3.52.54 \(\int \frac {15-5 e^x-5 e^x x \log (x) \log (\log (x))+(27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x) \log (x) \log ^2(\log (x))}{(9 x-6 e^x x+e^{2 x} x) \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=26 \[ 4-e^x+3 (-1+x)+\frac {5}{\left (-3+e^x\right ) \log (\log (x))} \]

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Rubi [F]  time = 3.08, 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 {15-5 e^x-5 e^x x \log (x) \log (\log (x))+\left (27 x-27 e^x x+9 e^{2 x} x-e^{3 x} x\right ) \log (x) \log ^2(\log (x))}{\left (9 x-6 e^x x+e^{2 x} x\right ) \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(15 - 5*E^x - 5*E^x*x*Log[x]*Log[Log[x]] + (27*x - 27*E^x*x + 9*E^(2*x)*x - E^(3*x)*x)*Log[x]*Log[Log[x]]^
2)/((9*x - 6*E^x*x + E^(2*x)*x)*Log[x]*Log[Log[x]]^2),x]

[Out]

-E^x + 3*x - 5*Defer[Int][1/((-3 + E^x)*x*Log[x]*Log[Log[x]]^2), x] - 15*Defer[Int][1/((-3 + E^x)^2*Log[Log[x]
]), x] - 5*Defer[Int][1/((-3 + E^x)*Log[Log[x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5 \left (-3+e^x\right )-x \log (x) \log (\log (x)) \left (5 e^x+\left (-3+e^x\right )^3 \log (\log (x))\right )}{\left (3-e^x\right )^2 x \log (x) \log ^2(\log (x))} \, dx\\ &=\int \left (3-e^x-\frac {15}{\left (-3+e^x\right )^2 \log (\log (x))}-\frac {5 (1+x \log (x) \log (\log (x)))}{\left (-3+e^x\right ) x \log (x) \log ^2(\log (x))}\right ) \, dx\\ &=3 x-5 \int \frac {1+x \log (x) \log (\log (x))}{\left (-3+e^x\right ) x \log (x) \log ^2(\log (x))} \, dx-15 \int \frac {1}{\left (-3+e^x\right )^2 \log (\log (x))} \, dx-\int e^x \, dx\\ &=-e^x+3 x-5 \int \left (\frac {1}{\left (-3+e^x\right ) x \log (x) \log ^2(\log (x))}+\frac {1}{\left (-3+e^x\right ) \log (\log (x))}\right ) \, dx-15 \int \frac {1}{\left (-3+e^x\right )^2 \log (\log (x))} \, dx\\ &=-e^x+3 x-5 \int \frac {1}{\left (-3+e^x\right ) x \log (x) \log ^2(\log (x))} \, dx-5 \int \frac {1}{\left (-3+e^x\right ) \log (\log (x))} \, dx-15 \int \frac {1}{\left (-3+e^x\right )^2 \log (\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 23, normalized size = 0.88 \begin {gather*} -e^x+3 x+\frac {5}{\left (-3+e^x\right ) \log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 - 5*E^x - 5*E^x*x*Log[x]*Log[Log[x]] + (27*x - 27*E^x*x + 9*E^(2*x)*x - E^(3*x)*x)*Log[x]*Log[Lo
g[x]]^2)/((9*x - 6*E^x*x + E^(2*x)*x)*Log[x]*Log[Log[x]]^2),x]

[Out]

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

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fricas [A]  time = 0.46, size = 35, normalized size = 1.35 \begin {gather*} \frac {{\left (3 \, {\left (x + 1\right )} e^{x} - 9 \, x - e^{\left (2 \, x\right )}\right )} \log \left (\log \relax (x)\right ) + 5}{{\left (e^{x} - 3\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*log(x)*log(log(x))^2-5*x*exp(x)*log(x)*log(log(x))-5*ex
p(x)+15)/(x*exp(x)^2-6*exp(x)*x+9*x)/log(x)/log(log(x))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.15, size = 47, normalized size = 1.81 \begin {gather*} \frac {3 \, x e^{x} \log \left (\log \relax (x)\right ) - 9 \, x \log \left (\log \relax (x)\right ) - e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right ) + 3 \, e^{x} \log \left (\log \relax (x)\right ) + 5}{e^{x} \log \left (\log \relax (x)\right ) - 3 \, \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*log(x)*log(log(x))^2-5*x*exp(x)*log(x)*log(log(x))-5*ex
p(x)+15)/(x*exp(x)^2-6*exp(x)*x+9*x)/log(x)/log(log(x))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 22, normalized size = 0.85

method	result	size
risch	\(3 x -{\mathrm e}^{x}+\frac {5}{\ln \left (\ln \relax (x )\right ) \left ({\mathrm e}^{x}-3\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*ln(x)*ln(ln(x))^2-5*x*exp(x)*ln(x)*ln(ln(x))-5*exp(x)+15)/(x*
exp(x)^2-6*exp(x)*x+9*x)/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 35, normalized size = 1.35 \begin {gather*} \frac {{\left (3 \, {\left (x + 1\right )} e^{x} - 9 \, x - e^{\left (2 \, x\right )}\right )} \log \left (\log \relax (x)\right ) + 5}{{\left (e^{x} - 3\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)^3+9*x*exp(x)^2-27*exp(x)*x+27*x)*log(x)*log(log(x))^2-5*x*exp(x)*log(x)*log(log(x))-5*ex
p(x)+15)/(x*exp(x)^2-6*exp(x)*x+9*x)/log(x)/log(log(x))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.49, size = 25, normalized size = 0.96 \begin {gather*} 3\,x-{\mathrm {e}}^x-\frac {5}{3\,\ln \left (\ln \relax (x)\right )-\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(x) - log(log(x))^2*log(x)*(27*x + 9*x*exp(2*x) - x*exp(3*x) - 27*x*exp(x)) + 5*x*log(log(x))*exp(x
)*log(x) - 15)/(log(log(x))^2*log(x)*(9*x + x*exp(2*x) - 6*x*exp(x))),x)

[Out]

3*x - exp(x) - 5/(3*log(log(x)) - log(log(x))*exp(x))

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sympy [A]  time = 0.30, size = 22, normalized size = 0.85 \begin {gather*} 3 x - e^{x} + \frac {5}{e^{x} \log {\left (\log {\relax (x )} \right )} - 3 \log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(x)**3+9*x*exp(x)**2-27*exp(x)*x+27*x)*ln(x)*ln(ln(x))**2-5*x*exp(x)*ln(x)*ln(ln(x))-5*exp(x
)+15)/(x*exp(x)**2-6*exp(x)*x+9*x)/ln(x)/ln(ln(x))**2,x)

[Out]

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

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