Integrand size = 90, antiderivative size = 31 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {e^{2 x} \left (5+e^{3-e^x-\frac {2 x \log \left (x^2\right )}{\log (4)}}\right )}{x} \]
Time = 1.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.90 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {e^{2 x} \left (5 x \log (4)+\frac {e^{3-e^x} x \left (x^2\right )^{-\frac {2 x}{\log (4)}} \left (-4 \log (2)+e^x \log (4)\right )}{-2+e^x}\right )}{x^2 \log (4)} \]
Integrate[(E^(2*x)*(-5 + 10*x)*Log[4] + E^((3*Log[4] - E^x*Log[4] - 2*x*Lo g[x^2])/Log[4])*(-(E^(3*x)*x*Log[4]) + E^(2*x)*(-4*x + (-1 + 2*x)*Log[4]) - 2*E^(2*x)*x*Log[x^2]))/(x^2*Log[4]),x]
(E^(2*x)*(5*x*Log[4] + (E^(3 - E^x)*x*(-4*Log[2] + E^x*Log[4]))/((-2 + E^x )*(x^2)^((2*x)/Log[4]))))/(x^2*Log[4])
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.
\(\displaystyle \int \frac {\left (-2 e^{2 x} x \log \left (x^2\right )-e^{3 x} x \log (4)+e^{2 x} ((2 x-1) \log (4)-4 x)\right ) \exp \left (\frac {-2 x \log \left (x^2\right )-e^x \log (4)+3 \log (4)}{\log (4)}\right )+e^{2 x} (10 x-5) \log (4)}{x^2 \log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {4^{\frac {3}{\log (4)}-\frac {e^x}{\log (4)}} \left (2 e^{2 x} \log \left (x^2\right ) x+e^{3 x} \log (4) x+e^{2 x} (\log (4) (1-2 x)+4 x)\right ) \left (x^2\right )^{-\frac {2 x}{\log (4)}}+5 e^{2 x} (1-2 x) \log (4)}{x^2}dx}{\log (4)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {4^{\frac {3}{\log (4)}-\frac {e^x}{\log (4)}} \left (2 e^{2 x} \log \left (x^2\right ) x+e^{3 x} \log (4) x+e^{2 x} (\log (4) (1-2 x)+4 x)\right ) \left (x^2\right )^{-\frac {2 x}{\log (4)}}+5 e^{2 x} (1-2 x) \log (4)}{x^2}dx}{\log (4)}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {\int \left (e^{2 x-e^x+3} \left (x^2\right )^{-\frac {2 x}{\log (4)}-1} \left (2 \log \left (x^2\right ) x+e^x \log (4) x+4 (1-\log (2)) x+\log (4)\right )-\frac {5 e^{2 x} (2 x-1) \log (4)}{x^2}\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log (4) \int e^{2 x-e^x+3} \left (x^2\right )^{-\frac {2 x}{\log (4)}-1}dx+2 \log \left (x^2\right ) \int e^{2 x-e^x+3} x \left (x^2\right )^{-\frac {2 x}{\log (4)}-1}dx+4 (1-\log (2)) \int e^{2 x-e^x+3} x \left (x^2\right )^{-\frac {2 x}{\log (4)}-1}dx+\log (4) \int e^{3 x-e^x+3} x \left (x^2\right )^{-\frac {2 x}{\log (4)}-1}dx-4 \int \frac {\int \frac {e^{2 x-e^x+3} \left (x^2\right )^{-\frac {2 x}{\log (4)}}}{x}dx}{x}dx-\frac {5 e^{2 x} \log (4)}{x}}{\log (4)}\) |
Int[(E^(2*x)*(-5 + 10*x)*Log[4] + E^((3*Log[4] - E^x*Log[4] - 2*x*Log[x^2] )/Log[4])*(-(E^(3*x)*x*Log[4]) + E^(2*x)*(-4*x + (-1 + 2*x)*Log[4]) - 2*E^ (2*x)*x*Log[x^2]))/(x^2*Log[4]),x]
3.10.59.3.1 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[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 8.55 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61
method | result | size |
parallelrisch | \(\frac {4 \ln \left (2\right ) {\mathrm e}^{-\frac {{\mathrm e}^{x} \ln \left (2\right )+x \ln \left (x^{2}\right )-3 \ln \left (2\right )}{\ln \left (2\right )}} {\mathrm e}^{2 x}+20 \ln \left (2\right ) {\mathrm e}^{2 x}}{4 \ln \left (2\right ) x}\) | \(50\) |
risch | \(\frac {5 \,{\mathrm e}^{2 x}}{x}+\frac {x^{-\frac {2 x}{\ln \left (2\right )}} {\mathrm e}^{-\frac {-i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 \,{\mathrm e}^{x} \ln \left (2\right )-4 x \ln \left (2\right )-6 \ln \left (2\right )}{2 \ln \left (2\right )}}}{x}\) | \(99\) |
