3.105 \(\int (d x)^m \text{PolyLog}(2,a x^2) \, dx\)

Optimal. Leaf size=94 \[ \frac{4 a (d x)^{m+3} \text{Hypergeometric2F1}\left (1,\frac{m+3}{2},\frac{m+5}{2},a x^2\right )}{d^3 (m+1)^2 (m+3)}+\frac{(d x)^{m+1} \text{PolyLog}\left (2,a x^2\right )}{d (m+1)}+\frac{2 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^2} \]

[Out]

(4*a*(d*x)^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, a*x^2])/(d^3*(1 + m)^2*(3 + m)) + (2*(d*x)^(1 +
m)*Log[1 - a*x^2])/(d*(1 + m)^2) + ((d*x)^(1 + m)*PolyLog[2, a*x^2])/(d*(1 + m))

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Rubi [A]  time = 0.0549094, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {6591, 2455, 16, 364} \[ \frac{(d x)^{m+1} \text{PolyLog}\left (2,a x^2\right )}{d (m+1)}+\frac{4 a (d x)^{m+3} \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};a x^2\right )}{d^3 (m+1)^2 (m+3)}+\frac{2 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*PolyLog[2, a*x^2],x]

[Out]

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

Mathematica [A]  time = 0.0454423, size = 72, normalized size = 0.77 \[ \frac{x (d x)^m \left (4 a x^2 \text{Hypergeometric2F1}\left (1,\frac{m+3}{2},\frac{m+5}{2},a x^2\right )+(m+3) \left ((m+1) \text{PolyLog}\left (2,a x^2\right )+2 \log \left (1-a x^2\right )\right )\right )}{(m+1)^2 (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d*x)^m*(4*a*x^2*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, a*x^2] + (3 + m)*(2*Log[1 - a*x^2] + (1 + m)*Po
lyLog[2, a*x^2])))/((1 + m)^2*(3 + m))

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Maple [C]  time = 0.236, size = 177, normalized size = 1.9 \begin{align*} -{\frac{ \left ( dx \right ) ^{m}{x}^{-m}}{2} \left ( -a \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( 2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( -12-4\,m \right ) }{ \left ( 1+m \right ) ^{3} \left ( 3+m \right ) a}}-2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( -6-2\,m \right ) \ln \left ( -a{x}^{2}+1 \right ) }{ \left ( 1+m \right ) ^{2} \left ( 3+m \right ) a}}+2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2}{\it polylog} \left ( 2,a{x}^{2} \right ) }{ \left ( 1+m \right ) a}}+2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( 6+2\,m \right ){\it LerchPhi} \left ( a{x}^{2},1,1/2+m/2 \right ) }{ \left ( 1+m \right ) ^{2} \left ( 3+m \right ) a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a*d^m*integrate(x^2*x^m/((a*m^2 + 2*a*m + a)*x^2 - m^2 - 2*m - 1), x) + ((d^m*m + d^m)*x*x^m*dilog(a*x^2) +
2*d^m*x*x^m*log(-a*x^2 + 1))/(m^2 + 2*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 Li}_2\left (a x^{2}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x)^m*dilog(a*x^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m}{\rm Li}_2\left (a x^{2}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^m*dilog(a*x^2), x)