3.14.32 \(\int \frac {16+8 x+2 x^4-2 x^5+(16+16 x^3-16 x^4) \log (x)+(32 x^2-32 x^3) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx\)

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

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Rubi [F]  time = 0.65, 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+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16 + 8*x + 2*x^4 - 2*x^5 + (16 + 16*x^3 - 16*x^4)*Log[x] + (32*x^2 - 32*x^3)*Log[x]^2)/(x^4 + 8*x^3*Log[x
] + 16*x^2*Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^2 (x+4 \log (x))^2} \, dx\\ &=\int \left (-2 (-1+x)+\frac {4 (4+x)}{x^2 (x+4 \log (x))^2}+\frac {4}{x^2 (x+4 \log (x))}\right ) \, dx\\ &=-(1-x)^2+4 \int \frac {4+x}{x^2 (x+4 \log (x))^2} \, dx+4 \int \frac {1}{x^2 (x+4 \log (x))} \, dx\\ &=-(1-x)^2+4 \int \frac {1}{x^2 (x+4 \log (x))} \, dx+4 \int \left (\frac {4}{x^2 (x+4 \log (x))^2}+\frac {1}{x (x+4 \log (x))^2}\right ) \, dx\\ &=-(1-x)^2+4 \int \frac {1}{x (x+4 \log (x))^2} \, dx+4 \int \frac {1}{x^2 (x+4 \log (x))} \, dx+16 \int \frac {1}{x^2 (x+4 \log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 26, normalized size = 0.96 \begin {gather*} -2 \left (-x+\frac {x^2}{2}+\frac {2}{x (x+4 \log (x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + 8*x + 2*x^4 - 2*x^5 + (16 + 16*x^3 - 16*x^4)*Log[x] + (32*x^2 - 32*x^3)*Log[x]^2)/(x^4 + 8*x^3
*Log[x] + 16*x^2*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.57, size = 36, normalized size = 1.33 \begin {gather*} -\frac {x^{4} - 2 \, x^{3} + 4 \, {\left (x^{3} - 2 \, x^{2}\right )} \log \relax (x) + 4}{x^{2} + 4 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3+32*x^2)*log(x)^2+(-16*x^4+16*x^3+16)*log(x)-2*x^5+2*x^4+8*x+16)/(16*x^2*log(x)^2+8*x^3*log
(x)+x^4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 22, normalized size = 0.81 \begin {gather*} -x^{2} + 2 \, x - \frac {4}{x^{2} + 4 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3+32*x^2)*log(x)^2+(-16*x^4+16*x^3+16)*log(x)-2*x^5+2*x^4+8*x+16)/(16*x^2*log(x)^2+8*x^3*log
(x)+x^4),x, algorithm="giac")

[Out]

-x^2 + 2*x - 4/(x^2 + 4*x*log(x))

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maple [A]  time = 0.03, size = 23, normalized size = 0.85

method	result	size
risch	\(-x^{2}+2 x -\frac {4}{x \left (4 \ln \relax (x )+x \right )}\)	\(23\)
norman	\(\frac {-4+8 x^{2} \ln \relax (x )+2 x^{3}-x^{4}-4 x^{3} \ln \relax (x )}{x \left (4 \ln \relax (x )+x \right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x^3+32*x^2)*ln(x)^2+(-16*x^4+16*x^3+16)*ln(x)-2*x^5+2*x^4+8*x+16)/(16*x^2*ln(x)^2+8*x^3*ln(x)+x^4),x
,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.70, size = 36, normalized size = 1.33 \begin {gather*} -\frac {x^{4} - 2 \, x^{3} + 4 \, {\left (x^{3} - 2 \, x^{2}\right )} \log \relax (x) + 4}{x^{2} + 4 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^3+32*x^2)*log(x)^2+(-16*x^4+16*x^3+16)*log(x)-2*x^5+2*x^4+8*x+16)/(16*x^2*log(x)^2+8*x^3*log
(x)+x^4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.04, size = 22, normalized size = 0.81 \begin {gather*} 2\,x-\frac {4}{x\,\left (x+4\,\ln \relax (x)\right )}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + log(x)*(16*x^3 - 16*x^4 + 16) + log(x)^2*(32*x^2 - 32*x^3) + 2*x^4 - 2*x^5 + 16)/(8*x^3*log(x) + 16
*x^2*log(x)^2 + x^4),x)

[Out]

2*x - 4/(x*(x + 4*log(x))) - x^2

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sympy [A]  time = 0.12, size = 17, normalized size = 0.63 \begin {gather*} - x^{2} + 2 x - \frac {4}{x^{2} + 4 x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x**3+32*x**2)*ln(x)**2+(-16*x**4+16*x**3+16)*ln(x)-2*x**5+2*x**4+8*x+16)/(16*x**2*ln(x)**2+8*x
**3*ln(x)+x**4),x)

[Out]

-x**2 + 2*x - 4/(x**2 + 4*x*log(x))

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