Integrand size = 62, antiderivative size = 25 \[ \int -\frac {48 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}{\left (2 x+x \log \left (x^2\right )\right ) \log ^3\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )} \, dx=1+4 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}+\log (2) \]
Time = 0.87 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int -\frac {48 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}{\left (2 x+x \log \left (x^2\right )\right ) \log ^3\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )} \, dx=4 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}} \]
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]
Time = 0.87 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.20, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {27, 3039, 7281, 27, 3039, 7266, 2720, 2624}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int -\frac {48 \exp \left (e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}\right )}{\left (x \log \left (x^2\right )+2 x\right ) \log ^3\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -48 \int \frac {\exp \left (e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}\right )}{\left (\log \left (x^2\right ) x+2 x\right ) \log ^3\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}dx\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle -24 \int \frac {\exp \left (e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}\right )}{\left (\log \left (x^2\right )+2\right ) \log ^3\left (\frac {1}{4} \left (\log \left (x^2\right )+2\right )\right )}d\log \left (x^2\right )\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle -96 \int \frac {\exp \left (e^{\frac {3}{\log ^2\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}}+\frac {3}{\log ^2\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}\right )}{4 \left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right ) \log ^3\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}d\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -24 \int \frac {\exp \left (e^{\frac {3}{\log ^2\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}}+\frac {3}{\log ^2\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}\right )}{\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right ) \log ^3\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}d\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle -24 \int \frac {\exp \left (e^{\frac {3}{\log ^2\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}}+\frac {3}{\log ^2\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}\right )}{\log ^3\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}d\log \left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle 12 \int e^{x^6} x^6d\frac {1}{\log ^2\left (\frac {\log \left (x^2\right )}{4}+\frac {1}{2}\right )}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle 4 \int x^2dx^6\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle 4 x^2\) |
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]
3.20.93.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst [[3]] Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /; !FalseQ[lst]] /; NonsumQ[u]
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Time = 9.50 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(4 \,{\mathrm e}^{{\mathrm e}^{\frac {3}{{\ln \left (\frac {1}{2}+\frac {\ln \left (x^{2}\right )}{4}\right )}^{2}}}}\) | \(18\) |
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,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (21) = 42\).
Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int -\frac {48 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}{\left (2 x+x \log \left (x^2\right )\right ) \log ^3\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )} \, dx=4 \, e^{\left (\frac {e^{\left (\frac {3}{\log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2}}\right )} \log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2} + 3}{\log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2}} - \frac {3}{\log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2}}\right )} \]
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=\
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)
Time = 0.48 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int -\frac {48 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}{\left (2 x+x \log \left (x^2\right )\right ) \log ^3\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )} \, dx=4 e^{e^{\frac {3}{\log {\left (\frac {\log {\left (x^{2} \right )}}{4} + \frac {1}{2} \right )}^{2}}}} \]
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*ln(x**2))**3,x)
Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int -\frac {48 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}{\left (2 x+x \log \left (x^2\right )\right ) \log ^3\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )} \, dx=4 \, e^{\left (e^{\left (\frac {3}{\log \left (2\right )^{2} - 2 \, \log \left (2\right ) \log \left (\log \left (x\right ) + 1\right ) + \log \left (\log \left (x\right ) + 1\right )^{2}}\right )}\right )} \]
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=\
Time = 1.79 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int -\frac {48 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}{\left (2 x+x \log \left (x^2\right )\right ) \log ^3\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )} \, dx=4 \, e^{\left (e^{\left (\frac {3}{\log \left (\frac {1}{4} \, \log \left (x^{2}\right ) + \frac {1}{2}\right )^{2}}\right )}\right )} \]
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=\
Time = 10.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int -\frac {48 e^{e^{\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}+\frac {3}{\log ^2\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )}}}{\left (2 x+x \log \left (x^2\right )\right ) \log ^3\left (\frac {1}{4} \left (2+\log \left (x^2\right )\right )\right )} \, dx=4\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {3}{{\ln \left (\frac {\ln \left (x^2\right )}{4}+\frac {1}{2}\right )}^2}}} \]