3.5.89 \(\int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} (12 e^6+3 e x^2)-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx\)

Optimal. Leaf size=24 \[ 5+x-4 \log \left (-4-3 e^{1+\frac {e^5}{x}}+x+\log (5)\right ) \]

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Rubi [F]  time = 1.05, 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 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*x^2 - x^3 + E^(E^5/x)*(12*E^6 + 3*E*x^2) - x^2*Log[5])/(4*x^2 + 3*E^(1 + E^5/x)*x^2 - x^3 - x^2*Log[5])
,x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{3 e^{1+\frac {e^5}{x}} x^2-x^3+x^2 (4-\log (5))} \, dx\\ &=\int \frac {-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )+x^2 (8-\log (5))}{3 e^{1+\frac {e^5}{x}} x^2-x^3+x^2 (4-\log (5))} \, dx\\ &=\int \left (\frac {4 e^5+x^2}{x^2}+\frac {4 \left (e^5 x+x^2-e^5 (4-\log (5))\right )}{x^2 \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )}\right ) \, dx\\ &=4 \int \frac {e^5 x+x^2-e^5 (4-\log (5))}{x^2 \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )} \, dx+\int \frac {4 e^5+x^2}{x^2} \, dx\\ &=4 \int \left (\frac {1}{3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )}+\frac {e^5}{x \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )}+\frac {e^5 (-4+\log (5))}{x^2 \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )}\right ) \, dx+\int \left (1+\frac {4 e^5}{x^2}\right ) \, dx\\ &=-\frac {4 e^5}{x}+x+4 \int \frac {1}{3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )} \, dx+\left (4 e^5\right ) \int \frac {1}{x \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )} \, dx-\left (4 e^5 (4-\log (5))\right ) \int \frac {1}{x^2 \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 27, normalized size = 1.12 \begin {gather*} x-4 \log \left (4+3 e^{1+\frac {e^5}{x}}-x-\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x^2 - x^3 + E^(E^5/x)*(12*E^6 + 3*E*x^2) - x^2*Log[5])/(4*x^2 + 3*E^(1 + E^5/x)*x^2 - x^3 - x^2*L
og[5]),x]

[Out]

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

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fricas [A]  time = 0.82, size = 25, normalized size = 1.04 \begin {gather*} x - 4 \, \log \left (-x + 3 \, e^{\left (\frac {x + e^{5}}{x}\right )} - \log \relax (5) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*exp(1)*exp(5)+3*x^2*exp(1))*exp(exp(5)/x)-x^2*log(5)-x^3+8*x^2)/(3*x^2*exp(1)*exp(exp(5)/x)-x^2
*log(5)-x^3+4*x^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.61, size = 67, normalized size = 2.79 \begin {gather*} -x {\left (\frac {4 \, e^{10} \log \left (-\frac {e^{5} \log \relax (5)}{x} + \frac {4 \, e^{5}}{x} + \frac {3 \, e^{\left (\frac {e^{5}}{x} + 6\right )}}{x} - e^{5}\right )}{x} - \frac {4 \, e^{10} \log \left (\frac {e^{5}}{x}\right )}{x} - e^{10}\right )} e^{\left (-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*exp(1)*exp(5)+3*x^2*exp(1))*exp(exp(5)/x)-x^2*log(5)-x^3+8*x^2)/(3*x^2*exp(1)*exp(exp(5)/x)-x^2
*log(5)-x^3+4*x^2),x, algorithm="giac")

[Out]

-x*(4*e^10*log(-e^5*log(5)/x + 4*e^5/x + 3*e^(e^5/x + 6)/x - e^5)/x - 4*e^10*log(e^5/x)/x - e^10)*e^(-10)

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maple [A]  time = 0.40, size = 23, normalized size = 0.96

method	result	size
risch	\(x -4 \ln \left ({\mathrm e}^{\frac {{\mathrm e}^{5}}{x}}-\frac {\left (\ln \relax (5)-4+x \right ) {\mathrm e}^{-1}}{3}\right )\)	\(23\)
norman	\(x -4 \ln \left (3 \,{\mathrm e}^{\frac {{\mathrm e}^{5}}{x}} {\mathrm e}-\ln \relax (5)-x +4\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*exp(1)*exp(5)+3*x^2*exp(1))*exp(exp(5)/x)-x^2*ln(5)-x^3+8*x^2)/(3*x^2*exp(1)*exp(exp(5)/x)-x^2*ln(5)-
x^3+4*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.60, size = 25, normalized size = 1.04 \begin {gather*} x - 4 \, \log \left (-\frac {1}{3} \, {\left (x - 3 \, e^{\left (\frac {e^{5}}{x} + 1\right )} + \log \relax (5) - 4\right )} e^{\left (-1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*exp(1)*exp(5)+3*x^2*exp(1))*exp(exp(5)/x)-x^2*log(5)-x^3+8*x^2)/(3*x^2*exp(1)*exp(exp(5)/x)-x^2
*log(5)-x^3+4*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.89, size = 31, normalized size = 1.29 \begin {gather*} x+4\,\ln \left (\frac {1}{x}\right )-4\,\ln \left (\frac {x+\ln \relax (5)-3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^5}{x}}\,\mathrm {e}-4}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(5)/x)*(12*exp(6) + 3*x^2*exp(1)) - x^2*log(5) + 8*x^2 - x^3)/(x^2*log(5) - 4*x^2 + x^3 - 3*x^2*e
xp(exp(5)/x)*exp(1)),x)

[Out]

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

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sympy [A]  time = 0.24, size = 22, normalized size = 0.92 \begin {gather*} x - 4 \log {\left (\frac {- x - \log {\relax (5 )} + 4}{3 e} + e^{\frac {e^{5}}{x}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*exp(1)*exp(5)+3*x**2*exp(1))*exp(exp(5)/x)-x**2*ln(5)-x**3+8*x**2)/(3*x**2*exp(1)*exp(exp(5)/x)
-x**2*ln(5)-x**3+4*x**2),x)

[Out]

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

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