3.63.21 \(\int \frac {-10 x+e^{3/4} x-x^4+(50-50 x^3-x^6+e^{3/4} (-5+4 x^3)) \log (x)+(-20 x+2 e^{3/4} x+x^4) \log (x) \log (\log (x))}{(25 x^2-10 x^5+x^8) \log (x)+(-10 x^3+2 x^6) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx\) [6221]

Optimal. Leaf size=30 \[ \frac {\frac {10-e^{3/4}}{x^2}+x}{-\frac {5}{x}+x^2+\log (\log (x))} \]

[Out]

(x+(10-exp(3/4))/x^2)/(ln(ln(x))+x^2-5/x)

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Rubi [F]
time = 1.88, 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 {-10 x+e^{3/4} x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*x + E^(3/4)*x - x^4 + (50 - 50*x^3 - x^6 + E^(3/4)*(-5 + 4*x^3))*Log[x] + (-20*x + 2*E^(3/4)*x + x^4)
*Log[x]*Log[Log[x]])/((25*x^2 - 10*x^5 + x^8)*Log[x] + (-10*x^3 + 2*x^6)*Log[x]*Log[Log[x]] + x^4*Log[x]*Log[L
og[x]]^2),x]

[Out]

-5*(10 - E^(3/4))*Defer[Int][1/(x^2*(-5 + x^3 + x*Log[Log[x]])^2), x] - 5*Defer[Int][x/(-5 + x^3 + x*Log[Log[x
]])^2, x] - 2*(10 - E^(3/4))*Defer[Int][x/(-5 + x^3 + x*Log[Log[x]])^2, x] - 2*Defer[Int][x^4/(-5 + x^3 + x*Lo
g[Log[x]])^2, x] - (10 - E^(3/4))*Defer[Int][1/(x*Log[x]*(-5 + x^3 + x*Log[Log[x]])^2), x] - Defer[Int][x^2/(L
og[x]*(-5 + x^3 + x*Log[Log[x]])^2), x] - 2*(10 - E^(3/4))*Defer[Int][1/(x^2*(-5 + x^3 + x*Log[Log[x]])), x] +
Defer[Int][x/(-5 + x^3 + x*Log[Log[x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-10+e^{3/4}\right ) x-x^4+\left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )\right ) \log (x)+\left (-20 x+2 e^{3/4} x+x^4\right ) \log (x) \log (\log (x))}{\left (25 x^2-10 x^5+x^8\right ) \log (x)+\left (-10 x^3+2 x^6\right ) \log (x) \log (\log (x))+x^4 \log (x) \log ^2(\log (x))} \, dx\\ &=\int \frac {x \left (-10+e^{3/4}-x^3\right )+\log (x) \left (50-50 x^3-x^6+e^{3/4} \left (-5+4 x^3\right )+x \left (-20+2 e^{3/4}+x^3\right ) \log (\log (x))\right )}{x^2 \log (x) \left (5-x^3-x \log (\log (x))\right )^2} \, dx\\ &=\int \left (-\frac {\left (10-e^{3/4}+x^3\right ) \left (x+5 \log (x)+2 x^3 \log (x)\right )}{x^2 \log (x) \left (-5+x^3+x \log (\log (x))\right )^2}+\frac {-20+2 e^{3/4}+x^3}{x^2 \left (-5+x^3+x \log (\log (x))\right )}\right ) \, dx\\ &=-\int \frac {\left (10-e^{3/4}+x^3\right ) \left (x+5 \log (x)+2 x^3 \log (x)\right )}{x^2 \log (x) \left (-5+x^3+x \log (\log (x))\right )^2} \, dx+\int \frac {-20+2 e^{3/4}+x^3}{x^2 \left (-5+x^3+x \log (\log (x))\right )} \, dx\\ &=-\int \left (-\frac {\left (-10+e^{3/4}\right ) \left (x+5 \log (x)+2 x^3 \log (x)\right )}{x^2 \log (x) \left (-5+x^3+x \log (\log (x))\right )^2}+\frac {x \left (x+5 \log (x)+2 x^3 \log (x)\right )}{\log (x) \left (-5+x^3+x \log (\log (x))\right )^2}\right ) \, dx+\int \left (\frac {2 \left (-10+e^{3/4}\right )}{x^2 \left (-5+x^3+x \log (\log (x))\right )}+\frac {x}{-5+x^3+x \log (\log (x))}\right ) \, dx\\ &=-\left (\left (10-e^{3/4}\right ) \int \frac {x+5 \log (x)+2 x^3 \log (x)}{x^2 \log (x) \left (-5+x^3+x \log (\log (x))\right )^2} \, dx\right )-\left (2 \left (10-e^{3/4}\right )\right ) \int \frac {1}{x^2 \left (-5+x^3+x \log (\log (x))\right )} \, dx-\int \frac {x \left (x+5 \log (x)+2 x^3 \log (x)\right )}{\log (x) \left (-5+x^3+x \log (\log (x))\right )^2} \, dx+\int \frac {x}{-5+x^3+x \log (\log (x))} \, dx\\ &=-\left (\left (10-e^{3/4}\right ) \int \left (\frac {5}{x^2 \left (-5+x^3+x \log (\log (x))\right )^2}+\frac {2 x}{\left (-5+x^3+x \log (\log (x))\right )^2}+\frac {1}{x \log (x) \left (-5+x^3+x \log (\log (x))\right )^2}\right ) \, dx\right )-\left (2 \left (10-e^{3/4}\right )\right ) \int \frac {1}{x^2 \left (-5+x^3+x \log (\log (x))\right )} \, dx+\int \frac {x}{-5+x^3+x \log (\log (x))} \, dx-\int \left (\frac {5 x}{\left (-5+x^3+x \log (\log (x))\right )^2}+\frac {2 x^4}{\left (-5+x^3+x \log (\log (x))\right )^2}+\frac {x^2}{\log (x) \left (-5+x^3+x \log (\log (x))\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {x^4}{\left (-5+x^3+x \log (\log (x))\right )^2} \, dx\right )-5 \int \frac {x}{\left (-5+x^3+x \log (\log (x))\right )^2} \, dx-\left (10-e^{3/4}\right ) \int \frac {1}{x \log (x) \left (-5+x^3+x \log (\log (x))\right )^2} \, dx-\left (2 \left (10-e^{3/4}\right )\right ) \int \frac {x}{\left (-5+x^3+x \log (\log (x))\right )^2} \, dx-\left (2 \left (10-e^{3/4}\right )\right ) \int \frac {1}{x^2 \left (-5+x^3+x \log (\log (x))\right )} \, dx-\left (5 \left (10-e^{3/4}\right )\right ) \int \frac {1}{x^2 \left (-5+x^3+x \log (\log (x))\right )^2} \, dx-\int \frac {x^2}{\log (x) \left (-5+x^3+x \log (\log (x))\right )^2} \, dx+\int \frac {x}{-5+x^3+x \log (\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 28, normalized size = 0.93 \begin {gather*} \frac {10-e^{3/4}+x^3}{x \left (-5+x^3+x \log (\log (x))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x + E^(3/4)*x - x^4 + (50 - 50*x^3 - x^6 + E^(3/4)*(-5 + 4*x^3))*Log[x] + (-20*x + 2*E^(3/4)*x
+ x^4)*Log[x]*Log[Log[x]])/((25*x^2 - 10*x^5 + x^8)*Log[x] + (-10*x^3 + 2*x^6)*Log[x]*Log[Log[x]] + x^4*Log[x]
*Log[Log[x]]^2),x]

[Out]

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

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Maple [A]
time = 2.27, size = 27, normalized size = 0.90

method	result	size
risch	\(-\frac {-x^{3}+{\mathrm e}^{\frac {3}{4}}-10}{x \left (x^{3}+x \ln \left (\ln \left (x \right )\right )-5\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(3/4)+x^4-20*x)*ln(x)*ln(ln(x))+((4*x^3-5)*exp(3/4)-x^6-50*x^3+50)*ln(x)+x*exp(3/4)-x^4-10*x)/(x^
4*ln(x)*ln(ln(x))^2+(2*x^6-10*x^3)*ln(x)*ln(ln(x))+(x^8-10*x^5+25*x^2)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-(-x^3+exp(3/4)-10)/x/(x^3+x*ln(ln(x))-5)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(3/4)+x^4-20*x)*log(x)*log(log(x))+((4*x^3-5)*exp(3/4)-x^6-50*x^3+50)*log(x)+x*exp(3/4)-x^4
-10*x)/(x^4*log(x)*log(log(x))^2+(2*x^6-10*x^3)*log(x)*log(log(x))+(x^8-10*x^5+25*x^2)*log(x)),x, algorithm="m
axima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [A]
time = 0.35, size = 26, normalized size = 0.87 \begin {gather*} \frac {x^{3} - e^{\frac {3}{4}} + 10}{x^{4} + x^{2} \log \left (\log \left (x\right )\right ) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(3/4)+x^4-20*x)*log(x)*log(log(x))+((4*x^3-5)*exp(3/4)-x^6-50*x^3+50)*log(x)+x*exp(3/4)-x^4
-10*x)/(x^4*log(x)*log(log(x))^2+(2*x^6-10*x^3)*log(x)*log(log(x))+(x^8-10*x^5+25*x^2)*log(x)),x, algorithm="f
ricas")