int(1/2*((-2*x*exp(x)^2*ln(x^2)-2*x*ln(2)*exp(x)^3+(2*(-1+2*x)*ln(2)-4*x)* exp(x)^2)*exp(1/2*(-2*x*ln(x^2)-2*exp(x)*ln(2)+6*ln(2))/ln(2))+2*(10*x-5)* ln(2)*exp(x)^2)/x^2/ln(2),x,method=_RETURNVERBOSE)
1/4/ln(2)/x*(4*ln(2)*exp(-(exp(x)*ln(2)+x*ln(x^2)-3*ln(2))/ln(2))*exp(x)^2 +20*ln(2)*exp(x)^2)
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x - \frac {e^{x} \log \left (2\right ) + x \log \left (x^{2}\right ) - 3 \, \log \left (2\right )}{\log \left (2\right )}\right )}}{x} \]
integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(-1+2*x)*log (2)-4*x)*exp(x)^2)*exp(1/2*(-2*x*log(x^2)-2*exp(x)*log(2)+6*log(2))/log(2) )+2*(10*x-5)*log(2)*exp(x)^2)/x^2/log(2),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {e^{2 x} e^{\frac {- x \log {\left (x^{2} \right )} - e^{x} \log {\left (2 \right )} + 3 \log {\left (2 \right )}}{\log {\left (2 \right )}}}}{x} + \frac {5 e^{2 x}}{x} \]
integrate(1/2*((-2*x*exp(x)**2*ln(x**2)-2*x*ln(2)*exp(x)**3+(2*(-1+2*x)*ln (2)-4*x)*exp(x)**2)*exp(1/2*(-2*x*ln(x**2)-2*exp(x)*ln(2)+6*ln(2))/ln(2))+ 2*(10*x-5)*ln(2)*exp(x)**2)/x**2/ln(2),x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {10 \, {\rm Ei}\left (2 \, x\right ) \log \left (2\right ) - 10 \, \Gamma \left (-1, -2 \, x\right ) \log \left (2\right ) + \frac {e^{\left (2 \, x - \frac {2 \, x \log \left (x\right )}{\log \left (2\right )} - e^{x} + 3\right )} \log \left (2\right )}{x}}{\log \left (2\right )} \]
integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(-1+2*x)*log (2)-4*x)*exp(x)^2)*exp(1/2*(-2*x*log(x^2)-2*exp(x)*log(2)+6*log(2))/log(2) )+2*(10*x-5)*log(2)*exp(x)^2)/x^2/log(2),x, algorithm=\
(10*Ei(2*x)*log(2) - 10*gamma(-1, -2*x)*log(2) + e^(2*x - 2*x*log(x)/log(2 ) - e^x + 3)*log(2)/x)/log(2)
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\frac {5 \, e^{\left (2 \, x\right )} \log \left (2\right ) + e^{\left (\frac {2 \, x \log \left (2\right ) - e^{x} \log \left (2\right ) - x \log \left (x^{2}\right )}{\log \left (2\right )} + 3\right )} \log \left (2\right )}{x \log \left (2\right )} \]
integrate(1/2*((-2*x*exp(x)^2*log(x^2)-2*x*log(2)*exp(x)^3+(2*(-1+2*x)*log (2)-4*x)*exp(x)^2)*exp(1/2*(-2*x*log(x^2)-2*exp(x)*log(2)+6*log(2))/log(2) )+2*(10*x-5)*log(2)*exp(x)^2)/x^2/log(2),x, algorithm=\
(5*e^(2*x)*log(2) + e^((2*x*log(2) - e^x*log(2) - x*log(x^2))/log(2) + 3)* log(2))/(x*log(2))
Timed out. \[ \int \frac {e^{2 x} (-5+10 x) \log (4)+e^{\frac {3 \log (4)-e^x \log (4)-2 x \log \left (x^2\right )}{\log (4)}} \left (-e^{3 x} x \log (4)+e^{2 x} (-4 x+(-1+2 x) \log (4))-2 e^{2 x} x \log \left (x^2\right )\right )}{x^2 \log (4)} \, dx=\int -\frac {\frac {{\mathrm {e}}^{-\frac {x\,\ln \left (x^2\right )-3\,\ln \left (2\right )+{\mathrm {e}}^x\,\ln \left (2\right )}{\ln \left (2\right )}}\,\left ({\mathrm {e}}^{2\,x}\,\left (4\,x-2\,\ln \left (2\right )\,\left (2\,x-1\right )\right )+2\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{2\,x}+2\,x\,{\mathrm {e}}^{3\,x}\,\ln \left (2\right )\right )}{2}-{\mathrm {e}}^{2\,x}\,\ln \left (2\right )\,\left (10\,x-5\right )}{x^2\,\ln \left (2\right )} \,d x \]
int(-((exp(-(x*log(x^2) - 3*log(2) + exp(x)*log(2))/log(2))*(exp(2*x)*(4*x - 2*log(2)*(2*x - 1)) + 2*x*log(x^2)*exp(2*x) + 2*x*exp(3*x)*log(2)))/2 - exp(2*x)*log(2)*(10*x - 5))/(x^2*log(2)),x)