3.23.21 \(\int \frac {-e^{24+6 x-24 x^2-6 x^3}+2 e^{12+3 x-12 x^2-3 x^3}+e^{24+6 x-24 x^2-6 x^3} (-6 x+48 x^2+18 x^3) \log (x)+(-2+2 e^{12+3 x-12 x^2-3 x^3}+e^{12+3 x-12 x^2-3 x^3} (6 x-48 x^2-18 x^3) \log (x)) \log (\log (x))-\log ^2(\log (x))}{x} \, dx\)

Optimal. Leaf size=30 \[ \frac {1}{4}-\log (x) \left (e^{3 (4+x) \left (1-x^2\right )}-\log (\log (x))\right )^2 \]

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Rubi [B]  time = 0.40, antiderivative size = 125, normalized size of antiderivative = 4.17, number of steps used = 10, number of rules used = 5, integrand size = 143, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {14, 2288, 2360, 2295, 2296} \begin {gather*} -\frac {e^{-6 x^3-24 x^2+6 x+24} \left (-3 x^3 \log (x)-8 x^2 \log (x)+x \log (x)\right )}{x \left (-3 x^2-8 x+1\right )}+\frac {2 e^{-3 x^3-12 x^2+3 x+12} \left (-3 x^3 \log (x) \log (\log (x))-8 x^2 \log (x) \log (\log (x))+x \log (x) \log (\log (x))\right )}{x \left (-3 x^2-8 x+1\right )}-\log (x) \log ^2(\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^(24 + 6*x - 24*x^2 - 6*x^3) + 2*E^(12 + 3*x - 12*x^2 - 3*x^3) + E^(24 + 6*x - 24*x^2 - 6*x^3)*(-6*x +
48*x^2 + 18*x^3)*Log[x] + (-2 + 2*E^(12 + 3*x - 12*x^2 - 3*x^3) + E^(12 + 3*x - 12*x^2 - 3*x^3)*(6*x - 48*x^2
- 18*x^3)*Log[x])*Log[Log[x]] - Log[Log[x]]^2)/x,x]

[Out]

-((E^(24 + 6*x - 24*x^2 - 6*x^3)*(x*Log[x] - 8*x^2*Log[x] - 3*x^3*Log[x]))/(x*(1 - 8*x - 3*x^2))) - Log[x]*Log
[Log[x]]^2 + (2*E^(12 + 3*x - 12*x^2 - 3*x^3)*(x*Log[x]*Log[Log[x]] - 8*x^2*Log[x]*Log[Log[x]] - 3*x^3*Log[x]*
Log[Log[x]]))/(x*(1 - 8*x - 3*x^2))

