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

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

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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 = {} $$\begin{array}{c}\int \frac{16+8x+2x^4-2x^5+(16+16x^3-16x^4)\log (x)+(32x^2-32x^3){\log }^2(x)}{x^4+8x^3\log (x)+16x^2{\log }^2(x)}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{16+8x+2x^4-2x^5+(16+16x^3-16x^4)\log (x)+(32x^2-32x^3){\log }^2(x)}{x^2(x+4\log (x){)}^2}dx\\ & =\int (-2(-1+x)+\frac{4(4+x)}{x^2(x+4\log (x){)}^2}+\frac{4}{x^2(x+4\log (x))})dx\\ & =-(1-x{)}^2+4\int \frac{4+x}{x^2(x+4\log (x){)}^2}dx+4\int \frac{1}{x^2(x+4\log (x))}dx\\ & =-(1-x{)}^2+4\int \frac{1}{x^2(x+4\log (x))}dx+4\int (\frac{4}{x^2(x+4\log (x){)}^2}+\frac{1}{x(x+4\log (x){)}^2})dx\\ & =-(1-x{)}^2+4\int \frac{1}{x(x+4\log (x){)}^2}dx+4\int \frac{1}{x^2(x+4\log (x))}dx+16\int \frac{1}{x^2(x+4\log (x){)}^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.15, size = 26, normalized size = 0.96 $$\begin{array}{c}-2(-x+\frac{x^2}{2}+\frac{2}{x(x+4\log (x))})\end{array}$$

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 $$\begin{array}{c}-\frac{x^4-2x^3+4(x^3-2x^2)\log (x)+4}{x^2+4x\log (x)}\end{array}$$

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 $$\begin{array}{c}-x^2+2x-\frac{4}{x^2+4x\log (x)}\end{array}$$

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	$-x^2+2x-\frac{4}{x(4\ln (x)+x)}$	$23$
norman	$\frac{-4+8x^2\ln (x)+2x^3-x^4-4x^3\ln (x)}{x(4\ln (x)+x)}$	$39$

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 $$\begin{array}{c}-\frac{x^4-2x^3+4(x^3-2x^2)\log (x)+4}{x^2+4x\log (x)}\end{array}$$

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 $$\begin{array}{c}2x-\frac{4}{x(x+4\ln (x))}-x^2\end{array}$$

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 $$\begin{array}{c}-x^2+2x-\frac{4}{x^2+4x\log (x)}\end{array}$$

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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