3.28.51 \(\int \frac {2-5 x+4 x^2-x^3+(-x+4 x^2-2 x^3) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+(-20 x^2+50 x^3-40 x^4+10 x^5) \log (4)+(5 x^3-10 x^4+5 x^5) \log ^2(4)+(20 x-30 x^2+10 x^3+(-10 x^2+10 x^3) \log (4)) \log (x)+5 x \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {x}{5 \left (-x+x^2+\frac {\log (x)}{\frac {-2+x}{x}+\log (4)}\right )} \]

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Rubi [F]  time = 1.93, 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 {2-5 x+4 x^2-x^3+\left (-x+4 x^2-2 x^3\right ) \log (4)-x^3 \log ^2(4)+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 - 5*x + 4*x^2 - x^3 + (-x + 4*x^2 - 2*x^3)*Log[4] - x^3*Log[4]^2 + (x + x*Log[4])*Log[x])/(20*x - 60*x^
2 + 65*x^3 - 30*x^4 + 5*x^5 + (-20*x^2 + 50*x^3 - 40*x^4 + 10*x^5)*Log[4] + (5*x^3 - 10*x^4 + 5*x^5)*Log[4]^2
+ (20*x - 30*x^2 + 10*x^3 + (-10*x^2 + 10*x^3)*Log[4])*Log[x] + 5*x*Log[x]^2),x]

[Out]

-1/5*((7 + Log[64])*Defer[Int][((-1 + x)*(-2 + x + x*Log[4]) + Log[x])^(-2), x]) + (2*Defer[Int][1/(x*((-1 + x
)*(-2 + x + x*Log[4]) + Log[x])^2), x])/5 + ((1 + Log[4])*(7 + Log[4])*Defer[Int][x/((-1 + x)*(-2 + x + x*Log[
4]) + Log[x])^2, x])/5 - (2*(1 + Log[4])^2*Defer[Int][x^2/((-1 + x)*(-2 + x + x*Log[4]) + Log[x])^2, x])/5 + (
(1 + Log[4])*Defer[Int][((-1 + x)*(-2 + x + x*Log[4]) + Log[x])^(-1), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2-5 x+4 x^2+\left (-x+4 x^2-2 x^3\right ) \log (4)+x^3 \left (-1-\log ^2(4)\right )+(x+x \log (4)) \log (x)}{20 x-60 x^2+65 x^3-30 x^4+5 x^5+\left (-20 x^2+50 x^3-40 x^4+10 x^5\right ) \log (4)+\left (5 x^3-10 x^4+5 x^5\right ) \log ^2(4)+\left (20 x-30 x^2+10 x^3+\left (-10 x^2+10 x^3\right ) \log (4)\right ) \log (x)+5 x \log ^2(x)} \, dx\\ &=\int \frac {2+4 x^2 (1+\log (4))-x^3 (1+\log (4))^2-x (5+\log (4))+x (1+\log (4)) \log (x)}{5 x ((-1+x) (-2+x+x \log (4))+\log (x))^2} \, dx\\ &=\frac {1}{5} \int \frac {2+4 x^2 (1+\log (4))-x^3 (1+\log (4))^2-x (5+\log (4))+x (1+\log (4)) \log (x)}{x ((-1+x) (-2+x+x \log (4))+\log (x))^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {(1-2 x) \left (2-3 x (1+\log (4))+x^2 (1+\log (4))^2\right )}{x \left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2}+\frac {1+\log (4)}{2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {(1-2 x) \left (2-3 x (1+\log (4))+x^2 (1+\log (4))^2\right )}{x \left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2} \, dx+\frac {1}{5} (1+\log (4)) \int \frac {1}{2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)} \, dx\\ &=\frac {1}{5} \int \frac {(1-2 x) \left (2-3 x (1+\log (4))+x^2 (1+\log (4))^2\right )}{x ((-1+x) (-2+x+x \log (4))+\log (x))^2} \, dx+\frac {1}{5} (1+\log (4)) \int \frac {1}{(-1+x) (-2+x+x \log (4))+\log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {2}{x \left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2}-\frac {7 \left (1+\frac {6 \log (2)}{7}\right )}{\left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2}-\frac {2 x^2 (1+\log (4))^2}{\left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2}+\frac {x (1+\log (4)) (7+\log (4))}{\left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2}\right ) \, dx+\frac {1}{5} (1+\log (4)) \int \frac {1}{(-1+x) (-2+x+x \log (4))+\log (x)} \, dx\\ &=\frac {2}{5} \int \frac {1}{x \left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2} \, dx+\frac {1}{5} (1+\log (4)) \int \frac {1}{(-1+x) (-2+x+x \log (4))+\log (x)} \, dx-\frac {1}{5} \left (2 (1+\log (4))^2\right ) \int \frac {x^2}{\left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2} \, dx+\frac {1}{5} ((1+\log (4)) (7+\log (4))) \int \frac {x}{\left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2} \, dx+\frac {1}{5} (-7-\log (64)) \int \frac {1}{\left (2-3 x \left (1+\frac {2 \log (2)}{3}\right )+x^2 (1+\log (4))+\log (x)\right )^2} \, dx\\ &=\frac {2}{5} \int \frac {1}{x ((-1+x) (-2+x+x \log (4))+\log (x))^2} \, dx+\frac {1}{5} (1+\log (4)) \int \frac {1}{(-1+x) (-2+x+x \log (4))+\log (x)} \, dx-\frac {1}{5} \left (2 (1+\log (4))^2\right ) \int \frac {x^2}{((-1+x) (-2+x+x \log (4))+\log (x))^2} \, dx+\frac {1}{5} ((1+\log (4)) (7+\log (4))) \int \frac {x}{((-1+x) (-2+x+x \log (4))+\log (x))^2} \, dx+\frac {1}{5} (-7-\log (64)) \int \frac {1}{((-1+x) (-2+x+x \log (4))+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.91, size = 27, normalized size = 0.93 \begin {gather*} \frac {-2+x+x \log (4)}{5 ((-1+x) (-2+x+x \log (4))+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 5*x + 4*x^2 - x^3 + (-x + 4*x^2 - 2*x^3)*Log[4] - x^3*Log[4]^2 + (x + x*Log[4])*Log[x])/(20*x -
60*x^2 + 65*x^3 - 30*x^4 + 5*x^5 + (-20*x^2 + 50*x^3 - 40*x^4 + 10*x^5)*Log[4] + (5*x^3 - 10*x^4 + 5*x^5)*Log
[4]^2 + (20*x - 30*x^2 + 10*x^3 + (-10*x^2 + 10*x^3)*Log[4])*Log[x] + 5*x*Log[x]^2),x]

