Integrand size = 138, antiderivative size = 32 \[ \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx=\log \left (\frac {1}{2} e^{\frac {4}{x}+10 x} \log ^2\left (1+e^{-4+x}\right )-\log (\log (4))\right ) \] Output:
ln(1/2*ln(exp(-4+x)+1)^2*exp(2/x+5*x)^2-ln(2*ln(2)))
\[ \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx=\int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx \] Input:
Integrate[(2*E^(-4 + x + (2*(2 + 5*x^2))/x)*x^2*Log[1 + E^(-4 + x)] + E^(( 2*(2 + 5*x^2))/x)*(-4 + 10*x^2 + E^(-4 + x)*(-4 + 10*x^2))*Log[1 + E^(-4 + x)]^2)/(E^((2*(2 + 5*x^2))/x)*(x^2 + E^(-4 + x)*x^2)*Log[1 + E^(-4 + x)]^ 2 + (-2*x^2 - 2*E^(-4 + x)*x^2)*Log[Log[4]]),x]
Output:
Integrate[(2*E^(-4 + x + (2*(2 + 5*x^2))/x)*x^2*Log[1 + E^(-4 + x)] + E^(( 2*(2 + 5*x^2))/x)*(-4 + 10*x^2 + E^(-4 + x)*(-4 + 10*x^2))*Log[1 + E^(-4 + x)]^2)/(E^((2*(2 + 5*x^2))/x)*(x^2 + E^(-4 + x)*x^2)*Log[1 + E^(-4 + x)]^ 2 + (-2*x^2 - 2*E^(-4 + x)*x^2)*Log[Log[4]]), x]
Time = 2.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {7292, 27, 27, 7259, 16}
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.
| \(\displaystyle \int \frac {e^{\frac {2 \left (5 x^2+2\right )}{x}} \left (10 x^2+e^{x-4} \left (10 x^2-4\right )-4\right ) \log ^2\left (e^{x-4}+1\right )+2 e^{\frac {2 \left (5 x^2+2\right )}{x}+x-4} x^2 \log \left (e^{x-4}+1\right )}{e^{\frac {2 \left (5 x^2+2\right )}{x}} \left (e^{x-4} x^2+x^2\right ) \log ^2\left (e^{x-4}+1\right )+\left (-2 e^{x-4} x^2-2 x^2\right ) \log (\log (4))} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^4 \left (e^{\frac {2 \left (5 x^2+2\right )}{x}} \left (10 x^2+e^{x-4} \left (10 x^2-4\right )-4\right ) \log ^2\left (e^{x-4}+1\right )+2 e^{\frac {2 \left (5 x^2+2\right )}{x}+x-4} x^2 \log \left (e^{x-4}+1\right )\right )}{\left (e^x+e^4\right ) x^2 \left (e^{10 x+\frac {4}{x}} \log ^2\left (e^{x-4}+1\right )-2 \log (\log (4))\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^4 \int \frac {2 \left (e^{x-4+\frac {2 \left (5 x^2+2\right )}{x}} x^2 \log \left (1+e^{x-4}\right )-e^{\frac {2 \left (5 x^2+2\right )}{x}} \left (-5 x^2+e^{x-4} \left (2-5 x^2\right )+2\right ) \log ^2\left (1+e^{x-4}\right )\right )}{\left (e^4+e^x\right ) x^2 \left (e^{10 x+\frac {4}{x}} \log ^2\left (1+e^{x-4}\right )-2 \log (\log (4))\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 e^4 \int \frac {e^{x-4+\frac {2 \left (5 x^2+2\right )}{x}} x^2 \log \left (1+e^{x-4}\right )-e^{\frac {2 \left (5 x^2+2\right )}{x}} \left (-5 x^2+e^{x-4} \left (2-5 x^2\right )+2\right ) \log ^2\left (1+e^{x-4}\right )}{\left (e^4+e^x\right ) x^2 \left (e^{10 x+\frac {4}{x}} \log ^2\left (1+e^{x-4}\right )-2 \log (\log (4))\right )}dx\) |
| \(\Big \downarrow \) 7259 |
| \(\displaystyle \int \frac {1}{e^{10 x+\frac {4}{x}} \log ^2\left (e^{x-4}+1\right )-2 \log (\log (4))}d\left (e^{10 x+\frac {4}{x}} \log ^2\left (e^{x-4}+1\right )\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \log \left (e^{10 x+\frac {4}{x}} \log ^2\left (e^{x-4}+1\right )-2 \log (\log (4))\right )\) |
Input:
