Integrand size = 116, antiderivative size = 24 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=\log (5)-\log \left (5-\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right ) \] Output:
ln(5)-ln(5-ln(x+exp((2+x)^2))*ln(2*x))
Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (5-\log (2 x) \log \left (e^{4+4 x+x^2}+x\right )\right ) \] Input:
Integrate[((-x + E^(4 + 4*x + x^2)*(-4*x - 2*x^2))*Log[2*x] + (-E^(4 + 4*x + x^2) - x)*Log[E^(4 + 4*x + x^2) + x])/(-5*E^(4 + 4*x + x^2)*x - 5*x^2 + (E^(4 + 4*x + x^2)*x + x^2)*Log[2*x]*Log[E^(4 + 4*x + x^2) + x]),x]
Output:
-Log[5 - Log[2*x]*Log[E^(4 + 4*x + x^2) + x]]
Time = 1.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {7292, 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.
\(\displaystyle \int \frac {\left (e^{x^2+4 x+4} \left (-2 x^2-4 x\right )-x\right ) \log (2 x)+\left (-e^{x^2+4 x+4}-x\right ) \log \left (e^{x^2+4 x+4}+x\right )}{-5 x^2-5 e^{x^2+4 x+4} x+\left (x^2+e^{x^2+4 x+4} x\right ) \log (2 x) \log \left (e^{x^2+4 x+4}+x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (\left (e^{x^2+4 x+4} \left (-2 x^2-4 x\right )-x\right ) \log (2 x)\right )-\left (-e^{x^2+4 x+4}-x\right ) \log \left (e^{x^2+4 x+4}+x\right )}{x \left (x+e^{(x+2)^2}\right ) \left (5-\log (2 x) \log \left (x+e^{(x+2)^2}\right )\right )}dx\) |
\(\Big \downarrow \) 7259 |
\(\displaystyle \int \frac {1}{5-\log (2 x) \log \left (x+e^{(x+2)^2}\right )}d\left (\log (2 x) \log \left (x+e^{(x+2)^2}\right )\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle -\log \left (5-\log (2 x) \log \left (x+e^{(x+2)^2}\right )\right )\) |
Input:
Int[((-x + E^(4 + 4*x + x^2)*(-4*x - 2*x^2))*Log[2*x] + (-E^(4 + 4*x + x^2 ) - x)*Log[E^(4 + 4*x + x^2) + x])/(-5*E^(4 + 4*x + x^2)*x - 5*x^2 + (E^(4 + 4*x + x^2)*x + x^2)*Log[2*x]*Log[E^(4 + 4*x + x^2) + x]),x]
Output:
-Log[5 - Log[2*x]*Log[E^(2 + x)^2 + x]]
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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 = 0.98 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-\ln \left (\ln \left (2 x \right ) \ln \left ({\mathrm e}^{x^{2}+4 x +4}+x \right )-5\right )\) | \(23\) |
risch | \(-\ln \left (\ln \left (2 x \right )\right )-\ln \left (\ln \left (x +{\mathrm e}^{\left (2+x \right )^{2}}\right )-\frac {5}{\ln \left (2 x \right )}\right )\) | \(30\) |
Input:
int(((-exp(x^2+4*x+4)-x)*ln(exp(x^2+4*x+4)+x)+((-2*x^2-4*x)*exp(x^2+4*x+4) -x)*ln(2*x))/((x*exp(x^2+4*x+4)+x^2)*ln(2*x)*ln(exp(x^2+4*x+4)+x)-5*x*exp( x^2+4*x+4)-5*x^2),x,method=_RETURNVERBOSE)
Output:
-ln(ln(2*x)*ln(exp(x^2+4*x+4)+x)-5)
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\frac {\log \left (2 \, x\right ) \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5}{\log \left (2 \, x\right )}\right ) - \log \left (\log \left (2 \, x\right )\right ) \] Input:
