3.101.65 \(\int \frac {-240 x \log ^2(x)+(480 e^3-480 x) \log (x) \log (-e^3+x)}{e^3 x-x^2} \, dx\)

Optimal. Leaf size=25 \[ -e^2+240 \log ^2(x) \log \left (\left (1-\frac {e^3}{x}\right ) x\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 14, normalized size of antiderivative = 0.56, number of steps used = 10, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1593, 6742, 2317, 2374, 6589, 2375} \begin {gather*} 240 \log ^2(x) \log \left (x-e^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-240*x*Log[x]^2 + (480*E^3 - 480*x)*Log[x]*Log[-E^3 + x])/(E^3*x - x^2),x]

[Out]

240*Log[x]^2*Log[-E^3 + x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2317

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

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 2375

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-240 x \log ^2(x)+\left (480 e^3-480 x\right ) \log (x) \log \left (-e^3+x\right )}{\left (e^3-x\right ) x} \, dx\\ &=\int \left (-\frac {240 \log ^2(x)}{e^3-x}+\frac {480 \log (x) \log \left (-e^3+x\right )}{x}\right ) \, dx\\ &=-\left (240 \int \frac {\log ^2(x)}{e^3-x} \, dx\right )+480 \int \frac {\log (x) \log \left (-e^3+x\right )}{x} \, dx\\ &=240 \log ^2(x) \log \left (-e^3+x\right )+240 \log ^2(x) \log \left (1-\frac {x}{e^3}\right )-240 \int \frac {\log ^2(x)}{-e^3+x} \, dx-480 \int \frac {\log (x) \log \left (1-\frac {x}{e^3}\right )}{x} \, dx\\ &=240 \log ^2(x) \log \left (-e^3+x\right )+480 \log (x) \text {Li}_2\left (\frac {x}{e^3}\right )+480 \int \frac {\log (x) \log \left (1-\frac {x}{e^3}\right )}{x} \, dx-480 \int \frac {\text {Li}_2\left (\frac {x}{e^3}\right )}{x} \, dx\\ &=240 \log ^2(x) \log \left (-e^3+x\right )-480 \text {Li}_3\left (\frac {x}{e^3}\right )+480 \int \frac {\text {Li}_2\left (\frac {x}{e^3}\right )}{x} \, dx\\ &=240 \log ^2(x) \log \left (-e^3+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 14, normalized size = 0.56 \begin {gather*} 240 \log ^2(x) \log \left (-e^3+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-240*x*Log[x]^2 + (480*E^3 - 480*x)*Log[x]*Log[-E^3 + x])/(E^3*x - x^2),x]

[Out]

240*Log[x]^2*Log[-E^3 + x]

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fricas [A]  time = 1.08, size = 13, normalized size = 0.52 \begin {gather*} 240 \, \log \left (x - e^{3}\right ) \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-240*x*log(x)^2+(480*exp(3)-480*x)*log(x-exp(3))*log(x))/(x*exp(3)-x^2),x, algorithm="fricas")

[Out]

240*log(x - e^3)*log(x)^2

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giac [A]  time = 0.16, size = 13, normalized size = 0.52 \begin {gather*} 240 \, \log \left (x - e^{3}\right ) \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-240*x*log(x)^2+(480*exp(3)-480*x)*log(x-exp(3))*log(x))/(x*exp(3)-x^2),x, algorithm="giac")

[Out]

240*log(x - e^3)*log(x)^2

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maple [A]  time = 0.16, size = 14, normalized size = 0.56

method	result	size
risch	\(240 \ln \relax (x )^{2} \ln \left (x -{\mathrm e}^{3}\right )\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-240*x*ln(x)^2+(480*exp(3)-480*x)*ln(x-exp(3))*ln(x))/(x*exp(3)-x^2),x,method=_RETURNVERBOSE)

[Out]

240*ln(x)^2*ln(x-exp(3))

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maxima [A]  time = 0.43, size = 13, normalized size = 0.52 \begin {gather*} 240 \, \log \left (x - e^{3}\right ) \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-240*x*log(x)^2+(480*exp(3)-480*x)*log(x-exp(3))*log(x))/(x*exp(3)-x^2),x, algorithm="maxima")

[Out]

240*log(x - e^3)*log(x)^2

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mupad [B]  time = 9.20, size = 13, normalized size = 0.52 \begin {gather*} 240\,\ln \left (x-{\mathrm {e}}^3\right )\,{\ln \relax (x)}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(240*x*log(x)^2 + log(x - exp(3))*log(x)*(480*x - 480*exp(3)))/(x*exp(3) - x^2),x)

[Out]

240*log(x - exp(3))*log(x)^2

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sympy [A]  time = 0.42, size = 12, normalized size = 0.48 \begin {gather*} 240 \log {\relax (x )}^{2} \log {\left (x - e^{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-240*x*ln(x)**2+(480*exp(3)-480*x)*ln(x-exp(3))*ln(x))/(x*exp(3)-x**2),x)

[Out]

240*log(x)**2*log(x - exp(3))

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