3.72.20 \(\int \frac {(-20+6 x+(10-2 x+100 x^2-40 x^3+4 x^4) \log (\frac {-1-10 x^2+2 x^3}{-5 x^2+x^3})) \log (3 x \log (\frac {-1-10 x^2+2 x^3}{-5 x^2+x^3}))}{(5 x-x^2+50 x^3-20 x^4+2 x^5) \log (\frac {-1-10 x^2+2 x^3}{-5 x^2+x^3})} \, dx\)

Optimal. Leaf size=19 \[ \log ^2\left (3 x \log \left (2-\frac {1}{(-5+x) x^2}\right )\right ) \]

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Rubi [F]  time = 81.91, 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 {\left (-20+6 x+\left (10-2 x+100 x^2-40 x^3+4 x^4\right ) \log \left (\frac {-1-10 x^2+2 x^3}{-5 x^2+x^3}\right )\right ) \log \left (3 x \log \left (\frac {-1-10 x^2+2 x^3}{-5 x^2+x^3}\right )\right )}{\left (5 x-x^2+50 x^3-20 x^4+2 x^5\right ) \log \left (\frac {-1-10 x^2+2 x^3}{-5 x^2+x^3}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-20 + 6*x + (10 - 2*x + 100*x^2 - 40*x^3 + 4*x^4)*Log[(-1 - 10*x^2 + 2*x^3)/(-5*x^2 + x^3)])*Log[3*x*Log
[(-1 - 10*x^2 + 2*x^3)/(-5*x^2 + x^3)]])/((5*x - x^2 + 50*x^3 - 20*x^4 + 2*x^5)*Log[(-1 - 10*x^2 + 2*x^3)/(-5*
x^2 + x^3)]),x]

[Out]

2*Defer[Int][Log[3*x*Log[(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x^2)]]/x, x] - 2*Defer[Int][Log[3*x*Log[(-1 - 10*x^2
+ 2*x^3)/((-5 + x)*x^2)]]/((-5 + x)*Log[(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x^2)]), x] - 4*Defer[Int][Log[3*x*Log[
(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x^2)]]/(x*Log[(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x^2)]), x] - 40*Defer[Int][(x*Lo
g[3*x*Log[(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x^2)]])/((-1 - 10*x^2 + 2*x^3)*Log[(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x
^2)]), x] + 12*Defer[Int][(x^2*Log[3*x*Log[(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x^2)]])/((-1 - 10*x^2 + 2*x^3)*Log[
(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x^2)]), x]

Rubi steps

Aborted

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Mathematica [A]  time = 0.07, size = 28, normalized size = 1.47 \begin {gather*} \log ^2\left (3 x \log \left (\frac {-1-10 x^2+2 x^3}{(-5+x) x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-20 + 6*x + (10 - 2*x + 100*x^2 - 40*x^3 + 4*x^4)*Log[(-1 - 10*x^2 + 2*x^3)/(-5*x^2 + x^3)])*Log[3
*x*Log[(-1 - 10*x^2 + 2*x^3)/(-5*x^2 + x^3)]])/((5*x - x^2 + 50*x^3 - 20*x^4 + 2*x^5)*Log[(-1 - 10*x^2 + 2*x^3
)/(-5*x^2 + x^3)]),x]

[Out]

