3.80.20 \(\int \frac {20 x-4 x^4+4 x^5+(-10+2 x^3) \log (2)+(60 x^2-30 x \log (2)) \log (2 x-\log (2))+(60 x^3-30 x^2 \log (2)) \log ^2(2 x-\log (2))+(20 x^4-10 x^3 \log (2)) \log ^3(2 x-\log (2))}{-2 x^3+x^2 \log (2)+(-6 x^4+3 x^3 \log (2)) \log (2 x-\log (2))+(-6 x^5+3 x^4 \log (2)) \log ^2(2 x-\log (2))+(-2 x^6+x^5 \log (2)) \log ^3(2 x-\log (2))} \, dx\)

Optimal. Leaf size=21 \[ \frac {10}{x}+\frac {1}{\left (\frac {1}{x}+\log (2 x-\log (2))\right )^2} \]

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Rubi [F]  time = 1.41, 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 {20 x-4 x^4+4 x^5+\left (-10+2 x^3\right ) \log (2)+\left (60 x^2-30 x \log (2)\right ) \log (2 x-\log (2))+\left (60 x^3-30 x^2 \log (2)\right ) \log ^2(2 x-\log (2))+\left (20 x^4-10 x^3 \log (2)\right ) \log ^3(2 x-\log (2))}{-2 x^3+x^2 \log (2)+\left (-6 x^4+3 x^3 \log (2)\right ) \log (2 x-\log (2))+\left (-6 x^5+3 x^4 \log (2)\right ) \log ^2(2 x-\log (2))+\left (-2 x^6+x^5 \log (2)\right ) \log ^3(2 x-\log (2))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(20*x - 4*x^4 + 4*x^5 + (-10 + 2*x^3)*Log[2] + (60*x^2 - 30*x*Log[2])*Log[2*x - Log[2]] + (60*x^3 - 30*x^2
*Log[2])*Log[2*x - Log[2]]^2 + (20*x^4 - 10*x^3*Log[2])*Log[2*x - Log[2]]^3)/(-2*x^3 + x^2*Log[2] + (-6*x^4 +
3*x^3*Log[2])*Log[2*x - Log[2]] + (-6*x^5 + 3*x^4*Log[2])*Log[2*x - Log[2]]^2 + (-2*x^6 + x^5*Log[2])*Log[2*x
- Log[2]]^3),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-10 x+2 x^4-2 x^5+5 \log (2)-x^3 \log (2)-15 x (2 x-\log (2)) \log (2 x-\log (2))-15 x^2 (2 x-\log (2)) \log ^2(2 x-\log (2))-5 x^3 (2 x-\log (2)) \log ^3(2 x-\log (2))\right )}{x^2 (2 x-\log (2)) (1+x \log (2 x-\log (2)))^3} \, dx\\ &=2 \int \frac {-10 x+2 x^4-2 x^5+5 \log (2)-x^3 \log (2)-15 x (2 x-\log (2)) \log (2 x-\log (2))-15 x^2 (2 x-\log (2)) \log ^2(2 x-\log (2))-5 x^3 (2 x-\log (2)) \log ^3(2 x-\log (2))}{x^2 (2 x-\log (2)) (1+x \log (2 x-\log (2)))^3} \, dx\\ &=2 \int \left (-\frac {5}{x^2}-\frac {x \left (-2 x+2 x^2+\log (2)\right )}{(2 x-\log (2)) (1+x \log (2 x-\log (2)))^3}\right ) \, dx\\ &=\frac {10}{x}-2 \int \frac {x \left (-2 x+2 x^2+\log (2)\right )}{(2 x-\log (2)) (1+x \log (2 x-\log (2)))^3} \, dx\\ &=\frac {10}{x}-2 \int \left (\frac {x^2}{(1+x \log (2 x-\log (2)))^3}+\frac {x (-2+\log (2))}{2 (1+x \log (2 x-\log (2)))^3}+\frac {\log ^2(2)}{4 (1+x \log (2 x-\log (2)))^3}+\frac {\log ^3(2)}{4 (2 x-\log (2)) (1+x \log (2 x-\log (2)))^3}\right ) \, dx\\ &=\frac {10}{x}-2 \int \frac {x^2}{(1+x \log (2 x-\log (2)))^3} \, dx-(-2+\log (2)) \int \frac {x}{(1+x \log (2 x-\log (2)))^3} \, dx-\frac {1}{2} \log ^2(2) \int \frac {1}{(1+x \log (2 x-\log (2)))^3} \, dx-\frac {1}{2} \log ^3(2) \int \frac {1}{(2 x-\log (2)) (1+x \log (2 x-\log (2)))^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 30, normalized size = 1.43 \begin {gather*} -2 \left (-\frac {5}{x}-\frac {x^2}{2 (1+x \log (2 x-\log (2)))^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*x - 4*x^4 + 4*x^5 + (-10 + 2*x^3)*Log[2] + (60*x^2 - 30*x*Log[2])*Log[2*x - Log[2]] + (60*x^3 -
30*x^2*Log[2])*Log[2*x - Log[2]]^2 + (20*x^4 - 10*x^3*Log[2])*Log[2*x - Log[2]]^3)/(-2*x^3 + x^2*Log[2] + (-6*
x^4 + 3*x^3*Log[2])*Log[2*x - Log[2]] + (-6*x^5 + 3*x^4*Log[2])*Log[2*x - Log[2]]^2 + (-2*x^6 + x^5*Log[2])*Lo
g[2*x - Log[2]]^3),x]

