3.109 \(\int (d x)^m \text {Li}_3(a x^3) \, dx\)

Optimal. Leaf size=118 \[ -\frac {27 a (d x)^{m+4} \, _2F_1\left (1,\frac {m+4}{3};\frac {m+7}{3};a x^3\right )}{d^4 (m+1)^3 (m+4)}-\frac {3 \text {Li}_2\left (a x^3\right ) (d x)^{m+1}}{d (m+1)^2}+\frac {\text {Li}_3\left (a x^3\right ) (d x)^{m+1}}{d (m+1)}-\frac {9 \log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)^3} \]

[Out]

-27*a*(d*x)^(4+m)*hypergeom([1, 4/3+1/3*m],[7/3+1/3*m],a*x^3)/d^4/(1+m)^3/(4+m)-9*(d*x)^(1+m)*ln(-a*x^3+1)/d/(
1+m)^3-3*(d*x)^(1+m)*polylog(2,a*x^3)/d/(1+m)^2+(d*x)^(1+m)*polylog(3,a*x^3)/d/(1+m)

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Rubi [A]  time = 0.07, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6591, 2455, 16, 364} \[ -\frac {3 (d x)^{m+1} \text {PolyLog}\left (2,a x^3\right )}{d (m+1)^2}+\frac {(d x)^{m+1} \text {PolyLog}\left (3,a x^3\right )}{d (m+1)}-\frac {27 a (d x)^{m+4} \, _2F_1\left (1,\frac {m+4}{3};\frac {m+7}{3};a x^3\right )}{d^4 (m+1)^3 (m+4)}-\frac {9 \log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)^3} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*PolyLog[3, a*x^3],x]

[Out]

(-27*a*(d*x)^(4 + m)*Hypergeometric2F1[1, (4 + m)/3, (7 + m)/3, a*x^3])/(d^4*(1 + m)^3*(4 + m)) - (9*(d*x)^(1
+ m)*Log[1 - a*x^3])/(d*(1 + m)^3) - (3*(d*x)^(1 + m)*PolyLog[2, a*x^3])/(d*(1 + m)^2) + ((d*x)^(1 + m)*PolyLo
g[3, a*x^3])/(d*(1 + m))

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2455

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

Rule 6591

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\left (a x^3\right ) \, dx &=\frac {(d x)^{1+m} \text {Li}_3\left (a x^3\right )}{d (1+m)}-\frac {3 \int (d x)^m \text {Li}_2\left (a x^3\right ) \, dx}{1+m}\\ &=-\frac {3 (d x)^{1+m} \text {Li}_2\left (a x^3\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_3\left (a x^3\right )}{d (1+m)}-\frac {9 \int (d x)^m \log \left (1-a x^3\right ) \, dx}{(1+m)^2}\\ &=-\frac {9 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^3}-\frac {3 (d x)^{1+m} \text {Li}_2\left (a x^3\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_3\left (a x^3\right )}{d (1+m)}-\frac {(27 a) \int \frac {x^2 (d x)^{1+m}}{1-a x^3} \, dx}{d (1+m)^3}\\ &=-\frac {9 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^3}-\frac {3 (d x)^{1+m} \text {Li}_2\left (a x^3\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_3\left (a x^3\right )}{d (1+m)}-\frac {(27 a) \int \frac {(d x)^{3+m}}{1-a x^3} \, dx}{d^3 (1+m)^3}\\ &=-\frac {27 a (d x)^{4+m} \, _2F_1\left (1,\frac {4+m}{3};\frac {7+m}{3};a x^3\right )}{d^4 (1+m)^3 (4+m)}-\frac {9 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^3}-\frac {3 (d x)^{1+m} \text {Li}_2\left (a x^3\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_3\left (a x^3\right )}{d (1+m)}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 126, normalized size = 1.07 \[ -\frac {3 x \Gamma \left (\frac {m+4}{3}\right ) (d x)^m \left (3 a (m+1) x^3 \Gamma \left (\frac {m+1}{3}\right ) \, _2\tilde {F}_1\left (1,\frac {m+4}{3};\frac {m+7}{3};a x^3\right )-m^2 \text {Li}_3\left (a x^3\right )+3 (m+1) \text {Li}_2\left (a x^3\right )-2 m \text {Li}_3\left (a x^3\right )-\text {Li}_3\left (a x^3\right )+9 \log \left (1-a x^3\right )\right )}{(m+1)^4 \Gamma \left (\frac {m+1}{3}\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*x)^m*PolyLog[3, a*x^3],x]

