3.84.14 \(\int \frac {(-16 x+28 x^2-8 x^3) \log (x)+(8 x-4 x^2+(8 x-28 x^2+12 x^3) \log (x)) \log (e^{-2 x} (-2 x^2+x^3))+(-16 x+8 x^2) \log (x) \log (e^{-2 x} (-2 x^2+x^3)) \log (\frac {\log (e^{-2 x} (-2 x^2+x^3))}{x \log (x)})}{(-2+x) \log (x) \log (e^{-2 x} (-2 x^2+x^3))} \, dx\)

Optimal. Leaf size=29 \[ 4 x^2 \left (x+\log \left (\frac {\log \left (e^{-2 x} (-2+x) x^2\right )}{x \log (x)}\right )\right ) \]

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Rubi [C]  time = 1.42, antiderivative size = 63, normalized size of antiderivative = 2.17, number of steps used = 22, number of rules used = 11, integrand size = 142, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6688, 2309, 2178, 6742, 43, 30, 2555, 12, 14, 2366, 6482} \begin {gather*} -4 \text {Ei}(2 \log (x))-4 \log (x) \text {Ei}(2 \log (x))+4 (\log (x)+1) \text {Ei}(2 \log (x))+4 x^3+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-16*x + 28*x^2 - 8*x^3)*Log[x] + (8*x - 4*x^2 + (8*x - 28*x^2 + 12*x^3)*Log[x])*Log[(-2*x^2 + x^3)/E^(2*
x)] + (-16*x + 8*x^2)*Log[x]*Log[(-2*x^2 + x^3)/E^(2*x)]*Log[Log[(-2*x^2 + x^3)/E^(2*x)]/(x*Log[x])])/((-2 + x
)*Log[x]*Log[(-2*x^2 + x^3)/E^(2*x)]),x]

[Out]

4*x^3 - 4*ExpIntegralEi[2*Log[x]] - 4*ExpIntegralEi[2*Log[x]]*Log[x] + 4*ExpIntegralEi[2*Log[x]]*(1 + Log[x])
+ 4*x^2*Log[Log[-(((2 - x)*x^2)/E^(2*x))]/(x*Log[x])]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 6482

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, x]