[Out]

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

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fricas [B]  time = 0.58, size = 67, normalized size = 3.19 \begin {gather*} \frac {10 \, x^{2} \log \left (2 \, x - \log \relax (2)\right )^{2} + x^{3} + 20 \, x \log \left (2 \, x - \log \relax (2)\right ) + 10}{x^{3} \log \left (2 \, x - \log \relax (2)\right )^{2} + 2 \, x^{2} \log \left (2 \, x - \log \relax (2)\right ) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^3*log(2)+20*x^4)*log(2*x-log(2))^3+(-30*x^2*log(2)+60*x^3)*log(2*x-log(2))^2+(-30*x*log(2)+6
0*x^2)*log(2*x-log(2))+(2*x^3-10)*log(2)+4*x^5-4*x^4+20*x)/((x^5*log(2)-2*x^6)*log(2*x-log(2))^3+(3*x^4*log(2)
-6*x^5)*log(2*x-log(2))^2+(3*x^3*log(2)-6*x^4)*log(2*x-log(2))+x^2*log(2)-2*x^3),x, algorithm="fricas")

[Out]

(10*x^2*log(2*x - log(2))^2 + x^3 + 20*x*log(2*x - log(2)) + 10)/(x^3*log(2*x - log(2))^2 + 2*x^2*log(2*x - lo
g(2)) + x)

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giac [B]  time = 0.27, size = 128, normalized size = 6.10 \begin {gather*} \frac {2 \, x^{4} - 2 \, x^{3} + x^{2} \log \relax (2)}{2 \, x^{4} \log \left (2 \, x - \log \relax (2)\right )^{2} - 2 \, x^{3} \log \left (2 \, x - \log \relax (2)\right )^{2} + x^{2} \log \relax (2) \log \left (2 \, x - \log \relax (2)\right )^{2} + 4 \, x^{3} \log \left (2 \, x - \log \relax (2)\right ) - 4 \, x^{2} \log \left (2 \, x - \log \relax (2)\right ) + 2 \, x \log \relax (2) \log \left (2 \, x - \log \relax (2)\right ) + 2 \, x^{2} - 2 \, x + \log \relax (2)} + \frac {10}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^3*log(2)+20*x^4)*log(2*x-log(2))^3+(-30*x^2*log(2)+60*x^3)*log(2*x-log(2))^2+(-30*x*log(2)+6
0*x^2)*log(2*x-log(2))+(2*x^3-10)*log(2)+4*x^5-4*x^4+20*x)/((x^5*log(2)-2*x^6)*log(2*x-log(2))^3+(3*x^4*log(2)
-6*x^5)*log(2*x-log(2))^2+(3*x^3*log(2)-6*x^4)*log(2*x-log(2))+x^2*log(2)-2*x^3),x, algorithm="giac")

[Out]

(2*x^4 - 2*x^3 + x^2*log(2))/(2*x^4*log(2*x - log(2))^2 - 2*x^3*log(2*x - log(2))^2 + x^2*log(2)*log(2*x - log
(2))^2 + 4*x^3*log(2*x - log(2)) - 4*x^2*log(2*x - log(2)) + 2*x*log(2)*log(2*x - log(2)) + 2*x^2 - 2*x + log(
2)) + 10/x

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maple [A]  time = 0.06, size = 26, normalized size = 1.24

method	result	size
risch	\(\frac {10}{x}+\frac {x^{2}}{\left (x \ln \left (2 x -\ln \relax (2)\right )+1\right )^{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x^3*ln(2)+20*x^4)*ln(2*x-ln(2))^3+(-30*x^2*ln(2)+60*x^3)*ln(2*x-ln(2))^2+(-30*x*ln(2)+60*x^2)*ln(2*x
-ln(2))+(2*x^3-10)*ln(2)+4*x^5-4*x^4+20*x)/((x^5*ln(2)-2*x^6)*ln(2*x-ln(2))^3+(3*x^4*ln(2)-6*x^5)*ln(2*x-ln(2)
)^2+(3*x^3*ln(2)-6*x^4)*ln(2*x-ln(2))+x^2*ln(2)-2*x^3),x,method=_RETURNVERBOSE)

