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

Optimal. Leaf size=121 \[ \frac{a (d x)^{m+2} \text{Hypergeometric2F1}(1,m+2,m+3,a x)}{d^2 (m+1)^4 (m+2)}+\frac{(d x)^{m+1} \text{PolyLog}(2,a x)}{d (m+1)^3}-\frac{(d x)^{m+1} \text{PolyLog}(3,a x)}{d (m+1)^2}+\frac{(d x)^{m+1} \text{PolyLog}(4,a x)}{d (m+1)}+\frac{\log (1-a x) (d x)^{m+1}}{d (m+1)^4} \]

[Out]

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

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Rubi [A]  time = 0.0857405, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6591, 2395, 64} \[ \frac{(d x)^{m+1} \text{PolyLog}(2,a x)}{d (m+1)^3}-\frac{(d x)^{m+1} \text{PolyLog}(3,a x)}{d (m+1)^2}+\frac{(d x)^{m+1} \text{PolyLog}(4,a x)}{d (m+1)}+\frac{a (d x)^{m+2} \, _2F_1(1,m+2;m+3;a x)}{d^2 (m+1)^4 (m+2)}+\frac{\log (1-a x) (d x)^{m+1}}{d (m+1)^4} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*PolyLog[4, a*x],x]

[Out]

(a*(d*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + m, a*x])/(d^2*(1 + m)^4*(2 + m)) + ((d*x)^(1 + m)*Log[1 - a*x
])/(d*(1 + m)^4) + ((d*x)^(1 + m)*PolyLog[2, a*x])/(d*(1 + m)^3) - ((d*x)^(1 + m)*PolyLog[3, a*x])/(d*(1 + m)^
2) + ((d*x)^(1 + m)*PolyLog[4, a*x])/(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 2395

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 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), 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])))

Rubi steps

\begin{align*} \int (d x)^m \text{Li}_4(a x) \, dx &=\frac{(d x)^{1+m} \text{Li}_4(a x)}{d (1+m)}-\frac{\int (d x)^m \text{Li}_3(a x) \, dx}{1+m}\\ &=-\frac{(d x)^{1+m} \text{Li}_3(a x)}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4(a x)}{d (1+m)}+\frac{\int (d x)^m \text{Li}_2(a x) \, dx}{(1+m)^2}\\ &=\frac{(d x)^{1+m} \text{Li}_2(a x)}{d (1+m)^3}-\frac{(d x)^{1+m} \text{Li}_3(a x)}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4(a x)}{d (1+m)}+\frac{\int (d x)^m \log (1-a x) \, dx}{(1+m)^3}\\ &=\frac{(d x)^{1+m} \log (1-a x)}{d (1+m)^4}+\frac{(d x)^{1+m} \text{Li}_2(a x)}{d (1+m)^3}-\frac{(d x)^{1+m} \text{Li}_3(a x)}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4(a x)}{d (1+m)}+\frac{a \int \frac{(d x)^{1+m}}{1-a x} \, dx}{d (1+m)^4}\\ &=\frac{a (d x)^{2+m} \, _2F_1(1,2+m;3+m;a x)}{d^2 (1+m)^4 (2+m)}+\frac{(d x)^{1+m} \log (1-a x)}{d (1+m)^4}+\frac{(d x)^{1+m} \text{Li}_2(a x)}{d (1+m)^3}-\frac{(d x)^{1+m} \text{Li}_3(a x)}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4(a x)}{d (1+m)}\\ \end{align*}

Mathematica [C]  time = 0.0726869, size = 119, normalized size = 0.98 \[ \frac{x \text{Gamma}(m+2) (d x)^m \left (a (m+1) x \text{Gamma}(m+1) \, _2\tilde{F}_1(1,m+2;m+3;a x)+m^3 \text{PolyLog}(4,a x)-m^2 \text{PolyLog}(3,a x)+3 m^2 \text{PolyLog}(4,a x)-2 m \text{PolyLog}(3,a x)+3 m \text{PolyLog}(4,a x)+(m+1) \text{PolyLog}(2,a x)-\text{PolyLog}(3,a x)+\text{PolyLog}(4,a x)+\log (1-a x)\right )}{(m+1)^5 \text{Gamma}(m+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*x)^m*PolyLog[4, 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] + Log
[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] + PolyLog[4,
a*x] + 3*m*PolyLog[4, a*x] + 3*m^2*PolyLog[4, a*x] + m^3*PolyLog[4, a*x]))/((1 + m)^5*Gamma[1 + m])

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Maple [C]  time = 0.794, size = 198, normalized size = 1.6 \begin{align*}{\frac{ \left ( dx \right ) ^{m}{x}^{-m} \left ( -a \right ) ^{-m}}{a} \left ({\frac{{x}^{m} \left ( -a \right ) ^{m} \left ( -a{m}^{2}x-2\,amx-{m}^{2}-3\,m-2 \right ) }{ \left ( 1+m \right ) ^{5} \left ( 2+m \right ) m}}-{\frac{{x}^{1+m}a \left ( -a \right ) ^{m} \left ( -m-2 \right ) \ln \left ( -ax+1 \right ) }{ \left ( 1+m \right ) ^{4} \left ( 2+m \right ) }}+{\frac{{x}^{1+m}a \left ( -a \right ) ^{m}{\it polylog} \left ( 2,ax \right ) }{ \left ( 1+m \right ) ^{3}}}+{\frac{{x}^{1+m}a \left ( -a \right ) ^{m} \left ( -m-2 \right ){\it polylog} \left ( 3,ax \right ) }{ \left ( 1+m \right ) ^{2} \left ( 2+m \right ) }}+{\frac{{x}^{1+m}a \left ( -a \right ) ^{m}{\it polylog} \left ( 4,ax \right ) }{1+m}}+{\frac{{x}^{m} \left ( -a \right ) ^{m}{\it LerchPhi} \left ( ax,1,m \right ) }{ \left ( 1+m \right ) ^{4}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*polylog(4,a*x),x)

[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)^5/m-1/(2+m)*x^(1+m)*a*(-a)^m
*(-m-2)/(1+m)^4*ln(-a*x+1)+x^(1+m)*a*(-a)^m/(1+m)^3*polylog(2,a*x)+1/(2+m)*x^(1+m)*a*(-a)^m*(-m-2)/(1+m)^2*pol
ylog(3,a*x)+x^(1+m)*a*(-a)^m/(1+m)*polylog(4,a*x)+x^m*(-a)^m/(1+m)^4*LerchPhi(a*x,1,m))

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

[Out]

-a*d^m*integrate(-x*x^m/(m^4 + 4*m^3 + 6*m^2 - (a*m^4 + 4*a*m^3 + 6*a*m^2 + 4*a*m + a)*x + 4*m + 1), x) + ((d^
m*m + d^m)*x*x^m*dilog(a*x) + d^m*x*x^m*log(-a*x + 1) + (d^m*m^3 + 3*d^m*m^2 + 3*d^m*m + d^m)*x*x^m*polylog(4,
a*x) - (d^m*m^2 + 2*d^m*m + d^m)*x*x^m*polylog(3, a*x))/(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\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x)^m*polylog(4, a*x), 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\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*x)**m*polylog(4, a*x), 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)\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^m*polylog(4, a*x), x)