Integrand size = 76, antiderivative size = 27 \[ \int \frac {-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )+e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (8 x-8 x^2+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )-4 \log ^2\left (\frac {16}{x^2}\right )\right )}{3 x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=\frac {(4-4 x) \left (e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}}+\log (5)\right )}{3 x} \]
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )+e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (8 x-8 x^2+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )-4 \log ^2\left (\frac {16}{x^2}\right )\right )}{3 x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=-\frac {4 \left (e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} (-1+x)-\log (5)\right )}{3 x} \]
Integrate[(-4*Log[5]*Log[16/x^2]^2 + E^(x/Log[16/x^2])*(8*x - 8*x^2 + (4*x - 4*x^2)*Log[16/x^2] - 4*Log[16/x^2]^2))/(3*x^2*Log[16/x^2]^2),x]
Leaf count is larger than twice the leaf count of optimal. \(86\) vs. \(2(27)=54\).
Time = 0.82 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.19, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {27, 27, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (-8 x^2-4 \log ^2\left (\frac {16}{x^2}\right )+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )+8 x\right )-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )}{3 x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {4 \left (\log (5) \log ^2\left (\frac {16}{x^2}\right )-e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (-2 x^2+2 x-\log ^2\left (\frac {16}{x^2}\right )+\left (x-x^2\right ) \log \left (\frac {16}{x^2}\right )\right )\right )}{x^2 \log ^2\left (\frac {16}{x^2}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{3} \int \frac {\log (5) \log ^2\left (\frac {16}{x^2}\right )-e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (-2 x^2+2 x-\log ^2\left (\frac {16}{x^2}\right )+\left (x-x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^2 \log ^2\left (\frac {16}{x^2}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {4}{3} \int \left (\frac {e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (\log \left (\frac {16}{x^2}\right ) x^2+2 x^2-\log \left (\frac {16}{x^2}\right ) x-2 x+\log ^2\left (\frac {16}{x^2}\right )\right )}{x^2 \log ^2\left (\frac {16}{x^2}\right )}+\frac {\log (5)}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4}{3} \left (-\frac {e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (-2 x^2+x^2 \left (-\log \left (\frac {16}{x^2}\right )\right )+x \log \left (\frac {16}{x^2}\right )+2 x\right )}{x^2 \left (\frac {2}{\log ^2\left (\frac {16}{x^2}\right )}+\frac {1}{\log \left (\frac {16}{x^2}\right )}\right ) \log ^2\left (\frac {16}{x^2}\right )}-\frac {\log (5)}{x}\right )\) |
Int[(-4*Log[5]*Log[16/x^2]^2 + E^(x/Log[16/x^2])*(8*x - 8*x^2 + (4*x - 4*x ^2)*Log[16/x^2] - 4*Log[16/x^2]^2))/(3*x^2*Log[16/x^2]^2),x]
(-4*(-(Log[5]/x) - (E^(x/Log[16/x^2])*(2*x - 2*x^2 + x*Log[16/x^2] - x^2*L og[16/x^2]))/(x^2*(2/Log[16/x^2]^2 + Log[16/x^2]^(-1))*Log[16/x^2]^2)))/3
3.28.100.3.1 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]]
Time = 0.75 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\frac {4 \ln \left (5\right )}{3 x}-\frac {4 \left (-1+x \right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}}{3 x}\) | \(28\) |
parallelrisch | \(\frac {-8 \,{\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}} x +8 \ln \left (5\right )+8 \,{\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}}{6 x}\) | \(38\) |
norman | \(\frac {\frac {4 \ln \left (5\right ) \ln \left (\frac {16}{x^{2}}\right )}{3}+\frac {4 \ln \left (\frac {16}{x^{2}}\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}}{3}-\frac {4 \ln \left (\frac {16}{x^{2}}\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}} x}{3}}{x \ln \left (\frac {16}{x^{2}}\right )}\) | \(63\) |