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^{24+6 x-24 x^2-6 x^3} \left (-1-6 x \log (x)+48 x^2 \log (x)+18 x^3 \log (x)\right )}{x}-\frac {\log (\log (x)) (2+\log (\log (x)))}{x}-\frac {2 e^{12+3 x-12 x^2-3 x^3} \left (-1-\log (\log (x))-3 x \log (x) \log (\log (x))+24 x^2 \log (x) \log (\log (x))+9 x^3 \log (x) \log (\log (x))\right )}{x}\right ) \, dx\\ &=-\left (2 \int \frac {e^{12+3 x-12 x^2-3 x^3} \left (-1-\log (\log (x))-3 x \log (x) \log (\log (x))+24 x^2 \log (x) \log (\log (x))+9 x^3 \log (x) \log (\log (x))\right )}{x} \, dx\right )+\int \frac {e^{24+6 x-24 x^2-6 x^3} \left (-1-6 x \log (x)+48 x^2 \log (x)+18 x^3 \log (x)\right )}{x} \, dx-\int \frac {\log (\log (x)) (2+\log (\log (x)))}{x} \, dx\\ &=-\frac {e^{24+6 x-24 x^2-6 x^3} \left (x \log (x)-8 x^2 \log (x)-3 x^3 \log (x)\right )}{x \left (1-8 x-3 x^2\right )}+\frac {2 e^{12+3 x-12 x^2-3 x^3} \left (x \log (x) \log (\log (x))-8 x^2 \log (x) \log (\log (x))-3 x^3 \log (x) \log (\log (x))\right )}{x \left (1-8 x-3 x^2\right )}-\operatorname {Subst}(\int \log (x) (2+\log (x)) \, dx,x,\log (x))\\ &=-\frac {e^{24+6 x-24 x^2-6 x^3} \left (x \log (x)-8 x^2 \log (x)-3 x^3 \log (x)\right )}{x \left (1-8 x-3 x^2\right )}+\frac {2 e^{12+3 x-12 x^2-3 x^3} \left (x \log (x) \log (\log (x))-8 x^2 \log (x) \log (\log (x))-3 x^3 \log (x) \log (\log (x))\right )}{x \left (1-8 x-3 x^2\right )}-\operatorname {Subst}\left (\int \left (2 \log (x)+\log ^2(x)\right ) \, dx,x,\log (x)\right )\\ &=-\frac {e^{24+6 x-24 x^2-6 x^3} \left (x \log (x)-8 x^2 \log (x)-3 x^3 \log (x)\right )}{x \left (1-8 x-3 x^2\right )}+\frac {2 e^{12+3 x-12 x^2-3 x^3} \left (x \log (x) \log (\log (x))-8 x^2 \log (x) \log (\log (x))-3 x^3 \log (x) \log (\log (x))\right )}{x \left (1-8 x-3 x^2\right )}-2 \operatorname {Subst}(\int \log (x) \, dx,x,\log (x))-\operatorname {Subst}\left (\int \log ^2(x) \, dx,x,\log (x)\right )\\ &=2 \log (x)-\frac {e^{24+6 x-24 x^2-6 x^3} \left (x \log (x)-8 x^2 \log (x)-3 x^3 \log (x)\right )}{x \left (1-8 x-3 x^2\right )}-2 \log (x) \log (\log (x))-\log (x) \log ^2(\log (x))+\frac {2 e^{12+3 x-12 x^2-3 x^3} \left (x \log (x) \log (\log (x))-8 x^2 \log (x) \log (\log (x))-3 x^3 \log (x) \log (\log (x))\right )}{x \left (1-8 x-3 x^2\right )}+2 \operatorname {Subst}(\int \log (x) \, dx,x,\log (x))\\ &=-\frac {e^{24+6 x-24 x^2-6 x^3} \left (x \log (x)-8 x^2 \log (x)-3 x^3 \log (x)\right )}{x \left (1-8 x-3 x^2\right )}-\log (x) \log ^2(\log (x))+\frac {2 e^{12+3 x-12 x^2-3 x^3} \left (x \log (x) \log (\log (x))-8 x^2 \log (x) \log (\log (x))-3 x^3 \log (x) \log (\log (x))\right )}{x \left (1-8 x-3 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 39, normalized size = 1.30 \begin {gather*} -e^{-6 x^2 (4+x)} \log (x) \left (e^{3 (4+x)}-e^{3 x^2 (4+x)} \log (\log (x))\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^(24 + 6*x - 24*x^2 - 6*x^3) + 2*E^(12 + 3*x - 12*x^2 - 3*x^3) + E^(24 + 6*x - 24*x^2 - 6*x^3)*(-
6*x + 48*x^2 + 18*x^3)*Log[x] + (-2 + 2*E^(12 + 3*x - 12*x^2 - 3*x^3) + E^(12 + 3*x - 12*x^2 - 3*x^3)*(6*x - 4
8*x^2 - 18*x^3)*Log[x])*Log[Log[x]] - Log[Log[x]]^2)/x,x]

[Out]

-((Log[x]*(E^(3*(4 + x)) - E^(3*x^2*(4 + x))*Log[Log[x]])^2)/E^(6*x^2*(4 + x)))

