3.31.36 \(\int \frac {-15 x+(-300+100 x-15 x^2+5 x^3) \log ^2(\frac {3-x}{x})}{(-3 x^2+x^3) \log ^2(\frac {3-x}{x})} \, dx\)

Optimal. Leaf size=32 \[ 6 x-\frac {-x+(10+x)^2}{x}+\frac {5}{\log \left (\frac {3-x}{x}\right )} \]

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Rubi [F]  time = 0.53, 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 {-15 x+\left (-300+100 x-15 x^2+5 x^3\right ) \log ^2\left (\frac {3-x}{x}\right )}{\left (-3 x^2+x^3\right ) \log ^2\left (\frac {3-x}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-15*x + (-300 + 100*x - 15*x^2 + 5*x^3)*Log[(3 - x)/x]^2)/((-3*x^2 + x^3)*Log[(3 - x)/x]^2),x]

[Out]

-100/x + 5*x - 5*Defer[Int][1/((-3 + x)*Log[-1 + 3/x]^2), x] + 5*Defer[Int][1/(x*Log[-1 + 3/x]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-15 x+\left (-300+100 x-15 x^2+5 x^3\right ) \log ^2\left (\frac {3-x}{x}\right )}{(-3+x) x^2 \log ^2\left (\frac {3-x}{x}\right )} \, dx\\ &=\int \frac {5 \left (60-20 x+3 x^2-x^3+\frac {3 x}{\log ^2\left (-1+\frac {3}{x}\right )}\right )}{(3-x) x^2} \, dx\\ &=5 \int \frac {60-20 x+3 x^2-x^3+\frac {3 x}{\log ^2\left (-1+\frac {3}{x}\right )}}{(3-x) x^2} \, dx\\ &=5 \int \left (\frac {20+x^2}{x^2}-\frac {3}{(-3+x) x \log ^2\left (-1+\frac {3}{x}\right )}\right ) \, dx\\ &=5 \int \frac {20+x^2}{x^2} \, dx-15 \int \frac {1}{(-3+x) x \log ^2\left (-1+\frac {3}{x}\right )} \, dx\\ &=5 \int \left (1+\frac {20}{x^2}\right ) \, dx-15 \int \left (\frac {1}{3 (-3+x) \log ^2\left (-1+\frac {3}{x}\right )}-\frac {1}{3 x \log ^2\left (-1+\frac {3}{x}\right )}\right ) \, dx\\ &=-\frac {100}{x}+5 x-5 \int \frac {1}{(-3+x) \log ^2\left (-1+\frac {3}{x}\right )} \, dx+5 \int \frac {1}{x \log ^2\left (-1+\frac {3}{x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.02, size = 19, normalized size = 0.59 \begin {gather*} 5 \left (-\frac {20}{x}+x+\frac {1}{\log \left (-1+\frac {3}{x}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x + (-300 + 100*x - 15*x^2 + 5*x^3)*Log[(3 - x)/x]^2)/((-3*x^2 + x^3)*Log[(3 - x)/x]^2),x]

[Out]

5*(-20/x + x + Log[-1 + 3/x]^(-1))

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fricas [A]  time = 0.60, size = 33, normalized size = 1.03 \begin {gather*} \frac {5 \, {\left ({\left (x^{2} - 20\right )} \log \left (-\frac {x - 3}{x}\right ) + x\right )}}{x \log \left (-\frac {x - 3}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^3-15*x^2+100*x-300)*log((3-x)/x)^2-15*x)/(x^3-3*x^2)/log((3-x)/x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 35, normalized size = 1.09 \begin {gather*} \frac {100 \, {\left (x - 3\right )}}{3 \, x} - \frac {15}{\frac {x - 3}{x} - 1} + \frac {5}{\log \left (-\frac {x - 3}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^3-15*x^2+100*x-300)*log((3-x)/x)^2-15*x)/(x^3-3*x^2)/log((3-x)/x)^2,x, algorithm="giac")

[Out]

100/3*(x - 3)/x - 15/((x - 3)/x - 1) + 5/log(-(x - 3)/x)

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maple [A]  time = 0.46, size = 23, normalized size = 0.72




method result size



derivativedivides \(\frac {5}{\ln \left (\frac {3}{x}-1\right )}-\frac {100}{x}+\frac {100}{3}+5 x\) \(23\)
default \(\frac {5}{\ln \left (\frac {3}{x}-1\right )}-\frac {100}{x}+\frac {100}{3}+5 x\) \(23\)
risch \(\frac {5 x^{2}-100}{x}+\frac {5}{\ln \left (\frac {3-x}{x}\right )}\) \(26\)
norman \(\frac {5 x +5 x^{2} \ln \left (\frac {3-x}{x}\right )-100 \ln \left (\frac {3-x}{x}\right )}{x \ln \left (\frac {3-x}{x}\right )}\) \(48\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^3-15*x^2+100*x-300)*ln((3-x)/x)^2-15*x)/(x^3-3*x^2)/ln((3-x)/x)^2,x,method=_RETURNVERBOSE)

[Out]

5/ln(3/x-1)-100/x+100/3+5*x

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maxima [B]  time = 0.57, size = 114, normalized size = 3.56 \begin {gather*} \frac {200}{3} \, {\left (\log \relax (x) - \log \left (-x + 3\right )\right )} \log \left (-\log \relax (x) + \log \left (-x + 3\right )\right ) + \frac {200}{3} \, \log \left (\frac {3}{x} - 1\right ) \log \left (-\log \relax (x) + \log \left (-x + 3\right )\right ) + 5 \, x + \frac {100 \, \log \left (\frac {3}{x} - 1\right )^{2}}{3 \, {\left (\log \relax (x) - \log \left (-x + 3\right )\right )}} - \frac {100}{x} - \frac {5}{\log \relax (x) - \log \left (-x + 3\right )} - \frac {100}{3} \, \log \left (x - 3\right ) - \frac {100}{3} \, \log \relax (x) + \frac {200}{3} \, \log \left (-x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^3-15*x^2+100*x-300)*log((3-x)/x)^2-15*x)/(x^3-3*x^2)/log((3-x)/x)^2,x, algorithm="maxima")

[Out]

200/3*(log(x) - log(-x + 3))*log(-log(x) + log(-x + 3)) + 200/3*log(3/x - 1)*log(-log(x) + log(-x + 3)) + 5*x
+ 100/3*log(3/x - 1)^2/(log(x) - log(-x + 3)) - 100/x - 5/(log(x) - log(-x + 3)) - 100/3*log(x - 3) - 100/3*lo
g(x) + 200/3*log(-x + 3)

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mupad [B]  time = 1.96, size = 22, normalized size = 0.69 \begin {gather*} 5\,x-\frac {100}{x}+\frac {5}{\ln \left (-\frac {x-3}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x - log(-(x - 3)/x)^2*(100*x - 15*x^2 + 5*x^3 - 300))/(log(-(x - 3)/x)^2*(3*x^2 - x^3)),x)

[Out]

5*x - 100/x + 5/log(-(x - 3)/x)

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sympy [A]  time = 0.13, size = 14, normalized size = 0.44 \begin {gather*} 5 x + \frac {5}{\log {\left (\frac {3 - x}{x} \right )}} - \frac {100}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**3-15*x**2+100*x-300)*ln((3-x)/x)**2-15*x)/(x**3-3*x**2)/ln((3-x)/x)**2,x)

[Out]

5*x + 5/log((3 - x)/x) - 100/x

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