Integrand size = 70, antiderivative size = 25 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=e^{\frac {2}{x^3}} \left (-\frac {1}{\frac {8}{3}+x}\right )^{\frac {1}{2 x^3}} \] Output:
exp(1/2*ln(-exp(4)/(x+8/3))/x^3)
\[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=\int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx \] Input:
Integrate[(3^(1/(2*x^3))*E^(2/x^3)*(-(8 + 3*x)^(-1))^(1/(2*x^3))*(-3*x + ( -24 - 9*x)*Log[(-3*E^4)/(8 + 3*x)]))/(16*x^4 + 6*x^5),x]
Output:
Integrate[(3^(1/(2*x^3))*E^(2/x^3)*(-(8 + 3*x)^(-1))^(1/(2*x^3))*(-3*x + ( -24 - 9*x)*Log[(-3*E^4)/(8 + 3*x)]))/(16*x^4 + 6*x^5), 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.
\(\displaystyle \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}} \left ((-9 x-24) \log \left (-\frac {3 e^4}{3 x+8}\right )-3 x\right )}{6 x^5+16 x^4} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}} \left ((-9 x-24) \log \left (-\frac {3 e^4}{3 x+8}\right )-3 x\right )}{x^4 (6 x+16)}dx\) |
\(\Big \downarrow \) 2725 |
\(\displaystyle \int \frac {\left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}} e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left ((-9 x-24) \log \left (-\frac {3 e^4}{3 x+8}\right )-3 x\right )}{x^4 (6 x+16)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {3 \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}} e^{\frac {2+\frac {\log (3)}{2}}{x^3}}}{2 x^3 (3 x+8)}-\frac {3 \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}} e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \log \left (-\frac {3 e^4}{3 x+8}\right )}{2 x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{16} \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}}}{x^3}dx-\frac {27 \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}}}{x}dx}{1024}+\frac {81 \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}}}{3 x+8}dx}{1024}-\frac {3}{2} \log \left (-\frac {3 e^4}{3 x+8}\right ) \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}}}{x^4}dx+\frac {9}{2} \int \frac {\int \frac {e^{\frac {4+\log (3)}{2 x^3}} \left (\frac {1}{-3 x-8}\right )^{\frac {1}{2 x^3}}}{x^4}dx}{-3 x-8}dx+\frac {9}{128} \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{3 x+8}\right )^{\frac {1}{2 x^3}}}{x^2}dx\) |
Input:
Int[(3^(1/(2*x^3))*E^(2/x^3)*(-(8 + 3*x)^(-1))^(1/(2*x^3))*(-3*x + (-24 - 9*x)*Log[(-3*E^4)/(8 + 3*x)]))/(16*x^4 + 6*x^5),x]
Output:
$Aborted
Time = 1.75 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\left (-\frac {3 \,{\mathrm e}^{4}}{8+3 x}\right )^{\frac {1}{2 x^{3}}}\) | \(18\) |
parallelrisch | \({\mathrm e}^{\frac {\ln \left (-\frac {3 \,{\mathrm e}^{4}}{8+3 x}\right )}{2 x^{3}}}\) | \(19\) |
Input:
int(((-9*x-24)*ln(-3*exp(4)/(8+3*x))-3*x)*exp(1/2*ln(-3*exp(4)/(8+3*x))/x^ 3)/(6*x^5+16*x^4),x,method=_RETURNVERBOSE)
Output:
(-3*exp(4)/(8+3*x))^(1/2/x^3)
Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=\left (-\frac {3 \, e^{4}}{3 \, x + 8}\right )^{\frac {1}{2 \, x^{3}}} \] Input:
integrate(((-9*x-24)*log(-3*exp(4)/(8+3*x))-3*x)*exp(1/2*log(-3*exp(4)/(8+ 3*x))/x^3)/(6*x^5+16*x^4),x, algorithm="fricas")
Output:
(-3*e^4/(3*x + 8))^(1/2/x^3)
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=e^{\frac {\log {\left (- \frac {3 e^{4}}{3 x + 8} \right )}}{2 x^{3}}} \] Input:
integrate(((-9*x-24)*ln(-3*exp(4)/(8+3*x))-3*x)*exp(1/2*ln(-3*exp(4)/(8+3* x))/x**3)/(6*x**5+16*x**4),x)
Output:
exp(log(-3*exp(4)/(3*x + 8))/(2*x**3))
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=e^{\left (\frac {\log \left (3\right )}{2 \, x^{3}} - \frac {\log \left (-3 \, x - 8\right )}{2 \, x^{3}} + \frac {2}{x^{3}}\right )} \] Input:
integrate(((-9*x-24)*log(-3*exp(4)/(8+3*x))-3*x)*exp(1/2*log(-3*exp(4)/(8+ 3*x))/x^3)/(6*x^5+16*x^4),x, algorithm="maxima")
Output:
e^(1/2*log(3)/x^3 - 1/2*log(-3*x - 8)/x^3 + 2/x^3)
\[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=\int { -\frac {3 \, {\left ({\left (3 \, x + 8\right )} \log \left (-\frac {3 \, e^{4}}{3 \, x + 8}\right ) + x\right )} \left (-\frac {3 \, e^{4}}{3 \, x + 8}\right )^{\frac {1}{2 \, x^{3}}}}{2 \, {\left (3 \, x^{5} + 8 \, x^{4}\right )}} \,d x } \] Input:
integrate(((-9*x-24)*log(-3*exp(4)/(8+3*x))-3*x)*exp(1/2*log(-3*exp(4)/(8+ 3*x))/x^3)/(6*x^5+16*x^4),x, algorithm="giac")
Output:
integrate(-3/2*((3*x + 8)*log(-3*e^4/(3*x + 8)) + x)*(-3*e^4/(3*x + 8))^(1 /2/x^3)/(3*x^5 + 8*x^4), x)
Time = 3.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx={\mathrm {e}}^{\frac {2}{x^3}}\,{\left (-\frac {3}{3\,x+8}\right )}^{\frac {1}{2\,x^3}} \] Input:
int(-(exp(log(-(3*exp(4))/(3*x + 8))/(2*x^3))*(3*x + log(-(3*exp(4))/(3*x + 8))*(9*x + 24)))/(16*x^4 + 6*x^5),x)
Output:
exp(2/x^3)*(-3/(3*x + 8))^(1/(2*x^3))
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=e^{\frac {\mathrm {log}\left (-\frac {3 e^{4}}{3 x +8}\right )}{2 x^{3}}} \] Input:
int(((-9*x-24)*log(-3*exp(4)/(8+3*x))-3*x)*exp(1/2*log(-3*exp(4)/(8+3*x))/ x^3)/(6*x^5+16*x^4),x)
Output:
e**(log(( - 3*e**4)/(3*x + 8))/(2*x**3))