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fricas [B]  time = 0.87, size = 53, normalized size = 1.77 \begin {gather*} 2 \, e^{\left (-3 \, x^{3} - 12 \, x^{2} + 3 \, x + 12\right )} \log \relax (x) \log \left (\log \relax (x)\right ) - \log \relax (x) \log \left (\log \relax (x)\right )^{2} - e^{\left (-6 \, x^{3} - 24 \, x^{2} + 6 \, x + 24\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(log(x))^2+((-18*x^3-48*x^2+6*x)*exp(-3*x^3-12*x^2+3*x+12)*log(x)+2*exp(-3*x^3-12*x^2+3*x+12)-2
)*log(log(x))+(18*x^3+48*x^2-6*x)*exp(-3*x^3-12*x^2+3*x+12)^2*log(x)-exp(-3*x^3-12*x^2+3*x+12)^2+2*exp(-3*x^3-
12*x^2+3*x+12))/x,x, algorithm="fricas")

[Out]

2*e^(-3*x^3 - 12*x^2 + 3*x + 12)*log(x)*log(log(x)) - log(x)*log(log(x))^2 - e^(-6*x^3 - 24*x^2 + 6*x + 24)*lo
g(x)

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giac [B]  time = 0.39, size = 53, normalized size = 1.77 \begin {gather*} 2 \, e^{\left (-3 \, x^{3} - 12 \, x^{2} + 3 \, x + 12\right )} \log \relax (x) \log \left (\log \relax (x)\right ) - \log \relax (x) \log \left (\log \relax (x)\right )^{2} - e^{\left (-6 \, x^{3} - 24 \, x^{2} + 6 \, x + 24\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(log(x))^2+((-18*x^3-48*x^2+6*x)*exp(-3*x^3-12*x^2+3*x+12)*log(x)+2*exp(-3*x^3-12*x^2+3*x+12)-2
)*log(log(x))+(18*x^3+48*x^2-6*x)*exp(-3*x^3-12*x^2+3*x+12)^2*log(x)-exp(-3*x^3-12*x^2+3*x+12)^2+2*exp(-3*x^3-
12*x^2+3*x+12))/x,x, algorithm="giac")

[Out]

2*e^(-3*x^3 - 12*x^2 + 3*x + 12)*log(x)*log(log(x)) - log(x)*log(log(x))^2 - e^(-6*x^3 - 24*x^2 + 6*x + 24)*lo
g(x)

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maple [A]  time = 0.07, size = 46, normalized size = 1.53

method	result	size
risch	\(-{\mathrm e}^{-6 \left (x -1\right ) \left (4+x \right ) \left (x +1\right )} \ln \relax (x )+2 \ln \relax (x ) \ln \left (\ln \relax (x )\right ) {\mathrm e}^{-3 \left (x -1\right ) \left (4+x \right ) \left (x +1\right )}-\ln \relax (x ) \ln \left (\ln \relax (x )\right )^{2}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(ln(x))^2+((-18*x^3-48*x^2+6*x)*exp(-3*x^3-12*x^2+3*x+12)*ln(x)+2*exp(-3*x^3-12*x^2+3*x+12)-2)*ln(ln(x
))+(18*x^3+48*x^2-6*x)*exp(-3*x^3-12*x^2+3*x+12)^2*ln(x)-exp(-3*x^3-12*x^2+3*x+12)^2+2*exp(-3*x^3-12*x^2+3*x+1
2))/x,x,method=_RETURNVERBOSE)

[Out]

-exp(-6*(x-1)*(4+x)*(x+1))*ln(x)+2*ln(x)*ln(ln(x))*exp(-3*(x-1)*(4+x)*(x+1))-ln(x)*ln(ln(x))^2

