3.110 \(\int (d x)^m \text{PolyLog}(4,a x^3) \, dx\)

Optimal. Leaf size=142 \[ \frac{81 a (d x)^{m+4} \text{Hypergeometric2F1}\left (1,\frac{m+4}{3},\frac{m+7}{3},a x^3\right )}{d^4 (m+1)^4 (m+4)}+\frac{9 (d x)^{m+1} \text{PolyLog}\left (2,a x^3\right )}{d (m+1)^3}-\frac{3 (d x)^{m+1} \text{PolyLog}\left (3,a x^3\right )}{d (m+1)^2}+\frac{(d x)^{m+1} \text{PolyLog}\left (4,a x^3\right )}{d (m+1)}+\frac{27 \log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)^4} \]

[Out]

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

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Rubi [A]  time = 0.0943941, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {6591, 2455, 16, 364} \[ \frac{9 (d x)^{m+1} \text{PolyLog}\left (2,a x^3\right )}{d (m+1)^3}-\frac{3 (d x)^{m+1} \text{PolyLog}\left (3,a x^3\right )}{d (m+1)^2}+\frac{(d x)^{m+1} \text{PolyLog}\left (4,a x^3\right )}{d (m+1)}+\frac{81 a (d x)^{m+4} \, _2F_1\left (1,\frac{m+4}{3};\frac{m+7}{3};a x^3\right )}{d^4 (m+1)^4 (m+4)}+\frac{27 \log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)^4} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*PolyLog[4, a*x^3],x]

[Out]

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

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]

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 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])

Rubi steps

\begin{align*} \int (d x)^m \text{Li}_4\left (a x^3\right ) \, dx &=\frac{(d x)^{1+m} \text{Li}_4\left (a x^3\right )}{d (1+m)}-\frac{3 \int (d x)^m \text{Li}_3\left (a x^3\right ) \, dx}{1+m}\\ &=-\frac{3 (d x)^{1+m} \text{Li}_3\left (a x^3\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^3\right )}{d (1+m)}+\frac{9 \int (d x)^m \text{Li}_2\left (a x^3\right ) \, dx}{(1+m)^2}\\ &=\frac{9 (d x)^{1+m} \text{Li}_2\left (a x^3\right )}{d (1+m)^3}-\frac{3 (d x)^{1+m} \text{Li}_3\left (a x^3\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^3\right )}{d (1+m)}+\frac{27 \int (d x)^m \log \left (1-a x^3\right ) \, dx}{(1+m)^3}\\ &=\frac{27 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^4}+\frac{9 (d x)^{1+m} \text{Li}_2\left (a x^3\right )}{d (1+m)^3}-\frac{3 (d x)^{1+m} \text{Li}_3\left (a x^3\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^3\right )}{d (1+m)}+\frac{(81 a) \int \frac{x^2 (d x)^{1+m}}{1-a x^3} \, dx}{d (1+m)^4}\\ &=\frac{27 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^4}+\frac{9 (d x)^{1+m} \text{Li}_2\left (a x^3\right )}{d (1+m)^3}-\frac{3 (d x)^{1+m} \text{Li}_3\left (a x^3\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^3\right )}{d (1+m)}+\frac{(81 a) \int \frac{(d x)^{3+m}}{1-a x^3} \, dx}{d^3 (1+m)^4}\\ &=\frac{81 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)^4 (4+m)}+\frac{27 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^4}+\frac{9 (d x)^{1+m} \text{Li}_2\left (a x^3\right )}{d (1+m)^3}-\frac{3 (d x)^{1+m} \text{Li}_3\left (a x^3\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^3\right )}{d (1+m)}\\ \end{align*}

Mathematica [C]  time = 0.0967328, size = 166, normalized size = 1.17 \[ \frac{3 x \text{Gamma}\left (\frac{m+4}{3}\right ) (d x)^m \left (9 a (m+1) x^3 \text{Gamma}\left (\frac{m+1}{3}\right ) \, _2\tilde{F}_1\left (1,\frac{m+4}{3};\frac{m+7}{3};a x^3\right )+m^3 \text{PolyLog}\left (4,a x^3\right )-3 m^2 \text{PolyLog}\left (3,a x^3\right )+3 m^2 \text{PolyLog}\left (4,a x^3\right )-6 m \text{PolyLog}\left (3,a x^3\right )+3 m \text{PolyLog}\left (4,a x^3\right )+9 (m+1) \text{PolyLog}\left (2,a x^3\right )-3 \text{PolyLog}\left (3,a x^3\right )+\text{PolyLog}\left (4,a x^3\right )+27 \log \left (1-a x^3\right )\right )}{(m+1)^5 \text{Gamma}\left (\frac{m+1}{3}\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*x)^m*PolyLog[4, a*x^3],x]

[Out]

(3*x*(d*x)^m*Gamma[(4 + m)/3]*(9*a*(1 + m)*x^3*Gamma[(1 + m)/3]*HypergeometricPFQRegularized[{1, (4 + m)/3}, {
(7 + m)/3}, a*x^3] + 27*Log[1 - a*x^3] + 9*(1 + m)*PolyLog[2, a*x^3] - 3*PolyLog[3, a*x^3] - 6*m*PolyLog[3, a*
x^3] - 3*m^2*PolyLog[3, a*x^3] + PolyLog[4, a*x^3] + 3*m*PolyLog[4, a*x^3] + 3*m^2*PolyLog[4, a*x^3] + m^3*Pol
yLog[4, a*x^3]))/((1 + m)^5*Gamma[(1 + m)/3])

