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

Optimal. Leaf size=33 \[ \frac{p \text{PolyLog}\left (3,-e x^m\right )}{m^2}-\frac{\log \left (f x^p\right ) \text{PolyLog}\left (2,-e x^m\right )}{m} \]

[Out]

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

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Rubi [A]  time = 0.0260934, antiderivative size = 33, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0115781, size = 33, normalized size = 1. \[ \frac{p \text{PolyLog}\left (3,-e x^m\right )}{m^2}-\frac{\log \left (f x^p\right ) \text{PolyLog}\left (2,-e x^m\right )}{m} \]

Antiderivative was successfully verified.

[In]

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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Maple [C]  time = 2.384, size = 191, normalized size = 5.8 \begin{align*} -{\frac{p\ln \left ( x \right ){\it polylog} \left ( 2,-e{x}^{m} \right ) }{m}}+{\frac{p{\it polylog} \left ( 3,-e{x}^{m} \right ) }{{m}^{2}}}-{\frac{ \left ( \ln \left ({x}^{p} \right ) -p\ln \left ( x \right ) \right ){\it dilog} \left ( 1+e{x}^{m} \right ) }{m}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+e{x}^{m} \right ) \pi \,{\it csgn} \left ( if \right ){\it csgn} \left ( i{x}^{p} \right ){\it csgn} \left ( if{x}^{p} \right ) }{m}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+e{x}^{m} \right ) \pi \,{\it csgn} \left ( if \right ) \left ({\it csgn} \left ( if{x}^{p} \right ) \right ) ^{2}}{m}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+e{x}^{m} \right ) \pi \,{\it csgn} \left ( i{x}^{p} \right ) \left ({\it csgn} \left ( if{x}^{p} \right ) \right ) ^{2}}{m}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+e{x}^{m} \right ) \pi \, \left ({\it csgn} \left ( if{x}^{p} \right ) \right ) ^{3}}{m}}-{\frac{{\it dilog} \left ( 1+e{x}^{m} \right ) \ln \left ( f \right ) }{m}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \,{\left (p \log \left (x\right )^{2} - 2 \, \log \left (f\right ) \log \left (x\right ) - 2 \, \log \left (x\right ) \log \left (x^{p}\right )\right )} \log \left (e x^{m} + 1\right ) - \int \frac{2 \, e m x^{m} \log \left (x\right ) \log \left (x^{p}\right ) -{\left (e m p \log \left (x\right )^{2} - 2 \, e m \log \left (f\right ) \log \left (x\right )\right )} x^{m}}{2 \,{\left (e x x^{m} + x\right )}}\,{d x} \end{align*}

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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Fricas [C]  time = 1.56477, size = 93, normalized size = 2.82 \begin{align*} -\frac{{\left (m p \log \left (x\right ) + m \log \left (f\right )\right )}{\rm Li}_2\left (-e x^{m}\right ) - p{\rm polylog}\left (3, -e x^{m}\right )}{m^{2}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (e x^{m} + 1\right ) \log \left (f x^{p}\right )}{x}\,{d x} \end{align*}

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)