frac {\int (d x)^m \log (1-a x) \, dx}{(1+m)^2}\\ &=-\frac {(d x)^{1+m} \log (1-a x)}{d (1+m)^3}-\frac {(d x)^{1+m} \text {Li}_2(a x)}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_3(a x)}{d (1+m)}-\frac {a \int \frac {(d x)^{1+m}}{1-a x} \, dx}{d (1+m)^3}\\ &=-\frac {a (d x)^{2+m} \, _2F_1(1,2+m;3+m;a x)}{d^2 (1+m)^3 (2+m)}-\frac {(d x)^{1+m} \log (1-a x)}{d (1+m)^3}-\frac {(d x)^{1+m} \text {Li}_2(a x)}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_3(a x)}{d (1+m)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
time = 0.04, size = 88, normalized size = 0.86 \begin {gather*} -\frac {x (d x)^m \Gamma (2+m) \left (a (1+m) x \Gamma (1+m) \, _2\tilde {F}_1(1,2+m;3+m;a x)+\log (1-a x)+(1+m) \text {PolyLog}(2,a x)-\text {PolyLog}(3,a x)-2 m \text {PolyLog}(3,a x)-m^2 \text {PolyLog}(3,a x)\right )}{(1+m)^4 \Gamma (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m*PolyLog[3, a*x],x]

[Out]

-((x*(d*x)^m*Gamma[2 + m]*(a*(1 + m)*x*Gamma[1 + m]*HypergeometricPFQRegularized[{1, 2 + m}, {3 + m}, a*x] + L
og[1 - a*x] + (1 + m)*PolyLog[2, a*x] - PolyLog[3, a*x] - 2*m*PolyLog[3, a*x] - m^2*PolyLog[3, a*x]))/((1 + m)
^4*Gamma[1 + m]))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.22, size = 173, normalized size = 1.70

method	result	size
meijerg	\(\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-m} \left (\frac {x^{m} \left (-a \right )^{m} \left (a \,m^{2} x +2 a m x +m^{2}+3 m +2\right )}{\left (2+m \right ) \left (1+m \right )^{4} m}-\frac {x^{1+m} a \left (-a \right )^{m} \ln \left (-a x +1\right )}{\left (1+m \right )^{3}}+\frac {x^{1+m} a \left (-a \right )^{m} \left (-m -2\right ) \polylog \left (2, a x \right )}{\left (2+m \right ) \left (1+m \right )^{2}}+\frac {x^{1+m} a \left (-a \right )^{m} \polylog \left (3, a x \right )}{1+m}+\frac {x^{m} \left (-a \right )^{m} \left (-m -2\right ) \Phi \left (a x , 1, m\right )}{\left (2+m \right ) \left (1+m \right )^{3}}\right )}{a}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*polylog(3,a*x),x,method=_RETURNVERBOSE)

[Out]

(d*x)^m*x^(-m)*(-a)^(-m)/a*(1/(2+m)*x^m*(-a)^m*(a*m^2*x+2*a*m*x+m^2+3*m+2)/(1+m)^4/m-x^(1+m)*a*(-a)^m/(1+m)^3*
ln(-a*x+1)+1/(2+m)*x^(1+m)*a*(-a)^m*(-m-2)/(1+m)^2*polylog(2,a*x)+x^(1+m)*a*(-a)^m/(1+m)*polylog(3,a*x)+1/(2+m
)*x^m*(-a)^m*(-m-2)/(1+m)^3*LerchPhi(a*x,1,m))

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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((d*x)^m*polylog(3,a*x),x, algorithm="maxima")

[Out]

a*d^m*integrate(-x*x^m/(m^3 - (m^3 + 3*m^2 + 3*m + 1)*a*x + 3*m^2 + 3*m + 1), x) - (d^m*(m + 1)*x*x^m*dilog(a*
x) - (m^2 + 2*m + 1)*d^m*x*x^m*polylog(3, a*x) + d^m*x*x^m*log(-a*x + 1))/(m^3 + 3*m^2 + 3*m + 1)

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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((d*x)^m*polylog(3,a*x),x, algorithm="fricas")

[Out]

integral((d*x)^m*polylog(3, a*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{m} \operatorname {Li}_{3}\left (a x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*polylog(3,a*x),x)

[Out]

Integral((d*x)**m*polylog(3, a*x), 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((d*x)^m*polylog(3,a*x),x, algorithm="giac")

[Out]

integrate((d*x)^m*polylog(3, a*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,x\right )}^m\,\mathrm {polylog}\left (3,a\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*polylog(3, a*x),x)

[Out]

int((d*x)^m*polylog(3, a*x), x)

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