Integrand size = 57, antiderivative size = 24 \[ \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right ) \log (\log (\log (5)))}{x^2} \, dx=4^{\frac {1}{x}} \left (2+e^x+e^{(5-x) x}\right ) \log (\log (\log (5))) \] Output:
exp(2*ln(2)/x)*(exp(x)+exp(x*(5-x))+2)*ln(ln(ln(5)))
Time = 0.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right ) \log (\log (\log (5)))}{x^2} \, dx=4^{\frac {1}{x}} e^{-x^2} \left (e^{5 x}+2 e^{x^2}+e^{x+x^2}\right ) \log (\log (\log (5))) \] Input:
Integrate[(4^x^(-1)*(E^x*(x^2 - Log[4]) + E^(5*x - x^2)*(5*x^2 - 2*x^3 - L og[4]) - 2*Log[4])*Log[Log[Log[5]]])/x^2,x]
Output:
(4^x^(-1)*(E^(5*x) + 2*E^x^2 + E^(x + x^2))*Log[Log[Log[5]]])/E^x^2
Time = 0.83 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {27, 7293, 2009}
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.
\(\displaystyle \int \frac {4^{\frac {1}{x}} \log (\log (\log (5))) \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (-2 x^3+5 x^2-\log (4)\right )-2 \log (4)\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (\log (\log (5))) \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (-2 x^3+5 x^2-\log (4)\right )-2 \log (4)\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \log (\log (\log (5))) \int \left (\frac {4^{\frac {1}{x}} e^{5 x-x^2} \left (-2 x^3+5 x^2-\log (4)\right )}{x^2}+\frac {4^{\frac {1}{x}} \left (e^x x^2-e^x \log (4)-2 \log (4)\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (4^{\frac {1}{x}} e^{5 x-x^2}+4^{\frac {1}{x}} \left (e^x+2\right )\right ) \log (\log (\log (5)))\) |
Input:
Int[(4^x^(-1)*(E^x*(x^2 - Log[4]) + E^(5*x - x^2)*(5*x^2 - 2*x^3 - Log[4]) - 2*Log[4])*Log[Log[Log[5]]])/x^2,x]
Output:
(4^x^(-1)*E^(5*x - x^2) + 4^x^(-1)*(2 + E^x))*Log[Log[Log[5]]]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 4.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\ln \left (\ln \left (\ln \left (5\right )\right )\right ) \left (2+{\mathrm e}^{x}+{\mathrm e}^{-\left (-5+x \right ) x}\right ) 4^{\frac {1}{x}}\) | \(22\) |
parallelrisch | \(\ln \left (\ln \left (\ln \left (5\right )\right )\right ) \left ({\mathrm e}^{x} {\mathrm e}^{\frac {2 \ln \left (2\right )}{x}}+{\mathrm e}^{-x^{2}+5 x} {\mathrm e}^{\frac {2 \ln \left (2\right )}{x}}+2 \,{\mathrm e}^{\frac {2 \ln \left (2\right )}{x}}\right )\) | \(47\) |
parts | \(\frac {\ln \left (\ln \left (\ln \left (5\right )\right )\right ) x \,{\mathrm e}^{-x^{2}+5 x} {\mathrm e}^{\frac {2 \ln \left (2\right )}{x}}+2 \ln \left (\ln \left (\ln \left (5\right )\right )\right ) x \,{\mathrm e}^{\frac {2 \ln \left (2\right )}{x}}}{x}+\ln \left (\ln \left (\ln \left (5\right )\right )\right ) {\mathrm e}^{x} {\mathrm e}^{\frac {2 \ln \left (2\right )}{x}}\) | \(61\) |
Input:
int(((x^2-2*ln(2))*exp(x)+(-2*ln(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*ln(2))*ex p(2*ln(2)/x)*ln(ln(ln(5)))/x^2,x,method=_RETURNVERBOSE)
Output:
ln(ln(ln(5)))*(2+exp(x)+exp(-(-5+x)*x))*4^(1/x)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right ) \log (\log (\log (5)))}{x^2} \, dx=2^{\frac {2}{x}} {\left (e^{\left (-x^{2} + 5 \, x\right )} + e^{x} + 2\right )} \log \left (\log \left (\log \left (5\right )\right )\right ) \] Input:
integrate(((x^2-2*log(2))*exp(x)+(-2*log(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*l og(2))*exp(2*log(2)/x)*log(log(log(5)))/x^2,x, algorithm="fricas")
Output:
2^(2/x)*(e^(-x^2 + 5*x) + e^x + 2)*log(log(log(5)))