Rule 6688

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

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 \left (-\frac {4 x}{\log (x)}+\frac {4 x \left (-4+7 x-2 x^2+(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right ) \left (-1+3 x+2 \log \left (\frac {\log \left (e^{-2 x} (-2+x) x^2\right )}{x \log (x)}\right )\right )\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}\right ) \, dx\\ &=-\left (4 \int \frac {x}{\log (x)} \, dx\right )+4 \int \frac {x \left (-4+7 x-2 x^2+(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right ) \left (-1+3 x+2 \log \left (\frac {\log \left (e^{-2 x} (-2+x) x^2\right )}{x \log (x)}\right )\right )\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )} \, dx\\ &=4 \int \left (\frac {x \left (-4+7 x-2 x^2+2 \log \left (e^{-2 x} (-2+x) x^2\right )-7 x \log \left (e^{-2 x} (-2+x) x^2\right )+3 x^2 \log \left (e^{-2 x} (-2+x) x^2\right )\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}+2 x \log \left (\frac {\log \left (e^{-2 x} (-2+x) x^2\right )}{x \log (x)}\right )\right ) \, dx-4 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-4 \text {Ei}(2 \log (x))+4 \int \frac {x \left (-4+7 x-2 x^2+2 \log \left (e^{-2 x} (-2+x) x^2\right )-7 x \log \left (e^{-2 x} (-2+x) x^2\right )+3 x^2 \log \left (e^{-2 x} (-2+x) x^2\right )\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )} \, dx+8 \int x \log \left (\frac {\log \left (e^{-2 x} (-2+x) x^2\right )}{x \log (x)}\right ) \, dx\\ &=-4 \text {Ei}(2 \log (x))+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right )+4 \int \frac {x \left (4-7 x+2 x^2-\left (2-7 x+3 x^2\right ) \log \left (e^{-2 x} (-2+x) x^2\right )\right )}{(2-x) \log \left (e^{-2 x} (-2+x) x^2\right )} \, dx-8 \int -\frac {1}{2} x \left (\frac {1}{\log (x)}+\frac {4-7 x+2 x^2+(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}\right ) \, dx\\ &=-4 \text {Ei}(2 \log (x))+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right )+4 \int \left (x (-1+3 x)-\frac {x \left (4-7 x+2 x^2\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}\right ) \, dx+4 \int x \left (\frac {1}{\log (x)}+\frac {4-7 x+2 x^2+(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}\right ) \, dx\\ &=-4 \text {Ei}(2 \log (x))+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right )+4 \int x (-1+3 x) \, dx+4 \int \left (\frac {x (1+\log (x))}{\log (x)}+\frac {x \left (4-7 x+2 x^2\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}\right ) \, dx-4 \int \frac {x \left (4-7 x+2 x^2\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )} \, dx\\ &=-4 \text {Ei}(2 \log (x))+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right )+4 \int \left (-x+3 x^2\right ) \, dx+4 \int \frac {x (1+\log (x))}{\log (x)} \, dx-4 \int \left (-\frac {2}{\log \left (e^{-2 x} (-2+x) x^2\right )}-\frac {4}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}-\frac {3 x}{\log \left (e^{-2 x} (-2+x) x^2\right )}+\frac {2 x^2}{\log \left (e^{-2 x} (-2+x) x^2\right )}\right ) \, dx+4 \int \frac {x \left (4-7 x+2 x^2\right )}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )} \, dx\\ &=-2 x^2+4 x^3-4 \text {Ei}(2 \log (x))+4 \text {Ei}(2 \log (x)) (1+\log (x))+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right )-4 \int \frac {\text {Ei}(2 \log (x))}{x} \, dx+4 \int \left (-\frac {2}{\log \left (e^{-2 x} (-2+x) x^2\right )}-\frac {4}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )}-\frac {3 x}{\log \left (e^{-2 x} (-2+x) x^2\right )}+\frac {2 x^2}{\log \left (e^{-2 x} (-2+x) x^2\right )}\right ) \, dx+8 \int \frac {1}{\log \left (e^{-2 x} (-2+x) x^2\right )} \, dx-8 \int \frac {x^2}{\log \left (e^{-2 x} (-2+x) x^2\right )} \, dx+12 \int \frac {x}{\log \left (e^{-2 x} (-2+x) x^2\right )} \, dx+16 \int \frac {1}{(-2+x) \log \left (e^{-2 x} (-2+x) x^2\right )} \, dx\\ &=-2 x^2+4 x^3-4 \text {Ei}(2 \log (x))+4 \text {Ei}(2 \log (x)) (1+\log (x))+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right )-4 \operatorname {Subst}(\int \text {Ei}(2 x) \, dx,x,\log (x))\\ &=4 x^3-4 \text {Ei}(2 \log (x))-4 \text {Ei}(2 \log (x)) \log (x)+4 \text {Ei}(2 \log (x)) (1+\log (x))+4 x^2 \log \left (\frac {\log \left (-e^{-2 x} (2-x) x^2\right )}{x \log (x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 33, normalized size = 1.14 \begin {gather*} 4 x^3+4 x^2 \log \left (\frac {\log \left (e^{-2 x} (-2+x) x^2\right )}{x \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-16*x + 28*x^2 - 8*x^3)*Log[x] + (8*x - 4*x^2 + (8*x - 28*x^2 + 12*x^3)*Log[x])*Log[(-2*x^2 + x^3)
/E^(2*x)] + (-16*x + 8*x^2)*Log[x]*Log[(-2*x^2 + x^3)/E^(2*x)]*Log[Log[(-2*x^2 + x^3)/E^(2*x)]/(x*Log[x])])/((
-2 + x)*Log[x]*Log[(-2*x^2 + x^3)/E^(2*x)]),x]

[Out]

4*x^3 + 4*x^2*Log[Log[((-2 + x)*x^2)/E^(2*x)]/(x*Log[x])]

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fricas [A]  time = 1.36, size = 35, normalized size = 1.21 \begin {gather*} 4 \, x^{3} + 4 \, x^{2} \log \left (\frac {\log \left ({\left (x^{3} - 2 \, x^{2}\right )} e^{\left (-2 \, x\right )}\right )}{x \log \relax (x)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-16*x)*log(x)*log((x^3-2*x^2)/exp(x)^2)*log(log((x^3-2*x^2)/exp(x)^2)/x/log(x))+((12*x^3-28*x
^2+8*x)*log(x)-4*x^2+8*x)*log((x^3-2*x^2)/exp(x)^2)+(-8*x^3+28*x^2-16*x)*log(x))/(x-2)/log(x)/log((x^3-2*x^2)/
exp(x)^2),x, algorithm="fricas")

[Out]

4*x^3 + 4*x^2*log(log((x^3 - 2*x^2)*e^(-2*x))/(x*log(x)))

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giac [A]  time = 1.98, size = 39, normalized size = 1.34 \begin {gather*} 4 \, x^{3} - 4 \, x^{2} \log \relax (x) + 4 \, x^{2} \log \left (-2 \, x + \log \left (x - 2\right ) + 2 \, \log \relax (x)\right ) - 4 \, x^{2} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-16*x)*log(x)*log((x^3-2*x^2)/exp(x)^2)*log(log((x^3-2*x^2)/exp(x)^2)/x/log(x))+((12*x^3-28*x
^2+8*x)*log(x)-4*x^2+8*x)*log((x^3-2*x^2)/exp(x)^2)+(-8*x^3+28*x^2-16*x)*log(x))/(x-2)/log(x)/log((x^3-2*x^2)/
exp(x)^2),x, algorithm="giac")

[Out]

4*x^3 - 4*x^2*log(x) + 4*x^2*log(-2*x + log(x - 2) + 2*log(x)) - 4*x^2*log(log(x))

