3.20.93 \(\int -\frac {48 e^{e^{\frac {3}{\log ^2(\frac {1}{4} (2+\log (x^2)))}}+\frac {3}{\log ^2(\frac {1}{4} (2+\log (x^2)))}}}{(2 x+x \log (x^2)) \log ^3(\frac {1}{4} (2+\log (x^2)))} \, dx\)

Optimal. Leaf size=25 \[ 1+4 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}+\log (2) \]

________________________________________________________________________________________

Rubi [A]  time = 0.78, antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 8, number of rules used = 4, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 6715, 2282, 2194} \begin {gather*} 4 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48*E^(E^(3/Log[(2 + Log[x^2])/4]^2) + 3/Log[(2 + Log[x^2])/4]^2))/((2*x + x*Log[x^2])*Log[(2 + Log[x^2])
/4]^3),x]

[Out]

4*E^E^(3/Log[(2 + Log[x^2])/4]^2)

Rule 12

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (48 \int \frac {\exp \left (e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}\right )}{\left (2 x+x \log \left (x^2\right )\right ) \log ^3\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )} \, dx\right )\\ &=-\left (24 \operatorname {Subst}\left (\int \frac {e^{e^{\frac {3}{\log ^2\left (\frac {2+x}{4}\right )}}+\frac {3}{\log ^2\left (\frac {2+x}{4}\right )}}}{(2+x) \log ^3\left (\frac {2+x}{4}\right )} \, dx,x,\log \left (x^2\right )\right )\right )\\ &=-\left (96 \operatorname {Subst}\left (\int \frac {e^{e^{\frac {3}{\log ^2(x)}}+\frac {3}{\log ^2(x)}}}{4 x \log ^3(x)} \, dx,x,\frac {1}{2}+\frac {\log \left (x^2\right )}{4}\right )\right )\\ &=-\left (24 \operatorname {Subst}\left (\int \frac {e^{e^{\frac {3}{\log ^2(x)}}+\frac {3}{\log ^2(x)}}}{x \log ^3(x)} \, dx,x,\frac {1}{2}+\frac {\log \left (x^2\right )}{4}\right )\right )\\ &=-\left (24 \operatorname {Subst}\left (\int \frac {e^{e^{\frac {3}{x^2}}+\frac {3}{x^2}}}{x^3} \, dx,x,\log \left (\frac {1}{2}+\frac {\log \left (x^2\right )}{4}\right )\right )\right )\\ &=12 \operatorname {Subst}\left (\int e^{e^{3 x}+3 x} \, dx,x,\frac {1}{\log ^2\left (\frac {1}{2}+\frac {\log \left (x^2\right )}{4}\right )}\right )\\ &=4 \operatorname {Subst}\left (\int e^x \, dx,x,e^{\frac {3}{\log ^2\left (\frac {1}{2}+\frac {\log \left (x^2\right )}{4}\right )}}\right )\\ &=4 e^{e^{\frac {3}{\log ^2\left (\frac {1}{2}+\frac {\log \left (x^2\right )}{4}\right )}}}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.85, size = 21, normalized size = 0.84 \begin {gather*} 4 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48*E^(E^(3/Log[(2 + Log[x^2])/4]^2) + 3/Log[(2 + Log[x^2])/4]^2))/((2*x + x*Log[x^2])*Log[(2 + Log
[x^2])/4]^3),x]

[Out]

4*E^E^(3/Log[(2 + Log[x^2])/4]^2)

________________________________________________________________________________________

fricas [B]  time = 1.02, size = 57, normalized size = 2.28 \begin {gather*} 4 \, e^{\left (\frac {e^{\left (\frac {3}{\log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2}}\right )} \log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2} + 3}{\log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2}} - \frac {3}{\log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-48*exp(3/log(1/2+1/4*log(x^2))^2)*exp(exp(3/log(1/2+1/4*log(x^2))^2))/(x*log(x^2)+2*x)/log(1/2+1/4*
log(x^2))^3,x, algorithm="fricas")

