\(\int \frac {-5 x^2+(8 x^2+x^2 \log (\frac {14}{5}))^{.\frac {1}{5}/x} (6-3 \log (8 x^2+x^2 \log (\frac {14}{5})))}{5 x^2} \, dx\) [7514]

Optimal result

Integrand size = 55, antiderivative size = 37 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=-x+3 \left (x^2\right )^{\left .\frac {1}{5}\right /x} \left (\frac {1}{2} \left (16+2 \log \left (\frac {14}{5}\right )\right )\right )^{\left .\frac {1}{5}\right /x} \]

[Out]

3*exp(1/5*ln(1/2*x^2*(2*ln(14/5)+16))/x)-x

Rubi [F]

\[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=\int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx \]

[In]

Int[(-5*x^2 + (8*x^2 + x^2*Log[14/5])^(1/(5*x))*(6 - 3*Log[8*x^2 + x^2*Log[14/5]]))/(5*x^2),x]

[Out]

-x + (6*Defer[Int][(x^2)^(-1 + 1/(5*x))*(8 + Log[14/5])^(1/(5*x)), x])/5 - (3*Log[x^2*(8 + Log[14/5])]*Defer[I
nt][(x^2)^(-1 + 1/(5*x))*(8 + Log[14/5])^(1/(5*x)), x])/5 + (6*Defer[Int][Defer[Int][(x^2)^(-1 + 1/(5*x))*(8 +
Log[14/5])^(1/(5*x)), x]/x, x])/5

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \left (-5-3 \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (-2+\log \left (x^2 \left (8+\log \left (\frac {14}{5}\right )\right )\right )\right )\right ) \, dx \\ & = -x-\frac {3}{5} \int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (-2+\log \left (x^2 \left (8+\log \left (\frac {14}{5}\right )\right )\right )\right ) \, dx \\ & = -x-\frac {3}{5} \int \left (-2 \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x}+\left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \log \left (x^2 \left (8+\log \left (\frac {14}{5}\right )\right )\right )\right ) \, dx \\ & = -x-\frac {3}{5} \int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \log \left (x^2 \left (8+\log \left (\frac {14}{5}\right )\right )\right ) \, dx+\frac {6}{5} \int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \, dx \\ & = -x+\frac {3}{5} \int \frac {2 \int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \, dx}{x} \, dx+\frac {6}{5} \int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \, dx-\frac {1}{5} \left (3 \log \left (x^2 \left (8+\log \left (\frac {14}{5}\right )\right )\right )\right ) \int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \, dx \\ & = -x+\frac {6}{5} \int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \, dx+\frac {6}{5} \int \frac {\int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \, dx}{x} \, dx-\frac {1}{5} \left (3 \log \left (x^2 \left (8+\log \left (\frac {14}{5}\right )\right )\right )\right ) \int \left (x^2\right )^{-1+\frac {1}{5 x}} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=\int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx \]

[In]

Integrate[(-5*x^2 + (8*x^2 + x^2*Log[14/5])^(1/(5*x))*(6 - 3*Log[8*x^2 + x^2*Log[14/5]]))/(5*x^2),x]

[Out]

Integrate[(-5*x^2 + (8*x^2 + x^2*Log[14/5])^(1/(5*x))*(6 - 3*Log[8*x^2 + x^2*Log[14/5]]))/x^2, x]/5

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.59

method	result	size
parallelrisch	\(-x +3 \,{\mathrm e}^{\frac {\ln \left (x^{2} \left (\ln \left (\frac {14}{5}\right )+8\right )\right )}{5 x}}\)	\(22\)
default	\(-x +3 \,{\mathrm e}^{\frac {\ln \left (x^{2} \ln \left (\frac {14}{5}\right )+8 x^{2}\right )}{5 x}}\)	\(26\)
parts	\(-x +3 \,{\mathrm e}^{\frac {\ln \left (x^{2} \ln \left (\frac {14}{5}\right )+8 x^{2}\right )}{5 x}}\)	\(26\)
risch	\(-x +3 \left (x^{2} \left (\ln \left (2\right )+\ln \left (7\right )-\ln \left (5\right )\right )+8 x^{2}\right )^{\frac {1}{5 x}}\)	\(32\)

[In]

int(1/5*((-3*ln(x^2*ln(14/5)+8*x^2)+6)*exp(1/5*ln(x^2*ln(14/5)+8*x^2)/x)-5*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-x+3*exp(1/5*ln(x^2*(ln(14/5)+8))/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.65 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=3 \, {\left (x^{2} \log \left (\frac {14}{5}\right ) + 8 \, x^{2}\right )}^{\frac {1}{5 \, x}} - x \]

[In]

integrate(1/5*((-3*log(x^2*log(14/5)+8*x^2)+6)*exp(1/5*log(x^2*log(14/5)+8*x^2)/x)-5*x^2)/x^2,x, algorithm="fr
icas")

[Out]

3*(x^2*log(14/5) + 8*x^2)^(1/5/x) - x

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.59 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=- x + 3 e^{\frac {\log {\left (x^{2} \log {\left (\frac {14}{5} \right )} + 8 x^{2} \right )}}{5 x}} \]

[In]

integrate(1/5*((-3*ln(x**2*ln(14/5)+8*x**2)+6)*exp(1/5*ln(x**2*ln(14/5)+8*x**2)/x)-5*x**2)/x**2,x)

[Out]

-x + 3*exp(log(x**2*log(14/5) + 8*x**2)/(5*x))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=-x + 3 \, e^{\left (\frac {2 \, \log \left (x\right )}{5 \, x} + \frac {\log \left (\log \left (7\right ) - \log \left (5\right ) + \log \left (2\right ) + 8\right )}{5 \, x}\right )} \]

[In]

integrate(1/5*((-3*log(x^2*log(14/5)+8*x^2)+6)*exp(1/5*log(x^2*log(14/5)+8*x^2)/x)-5*x^2)/x^2,x, algorithm="ma
xima")

[Out]

-x + 3*e^(2/5*log(x)/x + 1/5*log(log(7) - log(5) + log(2) + 8)/x)

Giac [F]

\[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=\int { -\frac {5 \, x^{2} + 3 \, {\left (x^{2} \log \left (\frac {14}{5}\right ) + 8 \, x^{2}\right )}^{\frac {1}{5 \, x}} {\left (\log \left (x^{2} \log \left (\frac {14}{5}\right ) + 8 \, x^{2}\right ) - 2\right )}}{5 \, x^{2}} \,d x } \]

[In]

integrate(1/5*((-3*log(x^2*log(14/5)+8*x^2)+6)*exp(1/5*log(x^2*log(14/5)+8*x^2)/x)-5*x^2)/x^2,x, algorithm="gi
ac")

[Out]

integrate(-1/5*(5*x^2 + 3*(x^2*log(14/5) + 8*x^2)^(1/5/x)*(log(x^2*log(14/5) + 8*x^2) - 2))/x^2, x)

Mupad [B] (verification not implemented)

Time = 12.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=3\,{\left (x^2\,\ln \left (14\right )-x^2\,\ln \left (5\right )+8\,x^2\right )}^{\frac {1}{5\,x}}-x \]

[In]

int(-((exp(log(x^2*log(14/5) + 8*x^2)/(5*x))*(3*log(x^2*log(14/5) + 8*x^2) - 6))/5 + x^2)/x^2,x)

[Out]

3*(x^2*log(14) - x^2*log(5) + 8*x^2)^(1/(5*x)) - x