Log[3*x*Log[(-1 - 10*x^2 + 2*x^3)/((-5 + x)*x^2)]]^2

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fricas [A]  time = 0.83, size = 31, normalized size = 1.63 \begin {gather*} \log \left (3 \, x \log \left (\frac {2 \, x^{3} - 10 \, x^{2} - 1}{x^{3} - 5 \, x^{2}}\right )\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-40*x^3+100*x^2-2*x+10)*log((2*x^3-10*x^2-1)/(x^3-5*x^2))+6*x-20)*log(3*x*log((2*x^3-10*x^2-1
)/(x^3-5*x^2)))/(2*x^5-20*x^4+50*x^3-x^2+5*x)/log((2*x^3-10*x^2-1)/(x^3-5*x^2)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.00, size = 92, normalized size = 4.84 \begin {gather*} \log \relax (x)^{2} + 2 \, {\left (\log \relax (x) + \log \left (\log \left (2 \, x^{3} - 10 \, x^{2} - 1\right ) - \log \left (x - 5\right ) - 2 \, \log \relax (x)\right )\right )} \log \left (3 \, \log \left (\frac {2 \, x^{3} - 10 \, x^{2} - 1}{x^{3} - 5 \, x^{2}}\right )\right ) - \log \left (\log \left (2 \, x^{3} - 10 \, x^{2} - 1\right ) - \log \left (x - 5\right ) - 2 \, \log \relax (x)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-40*x^3+100*x^2-2*x+10)*log((2*x^3-10*x^2-1)/(x^3-5*x^2))+6*x-20)*log(3*x*log((2*x^3-10*x^2-1
)/(x^3-5*x^2)))/(2*x^5-20*x^4+50*x^3-x^2+5*x)/log((2*x^3-10*x^2-1)/(x^3-5*x^2)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (4 x^{4}-40 x^{3}+100 x^{2}-2 x +10\right ) \ln \left (\frac {2 x^{3}-10 x^{2}-1}{x^{3}-5 x^{2}}\right )+6 x -20\right ) \ln \left (3 x \ln \left (\frac {2 x^{3}-10 x^{2}-1}{x^{3}-5 x^{2}}\right )\right )}{\left (2 x^{5}-20 x^{4}+50 x^{3}-x^{2}+5 x \right ) \ln \left (\frac {2 x^{3}-10 x^{2}-1}{x^{3}-5 x^{2}}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^4-40*x^3+100*x^2-2*x+10)*ln((2*x^3-10*x^2-1)/(x^3-5*x^2))+6*x-20)*ln(3*x*ln((2*x^3-10*x^2-1)/(x^3-5*
x^2)))/(2*x^5-20*x^4+50*x^3-x^2+5*x)/ln((2*x^3-10*x^2-1)/(x^3-5*x^2)),x)

[Out]

int(((4*x^4-40*x^3+100*x^2-2*x+10)*ln((2*x^3-10*x^2-1)/(x^3-5*x^2))+6*x-20)*ln(3*x*ln((2*x^3-10*x^2-1)/(x^3-5*
x^2)))/(2*x^5-20*x^4+50*x^3-x^2+5*x)/ln((2*x^3-10*x^2-1)/(x^3-5*x^2)),x)

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maxima [B]  time = 0.44, size = 124, normalized size = 6.53 \begin {gather*} 2 \, {\left (\log \relax (x) + \log \left (\log \left (2 \, x^{3} - 10 \, x^{2} - 1\right ) - \log \left (x - 5\right ) - 2 \, \log \relax (x)\right )\right )} \log \left (3 \, x \log \left (\frac {2 \, x^{3} - 10 \, x^{2} - 1}{x^{3} - 5 \, x^{2}}\right )\right ) - \log \relax (x)^{2} - 2 \, \log \relax (x) \log \left (\log \left (2 \, x^{3} - 10 \, x^{2} - 1\right ) - \log \left (x - 5\right ) - 2 \, \log \relax (x)\right ) - \log \left (\log \left (2 \, x^{3} - 10 \, x^{2} - 1\right ) - \log \left (x - 5\right ) - 2 \, \log \relax (x)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-40*x^3+100*x^2-2*x+10)*log((2*x^3-10*x^2-1)/(x^3-5*x^2))+6*x-20)*log(3*x*log((2*x^3-10*x^2-1
)/(x^3-5*x^2)))/(2*x^5-20*x^4+50*x^3-x^2+5*x)/log((2*x^3-10*x^2-1)/(x^3-5*x^2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.85, size = 33, normalized size = 1.74 \begin {gather*} {\ln \left (3\,x\,\ln \left (\frac {-2\,x^3+10\,x^2+1}{5\,x^2-x^3}\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3*x*log((10*x^2 - 2*x^3 + 1)/(5*x^2 - x^3)))*(6*x + log((10*x^2 - 2*x^3 + 1)/(5*x^2 - x^3))*(100*x^2
- 2*x - 40*x^3 + 4*x^4 + 10) - 20))/(log((10*x^2 - 2*x^3 + 1)/(5*x^2 - x^3))*(5*x - x^2 + 50*x^3 - 20*x^4 + 2*
x^5)),x)

[Out]

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

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sympy [A]  time = 0.69, size = 27, normalized size = 1.42 \begin {gather*} \log {\left (3 x \log {\left (\frac {2 x^{3} - 10 x^{2} - 1}{x^{3} - 5 x^{2}} \right )} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**4-40*x**3+100*x**2-2*x+10)*ln((2*x**3-10*x**2-1)/(x**3-5*x**2))+6*x-20)*ln(3*x*ln((2*x**3-10*
x**2-1)/(x**3-5*x**2)))/(2*x**5-20*x**4+50*x**3-x**2+5*x)/ln((2*x**3-10*x**2-1)/(x**3-5*x**2)),x)

[Out]

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

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