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Maple [C]  time = 0.856, size = 259, normalized size = 1.8 \begin{align*} -{\frac{ \left ( dx \right ) ^{m}{x}^{-m}}{3} \left ( -a \right ) ^{-{\frac{1}{3}}-{\frac{m}{3}}} \left ( 3\,{\frac{{x}^{1+m} \left ( -a \right ) ^{4/3+m/3} \left ( -324-81\,m \right ) }{ \left ( 1+m \right ) ^{5} \left ( 4+m \right ) a}}-3\,{\frac{{x}^{1+m} \left ( -a \right ) ^{4/3+m/3} \left ( -108-27\,m \right ) \ln \left ( -{x}^{3}a+1 \right ) }{ \left ( 1+m \right ) ^{4} \left ( 4+m \right ) a}}+3\,{\frac{{x}^{1+m} \left ( -a \right ) ^{4/3+m/3} \left ( 36+9\,m \right ){\it polylog} \left ( 2,{x}^{3}a \right ) }{ \left ( 1+m \right ) ^{3} \left ( 4+m \right ) a}}+3\,{\frac{{x}^{1+m} \left ( -a \right ) ^{4/3+m/3} \left ( -12-3\,m \right ){\it polylog} \left ( 3,{x}^{3}a \right ) }{ \left ( 1+m \right ) ^{2} \left ( 4+m \right ) a}}+3\,{\frac{{x}^{1+m} \left ( -a \right ) ^{4/3+m/3}{\it polylog} \left ( 4,{x}^{3}a \right ) }{ \left ( 1+m \right ) a}}+3\,{\frac{{x}^{1+m} \left ( -a \right ) ^{4/3+m/3} \left ( 108+27\,m \right ){\it LerchPhi} \left ({x}^{3}a,1,m/3+1/3 \right ) }{ \left ( 1+m \right ) ^{4} \left ( 4+m \right ) a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*polylog(4,x^3*a),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)*(-324-81*m)/(1+m)^5/a-3/(4+m)*x^(1+m)*
(-a)^(4/3+1/3*m)*(-108-27*m)/(1+m)^4/a*ln(-a*x^3+1)+3/(4+m)*x^(1+m)*(-a)^(4/3+1/3*m)*(36+9*m)/(1+m)^3/a*polylo
g(2,x^3*a)+3/(4+m)*x^(1+m)*(-a)^(4/3+1/3*m)*(-12-3*m)/(1+m)^2*polylog(3,x^3*a)/a+3*x^(1+m)*(-a)^(4/3+1/3*m)/(1
+m)/a*polylog(4,x^3*a)+3/(4+m)*x^(1+m)*(-a)^(4/3+1/3*m)*(108+27*m)/(1+m)^4/a*LerchPhi(x^3*a,1,1/3*m+1/3))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -81 \, a d^{m} \int -\frac{x^{3} x^{m}}{m^{4} -{\left (a m^{4} + 4 \, a m^{3} + 6 \, a m^{2} + 4 \, a m + a\right )} x^{3} + 4 \, m^{3} + 6 \, m^{2} + 4 \, m + 1}\,{d x} + \frac{9 \,{\left (d^{m} m + d^{m}\right )} x x^{m}{\rm Li}_2\left (a x^{3}\right ) + 27 \, d^{m} x x^{m} \log \left (-a x^{3} + 1\right ) +{\left (d^{m} m^{3} + 3 \, d^{m} m^{2} + 3 \, d^{m} m + d^{m}\right )} x x^{m}{\rm Li}_{4}(a x^{3}) - 3 \,{\left (d^{m} m^{2} + 2 \, d^{m} m + d^{m}\right )} x x^{m}{\rm Li}_{3}(a x^{3})}{m^{4} + 4 \, m^{3} + 6 \, m^{2} + 4 \, m + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*polylog(4,a*x^3),x, algorithm="maxima")

[Out]

-81*a*d^m*integrate(-x^3*x^m/(m^4 - (a*m^4 + 4*a*m^3 + 6*a*m^2 + 4*a*m + a)*x^3 + 4*m^3 + 6*m^2 + 4*m + 1), x)
+ (9*(d^m*m + d^m)*x*x^m*dilog(a*x^3) + 27*d^m*x*x^m*log(-a*x^3 + 1) + (d^m*m^3 + 3*d^m*m^2 + 3*d^m*m + d^m)*
x*x^m*polylog(4, a*x^3) - 3*(d^m*m^2 + 2*d^m*m + d^m)*x*x^m*polylog(3, a*x^3))/(m^4 + 4*m^3 + 6*m^2 + 4*m + 1)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (d x\right )^{m}{\rm polylog}\left (4, a x^{3}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*polylog(4,a*x^3),x, algorithm="fricas")

[Out]

integral((d*x)^m*polylog(4, a*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \operatorname{Li}_{4}\left (a x^{3}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*polylog(4,a*x**3),x)

[Out]

Integral((d*x)**m*polylog(4, a*x**3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m}{\rm Li}_{4}(a x^{3})\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*polylog(4,a*x^3),x, algorithm="giac")

[Out]

integrate((d*x)^m*polylog(4, a*x^3), x)