Integrand size = 114, antiderivative size = 30 \begin {dmath*} \int \frac {50+20 x^2+2 x^4+\left (-100-20 x^2\right ) \log (5)+50 \log ^2(5)+e^{\frac {9 x+x^3-5 x \log (5)}{-5-x^2+5 \log (5)}} \left (-45-6 x^2-x^4+\left (70+10 x^2\right ) \log (5)-25 \log ^2(5)\right )}{25+10 x^2+x^4+\left (-50-10 x^2\right ) \log (5)+25 \log ^2(5)} \, dx=e^{-x+\frac {4}{-\frac {5}{x}-x+\frac {5 \log (5)}{x}}}+2 x \end {dmath*}
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \begin {dmath*} \int \frac {50+20 x^2+2 x^4+\left (-100-20 x^2\right ) \log (5)+50 \log ^2(5)+e^{\frac {9 x+x^3-5 x \log (5)}{-5-x^2+5 \log (5)}} \left (-45-6 x^2-x^4+\left (70+10 x^2\right ) \log (5)-25 \log ^2(5)\right )}{25+10 x^2+x^4+\left (-50-10 x^2\right ) \log (5)+25 \log ^2(5)} \, dx=e^{-x-\frac {4 x}{5+x^2-5 \log (5)}}+2 x \end {dmath*}
Integrate[(50 + 20*x^2 + 2*x^4 + (-100 - 20*x^2)*Log[5] + 50*Log[5]^2 + E^ ((9*x + x^3 - 5*x*Log[5])/(-5 - x^2 + 5*Log[5]))*(-45 - 6*x^2 - x^4 + (70 + 10*x^2)*Log[5] - 25*Log[5]^2))/(25 + 10*x^2 + x^4 + (-50 - 10*x^2)*Log[5 ] + 25*Log[5]^2),x]
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.
| \(\displaystyle \int \frac {2 x^4+20 x^2+\left (-20 x^2-100\right ) \log (5)+e^{\frac {x^3+9 x-5 x \log (5)}{-x^2-5+5 \log (5)}} \left (-x^4-6 x^2+\left (10 x^2+70\right ) \log (5)-45-25 \log ^2(5)\right )+50+50 \log ^2(5)}{x^4+10 x^2+\left (-10 x^2-50\right ) \log (5)+25+25 \log ^2(5)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 x^4+20 x^2+\left (-20 x^2-100\right ) \log (5)+e^{\frac {x^3+9 x-5 x \log (5)}{-x^2-5+5 \log (5)}} \left (-x^4-6 x^2+\left (10 x^2+70\right ) \log (5)-45-25 \log ^2(5)\right )+50 \left (1+\log ^2(5)\right )}{x^4+10 x^2 (1-\log (5))+25 (1-\log (5))^2}dx\) |
| \(\Big \downarrow \) 1380 |
| \(\displaystyle \int \frac {2 x^4+20 x^2-20 \left (x^2+5\right ) \log (5)-5^{\frac {5 x}{x^2+5-5 \log (5)}} e^{-\frac {x^3+9 x}{x^2+5 (1-\log (5))}} \left (x^4+6 x^2-10 \left (x^2+7\right ) \log (5)+5 \left (9+5 \log ^2(5)\right )\right )+50 \left (1+\log ^2(5)\right )}{\left (x^2+5 (1-\log (5))\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {50 \left (1+\log ^2(5)\right )}{\left (x^2+5-5 \log (5)\right )^2}+\frac {20 x^2}{\left (x^2+5-5 \log (5)\right )^2}-\frac {20 \left (x^2+5\right ) \log (5)}{\left (x^2+5-5 \log (5)\right )^2}+\frac {2 x^4}{\left (x^2+5-5 \log (5)\right )^2}+\frac {5^{\frac {5 x}{x^2+5-5 \log (5)}} e^{-\frac {x \left (x^2+9\right )}{x^2+5-5 \log (5)}} \left (-x^4-2 x^2 (3-5 \log (5))-5 (9-5 \log (5)) (1-\log (5))\right )}{\left (x^2+5-5 \log (5)\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int 5^{\frac {5 x}{x^2-5 \log (5)+5}} e^{-\frac {x \left (x^2+9\right )}{x^2-5 \log (5)+5}}dx+2 \int \frac {5^{\frac {5 x}{x^2-5 \log (5)+5}} e^{-\frac {x \left (x^2+9\right )}{x^2-5 \log (5)+5}}}{\left (\sqrt {5 (-1+\log (5))}-x\right )^2}dx+\frac {2 \int \frac {5^{\frac {5 x}{x^2-5 \log (5)+5}-\frac {1}{2}} e^{-\frac {x \left (x^2+9\right )}{x^2-5 \log (5)+5}}}{\sqrt {5 (-1+\log (5))}-x}dx}{\sqrt {\log (5)-1}}-\frac {2 \int \frac {5^{\frac {5 x}{x^2-5 \log (5)+5}+\frac {1}{2}} e^{-\frac {x \left (x^2+9\right )}{x^2-5 \log (5)+5}}}{\sqrt {5 (-1+\log (5))}-x}dx}{5 \sqrt {\log (5)-1}}+2 \int \frac {5^{\frac {5 x}{x^2-5 \log (5)+5}} e^{-\frac {x \left (x^2+9\right )}{x^2-5 \log (5)+5}}}{\left (x+\sqrt {5 (-1+\log (5))}\right )^2}dx+\frac {2 \int \frac {5^{\frac {5 x}{x^2-5 \log (5)+5}-\frac {1}{2}} e^{-\frac {x \left (x^2+9\right )}{x^2-5 \log (5)+5}}}{x+\sqrt {5 (-1+\log (5))}}dx}{\sqrt {\log (5)-1}}-\frac {2 \int \frac {5^{\frac {5 x}{x^2-5 \log (5)+5}+\frac {1}{2}} e^{-\frac {x \left (x^2+9\right )}{x^2-5 \log (5)+5}}}{x+\sqrt {5 (-1+\log (5))}}dx}{5 \sqrt {\log (5)-1}}+\frac {\sqrt {5} \left (1+\log ^2(5)\right ) \text {arctanh}\left (\frac {x}{\sqrt {5 (\log (5)-1)}}\right )}{(\log (5)-1)^{3/2}}-\frac {2 \sqrt {5} (2-\log (5)) \log (5) \text {arctanh}\left (\frac {x}{\sqrt {5 (\log (5)-1)}}\right )}{(\log (5)-1)^{3/2}}-3 \sqrt {5 (\log (5)-1)} \text {arctanh}\left (\frac {x}{\sqrt {5 (\log (5)-1)}}\right )-2 \sqrt {\frac {5}{\log (5)-1}} \text {arctanh}\left (\frac {x}{\sqrt {5 (\log (5)-1)}}\right )+\frac {5 x \left (1+\log ^2(5)\right )}{(1-\log (5)) \left (x^2+5 (1-\log (5))\right )}-\frac {10 x \log ^2(5)}{(1-\log (5)) \left (x^2+5 (1-\log (5))\right )}-\frac {10 x}{x^2+5 (1-\log (5))}-\frac {x^3}{x^2+5 (1-\log (5))}+3 x\) |
Int[(50 + 20*x^2 + 2*x^4 + (-100 - 20*x^2)*Log[5] + 50*Log[5]^2 + E^((9*x + x^3 - 5*x*Log[5])/(-5 - x^2 + 5*Log[5]))*(-45 - 6*x^2 - x^4 + (70 + 10*x ^2)*Log[5] - 25*Log[5]^2))/(25 + 10*x^2 + x^4 + (-50 - 10*x^2)*Log[5] + 25 *Log[5]^2),x]
3.3.7.3.1 Defintions of rubi rules used
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
| method | result | size |
| parallelrisch | \(2 x +{\mathrm e}^{\frac {-5 x \ln \left (5\right )+x^{3}+9 x}{5 \ln \left (5\right )-x^{2}-5}}\) | \(32\) |
