3.107 \(\int (d x)^m \text {Li}_4(a x^2) \, dx\)

Optimal. Leaf size=142 \[ \frac {16 a (d x)^{m+3} \, _2F_1\left (1,\frac {m+3}{2};\frac {m+5}{2};a x^2\right )}{d^3 (m+1)^4 (m+3)}+\frac {4 \text {Li}_2\left (a x^2\right ) (d x)^{m+1}}{d (m+1)^3}-\frac {2 \text {Li}_3\left (a x^2\right ) (d x)^{m+1}}{d (m+1)^2}+\frac {\text {Li}_4\left (a x^2\right ) (d x)^{m+1}}{d (m+1)}+\frac {8 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^4} \]

[Out]

16*a*(d*x)^(3+m)*hypergeom([1, 3/2+1/2*m],[5/2+1/2*m],a*x^2)/d^3/(1+m)^4/(3+m)+8*(d*x)^(1+m)*ln(-a*x^2+1)/d/(1
+m)^4+4*(d*x)^(1+m)*polylog(2,a*x^2)/d/(1+m)^3-2*(d*x)^(1+m)*polylog(3,a*x^2)/d/(1+m)^2+(d*x)^(1+m)*polylog(4,
a*x^2)/d/(1+m)

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Rubi [A]  time = 0.09, antiderivative size = 142, normalized size of antiderivative = 1.00, 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} \[ \frac {4 (d x)^{m+1} \text {PolyLog}\left (2,a x^2\right )}{d (m+1)^3}-\frac {2 (d x)^{m+1} \text {PolyLog}\left (3,a x^2\right )}{d (m+1)^2}+\frac {(d x)^{m+1} \text {PolyLog}\left (4,a x^2\right )}{d (m+1)}+\frac {16 a (d x)^{m+3} \, _2F_1\left (1,\frac {m+3}{2};\frac {m+5}{2};a x^2\right )}{d^3 (m+1)^4 (m+3)}+\frac {8 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^4} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*PolyLog[4, a*x^2],x]

[Out]

(16*a*(d*x)^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, a*x^2])/(d^3*(1 + m)^4*(3 + m)) + (8*(d*x)^(1 +
m)*Log[1 - a*x^2])/(d*(1 + m)^4) + (4*(d*x)^(1 + m)*PolyLog[2, a*x^2])/(d*(1 + m)^3) - (2*(d*x)^(1 + m)*PolyL
og[3, a*x^2])/(d*(1 + m)^2) + ((d*x)^(1 + m)*PolyLog[4, a*x^2])/(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}_4\left (a x^2\right ) \, dx &=\frac {(d x)^{1+m} \text {Li}_4\left (a x^2\right )}{d (1+m)}-\frac {2 \int (d x)^m \text {Li}_3\left (a x^2\right ) \, dx}{1+m}\\ &=-\frac {2 (d x)^{1+m} \text {Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_4\left (a x^2\right )}{d (1+m)}+\frac {4 \int (d x)^m \text {Li}_2\left (a x^2\right ) \, dx}{(1+m)^2}\\ &=\frac {4 (d x)^{1+m} \text {Li}_2\left (a x^2\right )}{d (1+m)^3}-\frac {2 (d x)^{1+m} \text {Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_4\left (a x^2\right )}{d (1+m)}+\frac {8 \int (d x)^m \log \left (1-a x^2\right ) \, dx}{(1+m)^3}\\ &=\frac {8 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^4}+\frac {4 (d x)^{1+m} \text {Li}_2\left (a x^2\right )}{d (1+m)^3}-\frac {2 (d x)^{1+m} \text {Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_4\left (a x^2\right )}{d (1+m)}+\frac {(16 a) \int \frac {x (d x)^{1+m}}{1-a x^2} \, dx}{d (1+m)^4}\\ &=\frac {8 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^4}+\frac {4 (d x)^{1+m} \text {Li}_2\left (a x^2\right )}{d (1+m)^3}-\frac {2 (d x)^{1+m} \text {Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_4\left (a x^2\right )}{d (1+m)}+\frac {(16 a) \int \frac {(d x)^{2+m}}{1-a x^2} \, dx}{d^2 (1+m)^4}\\ &=\frac {16 a (d x)^{3+m} \, _2F_1\left (1,\frac {3+m}{2};\frac {5+m}{2};a x^2\right )}{d^3 (1+m)^4 (3+m)}+\frac {8 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^4}+\frac {4 (d x)^{1+m} \text {Li}_2\left (a x^2\right )}{d (1+m)^3}-\frac {2 (d x)^{1+m} \text {Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_4\left (a x^2\right )}{d (1+m)}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 166, normalized size = 1.17 \[ \frac {2 x \Gamma \left (\frac {m+3}{2}\right ) (d x)^m \left (4 a (m+1) x^2 \Gamma \left (\frac {m+1}{2}\right ) \, _2\tilde {F}_1\left (1,\frac {m+3}{2};\frac {m+5}{2};a x^2\right )+m^3 \text {Li}_4\left (a x^2\right )-2 m^2 \text {Li}_3\left (a x^2\right )+3 m^2 \text {Li}_4\left (a x^2\right )-4 m \text {Li}_3\left (a x^2\right )+3 m \text {Li}_4\left (a x^2\right )+4 (m+1) \text {Li}_2\left (a x^2\right )-2 \text {Li}_3\left (a x^2\right )+\text {Li}_4\left (a x^2\right )+8 \log \left (1-a x^2\right )\right )}{(m+1)^5 \Gamma \left (\frac {m+1}{2}\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*x)^m*PolyLog[4, a*x^2],x]