[Out]

4*e^((e^(3/log(1/4*log(x^2) + 1/2)^2)*log(1/4*log(x^2) + 1/2)^2 + 3)/log(1/4*log(x^2) + 1/2)^2 - 3/log(1/4*log
(x^2) + 1/2)^2)

________________________________________________________________________________________

giac [A]  time = 1.31, size = 17, normalized size = 0.68 \begin {gather*} 4 \, e^{\left (e^{\left (\frac {3}{\log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-48*exp(3/log(1/2+1/4*log(x^2))^2)*exp(exp(3/log(1/2+1/4*log(x^2))^2))/(x*log(x^2)+2*x)/log(1/2+1/4*
log(x^2))^3,x, algorithm="giac")

[Out]

4*e^(e^(3/log(1/4*log(x^2) + 1/2)^2))

________________________________________________________________________________________

maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[\int -\frac {48 \,{\mathrm e}^{\frac {3}{\ln \left (\frac {1}{2}+\frac {\ln \left (x^{2}\right )}{4}\right )^{2}}} {\mathrm e}^{{\mathrm e}^{\frac {3}{\ln \left (\frac {1}{2}+\frac {\ln \left (x^{2}\right )}{4}\right )^{2}}}}}{\left (x \ln \left (x^{2}\right )+2 x \right ) \ln \left (\frac {1}{2}+\frac {\ln \left (x^{2}\right )}{4}\right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-48*exp(3/ln(1/2+1/4*ln(x^2))^2)*exp(exp(3/ln(1/2+1/4*ln(x^2))^2))/(x*ln(x^2)+2*x)/ln(1/2+1/4*ln(x^2))^3,x
)

[Out]

int(-48*exp(3/ln(1/2+1/4*ln(x^2))^2)*exp(exp(3/ln(1/2+1/4*ln(x^2))^2))/(x*ln(x^2)+2*x)/ln(1/2+1/4*ln(x^2))^3,x
)

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 29, normalized size = 1.16 \begin {gather*} 4 \, e^{\left (e^{\left (\frac {3}{\log \relax (2)^{2} - 2 \, \log \relax (2) \log \left (\log \relax (x) + 1\right ) + \log \left (\log \relax (x) + 1\right )^{2}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-48*exp(3/log(1/2+1/4*log(x^2))^2)*exp(exp(3/log(1/2+1/4*log(x^2))^2))/(x*log(x^2)+2*x)/log(1/2+1/4*
log(x^2))^3,x, algorithm="maxima")

[Out]

4*e^(e^(3/(log(2)^2 - 2*log(2)*log(log(x) + 1) + log(log(x) + 1)^2)))

________________________________________________________________________________________

mupad [B]  time = 1.62, size = 17, normalized size = 0.68 \begin {gather*} 4\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {3}{{\ln \left (\frac {\ln \left (x^2\right )}{4}+\frac {1}{2}\right )}^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(48*exp(3/log(log(x^2)/4 + 1/2)^2)*exp(exp(3/log(log(x^2)/4 + 1/2)^2)))/(log(log(x^2)/4 + 1/2)^3*(2*x + x
*log(x^2))),x)

[Out]

4*exp(exp(3/log(log(x^2)/4 + 1/2)^2))

________________________________________________________________________________________

sympy [A]  time = 0.94, size = 19, normalized size = 0.76 \begin {gather*} 4 e^{e^{\frac {3}{\log {\left (\frac {\log {\left (x^{2} \right )}}{4} + \frac {1}{2} \right )}^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-48*exp(3/ln(1/2+1/4*ln(x**2))**2)*exp(exp(3/ln(1/2+1/4*ln(x**2))**2))/(x*ln(x**2)+2*x)/ln(1/2+1/4*l
n(x**2))**3,x)

[Out]

4*exp(exp(3/log(log(x**2)/4 + 1/2)**2))

________________________________________________________________________________________