| risch | \(2 x +{\mathrm e}^{-\frac {x \left (-x^{2}+5 \ln \left (5\right )-9\right )}{5 \ln \left (5\right )-x^{2}-5}}\) | \(33\) |
| parts | \(2 x +\frac {\left (5 \ln \left (5\right )-5\right ) {\mathrm e}^{\frac {-5 x \ln \left (5\right )+x^{3}+9 x}{5 \ln \left (5\right )-x^{2}-5}}-x^{2} {\mathrm e}^{\frac {-5 x \ln \left (5\right )+x^{3}+9 x}{5 \ln \left (5\right )-x^{2}-5}}}{5 \ln \left (5\right )-x^{2}-5}\) | \(86\) |
| norman | \(\frac {\left (5 \ln \left (5\right )-5\right ) {\mathrm e}^{\frac {-5 x \ln \left (5\right )+x^{3}+9 x}{5 \ln \left (5\right )-x^{2}-5}}+\left (10 \ln \left (5\right )-10\right ) x -2 x^{3}-x^{2} {\mathrm e}^{\frac {-5 x \ln \left (5\right )+x^{3}+9 x}{5 \ln \left (5\right )-x^{2}-5}}}{5 \ln \left (5\right )-x^{2}-5}\) | \(95\) |
int(((-25*ln(5)^2+(10*x^2+70)*ln(5)-x^4-6*x^2-45)*exp((-5*x*ln(5)+x^3+9*x) /(5*ln(5)-x^2-5))+50*ln(5)^2+(-20*x^2-100)*ln(5)+2*x^4+20*x^2+50)/(25*ln(5 )^2+(-10*x^2-50)*ln(5)+x^4+10*x^2+25),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {50+20 x^2+2 x^4+\left (-100-20 x^2\right ) \log (5)+50 \log ^2(5)+e^{\frac {9 x+x^3-5 x \log (5)}{-5-x^2+5 \log (5)}} \left (-45-6 x^2-x^4+\left (70+10 x^2\right ) \log (5)-25 \log ^2(5)\right )}{25+10 x^2+x^4+\left (-50-10 x^2\right ) \log (5)+25 \log ^2(5)} \, dx=2 \, x + e^{\left (-\frac {x^{3} - 5 \, x \log \left (5\right ) + 9 \, x}{x^{2} - 5 \, \log \left (5\right ) + 5}\right )} \end {dmath*}
integrate(((-25*log(5)^2+(10*x^2+70)*log(5)-x^4-6*x^2-45)*exp((-5*x*log(5) +x^3+9*x)/(5*log(5)-x^2-5))+50*log(5)^2+(-20*x^2-100)*log(5)+2*x^4+20*x^2+ 50)/(25*log(5)^2+(-10*x^2-50)*log(5)+x^4+10*x^2+25),x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \begin {dmath*} \int \frac {50+20 x^2+2 x^4+\left (-100-20 x^2\right ) \log (5)+50 \log ^2(5)+e^{\frac {9 x+x^3-5 x \log (5)}{-5-x^2+5 \log (5)}} \left (-45-6 x^2-x^4+\left (70+10 x^2\right ) \log (5)-25 \log ^2(5)\right )}{25+10 x^2+x^4+\left (-50-10 x^2\right ) \log (5)+25 \log ^2(5)} \, dx=2 x + e^{\frac {x^{3} - 5 x \log {\left (5 \right )} + 9 x}{- x^{2} - 5 + 5 \log {\left (5 \right )}}} \end {dmath*}
integrate(((-25*ln(5)**2+(10*x**2+70)*ln(5)-x**4-6*x**2-45)*exp((-5*x*ln(5 )+x**3+9*x)/(5*ln(5)-x**2-5))+50*ln(5)**2+(-20*x**2-100)*ln(5)+2*x**4+20*x **2+50)/(25*ln(5)**2+(-10*x**2-50)*ln(5)+x**4+10*x**2+25),x)
Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (27) = 54\).
