3.99.39 \(\int \frac {-16-24 x-65 x^2-36 x^3-4 x^4+(8+2 x+16 x^2+8 x^3) \log (x)-\log ^2(x)}{16 x^2+8 x^3+x^4+(-8 x^2-2 x^3) \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {1+x+\frac {x (4+2 x)^2}{-4-x+\log (x)}}{x} \]

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Rubi [F]  time = 0.78, 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 {-16-24 x-65 x^2-36 x^3-4 x^4+\left (8+2 x+16 x^2+8 x^3\right ) \log (x)-\log ^2(x)}{16 x^2+8 x^3+x^4+\left (-8 x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-16 - 24*x - 65*x^2 - 36*x^3 - 4*x^4 + (8 + 2*x + 16*x^2 + 8*x^3)*Log[x] - Log[x]^2)/(16*x^2 + 8*x^3 + x^
4 + (-8*x^2 - 2*x^3)*Log[x] + x^2*Log[x]^2),x]

[Out]

x^(-1) - 16*Defer[Int][1/(x*(4 + x - Log[x])^2), x] + 12*Defer[Int][x/(4 + x - Log[x])^2, x] + 4*Defer[Int][x^
2/(4 + x - Log[x])^2, x] - 16*Defer[Int][(4 + x - Log[x])^(-1), x] - 8*Defer[Int][x/(4 + x - Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16-24 x-65 x^2-36 x^3-4 x^4+2 \left (4+x+8 x^2+4 x^3\right ) \log (x)-\log ^2(x)}{x^2 (4+x-\log (x))^2} \, dx\\ &=\int \left (-\frac {1}{x^2}+\frac {4 (-1+x) (2+x)^2}{x (4+x-\log (x))^2}-\frac {8 (2+x)}{4+x-\log (x)}\right ) \, dx\\ &=\frac {1}{x}+4 \int \frac {(-1+x) (2+x)^2}{x (4+x-\log (x))^2} \, dx-8 \int \frac {2+x}{4+x-\log (x)} \, dx\\ &=\frac {1}{x}+4 \int \left (-\frac {4}{x (4+x-\log (x))^2}+\frac {3 x}{(4+x-\log (x))^2}+\frac {x^2}{(4+x-\log (x))^2}\right ) \, dx-8 \int \left (\frac {2}{4+x-\log (x)}+\frac {x}{4+x-\log (x)}\right ) \, dx\\ &=\frac {1}{x}+4 \int \frac {x^2}{(4+x-\log (x))^2} \, dx-8 \int \frac {x}{4+x-\log (x)} \, dx+12 \int \frac {x}{(4+x-\log (x))^2} \, dx-16 \int \frac {1}{x (4+x-\log (x))^2} \, dx-16 \int \frac {1}{4+x-\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 20, normalized size = 0.80 \begin {gather*} \frac {1}{x}+\frac {4 (2+x)^2}{-4-x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 24*x - 65*x^2 - 36*x^3 - 4*x^4 + (8 + 2*x + 16*x^2 + 8*x^3)*Log[x] - Log[x]^2)/(16*x^2 + 8*x^
3 + x^4 + (-8*x^2 - 2*x^3)*Log[x] + x^2*Log[x]^2),x]

[Out]

x^(-1) + (4*(2 + x)^2)/(-4 - x + Log[x])

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fricas [A]  time = 0.60, size = 33, normalized size = 1.32 \begin {gather*} -\frac {4 \, x^{3} + 16 \, x^{2} + 15 \, x + \log \relax (x) - 4}{x^{2} - x \log \relax (x) + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2+(8*x^3+16*x^2+2*x+8)*log(x)-4*x^4-36*x^3-65*x^2-24*x-16)/(x^2*log(x)^2+(-2*x^3-8*x^2)*log
(x)+x^4+8*x^3+16*x^2),x, algorithm="fricas")

[Out]

-(4*x^3 + 16*x^2 + 15*x + log(x) - 4)/(x^2 - x*log(x) + 4*x)

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giac [A]  time = 0.15, size = 23, normalized size = 0.92 \begin {gather*} -\frac {4 \, {\left (x^{2} + 4 \, x + 4\right )}}{x - \log \relax (x) + 4} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2+(8*x^3+16*x^2+2*x+8)*log(x)-4*x^4-36*x^3-65*x^2-24*x-16)/(x^2*log(x)^2+(-2*x^3-8*x^2)*log
(x)+x^4+8*x^3+16*x^2),x, algorithm="giac")

[Out]

-4*(x^2 + 4*x + 4)/(x - log(x) + 4) + 1/x

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maple [A]  time = 0.04, size = 24, normalized size = 0.96

method	result	size
risch	\(\frac {1}{x}-\frac {4 \left (x^{2}+4 x +4\right )}{-\ln \relax (x )+4+x}\)	\(24\)
norman	\(\frac {4-16 x^{2}-15 x -4 x^{3}-\ln \relax (x )}{x \left (-\ln \relax (x )+4+x \right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)^2+(8*x^3+16*x^2+2*x+8)*ln(x)-4*x^4-36*x^3-65*x^2-24*x-16)/(x^2*ln(x)^2+(-2*x^3-8*x^2)*ln(x)+x^4+8*
x^3+16*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 33, normalized size = 1.32 \begin {gather*} -\frac {4 \, x^{3} + 16 \, x^{2} + 15 \, x + \log \relax (x) - 4}{x^{2} - x \log \relax (x) + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^2+(8*x^3+16*x^2+2*x+8)*log(x)-4*x^4-36*x^3-65*x^2-24*x-16)/(x^2*log(x)^2+(-2*x^3-8*x^2)*log
(x)+x^4+8*x^3+16*x^2),x, algorithm="maxima")

[Out]

-(4*x^3 + 16*x^2 + 15*x + log(x) - 4)/(x^2 - x*log(x) + 4*x)

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mupad [B]  time = 6.02, size = 37, normalized size = 1.48 \begin {gather*} \frac {81\,x-\ln \relax (x)-24\,x\,\ln \relax (x)+8\,x^2-4\,x^3+4}{x\,\left (x-\ln \relax (x)+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(24*x + log(x)^2 + 65*x^2 + 36*x^3 + 4*x^4 - log(x)*(2*x + 16*x^2 + 8*x^3 + 8) + 16)/(x^2*log(x)^2 - log(
x)*(8*x^2 + 2*x^3) + 16*x^2 + 8*x^3 + x^4),x)

[Out]

(81*x - log(x) - 24*x*log(x) + 8*x^2 - 4*x^3 + 4)/(x*(x - log(x) + 4))

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sympy [A]  time = 0.13, size = 19, normalized size = 0.76 \begin {gather*} \frac {4 x^{2} + 16 x + 16}{- x + \log {\relax (x )} - 4} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)**2+(8*x**3+16*x**2+2*x+8)*ln(x)-4*x**4-36*x**3-65*x**2-24*x-16)/(x**2*ln(x)**2+(-2*x**3-8*x*
*2)*ln(x)+x**4+8*x**3+16*x**2),x)

[Out]

(4*x**2 + 16*x + 16)/(-x + log(x) - 4) + 1/x

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