Int[(2*E^(-4 + x + (2*(2 + 5*x^2))/x)*x^2*Log[1 + E^(-4 + x)] + E^((2*(2 + 5*x^2))/x)*(-4 + 10*x^2 + E^(-4 + x)*(-4 + 10*x^2))*Log[1 + E^(-4 + x)]^2 )/(E^((2*(2 + 5*x^2))/x)*(x^2 + E^(-4 + x)*x^2)*Log[1 + E^(-4 + x)]^2 + (- 2*x^2 - 2*E^(-4 + x)*x^2)*Log[Log[4]]),x]
Output:
Log[E^(4/x + 10*x)*Log[1 + E^(-4 + x)]^2 - 2*Log[Log[4]]]
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])]}, Simp[c Subst[Int[(a + b*x^p)^m, x] , x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]
Time = 64.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06
| method | result | size |
| parallelrisch | \(\ln \left ({\mathrm e}^{\frac {10 x^{2}+4}{x}} \ln \left ({\mathrm e}^{x -4}+1\right )^{2}-2 \ln \left (2 \ln \left (2\right )\right )\right )\) | \(34\) |
| risch | \(\frac {10 x^{2}+4}{x}+\ln \left (\ln \left ({\mathrm e}^{x -4}+1\right )^{2}-2 \left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) {\mathrm e}^{-\frac {2 \left (5 x^{2}+2\right )}{x}}\right )\) | \(46\) |
Input:
int((((10*x^2-4)*exp(x-4)+10*x^2-4)*exp((5*x^2+2)/x)^2*ln(exp(x-4)+1)^2+2* x^2*exp(x-4)*exp((5*x^2+2)/x)^2*ln(exp(x-4)+1))/((x^2*exp(x-4)+x^2)*exp((5 *x^2+2)/x)^2*ln(exp(x-4)+1)^2+(-2*x^2*exp(x-4)-2*x^2)*ln(2*ln(2))),x,metho d=_RETURNVERBOSE)
Output:
ln(exp((5*x^2+2)/x)^2*ln(exp(x-4)+1)^2-2*ln(2*ln(2)))
Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (30) = 60\).
Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 3.00 \[ \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx=\frac {10 \, x^{2} + x \log \left ({\left (e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )} \log \left ({\left (e^{\left (\frac {11 \, x^{2} - 4 \, x + 4}{x}\right )} + e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )}\right )} e^{\left (-\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )}\right )^{2} - 2 \, \log \left (2 \, \log \left (2\right )\right )\right )} e^{\left (-\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )}\right ) + 4}{x} \] Input:
integrate((((10*x^2-4)*exp(-4+x)+10*x^2-4)*exp((5*x^2+2)/x)^2*log(exp(-4+x )+1)^2+2*x^2*exp(-4+x)*exp((5*x^2+2)/x)^2*log(exp(-4+x)+1))/((x^2*exp(-4+x )+x^2)*exp((5*x^2+2)/x)^2*log(exp(-4+x)+1)^2+(-2*x^2*exp(-4+x)-2*x^2)*log( 2*log(2))),x, algorithm="fricas")
Output:
(10*x^2 + x*log((e^(2*(5*x^2 + 2)/x)*log((e^((11*x^2 - 4*x + 4)/x) + e^(2* (5*x^2 + 2)/x))*e^(-2*(5*x^2 + 2)/x))^2 - 2*log(2*log(2)))*e^(-2*(5*x^2 + 2)/x)) + 4)/x
Time = 1.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx=10 x + \log {\left (\log {\left (e^{x - 4} + 1 \right )}^{2} + \left (- 2 \log {\left (2 \right )} - 2 \log {\left (\log {\left (2 \right )} \right )}\right ) e^{- \frac {2 \cdot \left (5 x^{2} + 2\right )}{x}} \right )} + \frac {4}{x} \] Input:
integrate((((10*x**2-4)*exp(-4+x)+10*x**2-4)*exp((5*x**2+2)/x)**2*ln(exp(- 4+x)+1)**2+2*x**2*exp(-4+x)*exp((5*x**2+2)/x)**2*ln(exp(-4+x)+1))/((x**2*e xp(-4+x)+x**2)*exp((5*x**2+2)/x)**2*ln(exp(-4+x)+1)**2+(-2*x**2*exp(-4+x)- 2*x**2)*ln(2*ln(2))),x)
Output:
10*x + log(log(exp(x - 4) + 1)**2 + (-2*log(2) - 2*log(log(2)))*exp(-2*(5* x**2 + 2)/x)) + 4/x
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (30) = 60\).