[Out]

(2*x*(d*x)^m*Gamma[(3 + m)/2]*(4*a*(1 + m)*x^2*Gamma[(1 + m)/2]*HypergeometricPFQRegularized[{1, (3 + m)/2}, {
(5 + m)/2}, a*x^2] + 8*Log[1 - a*x^2] + 4*(1 + m)*PolyLog[2, a*x^2] - 2*PolyLog[3, a*x^2] - 4*m*PolyLog[3, a*x
^2] - 2*m^2*PolyLog[3, a*x^2] + PolyLog[4, a*x^2] + 3*m*PolyLog[4, a*x^2] + 3*m^2*PolyLog[4, a*x^2] + m^3*Poly
Log[4, a*x^2]))/((1 + m)^5*Gamma[(1 + m)/2])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d x\right )^{m} {\rm polylog}\left (4, a x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*polylog(4,a*x^2),x, algorithm="fricas")

[Out]

integral((d*x)^m*polylog(4, a*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{m} {\rm Li}_{4}(a x^{2})\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*polylog(4,a*x^2),x, algorithm="giac")

[Out]

integrate((d*x)^m*polylog(4, a*x^2), x)

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maple [C]  time = 0.67, size = 259, normalized size = 1.82 \[ -\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (-16 m -48\right )}{\left (m +3\right ) \left (1+m \right )^{5} a}-\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (-8 m -24\right ) \ln \left (-a \,x^{2}+1\right )}{\left (m +3\right ) \left (1+m \right )^{4} a}+\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (4 m +12\right ) \polylog \left (2, a \,x^{2}\right )}{\left (m +3\right ) \left (1+m \right )^{3} a}+\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (-2 m -6\right ) \polylog \left (3, a \,x^{2}\right )}{\left (m +3\right ) \left (1+m \right )^{2} a}+\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \polylog \left (4, a \,x^{2}\right )}{\left (1+m \right ) a}+\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (8 m +24\right ) \Phi \left (a \,x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{\left (m +3\right ) \left (1+m \right )^{4} a}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*polylog(4,a*x^2),x)

[Out]

-1/2*(d*x)^m*x^(-m)*(-a)^(-1/2-1/2*m)*(2/(m+3)*x^(1+m)*(-a)^(3/2+1/2*m)*(-16*m-48)/(1+m)^5/a-2/(m+3)*x^(1+m)*(
-a)^(3/2+1/2*m)*(-8*m-24)/(1+m)^4/a*ln(-a*x^2+1)+2/(m+3)*x^(1+m)*(-a)^(3/2+1/2*m)*(4*m+12)/(1+m)^3*polylog(2,a
*x^2)/a+2/(m+3)*x^(1+m)*(-a)^(3/2+1/2*m)*(-2*m-6)/(1+m)^2/a*polylog(3,a*x^2)+2*x^(1+m)*(-a)^(3/2+1/2*m)/(1+m)/
a*polylog(4,a*x^2)+2/(m+3)*x^(1+m)*(-a)^(3/2+1/2*m)*(8*m+24)/(1+m)^4/a*LerchPhi(a*x^2,1,1/2+1/2*m))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -16 \, a d^{m} \int -\frac {x^{2} x^{m}}{m^{4} + 4 \, m^{3} - {\left (a m^{4} + 4 \, a m^{3} + 6 \, a m^{2} + 4 \, a m + a\right )} x^{2} + 6 \, m^{2} + 4 \, m + 1}\,{d x} + \frac {4 \, {\left (d^{m} m + d^{m}\right )} x x^{m} {\rm Li}_2\left (a x^{2}\right ) + 8 \, d^{m} x x^{m} \log \left (-a x^{2} + 1\right ) + {\left (d^{m} m^{3} + 3 \, d^{m} m^{2} + 3 \, d^{m} m + d^{m}\right )} x x^{m} {\rm Li}_{4}(a x^{2}) - 2 \, {\left (d^{m} m^{2} + 2 \, d^{m} m + d^{m}\right )} x x^{m} {\rm Li}_{3}(a x^{2})}{m^{4} + 4 \, m^{3} + 6 \, m^{2} + 4 \, m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*polylog(4,a*x^2),x, algorithm="maxima")

[Out]

-16*a*d^m*integrate(-x^2*x^m/(m^4 + 4*m^3 - (a*m^4 + 4*a*m^3 + 6*a*m^2 + 4*a*m + a)*x^2 + 6*m^2 + 4*m + 1), x)
+ (4*(d^m*m + d^m)*x*x^m*dilog(a*x^2) + 8*d^m*x*x^m*log(-a*x^2 + 1) + (d^m*m^3 + 3*d^m*m^2 + 3*d^m*m + d^m)*x
*x^m*polylog(4, a*x^2) - 2*(d^m*m^2 + 2*d^m*m + d^m)*x*x^m*polylog(3, a*x^2))/(m^4 + 4*m^3 + 6*m^2 + 4*m + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {polylog}\left (4,a\,x^2\right )\,{\left (d\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(4, a*x^2)*(d*x)^m,x)

[Out]

int(polylog(4, a*x^2)*(d*x)^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{m} \operatorname {Li}_{4}\left (a x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*polylog(4,a*x**2),x)

[Out]

Integral((d*x)**m*polylog(4, a*x**2), x)

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