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

Optimal. Leaf size=27 \[ 4+x+\frac {5}{2} x^2 \left (-4-x-\frac {x}{\log \left (\log \left (\frac {x}{3}\right )\right )}\right ) \]

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Rubi [F]  time = 0.24, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 30, normalized size = 1.11 \begin {gather*} x-10 x^2-\frac {5 x^3}{2}-\frac {5 x^3}{2 \log \left (\log \left (\frac {x}{3}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[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]])

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fricas [A]  time = 1.87, size = 35, normalized size = 1.30 \begin {gather*} -\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.25, size = 24, normalized size = 0.89 \begin {gather*} -\frac {5}{2} \, x^{3} - 10 \, x^{2} - \frac {5 \, x^{3}}{2 \, \log \left (\log \left (\frac {1}{3} \, x\right )\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.04, size = 25, normalized size = 0.93

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.48, size = 27, normalized size = 1.00 \begin {gather*} -\frac {5}{2} \, x^{3} - 10 \, x^{2} - \frac {5 \, x^{3}}{2 \, \log \left (-\log \relax (3) + \log \relax (x)\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.93, size = 27, normalized size = 1.00 \begin {gather*} x-\frac {5\,x^3}{2\,\ln \left (\ln \relax (x)-\ln \relax (3)\right )}-10\,x^2-\frac {5\,x^3}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.27, size = 26, normalized size = 0.96 \begin {gather*} - \frac {5 x^{3}}{2} - \frac {5 x^{3}}{2 \log {\left (\log {\left (\frac {x}{3} \right )} \right )}} - 10 x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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