default | \(-\frac {4 \left (x \left (\ln \left (\frac {16}{x^{2}}\right )+2 \ln \left (x \right )\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}-\left (\ln \left (\frac {16}{x^{2}}\right )+2 \ln \left (x \right )\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}+2 \ln \left (x \right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}-2 x \ln \left (x \right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}\right )}{3 x \ln \left (\frac {16}{x^{2}}\right )}+\frac {4 \ln \left (5\right )}{3 x}\) | \(102\) |
parts | \(-\frac {4 \left (x \left (\ln \left (\frac {16}{x^{2}}\right )+2 \ln \left (x \right )\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}-\left (\ln \left (\frac {16}{x^{2}}\right )+2 \ln \left (x \right )\right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}+2 \ln \left (x \right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}-2 x \ln \left (x \right ) {\mathrm e}^{\frac {x}{\ln \left (\frac {16}{x^{2}}\right )}}\right )}{3 x \ln \left (\frac {16}{x^{2}}\right )}+\frac {4 \ln \left (5\right )}{3 x}\) | \(102\) |
int(1/3*((-4*ln(16/x^2)^2+(-4*x^2+4*x)*ln(16/x^2)-8*x^2+8*x)*exp(x/ln(16/x ^2))-4*ln(5)*ln(16/x^2)^2)/x^2/ln(16/x^2)^2,x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )+e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (8 x-8 x^2+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )-4 \log ^2\left (\frac {16}{x^2}\right )\right )}{3 x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=-\frac {4 \, {\left ({\left (x - 1\right )} e^{\left (\frac {x}{\log \left (\frac {16}{x^{2}}\right )}\right )} - \log \left (5\right )\right )}}{3 \, x} \]
integrate(1/3*((-4*log(16/x^2)^2+(-4*x^2+4*x)*log(16/x^2)-8*x^2+8*x)*exp(x /log(16/x^2))-4*log(5)*log(16/x^2)^2)/x^2/log(16/x^2)^2,x, algorithm=\
Exception generated. \[ \int \frac {-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )+e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (8 x-8 x^2+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )-4 \log ^2\left (\frac {16}{x^2}\right )\right )}{3 x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=\text {Exception raised: TypeError} \]
integrate(1/3*((-4*ln(16/x**2)**2+(-4*x**2+4*x)*ln(16/x**2)-8*x**2+8*x)*ex p(x/ln(16/x**2))-4*ln(5)*ln(16/x**2)**2)/x**2/ln(16/x**2)**2,x)
Exception generated. \[ \int \frac {-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )+e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (8 x-8 x^2+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )-4 \log ^2\left (\frac {16}{x^2}\right )\right )}{3 x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=\text {Exception raised: RuntimeError} \]
integrate(1/3*((-4*log(16/x^2)^2+(-4*x^2+4*x)*log(16/x^2)-8*x^2+8*x)*exp(x /log(16/x^2))-4*log(5)*log(16/x^2)^2)/x^2/log(16/x^2)^2,x, algorithm=\
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
Time = 0.41 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )+e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (8 x-8 x^2+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )-4 \log ^2\left (\frac {16}{x^2}\right )\right )}{3 x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=-\frac {4 \, {\left (x e^{\left (\frac {x}{\log \left (\frac {16}{x^{2}}\right )}\right )} - e^{\left (\frac {x}{\log \left (\frac {16}{x^{2}}\right )}\right )} - \log \left (5\right )\right )}}{3 \, x} \]
integrate(1/3*((-4*log(16/x^2)^2+(-4*x^2+4*x)*log(16/x^2)-8*x^2+8*x)*exp(x /log(16/x^2))-4*log(5)*log(16/x^2)^2)/x^2/log(16/x^2)^2,x, algorithm=\
Time = 14.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-4 \log (5) \log ^2\left (\frac {16}{x^2}\right )+e^{\frac {x}{\log \left (\frac {16}{x^2}\right )}} \left (8 x-8 x^2+\left (4 x-4 x^2\right ) \log \left (\frac {16}{x^2}\right )-4 \log ^2\left (\frac {16}{x^2}\right )\right )}{3 x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx=\frac {\frac {4\,{\mathrm {e}}^{\frac {x}{\ln \left (\frac {16}{x^2}\right )}}}{3}+\ln \left ({625}^{1/3}\right )}{x}-\frac {4\,{\mathrm {e}}^{\frac {x}{\ln \left (\frac {16}{x^2}\right )}}}{3} \]