∫ 256−127x+16x2+(−128x+32x2) log(2x)____________________________

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Rubi [F] time = 0.85, 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 {256-127 x+16 x^2+\left (-128 x+32 x^2\right ) \log (2 x)}{x^2+\left (256 x-128 x^2+16 x^3\right ) \log (2 x)} \, dx \end {gather*}

Int[(256 - 127*x + 16*x^2 + (-128*x + 32*x^2)*Log[2*x])/(x^2 + (256*x - 128*x^2 + 16*x^3)*Log[2*x]),x]

2*Log[4 - x] - 129*Defer[Int][(x + 16*(-4 + x)^2*Log[2*x])^(-1), x] - 8*Defer[Int][1/((-4 + x)*(x + 16*(-4 + x

)^2*Log[2*x])), x] + 256*Defer[Int][1/(x*(x + 16*(-4 + x)^2*Log[2*x])), x] + 16*Defer[Int][x/(x + 16*(-4 + x)^

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2}{-4+x}+\frac {-1024+764 x-193 x^2+16 x^3}{(-4+x) x \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )}\right ) \, dx\\ &=2 \log (4-x)+\int \frac {-1024+764 x-193 x^2+16 x^3}{(-4+x) x \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )} \, dx\\ &=2 \log (4-x)+\int \left (-\frac {129}{x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)}-\frac {8}{(-4+x) \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )}+\frac {256}{x \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )}+\frac {16 x}{x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)}\right ) \, dx\\ &=2 \log (4-x)-8 \int \frac {1}{(-4+x) \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )} \, dx+16 \int \frac {x}{x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)} \, dx-129 \int \frac {1}{x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)} \, dx+256 \int \frac {1}{x \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right )} \, dx\\ &=2 \log (4-x)-8 \int \frac {1}{(-4+x) \left (x+16 (-4+x)^2 \log (2 x)\right )} \, dx+16 \int \frac {x}{x+16 (-4+x)^2 \log (2 x)} \, dx-129 \int \frac {1}{x+16 (-4+x)^2 \log (2 x)} \, dx+256 \int \frac {1}{x \left (x+16 (-4+x)^2 \log (2 x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.21, size = 25, normalized size = 1.39 \begin {gather*} \log \left (x+256 \log (2 x)-128 x \log (2 x)+16 x^2 \log (2 x)\right ) \end {gather*}

Integrate[(256 - 127*x + 16*x^2 + (-128*x + 32*x^2)*Log[2*x])/(x^2 + (256*x - 128*x^2 + 16*x^3)*Log[2*x]),x]

Log[x + 256*Log[2*x] - 128*x*Log[2*x] + 16*x^2*Log[2*x]]

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fricas [B] time = 0.64, size = 35, normalized size = 1.94 \begin {gather*} 2 \, \log \left (x - 4\right ) + \log \left (\frac {16 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (2 \, x\right ) + x}{x^{2} - 8 \, x + 16}\right ) \end {gather*}

integrate(((32*x^2-128*x)*log(2*x)+16*x^2-127*x+256)/((16*x^3-128*x^2+256*x)*log(2*x)+x^2),x, algorithm="frica

2*log(x - 4) + log((16*(x^2 - 8*x + 16)*log(2*x) + x)/(x^2 - 8*x + 16))

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giac [A] time = 0.44, size = 25, normalized size = 1.39 \begin {gather*} \log \left (16 \, x^{2} \log \left (2 \, x\right ) - 128 \, x \log \left (2 \, x\right ) + x + 256 \, \log \left (2 \, x\right )\right ) \end {gather*}

integrate(((32*x^2-128*x)*log(2*x)+16*x^2-127*x+256)/((16*x^3-128*x^2+256*x)*log(2*x)+x^2),x, algorithm="giac"

log(16*x^2*log(2*x) - 128*x*log(2*x) + x + 256*log(2*x))

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method	result	size

norman	\(\ln \left (16 x^{2} \ln \left (2 x \right )-128 x \ln \left (2 x \right )+256 \ln \left (2 x \right )+x \right )\)	\(26\)
risch	\(2 \ln \left (x -4\right )+\ln \left (\ln \left (2 x \right )+\frac {x}{16 x^{2}-128 x +256}\right )\)	\(27\)

int(((32*x^2-128*x)*ln(2*x)+16*x^2-127*x+256)/((16*x^3-128*x^2+256*x)*ln(2*x)+x^2),x,method=_RETURNVERBOSE)

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maxima [B] time = 0.48, size = 53, normalized size = 2.94 \begin {gather*} 2 \, \log \left (x - 4\right ) + \log \left (\frac {16 \, x^{2} \log \relax (2) - x {\left (128 \, \log \relax (2) - 1\right )} + 16 \, {\left (x^{2} - 8 \, x + 16\right )} \log \relax (x) + 256 \, \log \relax (2)}{16 \, {\left (x^{2} - 8 \, x + 16\right )}}\right ) \end {gather*}

integrate(((32*x^2-128*x)*log(2*x)+16*x^2-127*x+256)/((16*x^3-128*x^2+256*x)*log(2*x)+x^2),x, algorithm="maxim

2*log(x - 4) + log(1/16*(16*x^2*log(2) - x*(128*log(2) - 1) + 16*(x^2 - 8*x + 16)*log(x) + 256*log(2))/(x^2 -

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mupad [B] time = 5.33, size = 25, normalized size = 1.39 \begin {gather*} \ln \left (x+256\,\ln \left (2\,x\right )-128\,x\,\ln \left (2\,x\right )+16\,x^2\,\ln \left (2\,x\right )\right ) \end {gather*}

int(-(127*x + log(2*x)*(128*x - 32*x^2) - 16*x^2 - 256)/(log(2*x)*(256*x - 128*x^2 + 16*x^3) + x^2),x)

log(x + 256*log(2*x) - 128*x*log(2*x) + 16*x^2*log(2*x))

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sympy [A] time = 0.40, size = 24, normalized size = 1.33 \begin {gather*} 2 \log {\left (x - 4 \right )} + \log {\left (\frac {x}{16 x^{2} - 128 x + 256} + \log {\left (2 x \right )} \right )} \end {gather*}

integrate(((32*x**2-128*x)*ln(2*x)+16*x**2-127*x+256)/((16*x**3-128*x**2+256*x)*ln(2*x)+x**2),x)

2*log(x - 4) + log(x/(16*x**2 - 128*x + 256) + log(2*x))

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3.82.29 \(\int \frac {256-127 x+16 x^2+(-128 x+32 x^2) \log (2 x)}{x^2+(256 x-128 x^2+16 x^3) \log (2 x)} \, dx\)