\(\int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} (-\frac {1}{8+3 x})^{\frac {1}{2 x^3}} (-3 x+(-24-9 x) \log (-\frac {3 e^4}{8+3 x}))}{16 x^4+6 x^5} \, dx\) [7838]

Optimal result

Integrand size = 70, antiderivative size = 25 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=e^{\frac {2}{x^3}} \left (-\frac {1}{\frac {8}{3}+x}\right )^{\frac {1}{2 x^3}} \]

[Out]

exp(1/2*ln(-exp(4)/(x+8/3))/x^3)

Rubi [F]

\[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=\int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx \]

[In]

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]

[Out]

(-3*Log[(-3*E^4)/(8 + 3*x)]*Defer[Int][(E^((2 + Log[3]/2)/x^3)*(-(8 + 3*x)^(-1))^(1/(2*x^3)))/x^4, x])/2 - (3*
Defer[Int][(E^((2 + Log[3]/2)/x^3)*(-(8 + 3*x)^(-1))^(1/(2*x^3)))/x^3, x])/16 + (9*Defer[Int][(E^((2 + Log[3]/
2)/x^3)*(-(8 + 3*x)^(-1))^(1/(2*x^3)))/x^2, x])/128 - (27*Defer[Int][(E^((2 + Log[3]/2)/x^3)*(-(8 + 3*x)^(-1))
^(1/(2*x^3)))/x, x])/1024 + (81*Defer[Int][(E^((2 + Log[3]/2)/x^3)*(-(8 + 3*x)^(-1))^(1/(2*x^3)))/(8 + 3*x), x
])/1024 + (9*Defer[Int][Defer[Int][(E^((4 + Log[3])/(2*x^3))*((-8 - 3*x)^(-1))^(1/(2*x^3)))/x^4, x]/(-8 - 3*x)
, x])/2

Rubi steps \begin{align*} \text {integral}& = \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 )}{x^4 (16+6 x)} \, dx \\ & = \int \frac {e^{\frac {2+\frac {\log (3)}{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 )}{x^4 (16+6 x)} \, dx \\ & = \int \left (-\frac {3 e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{2 x^3 (8+3 x)}-\frac {3 e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \log \left (-\frac {3 e^4}{8+3 x}\right )}{2 x^4}\right ) \, dx \\ & = -\left (\frac {3}{2} \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{x^3 (8+3 x)} \, dx\right )-\frac {3}{2} \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \log \left (-\frac {3 e^4}{8+3 x}\right )}{x^4} \, dx \\ & = -\left (\frac {3}{2} \int \left (\frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{8 x^3}-\frac {3 e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{64 x^2}+\frac {9 e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{512 x}-\frac {27 e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{512 (8+3 x)}\right ) \, dx\right )+\frac {3}{2} \int \frac {3 \int \frac {e^{\frac {4+\log (3)}{2 x^3}} \left (\frac {1}{-8-3 x}\right )^{\frac {1}{2 x^3}}}{x^4} \, dx}{-8-3 x} \, dx-\frac {1}{2} \left (3 \log \left (-\frac {3 e^4}{8+3 x}\right )\right ) \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{x^4} \, dx \\ & = -\frac {27 \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{x} \, dx}{1024}+\frac {9}{128} \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{x^2} \, dx+\frac {81 \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{8+3 x} \, dx}{1024}-\frac {3}{16} \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{x^3} \, dx+\frac {9}{2} \int \frac {\int \frac {e^{\frac {4+\log (3)}{2 x^3}} \left (\frac {1}{-8-3 x}\right )^{\frac {1}{2 x^3}}}{x^4} \, dx}{-8-3 x} \, dx-\frac {1}{2} \left (3 \log \left (-\frac {3 e^4}{8+3 x}\right )\right ) \int \frac {e^{\frac {2+\frac {\log (3)}{2}}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}}}{x^4} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=\int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx \]

[In]

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]

[Out]

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]

Maple [A] (verified)

Time = 1.97 (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	\({\mathrm e}^{\frac {\ln \left (-\frac {3 \,{\mathrm e}^{4}}{8+3 x}\right )}{2 x^{3}}}\)	\(19\)

[In]

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=_RETURNVE
RBOSE)

[Out]

(-3*exp(4)/(8+3*x))^(1/2/x^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=\left (-\frac {3 \, e^{4}}{3 \, x + 8}\right )^{\frac {1}{2 \, x^{3}}} \]

[In]

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, algorit
hm="fricas")

[Out]

(-3*e^4/(3*x + 8))^(1/2/x^3)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=e^{\frac {\log {\left (- \frac {3 e^{4}}{3 x + 8} \right )}}{2 x^{3}}} \]

[In]

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)

[Out]

exp(log(-3*exp(4)/(3*x + 8))/(2*x**3))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=e^{\left (\frac {\log \left (3\right )}{2 \, x^{3}} - \frac {\log \left (-3 \, x - 8\right )}{2 \, x^{3}} + \frac {2}{x^{3}}\right )} \]

[In]

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, algorit
hm="maxima")

[Out]

e^(1/2*log(3)/x^3 - 1/2*log(-3*x - 8)/x^3 + 2/x^3)

Giac [F]

\[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx=\int { -\frac {3 \, {\left ({\left (3 \, x + 8\right )} \log \left (-\frac {3 \, e^{4}}{3 \, x + 8}\right ) + x\right )} \left (-\frac {3 \, e^{4}}{3 \, x + 8}\right )^{\frac {1}{2 \, x^{3}}}}{2 \, {\left (3 \, x^{5} + 8 \, x^{4}\right )}} \,d x } \]

[In]

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, algorit
hm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 13.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {3^{\frac {1}{2 x^3}} e^{\frac {2}{x^3}} \left (-\frac {1}{8+3 x}\right )^{\frac {1}{2 x^3}} \left (-3 x+(-24-9 x) \log \left (-\frac {3 e^4}{8+3 x}\right )\right )}{16 x^4+6 x^5} \, dx={\mathrm {e}}^{\frac {2}{x^3}}\,{\left (-\frac {3}{3\,x+8}\right )}^{\frac {1}{2\,x^3}} \]

[In]

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)

[Out]

exp(2/x^3)*(-3/(3*x + 8))^(1/(2*x^3))