3.24.74 \(\int \frac {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+(-128-112 x-40 x^2-8 x^3) \log (2)+(32+16 x) \log ^2(2)+(32 x+28 x^2+10 x^3+2 x^4+(-16 x-8 x^2) \log (2)) \log (6+3 x)+(2 x^2+x^3) \log ^2(6+3 x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {x}{2-\log (2)+\frac {1}{4} x (3+x+\log (3 (2+x)))} \]

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Rubi [F]  time = 0.79, 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 {64+32 x-12 x^2-4 x^3+(-32-16 x) \log (2)}{128+160 x+98 x^2+37 x^3+8 x^4+x^5+\left (-128-112 x-40 x^2-8 x^3\right ) \log (2)+(32+16 x) \log ^2(2)+\left (32 x+28 x^2+10 x^3+2 x^4+\left (-16 x-8 x^2\right ) \log (2)\right ) \log (6+3 x)+\left (2 x^2+x^3\right ) \log ^2(6+3 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(64 + 32*x - 12*x^2 - 4*x^3 + (-32 - 16*x)*Log[2])/(128 + 160*x + 98*x^2 + 37*x^3 + 8*x^4 + x^5 + (-128 -
112*x - 40*x^2 - 8*x^3)*Log[2] + (32 + 16*x)*Log[2]^2 + (32*x + 28*x^2 + 10*x^3 + 2*x^4 + (-16*x - 8*x^2)*Log[
2])*Log[6 + 3*x] + (2*x^2 + x^3)*Log[6 + 3*x]^2),x]

[Out]

8*(5 - Log[4])*Defer[Int][(3*x + x^2 + 8*(1 - Log[2]/2) + x*Log[3*(2 + x)])^(-2), x] + 16*Defer[Int][1/((-2 -
x)*(3*x + x^2 + 8*(1 - Log[2]/2) + x*Log[3*(2 + x)])^2), x] - 4*Defer[Int][x/(3*x + x^2 + 8*(1 - Log[2]/2) + x
*Log[3*(2 + x)])^2, x] - 4*Defer[Int][x^2/(3*x + x^2 + 8*(1 - Log[2]/2) + x*Log[3*(2 + x)])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (-3 x^2-x^3+8 (2-\log (2))+4 x (2-\log (2))\right )}{(2+x) \left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2} \, dx\\ &=4 \int \frac {-3 x^2-x^3+8 (2-\log (2))+4 x (2-\log (2))}{(2+x) \left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2} \, dx\\ &=4 \int \left (\frac {4}{(-2-x) \left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2}-\frac {x}{\left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2}-\frac {x^2}{\left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2}+\frac {2 (5-\log (4))}{\left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {x}{\left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2} \, dx\right )-4 \int \frac {x^2}{\left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2} \, dx+16 \int \frac {1}{(-2-x) \left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2} \, dx+(8 (5-\log (4))) \int \frac {1}{\left (3 x+x^2+8 \left (1-\frac {\log (2)}{2}\right )+x \log (3 (2+x))\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.22, size = 25, normalized size = 1.04 \begin {gather*} \frac {4 x}{8+3 x+x^2-4 \log (2)+x \log (3 (2+x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64 + 32*x - 12*x^2 - 4*x^3 + (-32 - 16*x)*Log[2])/(128 + 160*x + 98*x^2 + 37*x^3 + 8*x^4 + x^5 + (-
128 - 112*x - 40*x^2 - 8*x^3)*Log[2] + (32 + 16*x)*Log[2]^2 + (32*x + 28*x^2 + 10*x^3 + 2*x^4 + (-16*x - 8*x^2
)*Log[2])*Log[6 + 3*x] + (2*x^2 + x^3)*Log[6 + 3*x]^2),x]

[Out]

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

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fricas [A]  time = 0.66, size = 25, normalized size = 1.04 \begin {gather*} \frac {4 \, x}{x^{2} + x \log \left (3 \, x + 6\right ) + 3 \, x - 4 \, \log \relax (2) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x-32)*log(2)-4*x^3-12*x^2+32*x+64)/((x^3+2*x^2)*log(6+3*x)^2+((-8*x^2-16*x)*log(2)+2*x^4+10*x^
3+28*x^2+32*x)*log(6+3*x)+(16*x+32)*log(2)^2+(-8*x^3-40*x^2-112*x-128)*log(2)+x^5+8*x^4+37*x^3+98*x^2+160*x+12
8),x, algorithm="fricas")

[Out]

4*x/(x^2 + x*log(3*x + 6) + 3*x - 4*log(2) + 8)

