3.37.56 \(\int \frac {-119+61 x+e^x (-x-121 x^2+60 x^3)+(60+60 e^x x^2) \log (x)}{300 x-150 x^2+e^x (-300 x^2+150 x^3)+(-150 x+150 e^x x^2) \log (x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {2}{5} \left (\log \left (-e^x+\frac {1}{x}\right )-\frac {1}{60} \log (2-x-\log (x))\right ) \]

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Rubi [F]  time = 1.36, 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 {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{300 x-150 x^2+e^x \left (-300 x^2+150 x^3\right )+\left (-150 x+150 e^x x^2\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-119 + 61*x + E^x*(-x - 121*x^2 + 60*x^3) + (60 + 60*E^x*x^2)*Log[x])/(300*x - 150*x^2 + E^x*(-300*x^2 +
150*x^3) + (-150*x + 150*E^x*x^2)*Log[x]),x]

[Out]

(2*x)/5 - Log[2 - x - Log[x]]/150 + (2*Defer[Int][(-1 + E^x*x)^(-1), x])/5 + (2*Defer[Int][1/(x*(-1 + E^x*x)),
x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{150 x \left (1-e^x x\right ) (2-x-\log (x))} \, dx\\ &=\frac {1}{150} \int \frac {-119+61 x+e^x \left (-x-121 x^2+60 x^3\right )+\left (60+60 e^x x^2\right ) \log (x)}{x \left (1-e^x x\right ) (2-x-\log (x))} \, dx\\ &=\frac {1}{150} \int \left (\frac {60 (1+x)}{x \left (-1+e^x x\right )}+\frac {-1-121 x+60 x^2+60 x \log (x)}{x (-2+x+\log (x))}\right ) \, dx\\ &=\frac {1}{150} \int \frac {-1-121 x+60 x^2+60 x \log (x)}{x (-2+x+\log (x))} \, dx+\frac {2}{5} \int \frac {1+x}{x \left (-1+e^x x\right )} \, dx\\ &=\frac {1}{150} \int \left (60+\frac {-1-x}{x (-2+x+\log (x))}\right ) \, dx+\frac {2}{5} \int \left (\frac {1}{-1+e^x x}+\frac {1}{x \left (-1+e^x x\right )}\right ) \, dx\\ &=\frac {2 x}{5}+\frac {1}{150} \int \frac {-1-x}{x (-2+x+\log (x))} \, dx+\frac {2}{5} \int \frac {1}{-1+e^x x} \, dx+\frac {2}{5} \int \frac {1}{x \left (-1+e^x x\right )} \, dx\\ &=\frac {2 x}{5}-\frac {1}{150} \log (2-x-\log (x))+\frac {2}{5} \int \frac {1}{-1+e^x x} \, dx+\frac {2}{5} \int \frac {1}{x \left (-1+e^x x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.37, size = 32, normalized size = 1.10 \begin {gather*} \frac {1}{150} \left (-60 \log (x)+60 \log \left (1-e^x x\right )-\log (2-x-\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-119 + 61*x + E^x*(-x - 121*x^2 + 60*x^3) + (60 + 60*E^x*x^2)*Log[x])/(300*x - 150*x^2 + E^x*(-300*
x^2 + 150*x^3) + (-150*x + 150*E^x*x^2)*Log[x]),x]

[Out]

(-60*Log[x] + 60*Log[1 - E^x*x] - Log[2 - x - Log[x]])/150

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fricas [A]  time = 0.67, size = 22, normalized size = 0.76 \begin {gather*} -\frac {1}{150} \, \log \left (x + \log \relax (x) - 2\right ) + \frac {2}{5} \, \log \left (\frac {x e^{x} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*exp(x)*x^2+60)*log(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/((150*exp(x)*x^2-150*x)*log(x)+(150*x
^3-300*x^2)*exp(x)-150*x^2+300*x),x, algorithm="fricas")

[Out]

-1/150*log(x + log(x) - 2) + 2/5*log((x*e^x - 1)/x)

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giac [A]  time = 0.16, size = 22, normalized size = 0.76 \begin {gather*} \frac {2}{5} \, \log \left (x e^{x} - 1\right ) - \frac {1}{150} \, \log \left (x + \log \relax (x) - 2\right ) - \frac {2}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*exp(x)*x^2+60)*log(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/((150*exp(x)*x^2-150*x)*log(x)+(150*x
^3-300*x^2)*exp(x)-150*x^2+300*x),x, algorithm="giac")

[Out]

2/5*log(x*e^x - 1) - 1/150*log(x + log(x) - 2) - 2/5*log(x)

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maple [A]  time = 0.06, size = 21, normalized size = 0.72

method	result	size
risch	\(\frac {2 \ln \left ({\mathrm e}^{x}-\frac {1}{x}\right )}{5}-\frac {\ln \left (x +\ln \relax (x )-2\right )}{150}\)	\(21\)
norman	\(-\frac {2 \ln \relax (x )}{5}-\frac {\ln \left (x +\ln \relax (x )-2\right )}{150}+\frac {2 \ln \left ({\mathrm e}^{x} x -1\right )}{5}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((60*exp(x)*x^2+60)*ln(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/((150*exp(x)*x^2-150*x)*ln(x)+(150*x^3-300*x
^2)*exp(x)-150*x^2+300*x),x,method=_RETURNVERBOSE)

[Out]

2/5*ln(exp(x)-1/x)-1/150*ln(x+ln(x)-2)

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maxima [A]  time = 0.43, size = 22, normalized size = 0.76 \begin {gather*} -\frac {1}{150} \, \log \left (x + \log \relax (x) - 2\right ) + \frac {2}{5} \, \log \left (\frac {x e^{x} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*exp(x)*x^2+60)*log(x)+(60*x^3-121*x^2-x)*exp(x)+61*x-119)/((150*exp(x)*x^2-150*x)*log(x)+(150*x
^3-300*x^2)*exp(x)-150*x^2+300*x),x, algorithm="maxima")

[Out]

-1/150*log(x + log(x) - 2) + 2/5*log((x*e^x - 1)/x)

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mupad [B]  time = 2.29, size = 22, normalized size = 0.76 \begin {gather*} \frac {2\,\ln \left (\frac {x\,{\mathrm {e}}^x-1}{x}\right )}{5}-\frac {\ln \left (x+\ln \relax (x)-2\right )}{150} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(61*x - exp(x)*(x + 121*x^2 - 60*x^3) + log(x)*(60*x^2*exp(x) + 60) - 119)/(exp(x)*(300*x^2 - 150*x^3) -
300*x + log(x)*(150*x - 150*x^2*exp(x)) + 150*x^2),x)

[Out]

(2*log((x*exp(x) - 1)/x))/5 - log(x + log(x) - 2)/150

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sympy [A]  time = 0.39, size = 20, normalized size = 0.69 \begin {gather*} \frac {2 \log {\left (e^{x} - \frac {1}{x} \right )}}{5} - \frac {\log {\left (x + \log {\relax (x )} - 2 \right )}}{150} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*exp(x)*x**2+60)*ln(x)+(60*x**3-121*x**2-x)*exp(x)+61*x-119)/((150*exp(x)*x**2-150*x)*ln(x)+(150
*x**3-300*x**2)*exp(x)-150*x**2+300*x),x)

[Out]

2*log(exp(x) - 1/x)/5 - log(x + log(x) - 2)/150

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