Integrand size = 107, antiderivative size = 31 \[ \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx=\frac {1}{5} e^{\frac {2 x}{\log (x)}} \left (-e^{-15+x-x \log (2 x)}+x\right )^2 \] Output:
1/5*exp(x/ln(x))^2*(x-exp(-x*ln(2*x)+x-15))^2
\[ \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx=\int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx \] Input:
Integrate[(E^((2*x)/Log[x])*(-2*x^2 + 2*x^2*Log[x] + 2*x*Log[x]^2 + E^(-30 + 2*x - 2*x*Log[2*x])*(-2 + 2*Log[x] - 2*Log[x]^2*Log[2*x]) + E^(-15 + x - x*Log[2*x])*(4*x - 4*x*Log[x] - 2*Log[x]^2 + 2*x*Log[x]^2*Log[2*x])))/(5 *Log[x]^2),x]
Output:
Integrate[(E^((2*x)/Log[x])*(-2*x^2 + 2*x^2*Log[x] + 2*x*Log[x]^2 + E^(-30 + 2*x - 2*x*Log[2*x])*(-2 + 2*Log[x] - 2*Log[x]^2*Log[2*x]) + E^(-15 + x - x*Log[2*x])*(4*x - 4*x*Log[x] - 2*Log[x]^2 + 2*x*Log[x]^2*Log[2*x])))/Lo g[x]^2, x]/5
Leaf count is larger than twice the leaf count of optimal. \(97\) vs. \(2(31)=62\).
Time = 8.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.13, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {27, 25, 7292, 2726}
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.
| \(\displaystyle \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{2 x-2 x \log (2 x)-30} \left (-2 \log (2 x) \log ^2(x)+2 \log (x)-2\right )+e^{x+x (-\log (2 x))-15} \left (4 x+2 x \log (2 x) \log ^2(x)-2 \log ^2(x)-4 x \log (x)\right )\right )}{5 \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int -\frac {e^{\frac {2 x}{\log (x)}} \left (2^{1-2 x} e^{2 x-30} \left (\log (2 x) \log ^2(x)-\log (x)+1\right ) x^{-2 x}-2^{1-x} e^{x-15} \left (x \log (2 x) \log ^2(x)-\log ^2(x)-2 x \log (x)+2 x\right ) x^{-x}-2 \log (x) x^2+2 x^2-2 \log ^2(x) x\right )}{\log ^2(x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{5} \int \frac {e^{\frac {2 x}{\log (x)}} \left (2^{1-2 x} e^{2 x-30} \left (\log (2 x) \log ^2(x)-\log (x)+1\right ) x^{-2 x}-2^{1-x} e^{x-15} \left (x \log (2 x) \log ^2(x)-\log ^2(x)-2 x \log (x)+2 x\right ) x^{-x}-2 \log (x) x^2+2 x^2-2 \log ^2(x) x\right )}{\log ^2(x)}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\frac {1}{5} \int \frac {2^{1-2 x} e^{\frac {2 x}{\log (x)}-30} x^{-2 x} \left (e^x-2^x e^{15} x^{x+1}\right ) \left (2^x e^{15} \log ^2(x) x^x+2^x e^{15} \log (x) x^{x+1}-2^x e^{15} x^{x+1}+e^x-e^x \log (x)+e^x \log ^2(x) \log (2 x)\right )}{\log ^2(x)}dx\) |
| \(\Big \downarrow \) 2726 |
| \(\displaystyle \frac {2^{-2 x} x^{-2 x} \left (e^x-e^{15} 2^x x^{x+1}\right ) e^{\frac {2 x}{\log (x)}-30} \left (-e^{15} 2^x x^{x+1}+e^{15} 2^x x^{x+1} \log (x)+e^x-e^x \log (x)\right )}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}\) |
Input:
Int[(E^((2*x)/Log[x])*(-2*x^2 + 2*x^2*Log[x] + 2*x*Log[x]^2 + E^(-30 + 2*x - 2*x*Log[2*x])*(-2 + 2*Log[x] - 2*Log[x]^2*Log[2*x]) + E^(-15 + x - x*Lo g[2*x])*(4*x - 4*x*Log[x] - 2*Log[x]^2 + 2*x*Log[x]^2*Log[2*x])))/(5*Log[x ]^2),x]
Output:
(E^(-30 + (2*x)/Log[x])*(E^x - 2^x*E^15*x^(1 + x))*(E^x - 2^x*E^15*x^(1 + x) - E^x*Log[x] + 2^x*E^15*x^(1 + x)*Log[x]))/(5*2^(2*x)*x^(2*x)*(Log[x]^( -2) - Log[x]^(-1))*Log[x]^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Leaf count of result is larger than twice the leaf count of optimal. \(62\) vs. \(2(28)=56\).
