3.33.15 $\int \frac{5+5x+5x^2+5x^3+(10x-10x^2)\log (x)}{(x+3x^2+3x^3+x^4){\log }^2(x)}dx$

Optimal. Leaf size=17 $$\frac{-5-5x^2}{(1+x{)}^2\log (x)}$$

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Rubi [F]  time = 0.50, 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 = {} $$\begin{array}{c}\int \frac{5+5x+5x^2+5x^3+(10x-10x^2)\log (x)}{(x+3x^2+3x^3+x^4){\log }^2(x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(5 + 5*x + 5*x^2 + 5*x^3 + (10*x - 10*x^2)*Log[x])/((x + 3*x^2 + 3*x^3 + x^4)*Log[x]^2),x]

[Out]

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{5(1+x+x^2+x^3-2(-1+x)x\log (x))}{x(1+x{)}^3{\log }^2(x)}dx\\ & =5\int \frac{1+x+x^2+x^3-2(-1+x)x\log (x)}{x(1+x{)}^3{\log }^2(x)}dx\\ & =5\int (\frac{1+x^2}{x(1+x{)}^2{\log }^2(x)}-\frac{2(-1+x)}{(1+x{)}^3\log (x)})dx\\ & =5\int \frac{1+x^2}{x(1+x{)}^2{\log }^2(x)}dx-10\int \frac{-1+x}{(1+x{)}^3\log (x)}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.08, size = 18, normalized size = 1.06 $$\begin{array}{c}\frac{5(-1-x^2)}{(1+x{)}^2\log (x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(5 + 5*x + 5*x^2 + 5*x^3 + (10*x - 10*x^2)*Log[x])/((x + 3*x^2 + 3*x^3 + x^4)*Log[x]^2),x]

[Out]

(5*(-1 - x^2))/((1 + x)^2*Log[x])

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fricas [A]  time = 0.52, size = 21, normalized size = 1.24 $$\begin{array}{c}-\frac{5(x^2+1)}{(x^2+2x+1)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+10*x)*log(x)+5*x^3+5*x^2+5*x+5)/(x^4+3*x^3+3*x^2+x)/log(x)^2,x, algorithm="fricas")

[Out]

-5*(x^2 + 1)/((x^2 + 2*x + 1)*log(x))

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giac [A]  time = 0.20, size = 23, normalized size = 1.35 $$\begin{array}{c}-\frac{5(x^2+1)}{x^2\log (x)+2x\log (x)+\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+10*x)*log(x)+5*x^3+5*x^2+5*x+5)/(x^4+3*x^3+3*x^2+x)/log(x)^2,x, algorithm="giac")

[Out]

-5*(x^2 + 1)/(x^2*log(x) + 2*x*log(x) + log(x))

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maple [A]  time = 0.05, size = 18, normalized size = 1.06

method	result	size
norman	$\frac{-5x^2-5}{\ln (x){(x+1)}^2}$	$18$
risch	$-\frac{5(x^2+1)}{(x^2+2x+1)\ln (x)}$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x^2+10*x)*ln(x)+5*x^3+5*x^2+5*x+5)/(x^4+3*x^3+3*x^2+x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.68, size = 21, normalized size = 1.24 $$\begin{array}{c}-\frac{5(x^2+1)}{(x^2+2x+1)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+10*x)*log(x)+5*x^3+5*x^2+5*x+5)/(x^4+3*x^3+3*x^2+x)/log(x)^2,x, algorithm="maxima")

[Out]

-5*(x^2 + 1)/((x^2 + 2*x + 1)*log(x))

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mupad [B]  time = 1.99, size = 16, normalized size = 0.94 $$\begin{array}{c}-\frac{5(x^2+1)}{\ln (x){(x+1)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + log(x)*(10*x - 10*x^2) + 5*x^2 + 5*x^3 + 5)/(log(x)^2*(x + 3*x^2 + 3*x^3 + x^4)),x)

[Out]

-(5*(x^2 + 1))/(log(x)*(x + 1)^2)

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sympy [A]  time = 0.16, size = 19, normalized size = 1.12 $$\begin{array}{c}\frac{-5x^2-5}{(x^2+2x+1)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x**2+10*x)*ln(x)+5*x**3+5*x**2+5*x+5)/(x**4+3*x**3+3*x**2+x)/ln(x)**2,x)

[Out]

(-5*x**2 - 5)/((x**2 + 2*x + 1)*log(x))

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