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giac [A]  time = 0.47, size = 25, normalized size = 1.04 \begin {gather*} \frac {4 \, x}{x^{2} + x \log \left (3 \, x + 6\right ) + 3 \, x - 4 \, \log \relax (2) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x-32)*log(2)-4*x^3-12*x^2+32*x+64)/((x^3+2*x^2)*log(6+3*x)^2+((-8*x^2-16*x)*log(2)+2*x^4+10*x^
3+28*x^2+32*x)*log(6+3*x)+(16*x+32)*log(2)^2+(-8*x^3-40*x^2-112*x-128)*log(2)+x^5+8*x^4+37*x^3+98*x^2+160*x+12
8),x, algorithm="giac")

[Out]

4*x/(x^2 + x*log(3*x + 6) + 3*x - 4*log(2) + 8)

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maple [A]  time = 0.31, size = 29, normalized size = 1.21

method	result	size
norman	\(-\frac {4 x}{-\ln \left (6+3 x \right ) x -x^{2}+4 \ln \relax (2)-3 x -8}\)	\(29\)
risch	\(-\frac {4 x}{-\ln \left (6+3 x \right ) x -x^{2}+4 \ln \relax (2)-3 x -8}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x-32)*ln(2)-4*x^3-12*x^2+32*x+64)/((x^3+2*x^2)*ln(6+3*x)^2+((-8*x^2-16*x)*ln(2)+2*x^4+10*x^3+28*x^2+
32*x)*ln(6+3*x)+(16*x+32)*ln(2)^2+(-8*x^3-40*x^2-112*x-128)*ln(2)+x^5+8*x^4+37*x^3+98*x^2+160*x+128),x,method=
_RETURNVERBOSE)

[Out]

-4*x/(-ln(6+3*x)*x-x^2+4*ln(2)-3*x-8)

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maxima [A]  time = 0.52, size = 26, normalized size = 1.08 \begin {gather*} \frac {4 \, x}{x^{2} + x {\left (\log \relax (3) + 3\right )} + x \log \left (x + 2\right ) - 4 \, \log \relax (2) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x-32)*log(2)-4*x^3-12*x^2+32*x+64)/((x^3+2*x^2)*log(6+3*x)^2+((-8*x^2-16*x)*log(2)+2*x^4+10*x^
3+28*x^2+32*x)*log(6+3*x)+(16*x+32)*log(2)^2+(-8*x^3-40*x^2-112*x-128)*log(2)+x^5+8*x^4+37*x^3+98*x^2+160*x+12
8),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\ln \relax (2)\,\left (16\,x+32\right )-32\,x+12\,x^2+4\,x^3-64}{160\,x+\ln \left (3\,x+6\right )\,\left (32\,x-\ln \relax (2)\,\left (8\,x^2+16\,x\right )+28\,x^2+10\,x^3+2\,x^4\right )+{\ln \relax (2)}^2\,\left (16\,x+32\right )-\ln \relax (2)\,\left (8\,x^3+40\,x^2+112\,x+128\right )+{\ln \left (3\,x+6\right )}^2\,\left (x^3+2\,x^2\right )+98\,x^2+37\,x^3+8\,x^4+x^5+128} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)*(16*x + 32) - 32*x + 12*x^2 + 4*x^3 - 64)/(160*x + log(3*x + 6)*(32*x - log(2)*(16*x + 8*x^2) + 2
8*x^2 + 10*x^3 + 2*x^4) + log(2)^2*(16*x + 32) - log(2)*(112*x + 40*x^2 + 8*x^3 + 128) + log(3*x + 6)^2*(2*x^2
+ x^3) + 98*x^2 + 37*x^3 + 8*x^4 + x^5 + 128),x)

[Out]

int(-(log(2)*(16*x + 32) - 32*x + 12*x^2 + 4*x^3 - 64)/(160*x + log(3*x + 6)*(32*x - log(2)*(16*x + 8*x^2) + 2
8*x^2 + 10*x^3 + 2*x^4) + log(2)^2*(16*x + 32) - log(2)*(112*x + 40*x^2 + 8*x^3 + 128) + log(3*x + 6)^2*(2*x^2
+ x^3) + 98*x^2 + 37*x^3 + 8*x^4 + x^5 + 128), x)

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sympy [A]  time = 0.22, size = 24, normalized size = 1.00 \begin {gather*} \frac {4 x}{x^{2} + x \log {\left (3 x + 6 \right )} + 3 x - 4 \log {\relax (2 )} + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x-32)*ln(2)-4*x**3-12*x**2+32*x+64)/((x**3+2*x**2)*ln(6+3*x)**2+((-8*x**2-16*x)*ln(2)+2*x**4+1
0*x**3+28*x**2+32*x)*ln(6+3*x)+(16*x+32)*ln(2)**2+(-8*x**3-40*x**2-112*x-128)*ln(2)+x**5+8*x**4+37*x**3+98*x**
2+160*x+128),x)

[Out]

4*x/(x**2 + x*log(3*x + 6) + 3*x - 4*log(2) + 8)

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