[Out]

(-2 + x + x*Log[4])/(5*((-1 + x)*(-2 + x + x*Log[4]) + Log[x]))

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fricas [A]  time = 0.59, size = 33, normalized size = 1.14 \begin {gather*} \frac {2 \, x \log \relax (2) + x - 2}{5 \, {\left (x^{2} + 2 \, {\left (x^{2} - x\right )} \log \relax (2) - 3 \, x + \log \relax (x) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+2*x*log(2))*log(x)-4*x^3*log(2)^2+2*(-2*x^3+4*x^2-x)*log(2)-x^3+4*x^2-5*x+2)/(5*x*log(x)^2+(2*(1
0*x^3-10*x^2)*log(2)+10*x^3-30*x^2+20*x)*log(x)+4*(5*x^5-10*x^4+5*x^3)*log(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2
)*log(2)+5*x^5-30*x^4+65*x^3-60*x^2+20*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 34, normalized size = 1.17 \begin {gather*} \frac {2 \, x \log \relax (2) + x - 2}{5 \, {\left (2 \, x^{2} \log \relax (2) + x^{2} - 2 \, x \log \relax (2) - 3 \, x + \log \relax (x) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+2*x*log(2))*log(x)-4*x^3*log(2)^2+2*(-2*x^3+4*x^2-x)*log(2)-x^3+4*x^2-5*x+2)/(5*x*log(x)^2+(2*(1
0*x^3-10*x^2)*log(2)+10*x^3-30*x^2+20*x)*log(x)+4*(5*x^5-10*x^4+5*x^3)*log(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2
)*log(2)+5*x^5-30*x^4+65*x^3-60*x^2+20*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 35, normalized size = 1.21

method	result	size
risch	\(\frac {2 x \ln \relax (2)+x -2}{10 x^{2} \ln \relax (2)-10 x \ln \relax (2)+5 x^{2}+5 \ln \relax (x )-15 x +10}\)	\(35\)
norman	\(\frac {-\frac {2}{5}+\left (\frac {2 \ln \relax (2)}{5}+\frac {1}{5}\right ) x}{2 x^{2} \ln \relax (2)-2 x \ln \relax (2)+x^{2}+\ln \relax (x )-3 x +2}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+2*x*ln(2))*ln(x)-4*x^3*ln(2)^2+2*(-2*x^3+4*x^2-x)*ln(2)-x^3+4*x^2-5*x+2)/(5*x*ln(x)^2+(2*(10*x^3-10*x^
2)*ln(2)+10*x^3-30*x^2+20*x)*ln(x)+4*(5*x^5-10*x^4+5*x^3)*ln(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2)*ln(2)+5*x^5-
30*x^4+65*x^3-60*x^2+20*x),x,method=_RETURNVERBOSE)

