3.43.13 \(\int \frac {-27+9 x+e^{e^x} (-10 x-x^2+x^3+e^x (-5 x^2+x^3))}{-45+9 x+e^{e^x} (-5 x^2+x^3)} \, dx\)

Optimal. Leaf size=24 \[ -5+x+\log \left ((-5+x)^2\right )+\log \left (3+\frac {1}{3} e^{e^x} x^2\right ) \]

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Rubi [F]  time = 2.91, 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 {-27+9 x+e^{e^x} \left (-10 x-x^2+x^3+e^x \left (-5 x^2+x^3\right )\right )}{-45+9 x+e^{e^x} \left (-5 x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-27 + 9*x + E^E^x*(-10*x - x^2 + x^3 + E^x*(-5*x^2 + x^3)))/(-45 + 9*x + E^E^x*(-5*x^2 + x^3)),x]

[Out]

x + 10*Defer[Int][E^E^x/(9 + E^E^x*x^2), x] + 18*Defer[Int][1/((-5 + x)*(9 + E^E^x*x^2)), x] + 50*Defer[Int][E
^E^x/((-5 + x)*(9 + E^E^x*x^2)), x] + 4*Defer[Int][(E^E^x*x)/(9 + E^E^x*x^2), x] + Defer[Int][(E^(E^x + x)*x^2
)/(9 + E^E^x*x^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {27-9 x-e^{e^x} \left (-10 x-x^2+x^3+e^x \left (-5 x^2+x^3\right )\right )}{(5-x) \left (9+e^{e^x} x^2\right )} \, dx\\ &=\int \left (-\frac {27}{(-5+x) \left (9+e^{e^x} x^2\right )}+\frac {9 x}{(-5+x) \left (9+e^{e^x} x^2\right )}-\frac {10 e^{e^x} x}{(-5+x) \left (9+e^{e^x} x^2\right )}+\frac {e^{e^x+x} x^2}{9+e^{e^x} x^2}-\frac {e^{e^x} x^2}{(-5+x) \left (9+e^{e^x} x^2\right )}+\frac {e^{e^x} x^3}{(-5+x) \left (9+e^{e^x} x^2\right )}\right ) \, dx\\ &=9 \int \frac {x}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-10 \int \frac {e^{e^x} x}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-27 \int \frac {1}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+\int \frac {e^{e^x+x} x^2}{9+e^{e^x} x^2} \, dx-\int \frac {e^{e^x} x^2}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+\int \frac {e^{e^x} x^3}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx\\ &=9 \int \left (\frac {1}{9+e^{e^x} x^2}+\frac {5}{(-5+x) \left (9+e^{e^x} x^2\right )}\right ) \, dx-10 \int \left (\frac {e^{e^x}}{9+e^{e^x} x^2}+\frac {5 e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )}\right ) \, dx-27 \int \frac {1}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+\int \frac {e^{e^x+x} x^2}{9+e^{e^x} x^2} \, dx-\int \left (\frac {5 e^{e^x}}{9+e^{e^x} x^2}+\frac {25 e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )}+\frac {e^{e^x} x}{9+e^{e^x} x^2}\right ) \, dx+\int \left (\frac {25 e^{e^x}}{9+e^{e^x} x^2}+\frac {125 e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )}+\frac {5 e^{e^x} x}{9+e^{e^x} x^2}+\frac {e^{e^x} x^2}{9+e^{e^x} x^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx\right )+5 \int \frac {e^{e^x} x}{9+e^{e^x} x^2} \, dx+9 \int \frac {1}{9+e^{e^x} x^2} \, dx-10 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx+25 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx-25 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-27 \int \frac {1}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+45 \int \frac {1}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-50 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+125 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-\int \frac {e^{e^x} x}{9+e^{e^x} x^2} \, dx+\int \frac {e^{e^x} x^2}{9+e^{e^x} x^2} \, dx+\int \frac {e^{e^x+x} x^2}{9+e^{e^x} x^2} \, dx\\ &=-\left (5 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx\right )+5 \int \frac {e^{e^x} x}{9+e^{e^x} x^2} \, dx+9 \int \frac {1}{9+e^{e^x} x^2} \, dx-10 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx+25 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx-25 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-27 \int \frac {1}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+45 \int \frac {1}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-50 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+125 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-\int \frac {e^{e^x} x}{9+e^{e^x} x^2} \, dx+\int \frac {e^{e^x+x} x^2}{9+e^{e^x} x^2} \, dx+\int \left (1-\frac {9}{9+e^{e^x} x^2}\right ) \, dx\\ &=x-5 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx+5 \int \frac {e^{e^x} x}{9+e^{e^x} x^2} \, dx-10 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx+25 \int \frac {e^{e^x}}{9+e^{e^x} x^2} \, dx-25 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-27 \int \frac {1}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+45 \int \frac {1}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-50 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx+125 \int \frac {e^{e^x}}{(-5+x) \left (9+e^{e^x} x^2\right )} \, dx-\int \frac {e^{e^x} x}{9+e^{e^x} x^2} \, dx+\int \frac {e^{e^x+x} x^2}{9+e^{e^x} x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 22, normalized size = 0.92 \begin {gather*} x+2 \log (5-x)+\log \left (9+e^{e^x} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-27 + 9*x + E^E^x*(-10*x - x^2 + x^3 + E^x*(-5*x^2 + x^3)))/(-45 + 9*x + E^E^x*(-5*x^2 + x^3)),x]