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (8 x^{2}-16 x \right ) \ln \relax (x ) \ln \left (\left (x^{3}-2 x^{2}\right ) {\mathrm e}^{-2 x}\right ) \ln \left (\frac {\ln \left (\left (x^{3}-2 x^{2}\right ) {\mathrm e}^{-2 x}\right )}{x \ln \relax (x )}\right )+\left (\left (12 x^{3}-28 x^{2}+8 x \right ) \ln \relax (x )-4 x^{2}+8 x \right ) \ln \left (\left (x^{3}-2 x^{2}\right ) {\mathrm e}^{-2 x}\right )+\left (-8 x^{3}+28 x^{2}-16 x \right ) \ln \relax (x )}{\left (x -2\right ) \ln \relax (x ) \ln \left (\left (x^{3}-2 x^{2}\right ) {\mathrm e}^{-2 x}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2-16*x)*ln(x)*ln((x^3-2*x^2)/exp(x)^2)*ln(ln((x^3-2*x^2)/exp(x)^2)/x/ln(x))+((12*x^3-28*x^2+8*x)*ln(
x)-4*x^2+8*x)*ln((x^3-2*x^2)/exp(x)^2)+(-8*x^3+28*x^2-16*x)*ln(x))/(x-2)/ln(x)/ln((x^3-2*x^2)/exp(x)^2),x)

[Out]

int(((8*x^2-16*x)*ln(x)*ln((x^3-2*x^2)/exp(x)^2)*ln(ln((x^3-2*x^2)/exp(x)^2)/x/ln(x))+((12*x^3-28*x^2+8*x)*ln(
x)-4*x^2+8*x)*ln((x^3-2*x^2)/exp(x)^2)+(-8*x^3+28*x^2-16*x)*ln(x))/(x-2)/ln(x)/ln((x^3-2*x^2)/exp(x)^2),x)

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maxima [A]  time = 0.43, size = 39, normalized size = 1.34 \begin {gather*} 4 \, x^{3} - 4 \, x^{2} \log \relax (x) + 4 \, x^{2} \log \left (-2 \, x + \log \left (x - 2\right ) + 2 \, \log \relax (x)\right ) - 4 \, x^{2} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-16*x)*log(x)*log((x^3-2*x^2)/exp(x)^2)*log(log((x^3-2*x^2)/exp(x)^2)/x/log(x))+((12*x^3-28*x
^2+8*x)*log(x)-4*x^2+8*x)*log((x^3-2*x^2)/exp(x)^2)+(-8*x^3+28*x^2-16*x)*log(x))/(x-2)/log(x)/log((x^3-2*x^2)/
exp(x)^2),x, algorithm="maxima")

[Out]

4*x^3 - 4*x^2*log(x) + 4*x^2*log(-2*x + log(x - 2) + 2*log(x)) - 4*x^2*log(log(x))

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mupad [B]  time = 5.75, size = 34, normalized size = 1.17 \begin {gather*} 4\,x^2\,\left (x+\ln \left (\frac {\ln \left (-{\mathrm {e}}^{-2\,x}\,\left (2\,x^2-x^3\right )\right )}{x\,\ln \relax (x)}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(16*x - 28*x^2 + 8*x^3) - log(-exp(-2*x)*(2*x^2 - x^3))*(8*x - 4*x^2 + log(x)*(8*x - 28*x^2 + 12*
x^3)) + log(log(-exp(-2*x)*(2*x^2 - x^3))/(x*log(x)))*log(-exp(-2*x)*(2*x^2 - x^3))*log(x)*(16*x - 8*x^2))/(lo
g(-exp(-2*x)*(2*x^2 - x^3))*log(x)*(x - 2)),x)

[Out]

4*x^2*(x + log(log(-exp(-2*x)*(2*x^2 - x^3))/(x*log(x))))

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sympy [B]  time = 3.25, size = 70, normalized size = 2.41 \begin {gather*} 4 x^{3} + \left (4 x^{2} - \frac {8}{3}\right ) \log {\left (\frac {\log {\left (\left (x^{3} - 2 x^{2}\right ) e^{- 2 x} \right )}}{x \log {\relax (x )}} \right )} - \frac {8 \log {\relax (x )}}{3} - \frac {8 \log {\left (\log {\relax (x )} \right )}}{3} + \frac {8 \log {\left (\log {\left (\left (x^{3} - 2 x^{2}\right ) e^{- 2 x} \right )} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2-16*x)*ln(x)*ln((x**3-2*x**2)/exp(x)**2)*ln(ln((x**3-2*x**2)/exp(x)**2)/x/ln(x))+((12*x**3-2
8*x**2+8*x)*ln(x)-4*x**2+8*x)*ln((x**3-2*x**2)/exp(x)**2)+(-8*x**3+28*x**2-16*x)*ln(x))/(x-2)/ln(x)/ln((x**3-2
*x**2)/exp(x)**2),x)

[Out]

4*x**3 + (4*x**2 - 8/3)*log(log((x**3 - 2*x**2)*exp(-2*x))/(x*log(x))) - 8*log(x)/3 - 8*log(log(x))/3 + 8*log(
log((x**3 - 2*x**2)*exp(-2*x)))/3

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