[Out]

1/5*(2*x*ln(2)+x-2)/(2*x^2*ln(2)-2*x*ln(2)+x^2+ln(x)-3*x+2)

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maxima [A]  time = 0.90, size = 37, normalized size = 1.28 \begin {gather*} \frac {x {\left (2 \, \log \relax (2) + 1\right )} - 2}{5 \, {\left (x^{2} {\left (2 \, \log \relax (2) + 1\right )} - x {\left (2 \, \log \relax (2) + 3\right )} + \log \relax (x) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+2*x*log(2))*log(x)-4*x^3*log(2)^2+2*(-2*x^3+4*x^2-x)*log(2)-x^3+4*x^2-5*x+2)/(5*x*log(x)^2+(2*(1
0*x^3-10*x^2)*log(2)+10*x^3-30*x^2+20*x)*log(x)+4*(5*x^5-10*x^4+5*x^3)*log(2)^2+2*(10*x^5-40*x^4+50*x^3-20*x^2
)*log(2)+5*x^5-30*x^4+65*x^3-60*x^2+20*x),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {5\,x+4\,x^3\,{\ln \relax (2)}^2+2\,\ln \relax (2)\,\left (2\,x^3-4\,x^2+x\right )-\ln \relax (x)\,\left (x+2\,x\,\ln \relax (2)\right )-4\,x^2+x^3-2}{20\,x-2\,\ln \relax (2)\,\left (-10\,x^5+40\,x^4-50\,x^3+20\,x^2\right )+5\,x\,{\ln \relax (x)}^2+4\,{\ln \relax (2)}^2\,\left (5\,x^5-10\,x^4+5\,x^3\right )+\ln \relax (x)\,\left (20\,x-2\,\ln \relax (2)\,\left (10\,x^2-10\,x^3\right )-30\,x^2+10\,x^3\right )-60\,x^2+65\,x^3-30\,x^4+5\,x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + 4*x^3*log(2)^2 + 2*log(2)*(x - 4*x^2 + 2*x^3) - log(x)*(x + 2*x*log(2)) - 4*x^2 + x^3 - 2)/(20*x -
2*log(2)*(20*x^2 - 50*x^3 + 40*x^4 - 10*x^5) + 5*x*log(x)^2 + 4*log(2)^2*(5*x^3 - 10*x^4 + 5*x^5) + log(x)*(2
0*x - 2*log(2)*(10*x^2 - 10*x^3) - 30*x^2 + 10*x^3) - 60*x^2 + 65*x^3 - 30*x^4 + 5*x^5),x)

[Out]

int(-(5*x + 4*x^3*log(2)^2 + 2*log(2)*(x - 4*x^2 + 2*x^3) - log(x)*(x + 2*x*log(2)) - 4*x^2 + x^3 - 2)/(20*x -
2*log(2)*(20*x^2 - 50*x^3 + 40*x^4 - 10*x^5) + 5*x*log(x)^2 + 4*log(2)^2*(5*x^3 - 10*x^4 + 5*x^5) + log(x)*(2
0*x - 2*log(2)*(10*x^2 - 10*x^3) - 30*x^2 + 10*x^3) - 60*x^2 + 65*x^3 - 30*x^4 + 5*x^5), x)

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sympy [A]  time = 0.20, size = 39, normalized size = 1.34 \begin {gather*} \frac {x + 2 x \log {\relax (2 )} - 2}{5 x^{2} + 10 x^{2} \log {\relax (2 )} - 15 x - 10 x \log {\relax (2 )} + 5 \log {\relax (x )} + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+2*x*ln(2))*ln(x)-4*x**3*ln(2)**2+2*(-2*x**3+4*x**2-x)*ln(2)-x**3+4*x**2-5*x+2)/(5*x*ln(x)**2+(2*
(10*x**3-10*x**2)*ln(2)+10*x**3-30*x**2+20*x)*ln(x)+4*(5*x**5-10*x**4+5*x**3)*ln(2)**2+2*(10*x**5-40*x**4+50*x
**3-20*x**2)*ln(2)+5*x**5-30*x**4+65*x**3-60*x**2+20*x),x)

[Out]

(x + 2*x*log(2) - 2)/(5*x**2 + 10*x**2*log(2) - 15*x - 10*x*log(2) + 5*log(x) + 10)

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