3.2.11 \(\int (d x)^m \text {PolyLog}(2,a x^q) \, dx\) [111]

Optimal. Leaf size=101 \[ \frac {a q^2 x^{1+q} (d x)^m \, _2F_1\left (1,\frac {1+m+q}{q};\frac {1+m+2 q}{q};a x^q\right )}{(1+m)^2 (1+m+q)}+\frac {q (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {PolyLog}\left (2,a x^q\right )}{d (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 101, 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, 20, 371} \begin {gather*} \frac {a q^2 x^{q+1} (d x)^m \, _2F_1\left (1,\frac {m+q+1}{q};\frac {m+2 q+1}{q};a x^q\right )}{(m+1)^2 (m+q+1)}+\frac {(d x)^{m+1} \text {Li}_2\left (a x^q\right )}{d (m+1)}+\frac {q (d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*PolyLog[2, a*x^q],x]

[Out]

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

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[b^IntPart[n]*((b*v)^FracPart[n]/(a^IntPart[n]
*(a*v)^FracPart[n])), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

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^q\right ) \, dx &=\frac {(d x)^{1+m} \text {Li}_2\left (a x^q\right )}{d (1+m)}+\frac {q \int (d x)^m \log \left (1-a x^q\right ) \, dx}{1+m}\\ &=\frac {q (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^q\right )}{d (1+m)}+\frac {\left (a q^2\right ) \int \frac {x^{-1+q} (d x)^{1+m}}{1-a x^q} \, dx}{d (1+m)^2}\\ &=\frac {q (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^q\right )}{d (1+m)}+\frac {\left (a q^2 x^{-m} (d x)^m\right ) \int \frac {x^{m+q}}{1-a x^q} \, dx}{(1+m)^2}\\ &=\frac {a q^2 x^{1+q} (d x)^m \, _2F_1\left (1,\frac {1+m+q}{q};\frac {1+m+2 q}{q};a x^q\right )}{(1+m)^2 (1+m+q)}+\frac {q (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \text {Li}_2\left (a x^q\right )}{d (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 80, normalized size = 0.79 \begin {gather*} \frac {x (d x)^m \left (a q^2 x^q \, _2F_1\left (1,\frac {1+m+q}{q};\frac {1+m+2 q}{q};a x^q\right )+(1+m+q) \left (q \log \left (1-a x^q\right )+(1+m) \text {PolyLog}\left (2,a x^q\right )\right )\right )}{(1+m)^2 (1+m+q)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.13, size = 148, normalized size = 1.47

method	result	size
meijerg	\(-\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {m}{q}-\frac {1}{q}} \left (-\frac {q^{2} x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \ln \left (1-a \,x^{q}\right )}{\left (1+m \right )^{2}}-\frac {q \,x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \polylog \left (2, a \,x^{q}\right )}{1+m}-\frac {q^{2} x^{1+m +q} a \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \Phi \left (a \,x^{q}, 1, \frac {1+m +q}{q}\right )}{\left (1+m \right )^{2}}\right )}{q}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x)^m*x^(-m)*(-a)^(-m/q-1/q)/q*(-q^2*x^(1+m)*(-a)^(m/q+1/q)/(1+m)^2*ln(1-a*x^q)-q*x^(1+m)*(-a)^(m/q+1/q)/(1
+m)*polylog(2,a*x^q)-q^2*x^(1+m+q)*a*(-a)^(m/q+1/q)/(1+m)^2*LerchPhi(a*x^q,1,(1+m+q)/q))

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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^q),x, algorithm="maxima")

[Out]

-d^m*q^2*integrate(-x^m/(m^2 - (a*m^2 + 2*a*m + a)*x^q + 2*m + 1), x) - (d^m*q^2*x*x^m - (d^m*m + d^m)*q*x*x^m
*log(-a*x^q + 1) - (d^m*m^2 + 2*d^m*m + d^m)*x*x^m*dilog(a*x^q))/(m^3 + 3*m^2 + 3*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^q),x, algorithm="fricas")

[Out]

integral((d*x)^m*dilog(a*x^q), 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^{q}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((d*x)^m*dilog(a*x^q), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*polylog(2, a*x^q),x)

[Out]

int((d*x)^m*polylog(2, a*x^q), x)

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