integrate(((-exp(x^2+4*x+4)-x)*log(exp(x^2+4*x+4)+x)+((-2*x^2-4*x)*exp(x^2 +4*x+4)-x)*log(2*x))/((x*exp(x^2+4*x+4)+x^2)*log(2*x)*log(exp(x^2+4*x+4)+x )-5*x*exp(x^2+4*x+4)-5*x^2),x, algorithm="fricas")
Output:
-log((log(2*x)*log(x + e^(x^2 + 4*x + 4)) - 5)/log(2*x)) - log(log(2*x))
Time = 0.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=- \log {\left (\log {\left (x + e^{x^{2} + 4 x + 4} \right )} - \frac {5}{\log {\left (2 x \right )}} \right )} - \log {\left (\log {\left (2 x \right )} \right )} \] Input:
integrate(((-exp(x**2+4*x+4)-x)*ln(exp(x**2+4*x+4)+x)+((-2*x**2-4*x)*exp(x **2+4*x+4)-x)*ln(2*x))/((x*exp(x**2+4*x+4)+x**2)*ln(2*x)*ln(exp(x**2+4*x+4 )+x)-5*x*exp(x**2+4*x+4)-5*x**2),x)
Output:
-log(log(x + exp(x**2 + 4*x + 4)) - 5/log(2*x)) - log(log(2*x))
Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\frac {{\left (\log \left (2\right ) + \log \left (x\right )\right )} \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5}{\log \left (2\right ) + \log \left (x\right )}\right ) - \log \left (\log \left (2\right ) + \log \left (x\right )\right ) \] Input:
integrate(((-exp(x^2+4*x+4)-x)*log(exp(x^2+4*x+4)+x)+((-2*x^2-4*x)*exp(x^2 +4*x+4)-x)*log(2*x))/((x*exp(x^2+4*x+4)+x^2)*log(2*x)*log(exp(x^2+4*x+4)+x )-5*x*exp(x^2+4*x+4)-5*x^2),x, algorithm="maxima")
Output:
-log(((log(2) + log(x))*log(x + e^(x^2 + 4*x + 4)) - 5)/(log(2) + log(x))) - log(log(2) + log(x))
Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\log \left (2 \, x\right ) \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5\right ) \] Input:
integrate(((-exp(x^2+4*x+4)-x)*log(exp(x^2+4*x+4)+x)+((-2*x^2-4*x)*exp(x^2 +4*x+4)-x)*log(2*x))/((x*exp(x^2+4*x+4)+x^2)*log(2*x)*log(exp(x^2+4*x+4)+x )-5*x*exp(x^2+4*x+4)-5*x^2),x, algorithm="giac")
Output:
-log(log(2*x)*log(x + e^(x^2 + 4*x + 4)) - 5)
Time = 4.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\ln \left (\frac {\ln \left (2\,x\right )\,\ln \left (x+{\mathrm {e}}^{{\left (x+2\right )}^2}\right )-5}{\ln \left (2\,x\right )}\right )-\ln \left (\ln \left (2\,x\right )\right ) \] Input:
int((log(x + exp(4*x + x^2 + 4))*(x + exp(4*x + x^2 + 4)) + log(2*x)*(x + exp(4*x + x^2 + 4)*(4*x + 2*x^2)))/(5*x*exp(4*x + x^2 + 4) + 5*x^2 - log(2 *x)*log(x + exp(4*x + x^2 + 4))*(x*exp(4*x + x^2 + 4) + x^2)),x)
Output:
- log((log(2*x)*log(x + exp((x + 2)^2)) - 5)/log(2*x)) - log(log(2*x))
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\mathrm {log}\left (\mathrm {log}\left (e^{x^{2}+4 x} e^{4}+x \right ) \mathrm {log}\left (2 x \right )-5\right ) \] Input:
int(((-exp(x^2+4*x+4)-x)*log(exp(x^2+4*x+4)+x)+((-2*x^2-4*x)*exp(x^2+4*x+4 )-x)*log(2*x))/((x*exp(x^2+4*x+4)+x^2)*log(2*x)*log(exp(x^2+4*x+4)+x)-5*x* exp(x^2+4*x+4)-5*x^2),x)
Output:
- log(log(e**(x**2 + 4*x)*e**4 + x)*log(2*x) - 5)