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.62 \[ \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx=\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x} + \log \left ({\left (e^{\left (10 \, x + \frac {4}{x}\right )} \log \left (e^{4} + e^{x}\right )^{2} - 8 \, e^{\left (10 \, x + \frac {4}{x}\right )} \log \left (e^{4} + e^{x}\right ) + 16 \, e^{\left (10 \, x + \frac {4}{x}\right )} - 2 \, \log \left (2\right ) - 2 \, \log \left (\log \left (2\right )\right )\right )} e^{\left (-10 \, x - \frac {4}{x}\right )}\right ) \] Input:
integrate((((10*x^2-4)*exp(-4+x)+10*x^2-4)*exp((5*x^2+2)/x)^2*log(exp(-4+x )+1)^2+2*x^2*exp(-4+x)*exp((5*x^2+2)/x)^2*log(exp(-4+x)+1))/((x^2*exp(-4+x )+x^2)*exp((5*x^2+2)/x)^2*log(exp(-4+x)+1)^2+(-2*x^2*exp(-4+x)-2*x^2)*log( 2*log(2))),x, algorithm="maxima")
Output:
2*(5*x^2 + 2)/x + log((e^(10*x + 4/x)*log(e^4 + e^x)^2 - 8*e^(10*x + 4/x)* log(e^4 + e^x) + 16*e^(10*x + 4/x) - 2*log(2) - 2*log(log(2)))*e^(-10*x - 4/x))
\[ \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx=\int { \frac {2 \, {\left (x^{2} e^{\left (x + \frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x} - 4\right )} \log \left (e^{\left (x - 4\right )} + 1\right ) + {\left (5 \, x^{2} + {\left (5 \, x^{2} - 2\right )} e^{\left (x - 4\right )} - 2\right )} e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )} \log \left (e^{\left (x - 4\right )} + 1\right )^{2}\right )}}{{\left (x^{2} e^{\left (x - 4\right )} + x^{2}\right )} e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )} \log \left (e^{\left (x - 4\right )} + 1\right )^{2} - 2 \, {\left (x^{2} e^{\left (x - 4\right )} + x^{2}\right )} \log \left (2 \, \log \left (2\right )\right )} \,d x } \] Input:
integrate((((10*x^2-4)*exp(-4+x)+10*x^2-4)*exp((5*x^2+2)/x)^2*log(exp(-4+x )+1)^2+2*x^2*exp(-4+x)*exp((5*x^2+2)/x)^2*log(exp(-4+x)+1))/((x^2*exp(-4+x )+x^2)*exp((5*x^2+2)/x)^2*log(exp(-4+x)+1)^2+(-2*x^2*exp(-4+x)-2*x^2)*log( 2*log(2))),x, algorithm="giac")
Output:
integrate(2*(x^2*e^(x + 2*(5*x^2 + 2)/x - 4)*log(e^(x - 4) + 1) + (5*x^2 + (5*x^2 - 2)*e^(x - 4) - 2)*e^(2*(5*x^2 + 2)/x)*log(e^(x - 4) + 1)^2)/((x^ 2*e^(x - 4) + x^2)*e^(2*(5*x^2 + 2)/x)*log(e^(x - 4) + 1)^2 - 2*(x^2*e^(x - 4) + x^2)*log(2*log(2))), x)
Time = 2.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx=\ln \left ({\mathrm {e}}^{10\,x+\frac {4}{x}}\,{\ln \left ({\mathrm {e}}^{-4}\,{\mathrm {e}}^x+1\right )}^2+\ln \left (\frac {1}{4\,{\ln \left (2\right )}^2}\right )\right ) \] Input:
int(-(log(exp(x - 4) + 1)^2*exp((2*(5*x^2 + 2))/x)*(exp(x - 4)*(10*x^2 - 4 ) + 10*x^2 - 4) + 2*x^2*log(exp(x - 4) + 1)*exp((2*(5*x^2 + 2))/x)*exp(x - 4))/(log(2*log(2))*(2*x^2*exp(x - 4) + 2*x^2) - log(exp(x - 4) + 1)^2*exp ((2*(5*x^2 + 2))/x)*(x^2*exp(x - 4) + x^2)),x)
Output:
log(log(1/(4*log(2)^2)) + exp(10*x + 4/x)*log(exp(-4)*exp(x) + 1)^2)
Time = 8.45 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.78 \[ \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx=\mathrm {log}\left (-e^{5 x} \sqrt {\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )}\, \sqrt {2}+e^{\frac {10 x^{2}+2}{x}} \mathrm {log}\left (\frac {e^{x}+e^{4}}{e^{4}}\right )\right )+\mathrm {log}\left (e^{5 x} \sqrt {\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )}\, \sqrt {2}+e^{\frac {10 x^{2}+2}{x}} \mathrm {log}\left (\frac {e^{x}+e^{4}}{e^{4}}\right )\right )-10 x \] Input:
int((((10*x^2-4)*exp(-4+x)+10*x^2-4)*exp((5*x^2+2)/x)^2*log(exp(-4+x)+1)^2 +2*x^2*exp(-4+x)*exp((5*x^2+2)/x)^2*log(exp(-4+x)+1))/((x^2*exp(-4+x)+x^2) *exp((5*x^2+2)/x)^2*log(exp(-4+x)+1)^2+(-2*x^2*exp(-4+x)-2*x^2)*log(2*log( 2))),x)
Output:
log( - e**(5*x)*sqrt(log(2*log(2)))*sqrt(2) + e**((10*x**2 + 2)/x)*log((e* *x + e**4)/e**4)) + log(e**(5*x)*sqrt(log(2*log(2)))*sqrt(2) + e**((10*x** 2 + 2)/x)*log((e**x + e**4)/e**4)) - 10*x