[Out]

10/x+x^2/(x*ln(2*x-ln(2))+1)^2

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maxima [B]  time = 0.48, size = 67, normalized size = 3.19 \begin {gather*} \frac {10 \, x^{2} \log \left (2 \, x - \log \relax (2)\right )^{2} + x^{3} + 20 \, x \log \left (2 \, x - \log \relax (2)\right ) + 10}{x^{3} \log \left (2 \, x - \log \relax (2)\right )^{2} + 2 \, x^{2} \log \left (2 \, x - \log \relax (2)\right ) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^3*log(2)+20*x^4)*log(2*x-log(2))^3+(-30*x^2*log(2)+60*x^3)*log(2*x-log(2))^2+(-30*x*log(2)+6
0*x^2)*log(2*x-log(2))+(2*x^3-10)*log(2)+4*x^5-4*x^4+20*x)/((x^5*log(2)-2*x^6)*log(2*x-log(2))^3+(3*x^4*log(2)
-6*x^5)*log(2*x-log(2))^2+(3*x^3*log(2)-6*x^4)*log(2*x-log(2))+x^2*log(2)-2*x^3),x, algorithm="maxima")

[Out]

(10*x^2*log(2*x - log(2))^2 + x^3 + 20*x*log(2*x - log(2)) + 10)/(x^3*log(2*x - log(2))^2 + 2*x^2*log(2*x - lo
g(2)) + x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {{\ln \left (2\,x-\ln \relax (2)\right )}^3\,\left (10\,x^3\,\ln \relax (2)-20\,x^4\right )-\ln \relax (2)\,\left (2\,x^3-10\right )-20\,x+{\ln \left (2\,x-\ln \relax (2)\right )}^2\,\left (30\,x^2\,\ln \relax (2)-60\,x^3\right )+\ln \left (2\,x-\ln \relax (2)\right )\,\left (30\,x\,\ln \relax (2)-60\,x^2\right )+4\,x^4-4\,x^5}{\ln \left (2\,x-\ln \relax (2)\right )\,\left (3\,x^3\,\ln \relax (2)-6\,x^4\right )+{\ln \left (2\,x-\ln \relax (2)\right )}^3\,\left (x^5\,\ln \relax (2)-2\,x^6\right )+{\ln \left (2\,x-\ln \relax (2)\right )}^2\,\left (3\,x^4\,\ln \relax (2)-6\,x^5\right )+x^2\,\ln \relax (2)-2\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*x - log(2))^3*(10*x^3*log(2) - 20*x^4) - log(2)*(2*x^3 - 10) - 20*x + log(2*x - log(2))^2*(30*x^2*
log(2) - 60*x^3) + log(2*x - log(2))*(30*x*log(2) - 60*x^2) + 4*x^4 - 4*x^5)/(log(2*x - log(2))*(3*x^3*log(2)
- 6*x^4) + log(2*x - log(2))^3*(x^5*log(2) - 2*x^6) + log(2*x - log(2))^2*(3*x^4*log(2) - 6*x^5) + x^2*log(2)
- 2*x^3),x)

[Out]

int(-(log(2*x - log(2))^3*(10*x^3*log(2) - 20*x^4) - log(2)*(2*x^3 - 10) - 20*x + log(2*x - log(2))^2*(30*x^2*
log(2) - 60*x^3) + log(2*x - log(2))*(30*x*log(2) - 60*x^2) + 4*x^4 - 4*x^5)/(log(2*x - log(2))*(3*x^3*log(2)
- 6*x^4) + log(2*x - log(2))^3*(x^5*log(2) - 2*x^6) + log(2*x - log(2))^2*(3*x^4*log(2) - 6*x^5) + x^2*log(2)
- 2*x^3), x)

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sympy [A]  time = 0.21, size = 32, normalized size = 1.52 \begin {gather*} \frac {x^{2}}{x^{2} \log {\left (2 x - \log {\relax (2 )} \right )}^{2} + 2 x \log {\left (2 x - \log {\relax (2 )} \right )} + 1} + \frac {10}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x**3*ln(2)+20*x**4)*ln(2*x-ln(2))**3+(-30*x**2*ln(2)+60*x**3)*ln(2*x-ln(2))**2+(-30*x*ln(2)+60
*x**2)*ln(2*x-ln(2))+(2*x**3-10)*ln(2)+4*x**5-4*x**4+20*x)/((x**5*ln(2)-2*x**6)*ln(2*x-ln(2))**3+(3*x**4*ln(2)
-6*x**5)*ln(2*x-ln(2))**2+(3*x**3*ln(2)-6*x**4)*ln(2*x-ln(2))+x**2*ln(2)-2*x**3),x)

[Out]

x**2/(x**2*log(2*x - log(2))**2 + 2*x*log(2*x - log(2)) + 1) + 10/x

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