3.207 \(\int (\frac{q \text{PolyLog}(-1+k,e x^q)}{b n x (a+b \log (c x^n))}-\frac{\text{PolyLog}(k,e x^q)}{x (a+b \log (c x^n))^2}) \, dx\)

Optimal. Leaf size=26 \[ \frac{\text{PolyLog}\left (k,e x^q\right )}{b n \left (a+b \log \left (c x^n\right )\right )} \]

[Out]

PolyLog[k, e*x^q]/(b*n*(a + b*Log[c*x^n]))

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Rubi [A]  time = 0.109864, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 57, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.018, Rules used = {2384} \[ \frac{\text{PolyLog}\left (k,e x^q\right )}{b n \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[(q*PolyLog[-1 + k, e*x^q])/(b*n*x*(a + b*Log[c*x^n])) - PolyLog[k, e*x^q]/(x*(a + b*Log[c*x^n])^2),x]

[Out]

PolyLog[k, e*x^q]/(b*n*(a + b*Log[c*x^n]))

Rule 2384

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[(PolyL
og[k, e*x^q]*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] - Dist[q/(b*n*(p + 1)), Int[(PolyLog[k - 1, e*x^q]*
(a + b*Log[c*x^n])^(p + 1))/x, x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && LtQ[p, -1]

Rubi steps

\begin{align*} \int \left (\frac{q \text{Li}_{-1+k}\left (e x^q\right )}{b n x \left (a+b \log \left (c x^n\right )\right )}-\frac{\text{Li}_k\left (e x^q\right )}{x \left (a+b \log \left (c x^n\right )\right )^2}\right ) \, dx &=\frac{q \int \frac{\text{Li}_{-1+k}\left (e x^q\right )}{x \left (a+b \log \left (c x^n\right )\right )} \, dx}{b n}-\int \frac{\text{Li}_k\left (e x^q\right )}{x \left (a+b \log \left (c x^n\right )\right )^2} \, dx\\ &=\frac{\text{Li}_k\left (e x^q\right )}{b n \left (a+b \log \left (c x^n\right )\right )}\\ \end{align*}

Mathematica [F]  time = 0.147066, size = 0, normalized size = 0. \[ \int \left (\frac{q \text{PolyLog}\left (-1+k,e x^q\right )}{b n x \left (a+b \log \left (c x^n\right )\right )}-\frac{\text{PolyLog}\left (k,e x^q\right )}{x \left (a+b \log \left (c x^n\right )\right )^2}\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(q*PolyLog[-1 + k, e*x^q])/(b*n*x*(a + b*Log[c*x^n])) - PolyLog[k, e*x^q]/(x*(a + b*Log[c*x^n])^2),x
]

[Out]

Integrate[(q*PolyLog[-1 + k, e*x^q])/(b*n*x*(a + b*Log[c*x^n])) - PolyLog[k, e*x^q]/(x*(a + b*Log[c*x^n])^2),
x]

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{\frac{q{\it polylog} \left ( -1+k,e{x}^{q} \right ) }{bnx \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }}-{\frac{{\it polylog} \left ( k,e{x}^{q} \right ) }{x \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(q*polylog(-1+k,e*x^q)/b/n/x/(a+b*ln(c*x^n))-polylog(k,e*x^q)/x/(a+b*ln(c*x^n))^2,x)

[Out]

int(q*polylog(-1+k,e*x^q)/b/n/x/(a+b*ln(c*x^n))-polylog(k,e*x^q)/x/(a+b*ln(c*x^n))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{q{\rm Li}_{k - 1}(e x^{q})}{{\left (b \log \left (c x^{n}\right ) + a\right )} b n x} - \frac{{\rm Li}_{k}(e x^{q})}{{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(q*polylog(-1+k,e*x^q)/b/n/x/(a+b*log(c*x^n))-polylog(k,e*x^q)/x/(a+b*log(c*x^n))^2,x, algorithm="max
ima")

[Out]

integrate(q*polylog(k - 1, e*x^q)/((b*log(c*x^n) + a)*b*n*x) - polylog(k, e*x^q)/((b*log(c*x^n) + a)^2*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b n{\rm polylog}\left (k, e x^{q}\right ) -{\left (b q \log \left (c x^{n}\right ) + a q\right )}{\rm polylog}\left (k - 1, e x^{q}\right )}{b^{3} n x \log \left (c x^{n}\right )^{2} + 2 \, a b^{2} n x \log \left (c x^{n}\right ) + a^{2} b n x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(q*polylog(-1+k,e*x^q)/b/n/x/(a+b*log(c*x^n))-polylog(k,e*x^q)/x/(a+b*log(c*x^n))^2,x, algorithm="fri
cas")

[Out]

integral(-(b*n*polylog(k, e*x^q) - (b*q*log(c*x^n) + a*q)*polylog(k - 1, e*x^q))/(b^3*n*x*log(c*x^n)^2 + 2*a*b
^2*n*x*log(c*x^n) + a^2*b*n*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a q \operatorname{Li}_{k - 1}\left (e x^{q}\right )}{a^{2} x + 2 a b x \log{\left (c x^{n} \right )} + b^{2} x \log{\left (c x^{n} \right )}^{2}}\, dx + \int - \frac{b n \operatorname{Li}_{k}\left (e x^{q}\right )}{a^{2} x + 2 a b x \log{\left (c x^{n} \right )} + b^{2} x \log{\left (c x^{n} \right )}^{2}}\, dx + \int \frac{b q \log{\left (c x^{n} \right )} \operatorname{Li}_{k - 1}\left (e x^{q}\right )}{a^{2} x + 2 a b x \log{\left (c x^{n} \right )} + b^{2} x \log{\left (c x^{n} \right )}^{2}}\, dx}{b n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(q*polylog(-1+k,e*x**q)/b/n/x/(a+b*ln(c*x**n))-polylog(k,e*x**q)/x/(a+b*ln(c*x**n))**2,x)

[Out]

(Integral(a*q*polylog(k - 1, e*x**q)/(a**2*x + 2*a*b*x*log(c*x**n) + b**2*x*log(c*x**n)**2), x) + Integral(-b*
n*polylog(k, e*x**q)/(a**2*x + 2*a*b*x*log(c*x**n) + b**2*x*log(c*x**n)**2), x) + Integral(b*q*log(c*x**n)*pol
ylog(k - 1, e*x**q)/(a**2*x + 2*a*b*x*log(c*x**n) + b**2*x*log(c*x**n)**2), x))/(b*n)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{q{\rm Li}_{k - 1}(e x^{q})}{{\left (b \log \left (c x^{n}\right ) + a\right )} b n x} - \frac{{\rm Li}_{k}(e x^{q})}{{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(q*polylog(-1+k,e*x^q)/b/n/x/(a+b*log(c*x^n))-polylog(k,e*x^q)/x/(a+b*log(c*x^n))^2,x, algorithm="gia
c")

[Out]

integrate(q*polylog(k - 1, e*x^q)/((b*log(c*x^n) + a)*b*n*x) - polylog(k, e*x^q)/((b*log(c*x^n) + a)^2*x), x)