Time = 3.66 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.03
| method | result | size |
| parallelrisch | \(\frac {x^{2} {\mathrm e}^{\frac {2 x}{\ln \left (x \right )}}}{5}+\frac {{\mathrm e}^{-2 x \ln \left (2 x \right )+2 x -30} {\mathrm e}^{\frac {2 x}{\ln \left (x \right )}}}{5}-\frac {2 \,{\mathrm e}^{-x \ln \left (2 x \right )+x -15} {\mathrm e}^{\frac {2 x}{\ln \left (x \right )}} x}{5}\) | \(63\) |
| risch | \(\frac {x^{2} {\mathrm e}^{\frac {2 x}{\ln \left (x \right )}}}{5}-\frac {2 x^{-x} \left (\frac {1}{2}\right )^{x} x \,{\mathrm e}^{\frac {x \ln \left (x \right )-15 \ln \left (x \right )+2 x}{\ln \left (x \right )}}}{5}+\frac {x^{-2 x} 2^{-2 x} {\mathrm e}^{\frac {2 x \ln \left (x \right )-30 \ln \left (x \right )+2 x}{\ln \left (x \right )}}}{5}\) | \(75\) |
Input:
int(1/5*((-2*ln(x)^2*ln(2*x)+2*ln(x)-2)*exp(-x*ln(2*x)+x-15)^2+(2*x*ln(x)^ 2*ln(2*x)-2*ln(x)^2-4*x*ln(x)+4*x)*exp(-x*ln(2*x)+x-15)+2*x*ln(x)^2+2*x^2* ln(x)-2*x^2)*exp(x/ln(x))^2/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
1/5*x^2*exp(x/ln(x))^2+1/5*exp(-x*ln(2*x)+x-15)^2*exp(x/ln(x))^2-2/5*exp(- x*ln(2*x)+x-15)*exp(x/ln(x))^2*x
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx=\frac {1}{5} \, {\left (x^{2} - 2 \, x e^{\left (-x \log \left (2\right ) - x \log \left (x\right ) + x - 15\right )} + e^{\left (-2 \, x \log \left (2\right ) - 2 \, x \log \left (x\right ) + 2 \, x - 30\right )}\right )} e^{\left (\frac {2 \, x}{\log \left (x\right )}\right )} \] Input:
integrate(1/5*((-2*log(x)^2*log(2*x)+2*log(x)-2)*exp(-x*log(2*x)+x-15)^2+( 2*x*log(x)^2*log(2*x)-2*log(x)^2-4*x*log(x)+4*x)*exp(-x*log(2*x)+x-15)+2*x *log(x)^2+2*x^2*log(x)-2*x^2)*exp(x/log(x))^2/log(x)^2,x, algorithm="frica s")
Output:
1/5*(x^2 - 2*x*e^(-x*log(2) - x*log(x) + x - 15) + e^(-2*x*log(2) - 2*x*lo g(x) + 2*x - 30))*e^(2*x/log(x))
Exception generated. \[ \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/5*((-2*ln(x)**2*ln(2*x)+2*ln(x)-2)*exp(-x*ln(2*x)+x-15)**2+(2* x*ln(x)**2*ln(2*x)-2*ln(x)**2-4*x*ln(x)+4*x)*exp(-x*ln(2*x)+x-15)+2*x*ln(x )**2+2*x**2*ln(x)-2*x**2)*exp(x/ln(x))**2/ln(x)**2,x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Exception generated. \[ \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/5*((-2*log(x)^2*log(2*x)+2*log(x)-2)*exp(-x*log(2*x)+x-15)^2+( 2*x*log(x)^2*log(2*x)-2*log(x)^2-4*x*log(x)+4*x)*exp(-x*log(2*x)+x-15)+2*x *log(x)^2+2*x^2*log(x)-2*x^2)*exp(x/log(x))^2/log(x)^2,x, algorithm="maxim a")
Output:
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (27) = 54\).
Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.68 \[ \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx=\frac {1}{5} \, x^{2} e^{\left (\frac {2 \, x}{\log \left (x\right )}\right )} - \frac {2}{5} \, x e^{\left (-\frac {x \log \left (2\right ) \log \left (x\right ) + x \log \left (x\right )^{2} - x \log \left (x\right ) - 2 \, x + 15 \, \log \left (x\right )}{\log \left (x\right )}\right )} + \frac {1}{5} \, e^{\left (-\frac {2 \, {\left (x \log \left (2\right ) \log \left (x\right ) + x \log \left (x\right )^{2} - x \log \left (x\right ) - x + 15 \, \log \left (x\right )\right )}}{\log \left (x\right )}\right )} \] Input:
integrate(1/5*((-2*log(x)^2*log(2*x)+2*log(x)-2)*exp(-x*log(2*x)+x-15)^2+( 2*x*log(x)^2*log(2*x)-2*log(x)^2-4*x*log(x)+4*x)*exp(-x*log(2*x)+x-15)+2*x *log(x)^2+2*x^2*log(x)-2*x^2)*exp(x/log(x))^2/log(x)^2,x, algorithm="giac" )
Output:
1/5*x^2*e^(2*x/log(x)) - 2/5*x*e^(-(x*log(2)*log(x) + x*log(x)^2 - x*log(x ) - 2*x + 15*log(x))/log(x)) + 1/5*e^(-2*(x*log(2)*log(x) + x*log(x)^2 - x *log(x) - x + 15*log(x))/log(x))
Time = 3.86 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx={\mathrm {e}}^{\frac {2\,x}{\ln \left (x\right )}}\,\left (\frac {x^2}{5}+\frac {{\mathrm {e}}^{2\,x-30}}{5\,2^{2\,x}\,x^{2\,x}}-\frac {2\,x\,{\mathrm {e}}^{x-15}}{5\,2^x\,x^x}\right ) \] Input:
int((exp((2*x)/log(x))*(2*x*log(x)^2 + 2*x^2*log(x) - exp(2*x - 2*x*log(2* x) - 30)*(2*log(2*x)*log(x)^2 - 2*log(x) + 2) + exp(x - x*log(2*x) - 15)*( 4*x - 2*log(x)^2 - 4*x*log(x) + 2*x*log(2*x)*log(x)^2) - 2*x^2))/(5*log(x) ^2),x)
Output:
exp((2*x)/log(x))*(x^2/5 + exp(2*x - 30)/(5*2^(2*x)*x^(2*x)) - (2*x*exp(x - 15))/(5*2^x*x^x))
\[ \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx=\frac {e^{\frac {2 x}{\mathrm {log}\left (x \right )}} e^{30} x^{2}-2 \left (\int \frac {e^{\frac {\mathrm {log}\left (x \right ) x +2 x}{\mathrm {log}\left (x \right )}}}{x^{x} 2^{x}}d x \right ) e^{15}-2 \left (\int \frac {e^{\frac {2 \,\mathrm {log}\left (x \right ) x +2 x}{\mathrm {log}\left (x \right )}}}{x^{2 x} 2^{2 x} \mathrm {log}\left (x \right )^{2}}d x \right )+2 \left (\int \frac {e^{\frac {2 \,\mathrm {log}\left (x \right ) x +2 x}{\mathrm {log}\left (x \right )}}}{x^{2 x} 2^{2 x} \mathrm {log}\left (x \right )}d x \right )+2 \left (\int \frac {e^{\frac {\mathrm {log}\left (x \right ) x +2 x}{\mathrm {log}\left (x \right )}} \mathrm {log}\left (2 x \right ) x}{x^{x} 2^{x}}d x \right ) e^{15}+4 \left (\int \frac {e^{\frac {\mathrm {log}\left (x \right ) x +2 x}{\mathrm {log}\left (x \right )}} x}{x^{x} 2^{x} \mathrm {log}\left (x \right )^{2}}d x \right ) e^{15}-4 \left (\int \frac {e^{\frac {\mathrm {log}\left (x \right ) x +2 x}{\mathrm {log}\left (x \right )}} x}{x^{x} 2^{x} \mathrm {log}\left (x \right )}d x \right ) e^{15}-2 \left (\int \frac {e^{\frac {2 \,\mathrm {log}\left (x \right ) x +2 x}{\mathrm {log}\left (x \right )}} \mathrm {log}\left (2 x \right )}{x^{2 x} 2^{2 x}}d x \right )}{5 e^{30}} \] Input:
int(1/5*((-2*log(x)^2*log(2*x)+2*log(x)-2)*exp(-x*log(2*x)+x-15)^2+(2*x*lo g(x)^2*log(2*x)-2*log(x)^2-4*x*log(x)+4*x)*exp(-x*log(2*x)+x-15)+2*x*log(x )^2+2*x^2*log(x)-2*x^2)*exp(x/log(x))^2/log(x)^2,x)
Output:
(e**((2*x)/log(x))*e**30*x**2 - 2*int(e**((log(x)*x + 2*x)/log(x))/(x**x*2 **x),x)*e**15 - 2*int(e**((2*log(x)*x + 2*x)/log(x))/(x**(2*x)*2**(2*x)*lo g(x)**2),x) + 2*int(e**((2*log(x)*x + 2*x)/log(x))/(x**(2*x)*2**(2*x)*log( x)),x) + 2*int((e**((log(x)*x + 2*x)/log(x))*log(2*x)*x)/(x**x*2**x),x)*e* *15 + 4*int((e**((log(x)*x + 2*x)/log(x))*x)/(x**x*2**x*log(x)**2),x)*e**1 5 - 4*int((e**((log(x)*x + 2*x)/log(x))*x)/(x**x*2**x*log(x)),x)*e**15 - 2 *int((e**((2*log(x)*x + 2*x)/log(x))*log(2*x))/(x**(2*x)*2**(2*x)),x))/(5* e**30)