3.68.34 \(\int \frac {-16 x^2+16 x^3+(12 x^2+4 x^3) \log (x \log (4))+8 x^2 \log ^2(x \log (4))+(-2+4 x+x^2-4 x^3+2 x^4) \log ^3(x \log (4))}{(-2 x+6 x^2-6 x^3+2 x^4) \log ^3(x \log (4))} \, dx\)

Optimal. Leaf size=27 \[ x+\log (x)-\frac {\left (x+\frac {4 x}{\log (x \log (4))}\right )^2}{4 (-1+x)^2} \]

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Rubi [F]  time = 0.94, 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 {-16 x^2+16 x^3+\left (12 x^2+4 x^3\right ) \log (x \log (4))+8 x^2 \log ^2(x \log (4))+\left (-2+4 x+x^2-4 x^3+2 x^4\right ) \log ^3(x \log (4))}{\left (-2 x+6 x^2-6 x^3+2 x^4\right ) \log ^3(x \log (4))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-16*x^2 + 16*x^3 + (12*x^2 + 4*x^3)*Log[x*Log[4]] + 8*x^2*Log[x*Log[4]]^2 + (-2 + 4*x + x^2 - 4*x^3 + 2*x
^4)*Log[x*Log[4]]^3)/((-2*x + 6*x^2 - 6*x^3 + 2*x^4)*Log[x*Log[4]]^3),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16 x^2-16 x^3-\left (12 x^2+4 x^3\right ) \log (x \log (4))-8 x^2 \log ^2(x \log (4))-\left (-2+4 x+x^2-4 x^3+2 x^4\right ) \log ^3(x \log (4))}{x \left (2-6 x+6 x^2-2 x^3\right ) \log ^3(x \log (4))} \, dx\\ &=\int \left (\frac {-2+4 x+x^2-4 x^3+2 x^4}{2 (-1+x)^3 x}+\frac {8 x}{(-1+x)^2 \log ^3(x \log (4))}+\frac {2 x (3+x)}{(-1+x)^3 \log ^2(x \log (4))}+\frac {4 x}{(-1+x)^3 \log (x \log (4))}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-2+4 x+x^2-4 x^3+2 x^4}{(-1+x)^3 x} \, dx+2 \int \frac {x (3+x)}{(-1+x)^3 \log ^2(x \log (4))} \, dx+4 \int \frac {x}{(-1+x)^3 \log (x \log (4))} \, dx+8 \int \frac {x}{(-1+x)^2 \log ^3(x \log (4))} \, dx\\ &=\frac {1}{2} \int \left (2+\frac {1}{(-1+x)^3}+\frac {1}{(-1+x)^2}+\frac {2}{x}\right ) \, dx+2 \int \frac {x (3+x)}{(-1+x)^3 \log ^2(x \log (4))} \, dx+4 \int \frac {x}{(-1+x)^3 \log (x \log (4))} \, dx+8 \int \frac {x}{(-1+x)^2 \log ^3(x \log (4))} \, dx\\ &=-\frac {1}{4 (1-x)^2}+\frac {1}{2 (1-x)}+x+\log (x)+2 \int \frac {x (3+x)}{(-1+x)^3 \log ^2(x \log (4))} \, dx+4 \int \frac {x}{(-1+x)^3 \log (x \log (4))} \, dx+8 \int \frac {x}{(-1+x)^2 \log ^3(x \log (4))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.29, size = 62, normalized size = 2.30 \begin {gather*} \frac {1}{2} \left (-\frac {1}{2 (-1+x)^2}-\frac {1}{-1+x}+2 x+2 \log (x)-\frac {8 x^2}{(-1+x)^2 \log ^2(x \log (4))}-\frac {4 x^2}{(-1+x)^2 \log (x \log (4))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^2 + 16*x^3 + (12*x^2 + 4*x^3)*Log[x*Log[4]] + 8*x^2*Log[x*Log[4]]^2 + (-2 + 4*x + x^2 - 4*x^3
+ 2*x^4)*Log[x*Log[4]]^3)/((-2*x + 6*x^2 - 6*x^3 + 2*x^4)*Log[x*Log[4]]^3),x]

[Out]

