3.14.44 \(\int \frac {1+3 x+x^2-x \log (x)}{-4 x+5 e^x x-x^2+x \log (x)} \, dx\)

Optimal. Leaf size=17 \[ -x+\log \left (-4+5 e^x-x+\log (x)\right ) \]

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Rubi [F]  time = 0.45, 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 {1+3 x+x^2-x \log (x)}{-4 x+5 e^x x-x^2+x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 17, normalized size = 1.00 \begin {gather*} -x+\log \left (4-5 e^x+x-\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x + Log[4 - 5*E^x + x - Log[x]]

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fricas [A]  time = 1.50, size = 16, normalized size = 0.94 \begin {gather*} -x + \log \left (-x + 5 \, e^{x} + \log \relax (x) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+x^2+3*x+1)/(x*log(x)+5*exp(x)*x-x^2-4*x),x, algorithm="fricas")

[Out]

-x + log(-x + 5*e^x + log(x) - 4)

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giac [A]  time = 0.39, size = 16, normalized size = 0.94 \begin {gather*} -x + \log \left (x - 5 \, e^{x} - \log \relax (x) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+x^2+3*x+1)/(x*log(x)+5*exp(x)*x-x^2-4*x),x, algorithm="giac")

[Out]

-x + log(x - 5*e^x - log(x) + 4)

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maple [A]  time = 0.02, size = 17, normalized size = 1.00

method	result	size
norman	\(-x +\ln \left (x -5 \,{\mathrm e}^{x}-\ln \relax (x )+4\right )\)	\(17\)
risch	\(\ln \left (\ln \relax (x )+5 \,{\mathrm e}^{x}-4-x \right )-x\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x)+x^2+3*x+1)/(x*ln(x)+5*exp(x)*x-x^2-4*x),x,method=_RETURNVERBOSE)

[Out]

-x+ln(x-5*exp(x)-ln(x)+4)

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maxima [A]  time = 0.51, size = 16, normalized size = 0.94 \begin {gather*} -x + \log \left (-\frac {1}{5} \, x + e^{x} + \frac {1}{5} \, \log \relax (x) - \frac {4}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+x^2+3*x+1)/(x*log(x)+5*exp(x)*x-x^2-4*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.07, size = 16, normalized size = 0.94 \begin {gather*} \ln \left (x-5\,{\mathrm {e}}^x-\ln \relax (x)+4\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 5*exp(x) - log(x) + 4) - x

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sympy [A]  time = 0.35, size = 17, normalized size = 1.00 \begin {gather*} - x + \log {\left (- \frac {x}{5} + e^{x} + \frac {\log {\relax (x )}}{5} - \frac {4}{5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x)+x**2+3*x+1)/(x*ln(x)+5*exp(x)*x-x**2-4*x),x)

[Out]

-x + log(-x/5 + exp(x) + log(x)/5 - 4/5)

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