3.50.42 \(\int \frac {e^{4-x+x^2} (5 x^2-x^3+(-15 x^2+7 x^3-11 x^4+2 x^5) \log (x) \log (\log (x)))}{(225-90 x+9 x^2) \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^{4-x+x^2} x^3}{9 (-5+x) \log (\log (x))} \]

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Rubi [F]  time = 2.99, 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 {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{\left (225-90 x+9 x^2\right ) \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(4 - x + x^2)*(5*x^2 - x^3 + (-15*x^2 + 7*x^3 - 11*x^4 + 2*x^5)*Log[x]*Log[Log[x]]))/((225 - 90*x + 9*x
^2)*Log[x]*Log[Log[x]]^2),x]

[Out]

(-5*Defer[Int][E^(4 - x + x^2)/(Log[x]*Log[Log[x]]^2), x])/9 - (25*Defer[Int][E^(4 - x + x^2)/((-5 + x)*Log[x]
*Log[Log[x]]^2), x])/9 - Defer[Int][(E^(4 - x + x^2)*x)/(Log[x]*Log[Log[x]]^2), x]/9 + (230*Defer[Int][E^(4 -
x + x^2)/Log[Log[x]], x])/9 - (125*Defer[Int][E^(4 - x + x^2)/((-5 + x)^2*Log[Log[x]]), x])/9 + 125*Defer[Int]
[E^(4 - x + x^2)/((-5 + x)*Log[Log[x]]), x] + (47*Defer[Int][(E^(4 - x + x^2)*x)/Log[Log[x]], x])/9 + Defer[In
t][(E^(4 - x + x^2)*x^2)/Log[Log[x]], x] + (2*Defer[Int][(E^(4 - x + x^2)*x^3)/Log[Log[x]], x])/9

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{9 (-5+x)^2 \log (x) \log ^2(\log (x))} \, dx\\ &=\frac {1}{9} \int \frac {e^{4-x+x^2} \left (5 x^2-x^3+\left (-15 x^2+7 x^3-11 x^4+2 x^5\right ) \log (x) \log (\log (x))\right )}{(-5+x)^2 \log (x) \log ^2(\log (x))} \, dx\\ &=\frac {1}{9} \int \left (-\frac {e^{4-x+x^2} x^2}{(-5+x) \log (x) \log ^2(\log (x))}+\frac {e^{4-x+x^2} x^2 \left (-15+7 x-11 x^2+2 x^3\right )}{(-5+x)^2 \log (\log (x))}\right ) \, dx\\ &=-\left (\frac {1}{9} \int \frac {e^{4-x+x^2} x^2}{(-5+x) \log (x) \log ^2(\log (x))} \, dx\right )+\frac {1}{9} \int \frac {e^{4-x+x^2} x^2 \left (-15+7 x-11 x^2+2 x^3\right )}{(-5+x)^2 \log (\log (x))} \, dx\\ &=-\left (\frac {1}{9} \int \left (\frac {5 e^{4-x+x^2}}{\log (x) \log ^2(\log (x))}+\frac {25 e^{4-x+x^2}}{(-5+x) \log (x) \log ^2(\log (x))}+\frac {e^{4-x+x^2} x}{\log (x) \log ^2(\log (x))}\right ) \, dx\right )+\frac {1}{9} \int \left (\frac {230 e^{4-x+x^2}}{\log (\log (x))}-\frac {125 e^{4-x+x^2}}{(-5+x)^2 \log (\log (x))}+\frac {1125 e^{4-x+x^2}}{(-5+x) \log (\log (x))}+\frac {47 e^{4-x+x^2} x}{\log (\log (x))}+\frac {9 e^{4-x+x^2} x^2}{\log (\log (x))}+\frac {2 e^{4-x+x^2} x^3}{\log (\log (x))}\right ) \, dx\\ &=-\left (\frac {1}{9} \int \frac {e^{4-x+x^2} x}{\log (x) \log ^2(\log (x))} \, dx\right )+\frac {2}{9} \int \frac {e^{4-x+x^2} x^3}{\log (\log (x))} \, dx-\frac {5}{9} \int \frac {e^{4-x+x^2}}{\log (x) \log ^2(\log (x))} \, dx-\frac {25}{9} \int \frac {e^{4-x+x^2}}{(-5+x) \log (x) \log ^2(\log (x))} \, dx+\frac {47}{9} \int \frac {e^{4-x+x^2} x}{\log (\log (x))} \, dx-\frac {125}{9} \int \frac {e^{4-x+x^2}}{(-5+x)^2 \log (\log (x))} \, dx+\frac {230}{9} \int \frac {e^{4-x+x^2}}{\log (\log (x))} \, dx+125 \int \frac {e^{4-x+x^2}}{(-5+x) \log (\log (x))} \, dx+\int \frac {e^{4-x+x^2} x^2}{\log (\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.49, size = 27, normalized size = 1.00 \begin {gather*} \frac {e^{4-x+x^2} x^3}{9 (-5+x) \log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4 - x + x^2)*(5*x^2 - x^3 + (-15*x^2 + 7*x^3 - 11*x^4 + 2*x^5)*Log[x]*Log[Log[x]]))/((225 - 90*x
+ 9*x^2)*Log[x]*Log[Log[x]]^2),x]

