3.2.5 \(\int (d x)^m \text {PolyLog}(2,a x^2) \, dx\) [105]

Optimal. Leaf size=94 \[ \frac {4 a (d x)^{3+m} \, _2F_1\left (1,\frac {3+m}{2};\frac {5+m}{2};a x^2\right )}{d^3 (1+m)^2 (3+m)}+\frac {2 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {PolyLog}\left (2,a x^2\right )}{d (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, 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 = {6726, 2505, 16, 371} \begin {gather*} \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 {\text {Li}_2\left (a x^2\right ) (d x)^{m+1}}{d (m+1)}+\frac {2 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^2} \end {gather*}

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 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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2505

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 6726

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}_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*}

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Mathematica [A]
time = 0.04, size = 72, normalized size = 0.77 \begin {gather*} \frac {x (d x)^m \left (4 a x^2 \, _2F_1\left (1,\frac {3+m}{2};\frac {5+m}{2};a x^2\right )+(3+m) \left (2 \log \left (1-a x^2\right )+(1+m) \text {PolyLog}\left (2,a x^2\right )\right )\right )}{(1+m)^2 (3+m)} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.12, size = 177, normalized size = 1.88

method	result	size
meijerg	\(-\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (-12-4 m \right )}{\left (3+m \right ) \left (1+m \right )^{3} a}-\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (-6-2 m \right ) \ln \left (-a \,x^{2}+1\right )}{\left (3+m \right ) \left (1+m \right )^{2} a}+\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \polylog \left (2, a \,x^{2}\right )}{\left (1+m \right ) a}+\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (6+2 m \right ) \Phi \left (a \,x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{\left (3+m \right ) \left (1+m \right )^{2} a}\right )}{2}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)/a*polylog(2,a*x^2)+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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{m} \operatorname {Li}_{2}\left (a x^{2}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)*(d*x)^m,x)

[Out]

int(polylog(2, a*x^2)*(d*x)^m, x)

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