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

Optimal. Leaf size=18 \[ \frac {\log ^2\left (\frac {1}{2}+x\right )}{x \log ^2(-2+x)} \]

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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 {gather*} \int \frac {\left (-8 x+4 x^2\right ) \log (-2+x) \log \left (\frac {1}{2} (1+2 x)\right )+\left (-2 x-4 x^2+\left (2+3 x-2 x^2\right ) \log (-2+x)\right ) \log ^2\left (\frac {1}{2} (1+2 x)\right )}{\left (-2 x^2-3 x^3+2 x^4\right ) \log ^3(-2+x)} \, dx \end {gather*}

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

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

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 {gather*} \frac {\log \left (x + \frac {1}{2}\right )^{2}}{x \log \left (x - 2\right )^{2}} \end {gather*}

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 {gather*} \frac {\log \relax (2)^{2}}{x \log \left (x - 2\right )^{2}} - \frac {2 \, \log \relax (2) \log \left (2 \, x + 1\right )}{x \log \left (x - 2\right )^{2}} + \frac {\log \left (2 \, x + 1\right )^{2}}{x \log \left (x - 2\right )^{2}} \end {gather*}

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	\(\frac {\ln \left (\frac {1}{2}+x \right )^{2}}{x \ln \left (x -2\right )^{2}}\)	\(17\)

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 {gather*} \frac {\log \relax (2)^{2} - 2 \, \log \relax (2) \log \left (2 \, x + 1\right ) + \log \left (2 \, x + 1\right )^{2}}{x \log \left (x - 2\right )^{2}} \end {gather*}

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 {gather*} \frac {{\ln \left (x+\frac {1}{2}\right )}^2}{x\,{\ln \left (x-2\right )}^2} \end {gather*}

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 {gather*} \frac {\log {\left (x + \frac {1}{2} \right )}^{2}}{x \log {\left (x - 2 \right )}^{2}} \end {gather*}

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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