Time = 4.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right ) \log (\log (\log (5)))}{x^2} \, dx=\left (e^{x} \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )} + 2 \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )}\right ) e^{\frac {2 \log {\left (2 \right )}}{x}} + e^{\frac {2 \log {\left (2 \right )}}{x}} e^{- x^{2} + 5 x} \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )} \] Input:
integrate(((x**2-2*ln(2))*exp(x)+(-2*ln(2)-2*x**3+5*x**2)*exp(-x**2+5*x)-4 *ln(2))*exp(2*ln(2)/x)*ln(ln(ln(5)))/x**2,x)
Output:
(exp(x)*log(log(log(5))) + 2*log(log(log(5))))*exp(2*log(2)/x) + exp(2*log (2)/x)*exp(-x**2 + 5*x)*log(log(log(5)))
Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right ) \log (\log (\log (5)))}{x^2} \, dx={\left ({\left (e^{\left (x^{2} + x\right )} + e^{\left (5 \, x\right )}\right )} e^{\left (-x^{2} + \frac {2 \, \log \left (2\right )}{x}\right )} + 2^{\frac {2}{x} + 1}\right )} \log \left (\log \left (\log \left (5\right )\right )\right ) \] Input:
integrate(((x^2-2*log(2))*exp(x)+(-2*log(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*l og(2))*exp(2*log(2)/x)*log(log(log(5)))/x^2,x, algorithm="maxima")
Output:
((e^(x^2 + x) + e^(5*x))*e^(-x^2 + 2*log(2)/x) + 2^(2/x + 1))*log(log(log( 5)))
\[ \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right ) \log (\log (\log (5)))}{x^2} \, dx=\int { -\frac {{\left ({\left (2 \, x^{3} - 5 \, x^{2} + 2 \, \log \left (2\right )\right )} e^{\left (-x^{2} + 5 \, x\right )} - {\left (x^{2} - 2 \, \log \left (2\right )\right )} e^{x} + 4 \, \log \left (2\right )\right )} 2^{\frac {2}{x}} \log \left (\log \left (\log \left (5\right )\right )\right )}{x^{2}} \,d x } \] Input:
integrate(((x^2-2*log(2))*exp(x)+(-2*log(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*l og(2))*exp(2*log(2)/x)*log(log(log(5)))/x^2,x, algorithm="giac")
Output:
integrate(-((2*x^3 - 5*x^2 + 2*log(2))*e^(-x^2 + 5*x) - (x^2 - 2*log(2))*e ^x + 4*log(2))*2^(2/x)*log(log(log(5)))/x^2, x)
Time = 4.43 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right ) \log (\log (\log (5)))}{x^2} \, dx=2\,2^{2/x}\,\ln \left (\ln \left (\ln \left (5\right )\right )\right )+2^{2/x}\,\ln \left (\ln \left (\ln \left (5\right )\right )\right )\,{\mathrm {e}}^{5\,x-x^2}+2^{2/x}\,\ln \left (\ln \left (\ln \left (5\right )\right )\right )\,{\mathrm {e}}^x \] Input:
int(-(log(log(log(5)))*exp((2*log(2))/x)*(4*log(2) + exp(x)*(2*log(2) - x^ 2) + exp(5*x - x^2)*(2*log(2) - 5*x^2 + 2*x^3)))/x^2,x)
Output:
2*2^(2/x)*log(log(log(5))) + 2^(2/x)*log(log(log(5)))*exp(5*x - x^2) + 2^( 2/x)*log(log(log(5)))*exp(x)
\[ \int \frac {4^{\frac {1}{x}} \left (e^x \left (x^2-\log (4)\right )+e^{5 x-x^2} \left (5 x^2-2 x^3-\log (4)\right )-2 \log (4)\right ) \log (\log (\log (5)))}{x^2} \, dx=\int \frac {\left (\left (x^{2}-2 \,\mathrm {log}\left (2\right )\right ) {\mathrm e}^{x}+\left (-2 \,\mathrm {log}\left (2\right )-2 x^{3}+5 x^{2}\right ) {\mathrm e}^{-x^{2}+5 x}-4 \,\mathrm {log}\left (2\right )\right ) {\mathrm e}^{\frac {2 \,\mathrm {log}\left (2\right )}{x}} \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (5\right )\right )\right )}{x^{2}}d x \] Input:
int(((x^2-2*log(2))*exp(x)+(-2*log(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*log(2)) *exp(2*log(2)/x)*log(log(log(5)))/x^2,x)
Output:
int(((x^2-2*log(2))*exp(x)+(-2*log(2)-2*x^3+5*x^2)*exp(-x^2+5*x)-4*log(2)) *exp(2*log(2)/x)*log(log(log(5)))/x^2,x)