Integrand size = 55, antiderivative size = 37 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=-x+3 \left (x^2\right )^{\left .\frac {1}{5}\right /x} \left (\frac {1}{2} \left (16+2 \log \left (\frac {14}{5}\right )\right )\right )^{\left .\frac {1}{5}\right /x} \]
\[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=\int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx \]
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]
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
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.
\(\displaystyle \int \frac {\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )-5 x^2}{5 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {5 x^2-3 \left (\log \left (\frac {14}{5}\right ) x^2+8 x^2\right )^{\left .\frac {1}{5}\right /x} \left (2-\log \left (\log \left (\frac {14}{5}\right ) x^2+8 x^2\right )\right )}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {5 x^2-3 \left (\log \left (\frac {14}{5}\right ) x^2+8 x^2\right )^{\left .\frac {1}{5}\right /x} \left (2-\log \left (\log \left (\frac {14}{5}\right ) x^2+8 x^2\right )\right )}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{5} \int \left (3 \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (\log \left (x^2 \left (8+\log \left (\frac {14}{5}\right )\right )\right )-2\right ) \left (x^2\right )^{\frac {1}{5 x}-1}+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-3 \log \left (x^2 \left (8+\log \left (\frac {14}{5}\right )\right )\right ) \int \left (x^2\right )^{\frac {1}{5 x}-1} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x}dx+6 \int \left (x^2\right )^{\frac {1}{5 x}-1} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x}dx+6 \int \frac {\int \left (x^2\right )^{\frac {1}{5 x}-1} \left (8+\log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x}dx}{x}dx-5 x\right )\) |
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]
3.16.14.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
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 | \(-x +3 \left (x^{2} \left (\ln \left (2\right )+\ln \left (7\right )-\ln \left (5\right )\right )+8 x^{2}\right )^{\frac {1}{5 x}}\) | \(32\) |
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)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.65 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=3 \, {\left (x^{2} \log \left (\frac {14}{5}\right ) + 8 \, x^{2}\right )}^{\frac {1}{5 \, x}} - x \]
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=\
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.59 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=- x + 3 e^{\frac {\log {\left (x^{2} \log {\left (\frac {14}{5} \right )} + 8 x^{2} \right )}}{5 x}} \]
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)
Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=-x + 3 \, e^{\left (\frac {2 \, \log \left (x\right )}{5 \, x} + \frac {\log \left (\log \left (7\right ) - \log \left (5\right ) + \log \left (2\right ) + 8\right )}{5 \, x}\right )} \]
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=\
\[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=\int { -\frac {5 \, x^{2} + 3 \, {\left (x^{2} \log \left (\frac {14}{5}\right ) + 8 \, x^{2}\right )}^{\frac {1}{5 \, x}} {\left (\log \left (x^{2} \log \left (\frac {14}{5}\right ) + 8 \, x^{2}\right ) - 2\right )}}{5 \, x^{2}} \,d x } \]
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=\
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)
Time = 12.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {-5 x^2+\left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )^{\left .\frac {1}{5}\right /x} \left (6-3 \log \left (8 x^2+x^2 \log \left (\frac {14}{5}\right )\right )\right )}{5 x^2} \, dx=3\,{\left (x^2\,\ln \left (14\right )-x^2\,\ln \left (5\right )+8\,x^2\right )}^{\frac {1}{5\,x}}-x \]