[Out]

x + 2*Log[5 - x] + Log[9 + E^E^x*x^2]

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fricas [A]  time = 0.61, size = 26, normalized size = 1.08 \begin {gather*} x + 2 \, \log \left (x^{2} - 5 \, x\right ) + \log \left (\frac {x^{2} e^{\left (e^{x}\right )} + 9}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3-5*x^2)*exp(x)+x^3-x^2-10*x)*exp(exp(x))+9*x-27)/((x^3-5*x^2)*exp(exp(x))+9*x-45),x, algorithm
="fricas")

[Out]

x + 2*log(x^2 - 5*x) + log((x^2*e^(e^x) + 9)/x^2)

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giac [A]  time = 0.20, size = 22, normalized size = 0.92 \begin {gather*} \log \left (x^{2} e^{\left (x + e^{x}\right )} + 9 \, e^{x}\right ) + 2 \, \log \left (x - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3-5*x^2)*exp(x)+x^3-x^2-10*x)*exp(exp(x))+9*x-27)/((x^3-5*x^2)*exp(exp(x))+9*x-45),x, algorithm
="giac")

[Out]

log(x^2*e^(x + e^x) + 9*e^x) + 2*log(x - 5)

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maple [A]  time = 0.06, size = 19, normalized size = 0.79

method	result	size
norman	\(x +2 \ln \left (x -5\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x}} x^{2}+9\right )\)	\(19\)
risch	\(x +2 \ln \left (x^{2}-5 x \right )+\ln \left ({\mathrm e}^{{\mathrm e}^{x}}+\frac {9}{x^{2}}\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^3-5*x^2)*exp(x)+x^3-x^2-10*x)*exp(exp(x))+9*x-27)/((x^3-5*x^2)*exp(exp(x))+9*x-45),x,method=_RETURNVE
RBOSE)

[Out]

x+2*ln(x-5)+ln(exp(exp(x))*x^2+9)

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maxima [A]  time = 0.40, size = 26, normalized size = 1.08 \begin {gather*} x + 2 \, \log \left (x - 5\right ) + 2 \, \log \relax (x) + \log \left (\frac {x^{2} e^{\left (e^{x}\right )} + 9}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^3-5*x^2)*exp(x)+x^3-x^2-10*x)*exp(exp(x))+9*x-27)/((x^3-5*x^2)*exp(exp(x))+9*x-45),x, algorithm
="maxima")

[Out]

x + 2*log(x - 5) + 2*log(x) + log((x^2*e^(e^x) + 9)/x^2)

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mupad [B]  time = 0.13, size = 18, normalized size = 0.75 \begin {gather*} x+2\,\ln \left (x-5\right )+\ln \left (x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}+9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x))*(10*x + exp(x)*(5*x^2 - x^3) + x^2 - x^3) - 9*x + 27)/(exp(exp(x))*(5*x^2 - x^3) - 9*x + 45),
x)

[Out]

x + 2*log(x - 5) + log(x^2*exp(exp(x)) + 9)

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sympy [A]  time = 0.23, size = 22, normalized size = 0.92 \begin {gather*} x + 2 \log {\left (x^{2} - 5 x \right )} + \log {\left (e^{e^{x}} + \frac {9}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**3-5*x**2)*exp(x)+x**3-x**2-10*x)*exp(exp(x))+9*x-27)/((x**3-5*x**2)*exp(exp(x))+9*x-45),x)

[Out]

x + 2*log(x**2 - 5*x) + log(exp(exp(x)) + 9/x**2)

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