Integrand size = 110, antiderivative size = 27 \[ \int \frac {e^{\frac {x^2}{-3-x^2+4 x^3}} \left (6 x^2+4 x^5+e^{-\frac {x^2}{-3-x^2+4 x^3}} \left (-9-6 x^2+24 x^3-x^4+8 x^5-16 x^6\right )\right )}{9 x+6 x^3-24 x^4+x^5-8 x^6+16 x^7} \, dx=3-e^{-\frac {x}{\frac {3}{x}+x-4 x^2}}-\log (x) \] Output:
3-1/exp(x/(3/x-4*x^2+x))-ln(x)
Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^2}{-3-x^2+4 x^3}} \left (6 x^2+4 x^5+e^{-\frac {x^2}{-3-x^2+4 x^3}} \left (-9-6 x^2+24 x^3-x^4+8 x^5-16 x^6\right )\right )}{9 x+6 x^3-24 x^4+x^5-8 x^6+16 x^7} \, dx=-e^{\frac {x^2}{-3-x^2+4 x^3}}-\log (x) \] Input:
Integrate[(E^(x^2/(-3 - x^2 + 4*x^3))*(6*x^2 + 4*x^5 + (-9 - 6*x^2 + 24*x^ 3 - x^4 + 8*x^5 - 16*x^6)/E^(x^2/(-3 - x^2 + 4*x^3))))/(9*x + 6*x^3 - 24*x ^4 + x^5 - 8*x^6 + 16*x^7),x]
Output:
-E^(x^2/(-3 - x^2 + 4*x^3)) - Log[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.
\(\displaystyle \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}} \left (4 x^5+6 x^2+e^{-\frac {x^2}{4 x^3-x^2-3}} \left (-16 x^6+8 x^5-x^4+24 x^3-6 x^2-9\right )\right )}{16 x^7-8 x^6+x^5-24 x^4+6 x^3+9 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}} \left (4 x^5+6 x^2+e^{-\frac {x^2}{4 x^3-x^2-3}} \left (-16 x^6+8 x^5-x^4+24 x^3-6 x^2-9\right )\right )}{x \left (16 x^6-8 x^5+x^4-24 x^3+6 x^2+9\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (-\frac {11 e^{\frac {x^2}{4 x^3-x^2-3}} \left (4 x^5+6 x^2+e^{-\frac {x^2}{4 x^3-x^2-3}} \left (-16 x^6+8 x^5-x^4+24 x^3-6 x^2-9\right )\right )}{500 (x-1) x}+\frac {e^{\frac {x^2}{4 x^3-x^2-3}} \left (4 x^5+6 x^2+e^{-\frac {x^2}{4 x^3-x^2-3}} \left (-16 x^6+8 x^5-x^4+24 x^3-6 x^2-9\right )\right )}{100 (x-1)^2 x}+\frac {e^{\frac {x^2}{4 x^3-x^2-3}} (44 x+57) \left (4 x^5+6 x^2+e^{-\frac {x^2}{4 x^3-x^2-3}} \left (-16 x^6+8 x^5-x^4+24 x^3-6 x^2-9\right )\right )}{500 x \left (4 x^2+3 x+3\right )}+\frac {e^{\frac {x^2}{4 x^3-x^2-3}} (44 x+37) \left (4 x^5+6 x^2+e^{-\frac {x^2}{4 x^3-x^2-3}} \left (-16 x^6+8 x^5-x^4+24 x^3-6 x^2-9\right )\right )}{100 x \left (4 x^2+3 x+3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8}{65} \left (3-i \sqrt {39}\right ) \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{\left (-8 x+i \sqrt {39}-3\right )^2}dx+\frac {288}{65} \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{\left (-8 x+i \sqrt {39}-3\right )^2}dx-\frac {8}{5} i \sqrt {\frac {3}{13}} \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{-8 x+i \sqrt {39}-3}dx+\frac {1}{10} \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{(x-1)^2}dx-\frac {\left (2457+1381 i \sqrt {39}\right ) \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{8 x-i \sqrt {39}+3}dx}{5200}+\frac {3}{400} \left (63+19 i \sqrt {39}\right ) \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{8 x-i \sqrt {39}+3}dx+\frac {8}{65} \left (3+i \sqrt {39}\right ) \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{\left (8 x+i \sqrt {39}+3\right )^2}dx+\frac {288}{65} \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{\left (8 x+i \sqrt {39}+3\right )^2}dx-\frac {\left (2457-1381 i \sqrt {39}\right ) \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{8 x+i \sqrt {39}+3}dx}{5200}+\frac {3}{400} \left (63-19 i \sqrt {39}\right ) \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{8 x+i \sqrt {39}+3}dx-\frac {8}{5} i \sqrt {\frac {3}{13}} \int \frac {e^{\frac {x^2}{4 x^3-x^2-3}}}{8 x+i \sqrt {39}+3}dx+\frac {523 x^3}{1500}-\frac {523 x^2}{500}+\frac {523 (1-x)^3}{1500}+\frac {523 x}{500}-\log (x)\) |
Input:
Int[(E^(x^2/(-3 - x^2 + 4*x^3))*(6*x^2 + 4*x^5 + (-9 - 6*x^2 + 24*x^3 - x^ 4 + 8*x^5 - 16*x^6)/E^(x^2/(-3 - x^2 + 4*x^3))))/(9*x + 6*x^3 - 24*x^4 + x ^5 - 8*x^6 + 16*x^7),x]
Output:
$Aborted
Time = 1.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
risch | \(-\ln \left (x \right )-{\mathrm e}^{\frac {x^{2}}{\left (-1+x \right ) \left (4 x^{2}+3 x +3\right )}}\) | \(30\) |
norman | \(\frac {\left (-4 x^{3}+x^{2}+3\right ) {\mathrm e}^{\frac {x^{2}}{4 x^{3}-x^{2}-3}}}{4 x^{3}-x^{2}-3}-\ln \left (x \right )\) | \(53\) |
parts | \(\frac {\left (-4 x^{3}+x^{2}+3\right ) {\mathrm e}^{\frac {x^{2}}{4 x^{3}-x^{2}-3}}}{4 x^{3}-x^{2}-3}-\ln \left (x \right )\) | \(53\) |
parallelrisch | \(-\frac {\left (-2352+3136 \ln \left (x \right ) x^{3} {\mathrm e}^{-\frac {x^{2}}{4 x^{3}-x^{2}-3}}-784 \ln \left (x \right ) x^{2} {\mathrm e}^{-\frac {x^{2}}{4 x^{3}-x^{2}-3}}+3136 x^{3}-2352 \ln \left (x \right ) {\mathrm e}^{-\frac {x^{2}}{4 x^{3}-x^{2}-3}}-784 x^{2}\right ) {\mathrm e}^{\frac {x^{2}}{4 x^{3}-x^{2}-3}}}{784 \left (4 x^{3}-x^{2}-3\right )}\) | \(129\) |
Input:
int(((-16*x^6+8*x^5-x^4+24*x^3-6*x^2-9)*exp(-x^2/(4*x^3-x^2-3))+4*x^5+6*x^ 2)/(16*x^7-8*x^6+x^5-24*x^4+6*x^3+9*x)/exp(-x^2/(4*x^3-x^2-3)),x,method=_R ETURNVERBOSE)
Output:
-ln(x)-exp(x^2/(-1+x)/(4*x^2+3*x+3))
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {x^2}{-3-x^2+4 x^3}} \left (6 x^2+4 x^5+e^{-\frac {x^2}{-3-x^2+4 x^3}} \left (-9-6 x^2+24 x^3-x^4+8 x^5-16 x^6\right )\right )}{9 x+6 x^3-24 x^4+x^5-8 x^6+16 x^7} \, dx=-e^{\left (\frac {x^{2}}{4 \, x^{3} - x^{2} - 3}\right )} - \log \left (x\right ) \] Input:
integrate(((-16*x^6+8*x^5-x^4+24*x^3-6*x^2-9)*exp(-x^2/(4*x^3-x^2-3))+4*x^ 5+6*x^2)/(16*x^7-8*x^6+x^5-24*x^4+6*x^3+9*x)/exp(-x^2/(4*x^3-x^2-3)),x, al gorithm="fricas")
Output:
-e^(x^2/(4*x^3 - x^2 - 3)) - log(x)
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {x^2}{-3-x^2+4 x^3}} \left (6 x^2+4 x^5+e^{-\frac {x^2}{-3-x^2+4 x^3}} \left (-9-6 x^2+24 x^3-x^4+8 x^5-16 x^6\right )\right )}{9 x+6 x^3-24 x^4+x^5-8 x^6+16 x^7} \, dx=- e^{\frac {x^{2}}{4 x^{3} - x^{2} - 3}} - \log {\left (x \right )} \] Input:
integrate(((-16*x**6+8*x**5-x**4+24*x**3-6*x**2-9)*exp(-x**2/(4*x**3-x**2- 3))+4*x**5+6*x**2)/(16*x**7-8*x**6+x**5-24*x**4+6*x**3+9*x)/exp(-x**2/(4*x **3-x**2-3)),x)
Output:
-exp(x**2/(4*x**3 - x**2 - 3)) - log(x)
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (27) = 54\).