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

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fricas [B]  time = 0.60, size = 79, normalized size = 2.93 \begin {gather*} \frac {4 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (2 \, x \log \relax (2)\right )^{3} - 8 \, x^{2} \log \left (2 \, x \log \relax (2)\right ) + {\left (4 \, x^{3} - 8 \, x^{2} + 2 \, x + 1\right )} \log \left (2 \, x \log \relax (2)\right )^{2} - 16 \, x^{2}}{4 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (2 \, x \log \relax (2)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-4*x^3+x^2+4*x-2)*log(2*x*log(2))^3+8*x^2*log(2*x*log(2))^2+(4*x^3+12*x^2)*log(2*x*log(2))+16
*x^3-16*x^2)/(2*x^4-6*x^3+6*x^2-2*x)/log(2*x*log(2))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.27, size = 125, normalized size = 4.63 \begin {gather*} x - \frac {2 \, {\left (x^{2} \log \relax (2) + x^{2} \log \left (x \log \relax (2)\right ) + 2 \, x^{2}\right )}}{x^{2} \log \relax (2)^{2} + 2 \, x^{2} \log \relax (2) \log \left (x \log \relax (2)\right ) + x^{2} \log \left (x \log \relax (2)\right )^{2} - 2 \, x \log \relax (2)^{2} - 4 \, x \log \relax (2) \log \left (x \log \relax (2)\right ) - 2 \, x \log \left (x \log \relax (2)\right )^{2} + \log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (x \log \relax (2)\right ) + \log \left (x \log \relax (2)\right )^{2}} - \frac {2 \, x - 1}{4 \, {\left (x^{2} - 2 \, x + 1\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-4*x^3+x^2+4*x-2)*log(2*x*log(2))^3+8*x^2*log(2*x*log(2))^2+(4*x^3+12*x^2)*log(2*x*log(2))+16
*x^3-16*x^2)/(2*x^4-6*x^3+6*x^2-2*x)/log(2*x*log(2))^3,x, algorithm="giac")

[Out]

x - 2*(x^2*log(2) + x^2*log(x*log(2)) + 2*x^2)/(x^2*log(2)^2 + 2*x^2*log(2)*log(x*log(2)) + x^2*log(x*log(2))^
2 - 2*x*log(2)^2 - 4*x*log(2)*log(x*log(2)) - 2*x*log(x*log(2))^2 + log(2)^2 + 2*log(2)*log(x*log(2)) + log(x*
log(2))^2) - 1/4*(2*x - 1)/(x^2 - 2*x + 1) + log(x)

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maple [B]  time = 0.22, size = 70, normalized size = 2.59

method	result	size
norman	\(\frac {\frac {\ln \left (2 x \ln \relax (2)\right )^{2}}{2}+\ln \left (2 x \ln \relax (2)\right )^{2} x^{3}-\frac {7 x^{2} \ln \left (2 x \ln \relax (2)\right )^{2}}{4}-4 x^{2}-2 x^{2} \ln \left (2 x \ln \relax (2)\right )}{\left (x -1\right )^{2} \ln \left (2 x \ln \relax (2)\right )^{2}}+\ln \relax (x )\)	\(70\)
risch	\(\frac {4 x^{2} \ln \relax (x )+4 x^{3}-8 x \ln \relax (x )-8 x^{2}+4 \ln \relax (x )+2 x +1}{4 x^{2}-8 x +4}-\frac {2 \left (\ln \left (2 x \ln \relax (2)\right )+2\right ) x^{2}}{\left (x^{2}-2 x +1\right ) \ln \left (2 x \ln \relax (2)\right )^{2}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^4-4*x^3+x^2+4*x-2)*ln(2*x*ln(2))^3+8*x^2*ln(2*x*ln(2))^2+(4*x^3+12*x^2)*ln(2*x*ln(2))+16*x^3-16*x^2)
/(2*x^4-6*x^3+6*x^2-2*x)/ln(2*x*ln(2))^3,x,method=_RETURNVERBOSE)

[Out]

(1/2*ln(2*x*ln(2))^2+ln(2*x*ln(2))^2*x^3-7/4*x^2*ln(2*x*ln(2))^2-4*x^2-2*x^2*ln(2*x*ln(2)))/(x-1)^2/ln(2*x*ln(
2))^2+ln(x)

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maxima [B]  time = 0.52, size = 261, normalized size = 9.67 \begin {gather*} \frac {4 \, {\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )} x^{3} - 8 \, {\left ({\left (2 \, \log \left (\log \relax (2)\right ) + 1\right )} \log \relax (2) + \log \relax (2)^{2} + \log \left (\log \relax (2)\right )^{2} + \log \left (\log \relax (2)\right ) + 2\right )} x^{2} + {\left (4 \, x^{3} - 8 \, x^{2} + 2 \, x + 1\right )} \log \relax (x)^{2} + 2 \, {\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )} x + \log \relax (2)^{2} + 2 \, {\left (4 \, x^{3} {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} - 4 \, x^{2} {\left (2 \, \log \relax (2) + 2 \, \log \left (\log \relax (2)\right ) + 1\right )} + 2 \, x {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} + \log \relax (2) + \log \left (\log \relax (2)\right )\right )} \log \relax (x) + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}}{4 \, {\left ({\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )} x^{2} + {\left (x^{2} - 2 \, x + 1\right )} \log \relax (x)^{2} - 2 \, {\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )} x + \log \relax (2)^{2} + 2 \, {\left (x^{2} {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} - 2 \, x {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} + \log \relax (2) + \log \left (\log \relax (2)\right )\right )} \log \relax (x) + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4-4*x^3+x^2+4*x-2)*log(2*x*log(2))^3+8*x^2*log(2*x*log(2))^2+(4*x^3+12*x^2)*log(2*x*log(2))+16
*x^3-16*x^2)/(2*x^4-6*x^3+6*x^2-2*x)/log(2*x*log(2))^3,x, algorithm="maxima")