[Out]

(E^(4 - x + x^2)*x^3)/(9*(-5 + x)*Log[Log[x]])

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fricas [A]  time = 0.71, size = 25, normalized size = 0.93 \begin {gather*} \frac {x^{3} e^{\left (x^{2} - x - \log \left (\log \left (\log \relax (x)\right )\right ) + 4\right )}}{9 \, {\left (x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5-11*x^4+7*x^3-15*x^2)*log(x)*log(log(x))-x^3+5*x^2)*exp(-log(log(log(x)))+x^2-x+4)/(9*x^2-90*
x+225)/log(x)/log(log(x)),x, algorithm="fricas")

[Out]

1/9*x^3*e^(x^2 - x - log(log(log(x))) + 4)/(x - 5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5-11*x^4+7*x^3-15*x^2)*log(x)*log(log(x))-x^3+5*x^2)*exp(-log(log(log(x)))+x^2-x+4)/(9*x^2-90*
x+225)/log(x)/log(log(x)),x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.18, size = 25, normalized size = 0.93

method	result	size
risch	\(\frac {x^{3} {\mathrm e}^{x^{2}-x +4}}{9 \left (x -5\right ) \ln \left (\ln \relax (x )\right )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^5-11*x^4+7*x^3-15*x^2)*ln(x)*ln(ln(x))-x^3+5*x^2)*exp(-ln(ln(ln(x)))+x^2-x+4)/(9*x^2-90*x+225)/ln(x)
/ln(ln(x)),x,method=_RETURNVERBOSE)

[Out]

1/9*x^3/(x-5)/ln(ln(x))*exp(x^2-x+4)

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maxima [A]  time = 0.43, size = 24, normalized size = 0.89 \begin {gather*} \frac {x^{3} e^{\left (x^{2} - x + 4\right )}}{9 \, {\left (x - 5\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5-11*x^4+7*x^3-15*x^2)*log(x)*log(log(x))-x^3+5*x^2)*exp(-log(log(log(x)))+x^2-x+4)/(9*x^2-90*
x+225)/log(x)/log(log(x)),x, algorithm="maxima")

[Out]

1/9*x^3*e^(x^2 - x + 4)/((x - 5)*log(log(x)))

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mupad [B]  time = 3.79, size = 29, normalized size = 1.07 \begin {gather*} -\frac {x^3\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^4}{9\,\left (5\,\ln \left (\ln \relax (x)\right )-x\,\ln \left (\ln \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x^2 - log(log(log(x))) - x + 4)*(x^3 - 5*x^2 + log(log(x))*log(x)*(15*x^2 - 7*x^3 + 11*x^4 - 2*x^5))
)/(log(log(x))*log(x)*(9*x^2 - 90*x + 225)),x)

[Out]

-(x^3*exp(-x)*exp(x^2)*exp(4))/(9*(5*log(log(x)) - x*log(log(x))))

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sympy [A]  time = 0.45, size = 26, normalized size = 0.96 \begin {gather*} \frac {x^{3} e^{x^{2} - x + 4}}{9 x \log {\left (\log {\relax (x )} \right )} - 45 \log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**5-11*x**4+7*x**3-15*x**2)*ln(x)*ln(ln(x))-x**3+5*x**2)*exp(-ln(ln(ln(x)))+x**2-x+4)/(9*x**2-9
0*x+225)/ln(x)/ln(ln(x)),x)

[Out]

x**3*exp(x**2 - x + 4)/(9*x*log(log(x)) - 45*log(log(x)))

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