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

Optimal. Leaf size=25 \[ 5+x \log (x)+\frac {1}{4} \log \left (\frac {5}{x}-e^x x^2\right ) \]

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Rubi [F]  time = 0.80, 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 {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x/4 + Log[x]/2 + x*Log[x] + (5*Defer[Int][(-5 + E^x*x^3)^(-1), x])/4 + (15*Defer[Int][1/(x*(-5 + E^x*x^3)), x]
)/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5+20 x+20 x \log (x)-e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{4 x \left (5-e^x x^3\right )} \, dx\\ &=\frac {1}{4} \int \frac {-5+20 x+20 x \log (x)-e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{x \left (5-e^x x^3\right )} \, dx\\ &=\frac {1}{4} \int \left (\frac {5 (3+x)}{x \left (-5+e^x x^3\right )}+\frac {2+5 x+4 x \log (x)}{x}\right ) \, dx\\ &=\frac {1}{4} \int \frac {2+5 x+4 x \log (x)}{x} \, dx+\frac {5}{4} \int \frac {3+x}{x \left (-5+e^x x^3\right )} \, dx\\ &=\frac {1}{4} \int \left (\frac {2+5 x}{x}+4 \log (x)\right ) \, dx+\frac {5}{4} \int \left (\frac {1}{-5+e^x x^3}+\frac {3}{x \left (-5+e^x x^3\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {2+5 x}{x} \, dx+\frac {5}{4} \int \frac {1}{-5+e^x x^3} \, dx+\frac {15}{4} \int \frac {1}{x \left (-5+e^x x^3\right )} \, dx+\int \log (x) \, dx\\ &=-x+x \log (x)+\frac {1}{4} \int \left (5+\frac {2}{x}\right ) \, dx+\frac {5}{4} \int \frac {1}{-5+e^x x^3} \, dx+\frac {15}{4} \int \frac {1}{x \left (-5+e^x x^3\right )} \, dx\\ &=\frac {x}{4}+\frac {\log (x)}{2}+x \log (x)+\frac {5}{4} \int \frac {1}{-5+e^x x^3} \, dx+\frac {15}{4} \int \frac {1}{x \left (-5+e^x x^3\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{4} \left (-\log (x)+4 x \log (x)+\log \left (5-e^x x^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-Log[x] + 4*x*Log[x] + Log[5 - E^x*x^3])/4

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fricas [A]  time = 0.80, size = 28, normalized size = 1.12 \begin {gather*} \frac {1}{4} \, {\left (4 \, x + 1\right )} \log \relax (x) + \frac {1}{4} \, \log \left (\frac {x^{2} e^{\left (x + \log \relax (x)\right )} - 5}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x + 1)*log(x) + 1/4*log((x^2*e^(x + log(x)) - 5)/x^2)

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giac [A]  time = 0.29, size = 20, normalized size = 0.80 \begin {gather*} x \log \relax (x) + \frac {1}{4} \, \log \left (x^{3} e^{x} - 5\right ) - \frac {1}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(x) + 1/4*log(x^3*e^x - 5) - 1/4*log(x)

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maple [A]  time = 0.04, size = 23, normalized size = 0.92

method	result	size
risch	\(x \ln \relax (x )+\frac {\ln \relax (x )}{4}+\frac {\ln \left ({\mathrm e}^{x} x -\frac {5}{x^{2}}\right )}{4}\)	\(23\)
norman	\(x \ln \relax (x )-\frac {\ln \relax (x )}{4}+\frac {\ln \left (x^{2} {\mathrm e}^{x +\ln \relax (x )}-5\right )}{4}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.83, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{2} \, {\left (2 \, x + 1\right )} \log \relax (x) + \frac {1}{4} \, \log \left (\frac {x^{3} e^{x} - 5}{x^{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.31, size = 26, normalized size = 1.04 \begin {gather*} \frac {\ln \left (\frac {1}{x^2}\right )}{4}+\frac {\ln \left (x^3\,{\mathrm {e}}^x-5\right )}{4}+\frac {\ln \relax (x)}{4}+x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.32, size = 20, normalized size = 0.80 \begin {gather*} x \log {\relax (x )} + \frac {\log {\relax (x )}}{2} + \frac {\log {\left (e^{x} - \frac {5}{x^{3}} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(x) + log(x)/2 + log(exp(x) - 5/x**3)/4

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