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maxima [B]  time = 0.71, size = 60, normalized size = 2.00 \begin {gather*} -{\left (e^{\left (24 \, x^{2}\right )} \log \relax (x) \log \left (\log \relax (x)\right )^{2} - 2 \, e^{\left (-3 \, x^{3} + 12 \, x^{2} + 3 \, x + 12\right )} \log \relax (x) \log \left (\log \relax (x)\right ) + e^{\left (-6 \, x^{3} + 6 \, x + 24\right )} \log \relax (x)\right )} e^{\left (-24 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(log(x))^2+((-18*x^3-48*x^2+6*x)*exp(-3*x^3-12*x^2+3*x+12)*log(x)+2*exp(-3*x^3-12*x^2+3*x+12)-2
)*log(log(x))+(18*x^3+48*x^2-6*x)*exp(-3*x^3-12*x^2+3*x+12)^2*log(x)-exp(-3*x^3-12*x^2+3*x+12)^2+2*exp(-3*x^3-
12*x^2+3*x+12))/x,x, algorithm="maxima")

[Out]

-(e^(24*x^2)*log(x)*log(log(x))^2 - 2*e^(-3*x^3 + 12*x^2 + 3*x + 12)*log(x)*log(log(x)) + e^(-6*x^3 + 6*x + 24
)*log(x))*e^(-24*x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\ln \left (\ln \relax (x)\right )}^2+\left ({\mathrm {e}}^{-3\,x^3-12\,x^2+3\,x+12}\,\ln \relax (x)\,\left (18\,x^3+48\,x^2-6\,x\right )-2\,{\mathrm {e}}^{-3\,x^3-12\,x^2+3\,x+12}+2\right )\,\ln \left (\ln \relax (x)\right )-2\,{\mathrm {e}}^{-3\,x^3-12\,x^2+3\,x+12}+{\mathrm {e}}^{-6\,x^3-24\,x^2+6\,x+24}-{\mathrm {e}}^{-6\,x^3-24\,x^2+6\,x+24}\,\ln \relax (x)\,\left (18\,x^3+48\,x^2-6\,x\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(6*x - 24*x^2 - 6*x^3 + 24) - 2*exp(3*x - 12*x^2 - 3*x^3 + 12) + log(log(x))*(exp(3*x - 12*x^2 - 3*x^
3 + 12)*log(x)*(48*x^2 - 6*x + 18*x^3) - 2*exp(3*x - 12*x^2 - 3*x^3 + 12) + 2) + log(log(x))^2 - exp(6*x - 24*
x^2 - 6*x^3 + 24)*log(x)*(48*x^2 - 6*x + 18*x^3))/x,x)

[Out]

int(-(exp(6*x - 24*x^2 - 6*x^3 + 24) - 2*exp(3*x - 12*x^2 - 3*x^3 + 12) + log(log(x))*(exp(3*x - 12*x^2 - 3*x^
3 + 12)*log(x)*(48*x^2 - 6*x + 18*x^3) - 2*exp(3*x - 12*x^2 - 3*x^3 + 12) + 2) + log(log(x))^2 - exp(6*x - 24*
x^2 - 6*x^3 + 24)*log(x)*(48*x^2 - 6*x + 18*x^3))/x, x)

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sympy [B]  time = 35.66, size = 56, normalized size = 1.87 \begin {gather*} - e^{- 6 x^{3} - 24 x^{2} + 6 x + 24} \log {\relax (x )} + 2 e^{- 3 x^{3} - 12 x^{2} + 3 x + 12} \log {\relax (x )} \log {\left (\log {\relax (x )} \right )} - \log {\relax (x )} \log {\left (\log {\relax (x )} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(ln(x))**2+((-18*x**3-48*x**2+6*x)*exp(-3*x**3-12*x**2+3*x+12)*ln(x)+2*exp(-3*x**3-12*x**2+3*x+1
2)-2)*ln(ln(x))+(18*x**3+48*x**2-6*x)*exp(-3*x**3-12*x**2+3*x+12)**2*ln(x)-exp(-3*x**3-12*x**2+3*x+12)**2+2*ex
p(-3*x**3-12*x**2+3*x+12))/x,x)

[Out]

-exp(-6*x**3 - 24*x**2 + 6*x + 24)*log(x) + 2*exp(-3*x**3 - 12*x**2 + 3*x + 12)*log(x)*log(log(x)) - log(x)*lo
g(log(x))**2

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