3.69.94 \(\int \frac {-60 e^{\frac {19}{9}+\frac {20 e^2}{x}}+x^2}{3 \sqrt [9]{e} x^2+3 e^{\frac {1}{9}+\frac {20 e^2}{x}} x^2+x^3} \, dx\)

Optimal. Leaf size=23 \[ \log \left (\sqrt [9]{e} \left (3+3 e^{\frac {20 e^2}{x}}\right )+x\right ) \]

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Rubi [F]  time = 1.37, 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 {-60 e^{\frac {19}{9}+\frac {20 e^2}{x}}+x^2}{3 \sqrt [9]{e} x^2+3 e^{\frac {1}{9}+\frac {20 e^2}{x}} x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-60*E^(19/9 + (20*E^2)/x) + x^2)/(3*E^(1/9)*x^2 + 3*E^(1/9 + (20*E^2)/x)*x^2 + x^3),x]

[Out]

(20*E^2)/x + Defer[Int][(3*E^(1/9) + 3*E^(1/9 + (20*E^2)/x) + x)^(-1), x] + 60*E^(19/9)*Defer[Int][1/(x^2*(3*E
^(1/9) + 3*E^(1/9 + (20*E^2)/x) + x)), x] + 20*E^2*Defer[Int][1/(x*(3*E^(1/9) + 3*E^(1/9 + (20*E^2)/x) + x)),
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {20 e^2}{x^2}+\frac {60 e^{19/9}+20 e^2 x+x^2}{x^2 \left (3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x\right )}\right ) \, dx\\ &=\frac {20 e^2}{x}+\int \frac {60 e^{19/9}+20 e^2 x+x^2}{x^2 \left (3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x\right )} \, dx\\ &=\frac {20 e^2}{x}+\int \left (\frac {1}{3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x}+\frac {60 e^{19/9}}{x^2 \left (3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x\right )}+\frac {20 e^2}{x \left (3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x\right )}\right ) \, dx\\ &=\frac {20 e^2}{x}+\left (20 e^2\right ) \int \frac {1}{x \left (3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x\right )} \, dx+\left (60 e^{19/9}\right ) \int \frac {1}{x^2 \left (3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x\right )} \, dx+\int \frac {1}{3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 26, normalized size = 1.13 \begin {gather*} \log \left (3 \sqrt [9]{e}+3 e^{\frac {1}{9}+\frac {20 e^2}{x}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60*E^(19/9 + (20*E^2)/x) + x^2)/(3*E^(1/9)*x^2 + 3*E^(1/9 + (20*E^2)/x)*x^2 + x^3),x]

[Out]

Log[3*E^(1/9) + 3*E^(1/9 + (20*E^2)/x) + x]

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fricas [A]  time = 0.66, size = 26, normalized size = 1.13 \begin {gather*} \log \left (x e^{2} + 3 \, e^{\frac {19}{9}} + 3 \, e^{\left (\frac {19 \, x + 180 \, e^{2}}{9 \, x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*exp(1/9)*exp(2)*exp(20*exp(2)/x)+x^2)/(3*x^2*exp(1/9)*exp(20*exp(2)/x)+3*x^2*exp(1/9)+x^3),x, a
lgorithm="fricas")

[Out]

log(x*e^2 + 3*e^(19/9) + 3*e^(1/9*(19*x + 180*e^2)/x))

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giac [B]  time = 0.19, size = 44, normalized size = 1.91 \begin {gather*} {\left (e^{2} \log \left (\frac {3 \, e^{2}}{x} + \frac {3 \, e^{\left (\frac {20 \, e^{2}}{x} + 2\right )}}{x} + e^{\frac {17}{9}}\right ) - e^{2} \log \left (\frac {e^{2}}{x}\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*exp(1/9)*exp(2)*exp(20*exp(2)/x)+x^2)/(3*x^2*exp(1/9)*exp(20*exp(2)/x)+3*x^2*exp(1/9)+x^3),x, a
lgorithm="giac")

[Out]

(e^2*log(3*e^2/x + 3*e^(20*e^2/x + 2)/x + e^(17/9)) - e^2*log(e^2/x))*e^(-2)

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maple [A]  time = 0.24, size = 20, normalized size = 0.87

method	result	size
norman	\(\ln \left (3 \,{\mathrm e}^{\frac {1}{9}} {\mathrm e}^{\frac {20 \,{\mathrm e}^{2}}{x}}+3 \,{\mathrm e}^{\frac {1}{9}}+x \right )\)	\(20\)
risch	\(\ln \left ({\mathrm e}^{\frac {20 \,{\mathrm e}^{2}}{x}}+\frac {\left (3 \,{\mathrm e}^{\frac {1}{9}}+x \right ) {\mathrm e}^{-\frac {1}{9}}}{3}\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-60*exp(1/9)*exp(2)*exp(20*exp(2)/x)+x^2)/(3*x^2*exp(1/9)*exp(20*exp(2)/x)+3*x^2*exp(1/9)+x^3),x,method=_
RETURNVERBOSE)

[Out]

ln(3*exp(1/9)*exp(20*exp(2)/x)+3*exp(1/9)+x)

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maxima [A]  time = 0.38, size = 23, normalized size = 1.00 \begin {gather*} \log \left (\frac {1}{3} \, {\left (x + 3 \, e^{\frac {1}{9}} + 3 \, e^{\left (\frac {20 \, e^{2}}{x} + \frac {1}{9}\right )}\right )} e^{\left (-\frac {1}{9}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*exp(1/9)*exp(2)*exp(20*exp(2)/x)+x^2)/(3*x^2*exp(1/9)*exp(20*exp(2)/x)+3*x^2*exp(1/9)+x^3),x, a
lgorithm="maxima")

[Out]

log(1/3*(x + 3*e^(1/9) + 3*e^(20*e^2/x + 1/9))*e^(-1/9))

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mupad [B]  time = 4.69, size = 30, normalized size = 1.30 \begin {gather*} \ln \left (\frac {x+3\,{\mathrm {e}}^{1/9}+3\,{\mathrm {e}}^{\frac {20\,{\mathrm {e}}^2}{x}}\,{\mathrm {e}}^{1/9}}{x}\right )-\ln \left (\frac {1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(60*exp((20*exp(2))/x)*exp(19/9) - x^2)/(3*x^2*exp(1/9) + x^3 + 3*x^2*exp((20*exp(2))/x)*exp(1/9)),x)

[Out]

log((x + 3*exp(1/9) + 3*exp((20*exp(2))/x)*exp(1/9))/x) - log(1/x)

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sympy [A]  time = 0.20, size = 24, normalized size = 1.04 \begin {gather*} \log {\left (\frac {x + 3 e^{\frac {1}{9}}}{3 e^{\frac {1}{9}}} + e^{\frac {20 e^{2}}{x}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*exp(1/9)*exp(2)*exp(20*exp(2)/x)+x**2)/(3*x**2*exp(1/9)*exp(20*exp(2)/x)+3*x**2*exp(1/9)+x**3),
x)

[Out]

log((x + 3*exp(1/9))*exp(-1/9)/3 + exp(20*exp(2)/x))

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