[Out]

1/4*(4*(log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2)*x^3 - 8*((2*log(log(2)) + 1)*log(2) + log(2)^2 + log(
log(2))^2 + log(log(2)) + 2)*x^2 + (4*x^3 - 8*x^2 + 2*x + 1)*log(x)^2 + 2*(log(2)^2 + 2*log(2)*log(log(2)) + l
og(log(2))^2)*x + log(2)^2 + 2*(4*x^3*(log(2) + log(log(2))) - 4*x^2*(2*log(2) + 2*log(log(2)) + 1) + 2*x*(log
(2) + log(log(2))) + log(2) + log(log(2)))*log(x) + 2*log(2)*log(log(2)) + log(log(2))^2)/((log(2)^2 + 2*log(2
)*log(log(2)) + log(log(2))^2)*x^2 + (x^2 - 2*x + 1)*log(x)^2 - 2*(log(2)^2 + 2*log(2)*log(log(2)) + log(log(2
))^2)*x + log(2)^2 + 2*(x^2*(log(2) + log(log(2))) - 2*x*(log(2) + log(log(2))) + log(2) + log(log(2)))*log(x)
+ 2*log(2)*log(log(2)) + log(log(2))^2) + log(x)

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mupad [B]  time = 4.22, size = 203, normalized size = 7.52 \begin {gather*} x+\ln \relax (x)+\frac {\frac {x\,\left (5\,x-x^2\right )}{{\left (x-1\right )}^3}+\frac {2\,x^2\,{\ln \left (2\,x\,\ln \relax (2)\right )}^2\,\left (x+2\right )}{{\left (x-1\right )}^4}+\frac {2\,x\,\ln \left (2\,x\,\ln \relax (2)\right )\,\left (x^2+5\,x\right )}{{\left (x-1\right )}^4}}{\ln \left (2\,x\,\ln \relax (2)\right )}-\frac {\frac {x^3}{2}+\frac {43\,x^2}{4}+x-\frac {1}{4}}{x^4-4\,x^3+6\,x^2-4\,x+1}-\frac {\frac {4\,x^2}{{\left (x-1\right )}^2}+\frac {2\,x^2\,{\ln \left (2\,x\,\ln \relax (2)\right )}^2}{{\left (x-1\right )}^3}+\frac {x^2\,\ln \left (2\,x\,\ln \relax (2)\right )\,\left (x+3\right )}{{\left (x-1\right )}^3}}{{\ln \left (2\,x\,\ln \relax (2)\right )}^2}-\frac {\ln \left (2\,x\,\ln \relax (2)\right )\,\left (2\,x^3+4\,x^2\right )}{x^4-4\,x^3+6\,x^2-4\,x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x^2*log(2*x*log(2))^2 - 16*x^2 + 16*x^3 + log(2*x*log(2))*(12*x^2 + 4*x^3) + log(2*x*log(2))^3*(4*x +
x^2 - 4*x^3 + 2*x^4 - 2))/(log(2*x*log(2))^3*(2*x - 6*x^2 + 6*x^3 - 2*x^4)),x)

[Out]

x + log(x) + ((x*(5*x - x^2))/(x - 1)^3 + (2*x^2*log(2*x*log(2))^2*(x + 2))/(x - 1)^4 + (2*x*log(2*x*log(2))*(
5*x + x^2))/(x - 1)^4)/log(2*x*log(2)) - (x + (43*x^2)/4 + x^3/2 - 1/4)/(6*x^2 - 4*x - 4*x^3 + x^4 + 1) - ((4*
x^2)/(x - 1)^2 + (2*x^2*log(2*x*log(2))^2)/(x - 1)^3 + (x^2*log(2*x*log(2))*(x + 3))/(x - 1)^3)/log(2*x*log(2)
)^2 - (log(2*x*log(2))*(4*x^2 + 2*x^3))/(6*x^2 - 4*x - 4*x^3 + x^4 + 1)

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sympy [B]  time = 0.49, size = 58, normalized size = 2.15 \begin {gather*} x + \frac {1 - 2 x}{4 x^{2} - 8 x + 4} + \frac {- 2 x^{2} \log {\left (2 x \log {\relax (2 )} \right )} - 4 x^{2}}{\left (x^{2} - 2 x + 1\right ) \log {\left (2 x \log {\relax (2 )} \right )}^{2}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**4-4*x**3+x**2+4*x-2)*ln(2*x*ln(2))**3+8*x**2*ln(2*x*ln(2))**2+(4*x**3+12*x**2)*ln(2*x*ln(2))+
16*x**3-16*x**2)/(2*x**4-6*x**3+6*x**2-2*x)/ln(2*x*ln(2))**3,x)

[Out]

x + (1 - 2*x)/(4*x**2 - 8*x + 4) + (-2*x**2*log(2*x*log(2)) - 4*x**2)/((x**2 - 2*x + 1)*log(2*x*log(2))**2) +
log(x)

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