Integrand size = 141, antiderivative size = 25 \begin {dmath*} \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (2 x+\log ^2\left (4 \left (-3-e+e^5-e^x\right )\right )+\log (x)\right ) \end {dmath*}
Time = 1.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \begin {dmath*} \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (2 x+\log ^2\left (-4 \left (3+e-e^5+e^x\right )\right )+\log (x)\right ) \end {dmath*}
Integrate[(3 + E^5*(-1 - 2*x) + 6*x + E*(1 + 2*x) + E^x*(1 + 2*x) + 2*E^x* x*Log[-12 - 4*E + 4*E^5 - 4*E^x])/(6*x^2 + 2*E*x^2 - 2*E^5*x^2 + 2*E^x*x^2 + (3*x + E*x - E^5*x + E^x*x)*Log[-12 - 4*E + 4*E^5 - 4*E^x]^2 + (3*x + E *x - E^5*x + E^x*x)*Log[x]),x]
Time = 0.91 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6, 6, 7292, 7235}
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.
\(\displaystyle \int \frac {e^5 (-2 x-1)+6 x+e^x (2 x+1)+e (2 x+1)+2 e^x x \log \left (-4 e^x-12-4 e+4 e^5\right )+3}{2 e^x x^2-2 e^5 x^2+2 e x^2+6 x^2+\left (e^x x-e^5 x+e x+3 x\right ) \log ^2\left (-4 e^x-12-4 e+4 e^5\right )+\left (e^x x-e^5 x+e x+3 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^5 (-2 x-1)+6 x+e^x (2 x+1)+e (2 x+1)+2 e^x x \log \left (-4 e^x-12-4 e+4 e^5\right )+3}{2 e^x x^2+(6+2 e) x^2-2 e^5 x^2+\left (e^x x-e^5 x+e x+3 x\right ) \log ^2\left (-4 e^x-12-4 e+4 e^5\right )+\left (e^x x-e^5 x+e x+3 x\right ) \log (x)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^5 (-2 x-1)+6 x+e^x (2 x+1)+e (2 x+1)+2 e^x x \log \left (-4 e^x-12-4 e+4 e^5\right )+3}{2 e^x x^2+\left (6+2 e-2 e^5\right ) x^2+\left (e^x x-e^5 x+e x+3 x\right ) \log ^2\left (-4 e^x-12-4 e+4 e^5\right )+\left (e^x x-e^5 x+e x+3 x\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (1-\frac {1}{e^4}\right ) e^5 (-2 x-1)+6 x+e^x (2 x+1)+2 e^x x \log \left (-4 e^x-12-4 e+4 e^5\right )+3}{\left (e^x+3 \left (1-\frac {1}{3} e \left (e^4-1\right )\right )\right ) x \left (2 x+\log ^2\left (-4 \left (e^x+3+e-e^5\right )\right )+\log (x)\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \log \left (2 x+\log ^2\left (-4 \left (e^x+3+e-e^5\right )\right )+\log (x)\right )\) |
Int[(3 + E^5*(-1 - 2*x) + 6*x + E*(1 + 2*x) + E^x*(1 + 2*x) + 2*E^x*x*Log[ -12 - 4*E + 4*E^5 - 4*E^x])/(6*x^2 + 2*E*x^2 - 2*E^5*x^2 + 2*E^x*x^2 + (3* x + E*x - E^5*x + E^x*x)*Log[-12 - 4*E + 4*E^5 - 4*E^x]^2 + (3*x + E*x - E ^5*x + E^x*x)*Log[x]),x]
3.13.77.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 5.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\ln \left (\ln \left (x \right )+\ln \left (-4 \,{\mathrm e}^{x}+4 \,{\mathrm e}^{5}-4 \,{\mathrm e}-12\right )^{2}+2 x \right )\) | \(25\) |
parallelrisch | \(\ln \left (\frac {\ln \left (-4 \,{\mathrm e}^{x}+4 \,{\mathrm e}^{5}-4 \,{\mathrm e}-12\right )^{2}}{2}+x +\frac {\ln \left (x \right )}{2}\right )\) | \(27\) |
int((2*x*exp(x)*ln(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(1+2*x)*exp(x)+(-1-2*x) *exp(5)+(1+2*x)*exp(1)+6*x+3)/((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*ln(-4*exp( x)+4*exp(5)-4*exp(1)-12)^2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*ln(x)+2*exp(x) *x^2-2*x^2*exp(5)+2*x^2*exp(1)+6*x^2),x,method=_RETURNVERBOSE)
Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \begin {dmath*} \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (\log \left (4 \, e^{5} - 4 \, e - 4 \, e^{x} - 12\right )^{2} + 2 \, x + \log \left (x\right )\right ) \end {dmath*}
integrate((2*x*exp(x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(1+2*x)*exp(x)+( -1-2*x)*exp(5)+(1+2*x)*exp(1)+6*x+3)/((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log (-4*exp(x)+4*exp(5)-4*exp(1)-12)^2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(x) +2*exp(x)*x^2-2*x^2*exp(5)+2*x^2*exp(1)+6*x^2),x, algorithm=\
Time = 1.56 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log {\left (2 x + \log {\left (x \right )} + \log {\left (- 4 e^{x} - 12 - 4 e + 4 e^{5} \right )}^{2} \right )} \end {dmath*}
integrate((2*x*exp(x)*ln(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(1+2*x)*exp(x)+(- 1-2*x)*exp(5)+(1+2*x)*exp(1)+6*x+3)/((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*ln(- 4*exp(x)+4*exp(5)-4*exp(1)-12)**2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*ln(x)+2 *exp(x)*x**2-2*x**2*exp(5)+2*x**2*exp(1)+6*x**2),x)
Result contains complex when optimal does not.