[Out]

(x^3 - e^(3/4) + 10)/(x^4 + x^2*log(log(x)) - 5*x)

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Sympy [A]
time = 0.09, size = 24, normalized size = 0.80 \begin {gather*} \frac {x^{3} - e^{\frac {3}{4}} + 10}{x^{4} + x^{2} \log {\left (\log {\left (x \right )} \right )} - 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(3/4)+x**4-20*x)*ln(x)*ln(ln(x))+((4*x**3-5)*exp(3/4)-x**6-50*x**3+50)*ln(x)+x*exp(3/4)-x**
4-10*x)/(x**4*ln(x)*ln(ln(x))**2+(2*x**6-10*x**3)*ln(x)*ln(ln(x))+(x**8-10*x**5+25*x**2)*ln(x)),x)

[Out]

(x**3 - exp(3/4) + 10)/(x**4 + x**2*log(log(x)) - 5*x)

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Giac [A]
time = 0.43, size = 26, normalized size = 0.87 \begin {gather*} \frac {x^{3} - e^{\frac {3}{4}} + 10}{x^{4} + x^{2} \log \left (\log \left (x\right )\right ) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(3/4)+x^4-20*x)*log(x)*log(log(x))+((4*x^3-5)*exp(3/4)-x^6-50*x^3+50)*log(x)+x*exp(3/4)-x^4
-10*x)/(x^4*log(x)*log(log(x))^2+(2*x^6-10*x^3)*log(x)*log(log(x))+(x^8-10*x^5+25*x^2)*log(x)),x, algorithm="g
iac")

[Out]

(x^3 - e^(3/4) + 10)/(x^4 + x^2*log(log(x)) - 5*x)

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Mupad [B]
time = 4.42, size = 101, normalized size = 3.37 \begin {gather*} \frac {x^4\,\ln \left (x\right )-5\,{\mathrm {e}}^{3/4}\,{\ln \left (x\right )}^2+x\,\left (10\,\ln \left (x\right )-{\mathrm {e}}^{3/4}\,\ln \left (x\right )\right )-x^3\,\left (2\,{\mathrm {e}}^{3/4}\,{\ln \left (x\right )}^2-25\,{\ln \left (x\right )}^2\right )+50\,{\ln \left (x\right )}^2+2\,x^6\,{\ln \left (x\right )}^2}{x\,\left (2\,x^3\,{\ln \left (x\right )}^2+x\,\ln \left (x\right )+5\,{\ln \left (x\right )}^2\right )\,\left (x\,\ln \left (\ln \left (x\right )\right )+x^3-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(3/4) - 10*x + log(x)*(exp(3/4)*(4*x^3 - 5) - 50*x^3 - x^6 + 50) - x^4 + log(log(x))*log(x)*(2*x*exp
(3/4) - 20*x + x^4))/(log(x)*(25*x^2 - 10*x^5 + x^8) - log(log(x))*log(x)*(10*x^3 - 2*x^6) + x^4*log(log(x))^2
*log(x)),x)

[Out]

(x^4*log(x) - 5*exp(3/4)*log(x)^2 + x*(10*log(x) - exp(3/4)*log(x)) - x^3*(2*exp(3/4)*log(x)^2 - 25*log(x)^2)
+ 50*log(x)^2 + 2*x^6*log(x)^2)/(x*(5*log(x)^2 + 2*x^3*log(x)^2 + x*log(x))*(x*log(log(x)) + x^3 - 5))

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