Time = 0.40 (sec) , antiderivative size = 396, normalized size of antiderivative = 13.20 \begin {dmath*} \int \frac {50+20 x^2+2 x^4+\left (-100-20 x^2\right ) \log (5)+50 \log ^2(5)+e^{\frac {9 x+x^3-5 x \log (5)}{-5-x^2+5 \log (5)}} \left (-45-6 x^2-x^4+\left (70+10 x^2\right ) \log (5)-25 \log ^2(5)\right )}{25+10 x^2+x^4+\left (-50-10 x^2\right ) \log (5)+25 \log ^2(5)} \, dx=-\frac {5}{2} \, {\left (\frac {2 \, x}{x^{2} {\left (\log \left (5\right ) - 1\right )} - 5 \, \log \left (5\right )^{2} + 10 \, \log \left (5\right ) - 5} + \frac {\log \left (\frac {x - \sqrt {5 \, \log \left (5\right ) - 5}}{x + \sqrt {5 \, \log \left (5\right ) - 5}}\right )}{\sqrt {5 \, \log \left (5\right ) - 5} {\left (\log \left (5\right ) - 1\right )}}\right )} \log \left (5\right )^{2} - 5 \, {\left (\frac {\log \left (\frac {x - \sqrt {5 \, \log \left (5\right ) - 5}}{x + \sqrt {5 \, \log \left (5\right ) - 5}}\right )}{\sqrt {5 \, \log \left (5\right ) - 5}} - \frac {2 \, x}{x^{2} - 5 \, \log \left (5\right ) + 5}\right )} \log \left (5\right ) + 5 \, {\left (\frac {2 \, x}{x^{2} {\left (\log \left (5\right ) - 1\right )} - 5 \, \log \left (5\right )^{2} + 10 \, \log \left (5\right ) - 5} + \frac {\log \left (\frac {x - \sqrt {5 \, \log \left (5\right ) - 5}}{x + \sqrt {5 \, \log \left (5\right ) - 5}}\right )}{\sqrt {5 \, \log \left (5\right ) - 5} {\left (\log \left (5\right ) - 1\right )}}\right )} \log \left (5\right ) + \frac {15 \, {\left (\log \left (5\right ) - 1\right )} \log \left (\frac {x - \sqrt {5 \, \log \left (5\right ) - 5}}{x + \sqrt {5 \, \log \left (5\right ) - 5}}\right )}{2 \, \sqrt {5 \, \log \left (5\right ) - 5}} + 2 \, x - \frac {5 \, x {\left (\log \left (5\right ) - 1\right )}}{x^{2} - 5 \, \log \left (5\right ) + 5} + \frac {5 \, \log \left (\frac {x - \sqrt {5 \, \log \left (5\right ) - 5}}{x + \sqrt {5 \, \log \left (5\right ) - 5}}\right )}{\sqrt {5 \, \log \left (5\right ) - 5}} - \frac {5 \, x}{x^{2} {\left (\log \left (5\right ) - 1\right )} - 5 \, \log \left (5\right )^{2} + 10 \, \log \left (5\right ) - 5} - \frac {10 \, x}{x^{2} - 5 \, \log \left (5\right ) + 5} - \frac {5 \, \log \left (\frac {x - \sqrt {5 \, \log \left (5\right ) - 5}}{x + \sqrt {5 \, \log \left (5\right ) - 5}}\right )}{2 \, \sqrt {5 \, \log \left (5\right ) - 5} {\left (\log \left (5\right ) - 1\right )}} + e^{\left (-x - \frac {4 \, x}{x^{2} - 5 \, \log \left (5\right ) + 5}\right )} \end {dmath*}