Time = 0.36 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \begin {dmath*} \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (-\pi ^{2} + 4 i \, \pi \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 2 \, {\left (-i \, \pi - 2 \, \log \left (2\right )\right )} \log \left (-e^{5} + e + e^{x} + 3\right ) + \log \left (-e^{5} + e + e^{x} + 3\right )^{2} + 2 \, x + \log \left (x\right )\right ) \end {dmath*}
integrate((2*x*exp(x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(1+2*x)*exp(x)+( -1-2*x)*exp(5)+(1+2*x)*exp(1)+6*x+3)/((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log (-4*exp(x)+4*exp(5)-4*exp(1)-12)^2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(x) +2*exp(x)*x^2-2*x^2*exp(5)+2*x^2*exp(1)+6*x^2),x, algorithm=\
log(-pi^2 + 4*I*pi*log(2) + 4*log(2)^2 - 2*(-I*pi - 2*log(2))*log(-e^5 + e + e^x + 3) + log(-e^5 + e + e^x + 3)^2 + 2*x + log(x))
Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \begin {dmath*} \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\log \left (\log \left (4 \, e^{5} - 4 \, e - 4 \, e^{x} - 12\right )^{2} + 2 \, x + \log \left (x\right )\right ) \end {dmath*}
integrate((2*x*exp(x)*log(-4*exp(x)+4*exp(5)-4*exp(1)-12)+(1+2*x)*exp(x)+( -1-2*x)*exp(5)+(1+2*x)*exp(1)+6*x+3)/((exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log (-4*exp(x)+4*exp(5)-4*exp(1)-12)^2+(exp(x)*x-x*exp(5)+x*exp(1)+3*x)*log(x) +2*exp(x)*x^2-2*x^2*exp(5)+2*x^2*exp(1)+6*x^2),x, algorithm=\
Time = 16.84 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \begin {dmath*} \int \frac {3+e^5 (-1-2 x)+6 x+e (1+2 x)+e^x (1+2 x)+2 e^x x \log \left (-12-4 e+4 e^5-4 e^x\right )}{6 x^2+2 e x^2-2 e^5 x^2+2 e^x x^2+\left (3 x+e x-e^5 x+e^x x\right ) \log ^2\left (-12-4 e+4 e^5-4 e^x\right )+\left (3 x+e x-e^5 x+e^x x\right ) \log (x)} \, dx=\ln \left ({\ln \left (4\,{\mathrm {e}}^5-4\,\mathrm {e}-4\,{\mathrm {e}}^x-12\right )}^2+2\,x+\ln \left (x\right )\right ) \end {dmath*}
int((6*x + exp(x)*(2*x + 1) + exp(1)*(2*x + 1) - exp(5)*(2*x + 1) + 2*x*ex p(x)*log(4*exp(5) - 4*exp(1) - 4*exp(x) - 12) + 3)/(2*x^2*exp(x) + 2*x^2*e xp(1) - 2*x^2*exp(5) + log(x)*(3*x + x*exp(1) - x*exp(5) + x*exp(x)) + log (4*exp(5) - 4*exp(1) - 4*exp(x) - 12)^2*(3*x + x*exp(1) - x*exp(5) + x*exp (x)) + 6*x^2),x)