3.98.12 \(\int \frac {-60+16 x-601 x^2+140 x^3-10 x^4-4 x \log (x)}{5000 x^5-1000 x^6+50 x^7+(2000 x^3-400 x^4+20 x^5) \log (x)+(200 x-40 x^2+2 x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {-3+\frac {x}{2}}{(-10+x) \left (5 x^2+\log (x)\right )} \]

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Rubi [F]  time = 0.95, 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 {-60+16 x-601 x^2+140 x^3-10 x^4-4 x \log (x)}{5000 x^5-1000 x^6+50 x^7+\left (2000 x^3-400 x^4+20 x^5\right ) \log (x)+\left (200 x-40 x^2+2 x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-60 + 16*x - 601*x^2 + 140*x^3 - 10*x^4 - 4*x*Log[x])/(5000*x^5 - 1000*x^6 + 50*x^7 + (2000*x^3 - 400*x^4
 + 20*x^5)*Log[x] + (200*x - 40*x^2 + 2*x^3)*Log[x]^2),x]

[Out]

-20*Defer[Int][(5*x^2 + Log[x])^(-2), x] - (1001*Defer[Int][1/((-10 + x)*(5*x^2 + Log[x])^2), x])/5 - (3*Defer
[Int][1/(x*(5*x^2 + Log[x])^2), x])/10 - 5*Defer[Int][x/(5*x^2 + Log[x])^2, x] - 2*Defer[Int][1/((-10 + x)^2*(
5*x^2 + Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-60+16 x-601 x^2+140 x^3-10 x^4-4 x \log (x)}{2 (10-x)^2 x \left (5 x^2+\log (x)\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {-60+16 x-601 x^2+140 x^3-10 x^4-4 x \log (x)}{(10-x)^2 x \left (5 x^2+\log (x)\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {6-x+60 x^2-10 x^3}{(-10+x) x \left (5 x^2+\log (x)\right )^2}-\frac {4}{(-10+x)^2 \left (5 x^2+\log (x)\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {6-x+60 x^2-10 x^3}{(-10+x) x \left (5 x^2+\log (x)\right )^2} \, dx-2 \int \frac {1}{(-10+x)^2 \left (5 x^2+\log (x)\right )} \, dx\\ &=\frac {1}{2} \int \left (-\frac {40}{\left (5 x^2+\log (x)\right )^2}-\frac {2002}{5 (-10+x) \left (5 x^2+\log (x)\right )^2}-\frac {3}{5 x \left (5 x^2+\log (x)\right )^2}-\frac {10 x}{\left (5 x^2+\log (x)\right )^2}\right ) \, dx-2 \int \frac {1}{(-10+x)^2 \left (5 x^2+\log (x)\right )} \, dx\\ &=-\left (\frac {3}{10} \int \frac {1}{x \left (5 x^2+\log (x)\right )^2} \, dx\right )-2 \int \frac {1}{(-10+x)^2 \left (5 x^2+\log (x)\right )} \, dx-5 \int \frac {x}{\left (5 x^2+\log (x)\right )^2} \, dx-20 \int \frac {1}{\left (5 x^2+\log (x)\right )^2} \, dx-\frac {1001}{5} \int \frac {1}{(-10+x) \left (5 x^2+\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 24, normalized size = 1.04 \begin {gather*} -\frac {6-x}{2 (-10+x) \left (5 x^2+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60 + 16*x - 601*x^2 + 140*x^3 - 10*x^4 - 4*x*Log[x])/(5000*x^5 - 1000*x^6 + 50*x^7 + (2000*x^3 - 4
00*x^4 + 20*x^5)*Log[x] + (200*x - 40*x^2 + 2*x^3)*Log[x]^2),x]

[Out]

-1/2*(6 - x)/((-10 + x)*(5*x^2 + Log[x]))

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fricas [A]  time = 0.83, size = 24, normalized size = 1.04 \begin {gather*} \frac {x - 6}{2 \, {\left (5 \, x^{3} - 50 \, x^{2} + {\left (x - 10\right )} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(x)-10*x^4+140*x^3-601*x^2+16*x-60)/((2*x^3-40*x^2+200*x)*log(x)^2+(20*x^5-400*x^4+2000*x^3
)*log(x)+50*x^7-1000*x^6+5000*x^5),x, algorithm="fricas")

[Out]

1/2*(x - 6)/(5*x^3 - 50*x^2 + (x - 10)*log(x))

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giac [A]  time = 0.19, size = 26, normalized size = 1.13 \begin {gather*} \frac {x - 6}{2 \, {\left (5 \, x^{3} - 50 \, x^{2} + x \log \relax (x) - 10 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(x)-10*x^4+140*x^3-601*x^2+16*x-60)/((2*x^3-40*x^2+200*x)*log(x)^2+(20*x^5-400*x^4+2000*x^3
)*log(x)+50*x^7-1000*x^6+5000*x^5),x, algorithm="giac")

[Out]

1/2*(x - 6)/(5*x^3 - 50*x^2 + x*log(x) - 10*log(x))

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maple [A]  time = 0.09, size = 21, normalized size = 0.91




method result size



risch \(\frac {x -6}{2 \left (x -10\right ) \left (5 x^{2}+\ln \relax (x )\right )}\) \(21\)
norman \(\frac {\frac {x}{2}-3}{5 x^{3}+x \ln \relax (x )-50 x^{2}-10 \ln \relax (x )}\) \(28\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x*ln(x)-10*x^4+140*x^3-601*x^2+16*x-60)/((2*x^3-40*x^2+200*x)*ln(x)^2+(20*x^5-400*x^4+2000*x^3)*ln(x)+
50*x^7-1000*x^6+5000*x^5),x,method=_RETURNVERBOSE)

[Out]

1/2*(x-6)/(x-10)/(5*x^2+ln(x))

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maxima [A]  time = 0.39, size = 24, normalized size = 1.04 \begin {gather*} \frac {x - 6}{2 \, {\left (5 \, x^{3} - 50 \, x^{2} + {\left (x - 10\right )} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(x)-10*x^4+140*x^3-601*x^2+16*x-60)/((2*x^3-40*x^2+200*x)*log(x)^2+(20*x^5-400*x^4+2000*x^3
)*log(x)+50*x^7-1000*x^6+5000*x^5),x, algorithm="maxima")

[Out]

1/2*(x - 6)/(5*x^3 - 50*x^2 + (x - 10)*log(x))

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mupad [B]  time = 6.49, size = 20, normalized size = 0.87 \begin {gather*} \frac {x-6}{2\,\left (\ln \relax (x)+5\,x^2\right )\,\left (x-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x*log(x) - 16*x + 601*x^2 - 140*x^3 + 10*x^4 + 60)/(log(x)^2*(200*x - 40*x^2 + 2*x^3) + log(x)*(2000*x
^3 - 400*x^4 + 20*x^5) + 5000*x^5 - 1000*x^6 + 50*x^7),x)

[Out]

(x - 6)/(2*(log(x) + 5*x^2)*(x - 10))

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sympy [A]  time = 0.13, size = 20, normalized size = 0.87 \begin {gather*} \frac {x - 6}{10 x^{3} - 100 x^{2} + \left (2 x - 20\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*ln(x)-10*x**4+140*x**3-601*x**2+16*x-60)/((2*x**3-40*x**2+200*x)*ln(x)**2+(20*x**5-400*x**4+20
00*x**3)*ln(x)+50*x**7-1000*x**6+5000*x**5),x)

[Out]

(x - 6)/(10*x**3 - 100*x**2 + (2*x - 20)*log(x))

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