\(\int \frac {5 x^2-15 x^2 \log (\frac {x}{3}) \log (\log (\frac {x}{3}))+(2-40 x-15 x^2) \log (\frac {x}{3}) \log ^2(\log (\frac {x}{3}))}{2 \log (\frac {x}{3}) \log ^2(\log (\frac {x}{3}))} \, dx\) [7649]

Optimal result

Integrand size = 71, antiderivative size = 27 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=4+x+\frac {5}{2} x^2 \left (-4-x-\frac {x}{\log \left (\log \left (\frac {x}{3}\right )\right )}\right ) \]

[Out]

5/2*x^2*(-4-x/ln(ln(1/3*x))-x)+4+x

Rubi [F]

\[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=\int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx \]

[In]

Int[(5*x^2 - 15*x^2*Log[x/3]*Log[Log[x/3]] + (2 - 40*x - 15*x^2)*Log[x/3]*Log[Log[x/3]]^2)/(2*Log[x/3]*Log[Log
[x/3]]^2),x]

[Out]

x - 10*x^2 - (5*x^3)/2 + (5*Defer[Int][x^2/(Log[x/3]*Log[Log[x/3]]^2), x])/2 - (15*Defer[Int][x^2/Log[Log[x/3]
], x])/2

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{\log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx \\ & = \frac {1}{2} \int \left (2-40 x-15 x^2+\frac {5 x^2}{\log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}-\frac {15 x^2}{\log \left (\log \left (\frac {x}{3}\right )\right )}\right ) \, dx \\ & = x-10 x^2-\frac {5 x^3}{2}+\frac {5}{2} \int \frac {x^2}{\log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx-\frac {15}{2} \int \frac {x^2}{\log \left (\log \left (\frac {x}{3}\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=x-10 x^2-\frac {5 x^3}{2}-\frac {5 x^3}{2 \log \left (\log \left (\frac {x}{3}\right )\right )} \]

[In]

Integrate[(5*x^2 - 15*x^2*Log[x/3]*Log[Log[x/3]] + (2 - 40*x - 15*x^2)*Log[x/3]*Log[Log[x/3]]^2)/(2*Log[x/3]*L
og[Log[x/3]]^2),x]

[Out]

x - 10*x^2 - (5*x^3)/2 - (5*x^3)/(2*Log[Log[x/3]])

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {5 x^{3}}{2}-10 x^{2}+x -\frac {5 x^{3}}{2 \ln \left (\ln \left (\frac {x}{3}\right )\right )}\)	\(25\)
parallelrisch	\(-\frac {5 \ln \left (\ln \left (\frac {x}{3}\right )\right ) x^{3}+5 x^{3}+20 \ln \left (\ln \left (\frac {x}{3}\right )\right ) x^{2}-2 x \ln \left (\ln \left (\frac {x}{3}\right )\right )}{2 \ln \left (\ln \left (\frac {x}{3}\right )\right )}\)	\(44\)

[In]

int(1/2*((-15*x^2-40*x+2)*ln(1/3*x)*ln(ln(1/3*x))^2-15*x^2*ln(1/3*x)*ln(ln(1/3*x))+5*x^2)/ln(1/3*x)/ln(ln(1/3*
x))^2,x,method=_RETURNVERBOSE)

[Out]

-5/2*x^3-10*x^2+x-5/2*x^3/ln(ln(1/3*x))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=-\frac {5 \, x^{3} + {\left (5 \, x^{3} + 20 \, x^{2} - 2 \, x\right )} \log \left (\log \left (\frac {1}{3} \, x\right )\right )}{2 \, \log \left (\log \left (\frac {1}{3} \, x\right )\right )} \]

[In]

integrate(1/2*((-15*x^2-40*x+2)*log(1/3*x)*log(log(1/3*x))^2-15*x^2*log(1/3*x)*log(log(1/3*x))+5*x^2)/log(1/3*
x)/log(log(1/3*x))^2,x, algorithm="fricas")

[Out]

-1/2*(5*x^3 + (5*x^3 + 20*x^2 - 2*x)*log(log(1/3*x)))/log(log(1/3*x))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=- \frac {5 x^{3}}{2} - \frac {5 x^{3}}{2 \log {\left (\log {\left (\frac {x}{3} \right )} \right )}} - 10 x^{2} + x \]

[In]

integrate(1/2*((-15*x**2-40*x+2)*ln(1/3*x)*ln(ln(1/3*x))**2-15*x**2*ln(1/3*x)*ln(ln(1/3*x))+5*x**2)/ln(1/3*x)/
ln(ln(1/3*x))**2,x)

[Out]

-5*x**3/2 - 5*x**3/(2*log(log(x/3))) - 10*x**2 + x

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=-\frac {5}{2} \, x^{3} - 10 \, x^{2} - \frac {5 \, x^{3}}{2 \, \log \left (-\log \left (3\right ) + \log \left (x\right )\right )} + x \]

[In]

integrate(1/2*((-15*x^2-40*x+2)*log(1/3*x)*log(log(1/3*x))^2-15*x^2*log(1/3*x)*log(log(1/3*x))+5*x^2)/log(1/3*
x)/log(log(1/3*x))^2,x, algorithm="maxima")

[Out]

-5/2*x^3 - 10*x^2 - 5/2*x^3/log(-log(3) + log(x)) + x

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=-\frac {5}{2} \, x^{3} - 10 \, x^{2} - \frac {5 \, x^{3}}{2 \, \log \left (\log \left (\frac {1}{3} \, x\right )\right )} + x \]

[In]

integrate(1/2*((-15*x^2-40*x+2)*log(1/3*x)*log(log(1/3*x))^2-15*x^2*log(1/3*x)*log(log(1/3*x))+5*x^2)/log(1/3*
x)/log(log(1/3*x))^2,x, algorithm="giac")

[Out]

-5/2*x^3 - 10*x^2 - 5/2*x^3/log(log(1/3*x)) + x

Mupad [B] (verification not implemented)

Time = 13.91 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^2-15 x^2 \log \left (\frac {x}{3}\right ) \log \left (\log \left (\frac {x}{3}\right )\right )+\left (2-40 x-15 x^2\right ) \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )}{2 \log \left (\frac {x}{3}\right ) \log ^2\left (\log \left (\frac {x}{3}\right )\right )} \, dx=x-\frac {5\,x^3}{2\,\ln \left (\ln \left (x\right )-\ln \left (3\right )\right )}-10\,x^2-\frac {5\,x^3}{2} \]

[In]

int(-((log(x/3)*log(log(x/3))^2*(40*x + 15*x^2 - 2))/2 - (5*x^2)/2 + (15*x^2*log(x/3)*log(log(x/3)))/2)/(log(x
/3)*log(log(x/3))^2),x)

[Out]

x - (5*x^3)/(2*log(log(x) - log(3))) - 10*x^2 - (5*x^3)/2