3.10.100 \(\int \frac {8 e^4+4 e^2 x+(-8 e^4+e^2 (-4 x+4 x^2)) \log (x \log (2))+(e^4 (4-8 x)+e^2 (x-2 x^2+x^3)) \log ^2(x \log (2))}{4 e^2 x^2+e^2 (-4 x^2+4 x^3) \log (x \log (2))+e^2 (x^2-2 x^3+x^4) \log ^2(x \log (2))} \, dx\)

Optimal. Leaf size=25 \[ \log (x)+\frac {4 e^2}{x \left (-1+x+\frac {2}{\log (x \log (2))}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 2.04, 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 {8 e^4+4 e^2 x+\left (-8 e^4+e^2 \left (-4 x+4 x^2\right )\right ) \log (x \log (2))+\left (e^4 (4-8 x)+e^2 \left (x-2 x^2+x^3\right )\right ) \log ^2(x \log (2))}{4 e^2 x^2+e^2 \left (-4 x^2+4 x^3\right ) \log (x \log (2))+e^2 \left (x^2-2 x^3+x^4\right ) \log ^2(x \log (2))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*E^4 + 4*E^2*x + (-8*E^4 + E^2*(-4*x + 4*x^2))*Log[x*Log[2]] + (E^4*(4 - 8*x) + E^2*(x - 2*x^2 + x^3))*L
og[x*Log[2]]^2)/(4*E^2*x^2 + E^2*(-4*x^2 + 4*x^3)*Log[x*Log[2]] + E^2*(x^2 - 2*x^3 + x^4)*Log[x*Log[2]]^2),x]

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.55, size = 29, normalized size = 1.16 \begin {gather*} \log (x)+\frac {4 e^2 \log (x \log (2))}{x (2+(-1+x) \log (x \log (2)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.02, size = 49, normalized size = 1.96 \begin {gather*} \frac {{\left (x^{2} - x\right )} \log \left (x \log \relax (2)\right )^{2} + 2 \, {\left (x + 2 \, e^{2}\right )} \log \left (x \log \relax (2)\right )}{{\left (x^{2} - x\right )} \log \left (x \log \relax (2)\right ) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^2 - x)*log(x*log(2))^2 + 2*(x + 2*e^2)*log(x*log(2)))/((x^2 - x)*log(x*log(2)) + 2*x)

________________________________________________________________________________________

giac [B]  time = 0.45, size = 60, normalized size = 2.40 \begin {gather*} \frac {x^{2} \log \left (x \log \relax (2)\right ) \log \relax (x) - x \log \left (x \log \relax (2)\right ) \log \relax (x) + 4 \, e^{2} \log \left (x \log \relax (2)\right ) + 2 \, x \log \relax (x)}{x^{2} \log \left (x \log \relax (2)\right ) - x \log \left (x \log \relax (2)\right ) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.13, size = 57, normalized size = 2.28

method	result	size
risch	\(\frac {x^{2} \ln \relax (x )-x \ln \relax (x )+4 \,{\mathrm e}^{2}}{x \left (x -1\right )}-\frac {8 \,{\mathrm e}^{2}}{x \left (x -1\right ) \left (x \ln \left (x \ln \relax (2)\right )-\ln \left (x \ln \relax (2)\right )+2\right )}\)	\(57\)
norman	\(\frac {-\ln \left (x \ln \relax (2)\right )^{2} x +\ln \left (x \ln \relax (2)\right )^{2} x^{2}+2 x \ln \left (x \ln \relax (2)\right )+4 \,{\mathrm e}^{-2} {\mathrm e}^{4} \ln \left (x \ln \relax (2)\right )}{x \left (x \ln \left (x \ln \relax (2)\right )-\ln \left (x \ln \relax (2)\right )+2\right )}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x+4)*exp(4)+(x^3-2*x^2+x)*exp(2))*ln(x*ln(2))^2+(-8*exp(4)+(4*x^2-4*x)*exp(2))*ln(x*ln(2))+8*exp(4)+
4*exp(2)*x)/((x^4-2*x^3+x^2)*exp(2)*ln(x*ln(2))^2+(4*x^3-4*x^2)*exp(2)*ln(x*ln(2))+4*x^2*exp(2)),x,method=_RET
URNVERBOSE)

[Out]

(x^2*ln(x)-x*ln(x)+4*exp(2))/x/(x-1)-8*exp(2)/x/(x-1)/(x*ln(x*ln(2))-ln(x*ln(2))+2)

________________________________________________________________________________________

maxima [A]  time = 0.95, size = 45, normalized size = 1.80 \begin {gather*} \frac {4 \, {\left (e^{2} \log \relax (x) + e^{2} \log \left (\log \relax (2)\right )\right )}}{x^{2} \log \left (\log \relax (2)\right ) - x {\left (\log \left (\log \relax (2)\right ) - 2\right )} + {\left (x^{2} - x\right )} \log \relax (x)} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\left ({\mathrm {e}}^2\,\left (x^3-2\,x^2+x\right )-{\mathrm {e}}^4\,\left (8\,x-4\right )\right )\,{\ln \left (x\,\ln \relax (2)\right )}^2+\left (-8\,{\mathrm {e}}^4-{\mathrm {e}}^2\,\left (4\,x-4\,x^2\right )\right )\,\ln \left (x\,\ln \relax (2)\right )+8\,{\mathrm {e}}^4+4\,x\,{\mathrm {e}}^2}{4\,x^2\,{\mathrm {e}}^2+{\mathrm {e}}^2\,{\ln \left (x\,\ln \relax (2)\right )}^2\,\left (x^4-2\,x^3+x^2\right )-{\mathrm {e}}^2\,\ln \left (x\,\ln \relax (2)\right )\,\left (4\,x^2-4\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 0.36, size = 42, normalized size = 1.68 \begin {gather*} \log {\relax (x )} - \frac {8 e^{2}}{2 x^{2} - 2 x + \left (x^{3} - 2 x^{2} + x\right ) \log {\left (x \log {\relax (2 )} \right )}} + \frac {4 e^{2}}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x+4)*exp(4)+(x**3-2*x**2+x)*exp(2))*ln(x*ln(2))**2+(-8*exp(4)+(4*x**2-4*x)*exp(2))*ln(x*ln(2))
+8*exp(4)+4*exp(2)*x)/((x**4-2*x**3+x**2)*exp(2)*ln(x*ln(2))**2+(4*x**3-4*x**2)*exp(2)*ln(x*ln(2))+4*x**2*exp(
2)),x)

[Out]

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

________________________________________________________________________________________