3.3.7 \(\int (\frac {q \text {Li}_{-1+k}(e x^q)}{b n x (a+b \log (c x^n))}-\frac {\text {Li}_k(e x^q)}{x (a+b \log (c x^n))^2}) \, dx\) [207]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 26, normalized size of antiderivative = 1.00, 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 = {2431} \begin {gather*} \frac {\text {PolyLog}\left (k,e x^q\right )}{b n \left (a+b \log \left (c x^n\right )\right )} \end {gather*}

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 2431

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[PolyLo
g[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*}

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Mathematica [F]
time = 0.07, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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 = 0.03, size = 0, normalized size = 0.00 \[\int \frac {q \polylog \left (k -1, e \,x^{q}\right )}{b n x \left (a +b \ln \left (c \,x^{n}\right )\right )}-\frac {\polylog \left (k , e \,x^{q}\right )}{x \left (a +b \ln \left (c \,x^{n}\right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(q*polylog(k-1,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(k-1,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 = 0.00, 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(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, x^q*e)/((b*log(c*x^n) + a)*b*n*x) - polylog(k, x^q*e)/((b*log(c*x^n) + a)^2*x), x)

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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(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, x^q*e) - (b*q*log(c*x^n) + a*q)*polylog(k - 1, x^q*e))/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \left (- \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}}\right )\, 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 {gather*}

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 = 0.00, 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(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, x^q*e)/((b*log(c*x^n) + a)*b*n*x) - polylog(k, x^q*e)/((b*log(c*x^n) + a)^2*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {q\,\mathrm {polylog}\left (k-1,e\,x^q\right )}{b\,n\,x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}-\frac {\mathrm {polylog}\left (k,e\,x^q\right )}{x\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((q*polylog(k - 1, 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]

int((q*polylog(k - 1, 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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