3.17.61 $\int \frac{(-8x+4x^2)\log (-2+x)\log (\frac{1}{2}(1+2x))+(-2x-4x^2+(2+3x-2x^2)\log (-2+x)){\log }^2(\frac{1}{2}(1+2x))}{(-2x^2-3x^3+2x^4){\log }^3(-2+x)}dx$

Optimal. Leaf size=18 $$\frac{{\log }^2(\frac{1}{2}+x)}{x{\log }^2(-2+x)}$$

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Rubi [F]  time = 2.71, 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 = {} $$\begin{array}{c}\int \frac{(-8x+4x^2)\log (-2+x)\log (\frac{1}{2}(1+2x))+(-2x-4x^2+(2+3x-2x^2)\log (-2+x)){\log }^2(\frac{1}{2}(1+2x))}{(-2x^2-3x^3+2x^4){\log }^3(-2+x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[((-8*x + 4*x^2)*Log[-2 + x]*Log[(1 + 2*x)/2] + (-2*x - 4*x^2 + (2 + 3*x - 2*x^2)*Log[-2 + x])*Log[(1 + 2*x
)/2]^2)/((-2*x^2 - 3*x^3 + 2*x^4)*Log[-2 + x]^3),x]

[Out]

4*Defer[Int][Log[1/2 + x]/(x*Log[-2 + x]^2), x] - 8*Defer[Int][Log[1/2 + x]/((1 + 2*x)*Log[-2 + x]^2), x] - De
fer[Int][Log[1/2 + x]^2/((-2 + x)*Log[-2 + x]^3), x] + Defer[Int][Log[1/2 + x]^2/(x*Log[-2 + x]^3), x] - Defer
[Int][Log[1/2 + x]^2/(x^2*Log[-2 + x]^2), x]

Rubi steps

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

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Mathematica [F]  time = 0.29, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{(-8x+4x^2)\log (-2+x)\log (\frac{1}{2}(1+2x))+(-2x-4x^2+(2+3x-2x^2)\log (-2+x)){\log }^2(\frac{1}{2}(1+2x))}{(-2x^2-3x^3+2x^4){\log }^3(-2+x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Integrate[((-8*x + 4*x^2)*Log[-2 + x]*Log[(1 + 2*x)/2] + (-2*x - 4*x^2 + (2 + 3*x - 2*x^2)*Log[-2 + x])*Log[(1
+ 2*x)/2]^2)/((-2*x^2 - 3*x^3 + 2*x^4)*Log[-2 + x]^3),x]

[Out]

Integrate[((-8*x + 4*x^2)*Log[-2 + x]*Log[(1 + 2*x)/2] + (-2*x - 4*x^2 + (2 + 3*x - 2*x^2)*Log[-2 + x])*Log[(1
+ 2*x)/2]^2)/((-2*x^2 - 3*x^3 + 2*x^4)*Log[-2 + x]^3), x]

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fricas [A]  time = 0.88, size = 16, normalized size = 0.89 $$\begin{array}{c}\frac{\log {(x+\frac{1}{2})}^2}{x\log {(x-2)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+3*x+2)*log(x-2)-4*x^2-2*x)*log(1/2+x)^2+(4*x^2-8*x)*log(x-2)*log(1/2+x))/(2*x^4-3*x^3-2*x^
2)/log(x-2)^3,x, algorithm="fricas")

[Out]

log(x + 1/2)^2/(x*log(x - 2)^2)

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giac [B]  time = 0.34, size = 52, normalized size = 2.89 $$\begin{array}{c}\frac{\log (2{)}^2}{x\log {(x-2)}^2}-\frac{2\log (2)\log (2x+1)}{x\log {(x-2)}^2}+\frac{\log {(2x+1)}^2}{x\log {(x-2)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+3*x+2)*log(x-2)-4*x^2-2*x)*log(1/2+x)^2+(4*x^2-8*x)*log(x-2)*log(1/2+x))/(2*x^4-3*x^3-2*x^
2)/log(x-2)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 17, normalized size = 0.94

method	result	size
risch	$\frac{\ln {(\frac{1}{2}+x)}^2}{x\ln {(x-2)}^2}$	$17$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2+3*x+2)*ln(x-2)-4*x^2-2*x)*ln(1/2+x)^2+(4*x^2-8*x)*ln(x-2)*ln(1/2+x))/(2*x^4-3*x^3-2*x^2)/ln(x-2)
^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.70, size = 33, normalized size = 1.83 $$\begin{array}{c}\frac{\log (2{)}^2-2\log (2)\log (2x+1)+\log {(2x+1)}^2}{x\log {(x-2)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+3*x+2)*log(x-2)-4*x^2-2*x)*log(1/2+x)^2+(4*x^2-8*x)*log(x-2)*log(1/2+x))/(2*x^4-3*x^3-2*x^
2)/log(x-2)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.35, size = 16, normalized size = 0.89 $$\begin{array}{c}\frac{{\ln (x+\frac{1}{2})}^2}{x{\ln (x-2)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 1/2)^2*(2*x - log(x - 2)*(3*x - 2*x^2 + 2) + 4*x^2) + log(x - 2)*log(x + 1/2)*(8*x - 4*x^2))/(log
(x - 2)^3*(2*x^2 + 3*x^3 - 2*x^4)),x)

[Out]

log(x + 1/2)^2/(x*log(x - 2)^2)

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sympy [A]  time = 0.39, size = 15, normalized size = 0.83 $$\begin{array}{c}\frac{\log {(x+\frac{1}{2})}^2}{x\log {(x-2)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2+3*x+2)*ln(x-2)-4*x**2-2*x)*ln(1/2+x)**2+(4*x**2-8*x)*ln(x-2)*ln(1/2+x))/(2*x**4-3*x**3-2*
x**2)/ln(x-2)**3,x)

[Out]

log(x + 1/2)**2/(x*log(x - 2)**2)

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