[Out]

(-3*x*(d*x)^m*Gamma[(4 + m)/3]*(3*a*(1 + m)*x^3*Gamma[(1 + m)/3]*HypergeometricPFQRegularized[{1, (4 + m)/3},
{(7 + m)/3}, a*x^3] + 9*Log[1 - a*x^3] + 3*(1 + m)*PolyLog[2, a*x^3] - PolyLog[3, a*x^3] - 2*m*PolyLog[3, a*x^
3] - m^2*PolyLog[3, a*x^3]))/((1 + m)^4*Gamma[(1 + m)/3])

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d x\right )^{m} {\rm polylog}\left (3, a x^{3}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x)^m*polylog(3, a*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{m} {\rm Li}_{3}(a x^{3})\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.33, size = 218, normalized size = 1.85 \[ -\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {1}{3}-\frac {m}{3}} \left (\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (108+27 m \right )}{\left (4+m \right ) \left (1+m \right )^{4} a}-\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (36+9 m \right ) \ln \left (-a \,x^{3}+1\right )}{\left (4+m \right ) \left (1+m \right )^{3} a}+\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (-12-3 m \right ) \polylog \left (2, a \,x^{3}\right )}{\left (4+m \right ) \left (1+m \right )^{2} a}+\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \polylog \left (3, a \,x^{3}\right )}{\left (1+m \right ) a}+\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (-36-9 m \right ) \Phi \left (a \,x^{3}, 1, \frac {m}{3}+\frac {1}{3}\right )}{\left (4+m \right ) \left (1+m \right )^{3} a}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(d*x)^m*x^(-m)*(-a)^(-1/3-1/3*m)*(3/(4+m)*x^(1+m)*(-a)^(4/3+1/3*m)*(108+27*m)/(1+m)^4/a-3/(4+m)*x^(1+m)*(
-a)^(4/3+1/3*m)*(36+9*m)/(1+m)^3/a*ln(-a*x^3+1)+3/(4+m)*x^(1+m)*(-a)^(4/3+1/3*m)*(-12-3*m)/(1+m)^2*polylog(2,a
*x^3)/a+3*x^(1+m)*(-a)^(4/3+1/3*m)/(1+m)/a*polylog(3,a*x^3)+3/(4+m)*x^(1+m)*(-a)^(4/3+1/3*m)*(-36-9*m)/(1+m)^3
/a*LerchPhi(a*x^3,1,1/3*m+1/3))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 27 \, a d^{m} \int \frac {x^{3} x^{m}}{{\left (m^{3} + 3 \, m^{2} + 3 \, m + 1\right )} a x^{3} - m^{3} - 3 \, m^{2} - 3 \, m - 1}\,{d x} - \frac {3 \, d^{m} {\left (m + 1\right )} x x^{m} {\rm Li}_2\left (a x^{3}\right ) - {\left (m^{2} + 2 \, m + 1\right )} d^{m} x x^{m} {\rm Li}_{3}(a x^{3}) + 9 \, d^{m} x x^{m} \log \left (-a x^{3} + 1\right )}{m^{3} + 3 \, m^{2} + 3 \, m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27*a*d^m*integrate(x^3*x^m/((m^3 + 3*m^2 + 3*m + 1)*a*x^3 - m^3 - 3*m^2 - 3*m - 1), x) - (3*d^m*(m + 1)*x*x^m*
dilog(a*x^3) - (m^2 + 2*m + 1)*d^m*x*x^m*polylog(3, a*x^3) + 9*d^m*x*x^m*log(-a*x^3 + 1))/(m^3 + 3*m^2 + 3*m +
1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^3)*(d*x)^m,x)

[Out]

int(polylog(3, a*x^3)*(d*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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