3.54.35 \(\int \frac {-20-6 x-2 x^2-5 \log (\frac {4 e^{x^2}}{x})-5 \log (x)+(4+x+\log (\frac {4 e^{x^2}}{x})+\log (x)) \log (4+x+\log (\frac {4 e^{x^2}}{x})+\log (x))}{4 x^2+x^3+x^2 \log (\frac {4 e^{x^2}}{x})+x^2 \log (x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {5-\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x} \]

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Rubi [F]  time = 1.42, 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 {-20-6 x-2 x^2-5 \log \left (\frac {4 e^{x^2}}{x}\right )-5 \log (x)+\left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right ) \log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{4 x^2+x^3+x^2 \log \left (\frac {4 e^{x^2}}{x}\right )+x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20-6 x-2 x^2-5 \log \left (\frac {4 e^{x^2}}{x}\right )-5 \log (x)+\left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right ) \log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx\\ &=\int \left (-\frac {2}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)}-\frac {20}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {6}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {5 \log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {5 \log (x)}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}+\frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx\right )-5 \int \frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-5 \int \frac {\log (x)}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-20 \int \frac {1}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ &=-\left (2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx\right )-5 \int \frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-5 \int \left (\frac {1}{x^2}+\frac {-4-x-\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}\right ) \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-20 \int \frac {1}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ &=\frac {5}{x}-2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx-5 \int \frac {-4-x-\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-5 \int \frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-20 \int \frac {1}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ &=\frac {5}{x}-2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx-5 \int \frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-5 \int \left (-\frac {4}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}\right ) \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-20 \int \frac {1}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ &=\frac {5}{x}-2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx+5 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 28, normalized size = 1.12 \begin {gather*} \frac {5}{x}-\frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 6*x - 2*x^2 - 5*Log[(4*E^x^2)/x] - 5*Log[x] + (4 + x + Log[(4*E^x^2)/x] + Log[x])*Log[4 + x +
Log[(4*E^x^2)/x] + Log[x]])/(4*x^2 + x^3 + x^2*Log[(4*E^x^2)/x] + x^2*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.60, size = 18, normalized size = 0.72 \begin {gather*} -\frac {\log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right ) - 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(x^2 + x + 2*log(2) + 4) - 5)/x

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giac [A]  time = 0.15, size = 22, normalized size = 0.88 \begin {gather*} -\frac {\log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right )}{x} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x^2 + x + 2*log(2) + 4)/x + 5/x

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maple [A]  time = 0.16, size = 28, normalized size = 1.12

method	result	size
default	\(-\frac {\ln \left (\ln \left (\frac {4 \,{\mathrm e}^{x^{2}}}{x}\right )+4+x +\ln \relax (x )\right )}{x}+\frac {5}{x}\)	\(28\)
risch	\(-\frac {\ln \left (2 \ln \relax (2)+\ln \left ({\mathrm e}^{x^{2}}\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x^{2}}}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x^{2}}}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x^{2}}}{x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{x^{2}}\right )\right )}{2}+4+x \right )}{x}+\frac {5}{x}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x*ln(ln(4*exp(x^2)/x)+4+x+ln(x))+5/x

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maxima [B]  time = 0.66, size = 311, normalized size = 12.44 \begin {gather*} \frac {{\left (3 \, \log \relax (2) + 1\right )} \log \relax (x)}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} + \frac {{\left (4 \, \log \relax (2)^{2} - 7 \, \log \relax (2) - 25\right )} \arctan \left (\frac {2 \, x + 1}{\sqrt {8 \, \log \relax (2) + 15}}\right )}{{\left (\log \relax (2)^{2} + 4 \, \log \relax (2) + 4\right )} \sqrt {8 \, \log \relax (2) + 15}} - \frac {4 \, \arctan \left (\frac {2 \, x + 1}{\sqrt {8 \, \log \relax (2) + 15}}\right )}{\sqrt {8 \, \log \relax (2) + 15}} + \frac {5 \, {\left (4 \, \log \relax (2) + 7\right )} \arctan \left (\frac {2 \, x + 1}{\sqrt {8 \, \log \relax (2) + 15}}\right )}{{\left (\log \relax (2)^{2} + 4 \, \log \relax (2) + 4\right )} \sqrt {8 \, \log \relax (2) + 15}} - \frac {5 \, \log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right )}{2 \, {\left (\log \relax (2)^{2} + 4 \, \log \relax (2) + 4\right )}} + \frac {3 \, \log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right )}{2 \, {\left (\log \relax (2) + 2\right )}} + \frac {5 \, \log \relax (x)}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {3 \, \log \relax (x)}{\log \relax (2) + 2} + \frac {3 \, \arctan \left (\frac {2 \, x + 1}{\sqrt {8 \, \log \relax (2) + 15}}\right )}{\sqrt {8 \, \log \relax (2) + 15} {\left (\log \relax (2) + 2\right )}} + \frac {10 \, \log \relax (2)^{2} - {\left (x {\left (3 \, \log \relax (2) + 1\right )} + 2 \, \log \relax (2)^{2} + 8 \, \log \relax (2) + 8\right )} \log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right ) + 20 \, \log \relax (2)}{2 \, {\left (\log \relax (2)^{2} + 4 \, \log \relax (2) + 4\right )} x} + \frac {10}{x {\left (\log \relax (2) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*log(2) + 1)*log(x)/(log(2)^2 + 4*log(2) + 4) + (4*log(2)^2 - 7*log(2) - 25)*arctan((2*x + 1)/sqrt(8*log(2)
+ 15))/((log(2)^2 + 4*log(2) + 4)*sqrt(8*log(2) + 15)) - 4*arctan((2*x + 1)/sqrt(8*log(2) + 15))/sqrt(8*log(2)
+ 15) + 5*(4*log(2) + 7)*arctan((2*x + 1)/sqrt(8*log(2) + 15))/((log(2)^2 + 4*log(2) + 4)*sqrt(8*log(2) + 15)
) - 5/2*log(x^2 + x + 2*log(2) + 4)/(log(2)^2 + 4*log(2) + 4) + 3/2*log(x^2 + x + 2*log(2) + 4)/(log(2) + 2) +
5*log(x)/(log(2)^2 + 4*log(2) + 4) - 3*log(x)/(log(2) + 2) + 3*arctan((2*x + 1)/sqrt(8*log(2) + 15))/(sqrt(8*
log(2) + 15)*(log(2) + 2)) + 1/2*(10*log(2)^2 - (x*(3*log(2) + 1) + 2*log(2)^2 + 8*log(2) + 8)*log(x^2 + x + 2
*log(2) + 4) + 20*log(2))/((log(2)^2 + 4*log(2) + 4)*x) + 10/(x*(log(2) + 2))

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mupad [B]  time = 3.72, size = 22, normalized size = 0.88 \begin {gather*} -\frac {\ln \left (x+\ln \left (\frac {4}{x}\right )+\ln \relax (x)+x^2+4\right )-5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(x + log(4/x) + log(x) + x^2 + 4) - 5)/x

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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