3.619 \(\int \frac {\log (f x^p) \log (1+e x^m)}{x} \, dx\)

Optimal. Leaf size=33 \[ \frac {p \text {Li}_3\left (-e x^m\right )}{m^2}-\frac {\text {Li}_2\left (-e x^m\right ) \log \left (f x^p\right )}{m} \]

[Out]

-ln(f*x^p)*polylog(2,-e*x^m)/m+p*polylog(3,-e*x^m)/m^2

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Rubi [A]  time = 0.03, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2374, 6589} \[ \frac {p \text {PolyLog}\left (3,-e x^m\right )}{m^2}-\frac {\log \left (f x^p\right ) \text {PolyLog}\left (2,-e x^m\right )}{m} \]

Antiderivative was successfully verified.

[In]

Int[(Log[f*x^p]*Log[1 + e*x^m])/x,x]

[Out]

-((Log[f*x^p]*PolyLog[2, -(e*x^m)])/m) + (p*PolyLog[3, -(e*x^m)])/m^2

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\log \left (f x^p\right ) \log \left (1+e x^m\right )}{x} \, dx &=-\frac {\log \left (f x^p\right ) \text {Li}_2\left (-e x^m\right )}{m}+\frac {p \int \frac {\text {Li}_2\left (-e x^m\right )}{x} \, dx}{m}\\ &=-\frac {\log \left (f x^p\right ) \text {Li}_2\left (-e x^m\right )}{m}+\frac {p \text {Li}_3\left (-e x^m\right )}{m^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.00 \[ \frac {p \text {Li}_3\left (-e x^m\right )}{m^2}-\frac {\text {Li}_2\left (-e x^m\right ) \log \left (f x^p\right )}{m} \]

Antiderivative was successfully verified.

[In]

Integrate[(Log[f*x^p]*Log[1 + e*x^m])/x,x]

[Out]

-((Log[f*x^p]*PolyLog[2, -(e*x^m)])/m) + (p*PolyLog[3, -(e*x^m)])/m^2

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fricas [C]  time = 0.70, size = 35, normalized size = 1.06 \[ -\frac {{\left (m p \log \relax (x) + m \log \relax (f)\right )} {\rm Li}_2\left (-e x^{m}\right ) - p {\rm polylog}\left (3, -e x^{m}\right )}{m^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(f*x^p)*log(1+e*x^m)/x,x, algorithm="fricas")

[Out]

-((m*p*log(x) + m*log(f))*dilog(-e*x^m) - p*polylog(3, -e*x^m))/m^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (e x^{m} + 1\right ) \log \left (f x^{p}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(f*x^p)*log(1+e*x^m)/x,x, algorithm="giac")

[Out]

integrate(log(e*x^m + 1)*log(f*x^p)/x, x)

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maple [C]  time = 2.06, size = 191, normalized size = 5.79 \[ \frac {i \pi \dilog \left (e \,x^{m}+1\right ) \mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i x^{p}\right ) \mathrm {csgn}\left (i f \,x^{p}\right )}{2 m}-\frac {i \pi \dilog \left (e \,x^{m}+1\right ) \mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i f \,x^{p}\right )^{2}}{2 m}-\frac {i \pi \dilog \left (e \,x^{m}+1\right ) \mathrm {csgn}\left (i x^{p}\right ) \mathrm {csgn}\left (i f \,x^{p}\right )^{2}}{2 m}+\frac {i \pi \dilog \left (e \,x^{m}+1\right ) \mathrm {csgn}\left (i f \,x^{p}\right )^{3}}{2 m}-\frac {p \polylog \left (2, -e \,x^{m}\right ) \ln \relax (x )}{m}-\frac {\dilog \left (e \,x^{m}+1\right ) \ln \relax (f )}{m}-\frac {\left (-p \ln \relax (x )+\ln \left (x^{p}\right )\right ) \dilog \left (e \,x^{m}+1\right )}{m}+\frac {p \polylog \left (3, -e \,x^{m}\right )}{m^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(f*x^p)*ln(1+e*x^m)/x,x)

[Out]

-p/m*ln(x)*polylog(2,-e*x^m)+p*polylog(3,-e*x^m)/m^2-(ln(x^p)-p*ln(x))/m*dilog(1+e*x^m)+1/2*I/m*dilog(1+e*x^m)
*Pi*csgn(I*f)*csgn(I*x^p)*csgn(I*f*x^p)-1/2*I/m*dilog(1+e*x^m)*Pi*csgn(I*f)*csgn(I*f*x^p)^2-1/2*I/m*dilog(1+e*
x^m)*Pi*csgn(I*x^p)*csgn(I*f*x^p)^2+1/2*I/m*dilog(1+e*x^m)*Pi*csgn(I*f*x^p)^3-1/m*dilog(1+e*x^m)*ln(f)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (p \log \relax (x)^{2} - 2 \, \log \relax (f) \log \relax (x) - 2 \, \log \relax (x) \log \left (x^{p}\right )\right )} \log \left (e x^{m} + 1\right ) - \int \frac {2 \, e m x^{m} \log \relax (x) \log \left (x^{p}\right ) - {\left (e m p \log \relax (x)^{2} - 2 \, e m \log \relax (f) \log \relax (x)\right )} x^{m}}{2 \, {\left (e x x^{m} + x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(f*x^p)*log(1+e*x^m)/x,x, algorithm="maxima")

[Out]

-1/2*(p*log(x)^2 - 2*log(f)*log(x) - 2*log(x)*log(x^p))*log(e*x^m + 1) - integrate(1/2*(2*e*m*x^m*log(x)*log(x
^p) - (e*m*p*log(x)^2 - 2*e*m*log(f)*log(x))*x^m)/(e*x*x^m + x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\ln \left (f\,x^p\right )\,\ln \left (e\,x^m+1\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(f*x^p)*log(e*x^m + 1))/x,x)

[Out]

int((log(f*x^p)*log(e*x^m + 1))/x, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(f*x**p)*ln(1+e*x**m)/x,x)

[Out]

Timed out

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