Time = 0.20 (sec) , antiderivative size = 201, normalized size of antiderivative = 7.44 \[ \int \frac {e^{\frac {x^2}{-3-x^2+4 x^3}} \left (6 x^2+4 x^5+e^{-\frac {x^2}{-3-x^2+4 x^3}} \left (-9-6 x^2+24 x^3-x^4+8 x^5-16 x^6\right )\right )}{9 x+6 x^3-24 x^4+x^5-8 x^6+16 x^7} \, dx=\frac {181 \, x^{2} + 12 \, x + 327}{325 \, {\left (4 \, x^{3} - x^{2} - 3\right )}} - \frac {2 \, {\left (109 \, x^{2} + 18 \, x + 3\right )}}{325 \, {\left (4 \, x^{3} - x^{2} - 3\right )}} + \frac {3 \, {\left (72 \, x^{2} - 6 \, x - 1\right )}}{325 \, {\left (4 \, x^{3} - x^{2} - 3\right )}} - \frac {3 \, {\left (24 \, x^{2} - 2 \, x - 217\right )}}{650 \, {\left (4 \, x^{3} - x^{2} - 3\right )}} - \frac {12 \, {\left (12 \, x^{2} - x + 54\right )}}{325 \, {\left (4 \, x^{3} - x^{2} - 3\right )}} + \frac {2 \, x^{2} + 54 \, x + 9}{650 \, {\left (4 \, x^{3} - x^{2} - 3\right )}} - e^{\left (\frac {3 \, x}{5 \, {\left (4 \, x^{2} + 3 \, x + 3\right )}} + \frac {3}{10 \, {\left (4 \, x^{2} + 3 \, x + 3\right )}} + \frac {1}{10 \, {\left (x - 1\right )}}\right )} - \log \left (x\right ) \] Input:
integrate(((-16*x^6+8*x^5-x^4+24*x^3-6*x^2-9)*exp(-x^2/(4*x^3-x^2-3))+4*x^ 5+6*x^2)/(16*x^7-8*x^6+x^5-24*x^4+6*x^3+9*x)/exp(-x^2/(4*x^3-x^2-3)),x, al gorithm="maxima")
Output:
1/325*(181*x^2 + 12*x + 327)/(4*x^3 - x^2 - 3) - 2/325*(109*x^2 + 18*x + 3 )/(4*x^3 - x^2 - 3) + 3/325*(72*x^2 - 6*x - 1)/(4*x^3 - x^2 - 3) - 3/650*( 24*x^2 - 2*x - 217)/(4*x^3 - x^2 - 3) - 12/325*(12*x^2 - x + 54)/(4*x^3 - x^2 - 3) + 1/650*(2*x^2 + 54*x + 9)/(4*x^3 - x^2 - 3) - e^(3/5*x/(4*x^2 + 3*x + 3) + 3/10/(4*x^2 + 3*x + 3) + 1/10/(x - 1)) - log(x)
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {x^2}{-3-x^2+4 x^3}} \left (6 x^2+4 x^5+e^{-\frac {x^2}{-3-x^2+4 x^3}} \left (-9-6 x^2+24 x^3-x^4+8 x^5-16 x^6\right )\right )}{9 x+6 x^3-24 x^4+x^5-8 x^6+16 x^7} \, dx=-e^{\left (\frac {x^{2}}{4 \, x^{3} - x^{2} - 3}\right )} - \log \left (x\right ) \] Input:
integrate(((-16*x^6+8*x^5-x^4+24*x^3-6*x^2-9)*exp(-x^2/(4*x^3-x^2-3))+4*x^ 5+6*x^2)/(16*x^7-8*x^6+x^5-24*x^4+6*x^3+9*x)/exp(-x^2/(4*x^3-x^2-3)),x, al gorithm="giac")
Output:
-e^(x^2/(4*x^3 - x^2 - 3)) - log(x)
Time = 3.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {x^2}{-3-x^2+4 x^3}} \left (6 x^2+4 x^5+e^{-\frac {x^2}{-3-x^2+4 x^3}} \left (-9-6 x^2+24 x^3-x^4+8 x^5-16 x^6\right )\right )}{9 x+6 x^3-24 x^4+x^5-8 x^6+16 x^7} \, dx=-{\mathrm {e}}^{-\frac {x^2}{-4\,x^3+x^2+3}}-\ln \left (x\right ) \] Input:
int((exp(-x^2/(x^2 - 4*x^3 + 3))*(6*x^2 + 4*x^5 - exp(x^2/(x^2 - 4*x^3 + 3 ))*(6*x^2 - 24*x^3 + x^4 - 8*x^5 + 16*x^6 + 9)))/(9*x + 6*x^3 - 24*x^4 + x ^5 - 8*x^6 + 16*x^7),x)
Output:
- exp(-x^2/(x^2 - 4*x^3 + 3)) - log(x)
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x^2}{-3-x^2+4 x^3}} \left (6 x^2+4 x^5+e^{-\frac {x^2}{-3-x^2+4 x^3}} \left (-9-6 x^2+24 x^3-x^4+8 x^5-16 x^6\right )\right )}{9 x+6 x^3-24 x^4+x^5-8 x^6+16 x^7} \, dx=-e^{\frac {x^{2}}{4 x^{3}-x^{2}-3}}-\mathrm {log}\left (x \right ) \] Input:
int(((-16*x^6+8*x^5-x^4+24*x^3-6*x^2-9)*exp(-x^2/(4*x^3-x^2-3))+4*x^5+6*x^ 2)/(16*x^7-8*x^6+x^5-24*x^4+6*x^3+9*x)/exp(-x^2/(4*x^3-x^2-3)),x)
Output:
- (e**(x**2/(4*x**3 - x**2 - 3)) + log(x))