integrate(((-25*log(5)^2+(10*x^2+70)*log(5)-x^4-6*x^2-45)*exp((-5*x*log(5) +x^3+9*x)/(5*log(5)-x^2-5))+50*log(5)^2+(-20*x^2-100)*log(5)+2*x^4+20*x^2+ 50)/(25*log(5)^2+(-10*x^2-50)*log(5)+x^4+10*x^2+25),x, algorithm=\
-5/2*(2*x/(x^2*(log(5) - 1) - 5*log(5)^2 + 10*log(5) - 5) + log((x - sqrt( 5*log(5) - 5))/(x + sqrt(5*log(5) - 5)))/(sqrt(5*log(5) - 5)*(log(5) - 1)) )*log(5)^2 - 5*(log((x - sqrt(5*log(5) - 5))/(x + sqrt(5*log(5) - 5)))/sqr t(5*log(5) - 5) - 2*x/(x^2 - 5*log(5) + 5))*log(5) + 5*(2*x/(x^2*(log(5) - 1) - 5*log(5)^2 + 10*log(5) - 5) + log((x - sqrt(5*log(5) - 5))/(x + sqrt (5*log(5) - 5)))/(sqrt(5*log(5) - 5)*(log(5) - 1)))*log(5) + 15/2*(log(5) - 1)*log((x - sqrt(5*log(5) - 5))/(x + sqrt(5*log(5) - 5)))/sqrt(5*log(5) - 5) + 2*x - 5*x*(log(5) - 1)/(x^2 - 5*log(5) + 5) + 5*log((x - sqrt(5*log (5) - 5))/(x + sqrt(5*log(5) - 5)))/sqrt(5*log(5) - 5) - 5*x/(x^2*(log(5) - 1) - 5*log(5)^2 + 10*log(5) - 5) - 10*x/(x^2 - 5*log(5) + 5) - 5/2*log(( x - sqrt(5*log(5) - 5))/(x + sqrt(5*log(5) - 5)))/(sqrt(5*log(5) - 5)*(log (5) - 1)) + e^(-x - 4*x/(x^2 - 5*log(5) + 5))
Time = 0.41 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {50+20 x^2+2 x^4+\left (-100-20 x^2\right ) \log (5)+50 \log ^2(5)+e^{\frac {9 x+x^3-5 x \log (5)}{-5-x^2+5 \log (5)}} \left (-45-6 x^2-x^4+\left (70+10 x^2\right ) \log (5)-25 \log ^2(5)\right )}{25+10 x^2+x^4+\left (-50-10 x^2\right ) \log (5)+25 \log ^2(5)} \, dx=2 \, x + e^{\left (-\frac {x^{3} - 5 \, x \log \left (5\right ) + 9 \, x}{x^{2} - 5 \, \log \left (5\right ) + 5}\right )} \end {dmath*}
integrate(((-25*log(5)^2+(10*x^2+70)*log(5)-x^4-6*x^2-45)*exp((-5*x*log(5) +x^3+9*x)/(5*log(5)-x^2-5))+50*log(5)^2+(-20*x^2-100)*log(5)+2*x^4+20*x^2+ 50)/(25*log(5)^2+(-10*x^2-50)*log(5)+x^4+10*x^2+25),x, algorithm=\
Time = 14.71 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \begin {dmath*} \int \frac {50+20 x^2+2 x^4+\left (-100-20 x^2\right ) \log (5)+50 \log ^2(5)+e^{\frac {9 x+x^3-5 x \log (5)}{-5-x^2+5 \log (5)}} \left (-45-6 x^2-x^4+\left (70+10 x^2\right ) \log (5)-25 \log ^2(5)\right )}{25+10 x^2+x^4+\left (-50-10 x^2\right ) \log (5)+25 \log ^2(5)} \, dx=2\,x+5^{\frac {5\,x}{x^2-5\,\ln \left (5\right )+5}}\,{\mathrm {e}}^{-\frac {9\,x}{x^2-5\,\ln \left (5\right )+5}}\,{\mathrm {e}}^{-\frac {x^3}{x^2-5\,\ln \left (5\right )+5}} \end {dmath*}
int((50*log(5)^2 - log(5)*(20*x^2 + 100) - exp(-(9*x - 5*x*log(5) + x^3)/( x^2 - 5*log(5) + 5))*(25*log(5)^2 - log(5)*(10*x^2 + 70) + 6*x^2 + x^4 + 4 5) + 20*x^2 + 2*x^4 + 50)/(25*log(5)^2 - log(5)*(10*x^2 + 50) + 10*x^2 + x ^4 + 25),x)