3.19.25 \(\int \frac {8 x-6 x^2-2 x^3+\sqrt [5]{e} (2 x-2 x^2)+(15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} (-10 x^2-2 x^3)+(-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3) \log (x)) \log (-5-x+\log (x))+(2 x-2 x^2+(-10 x^2-2 x^3+2 x^2 \log (x)) \log (-5-x+\log (x))) \log (\log (-5-x+\log (x)))}{(-5 x^2-x^3+x^2 \log (x)) \log (-5-x+\log (x))} \, dx\)

Optimal. Leaf size=28 \[ \frac {3}{x}+2 x+\left (4+\sqrt [5]{e}+x+\log (\log (-5-x+\log (x)))\right )^2 \]

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Rubi [F]  time = 1.70, 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 {8 x-6 x^2-2 x^3+\sqrt [5]{e} \left (2 x-2 x^2\right )+\left (15+3 x-50 x^2-20 x^3-2 x^4+\sqrt [5]{e} \left (-10 x^2-2 x^3\right )+\left (-3+10 x^2+2 \sqrt [5]{e} x^2+2 x^3\right ) \log (x)\right ) \log (-5-x+\log (x))+\left (2 x-2 x^2+\left (-10 x^2-2 x^3+2 x^2 \log (x)\right ) \log (-5-x+\log (x))\right ) \log (\log (-5-x+\log (x)))}{\left (-5 x^2-x^3+x^2 \log (x)\right ) \log (-5-x+\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*x - 6*x^2 - 2*x^3 + E^(1/5)*(2*x - 2*x^2) + (15 + 3*x - 50*x^2 - 20*x^3 - 2*x^4 + E^(1/5)*(-10*x^2 - 2*
x^3) + (-3 + 10*x^2 + 2*E^(1/5)*x^2 + 2*x^3)*Log[x])*Log[-5 - x + Log[x]] + (2*x - 2*x^2 + (-10*x^2 - 2*x^3 +
2*x^2*Log[x])*Log[-5 - x + Log[x]])*Log[Log[-5 - x + Log[x]]])/((-5*x^2 - x^3 + x^2*Log[x])*Log[-5 - x + Log[x
]]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 \left (5+\sqrt [5]{e}\right )-\frac {3}{x^2}+2 x+2 \log (\log (-5-x+\log (x)))+\frac {2 (-1+x) \left (4 \left (1+\frac {\sqrt [5]{e}}{4}\right )+x+\log (\log (-5-x+\log (x)))\right )}{x (5+x-\log (x)) \log (-5-x+\log (x))}\right ) \, dx\\ &=\frac {3}{x}+2 \left (5+\sqrt [5]{e}\right ) x+x^2+2 \int \log (\log (-5-x+\log (x))) \, dx+2 \int \frac {(-1+x) \left (4 \left (1+\frac {\sqrt [5]{e}}{4}\right )+x+\log (\log (-5-x+\log (x)))\right )}{x (5+x-\log (x)) \log (-5-x+\log (x))} \, dx\\ &=\frac {3}{x}+2 \left (5+\sqrt [5]{e}\right ) x+x^2+2 \int \log (\log (-5-x+\log (x))) \, dx+2 \int \left (\frac {(-1+x) \left (4+\sqrt [5]{e}+x\right )}{x (5+x-\log (x)) \log (-5-x+\log (x))}+\frac {(-1+x) \log (\log (-5-x+\log (x)))}{x (5+x-\log (x)) \log (-5-x+\log (x))}\right ) \, dx\\ &=\frac {3}{x}+2 \left (5+\sqrt [5]{e}\right ) x+x^2+2 \int \frac {(-1+x) \left (4+\sqrt [5]{e}+x\right )}{x (5+x-\log (x)) \log (-5-x+\log (x))} \, dx+2 \int \log (\log (-5-x+\log (x))) \, dx+2 \int \frac {(-1+x) \log (\log (-5-x+\log (x)))}{x (5+x-\log (x)) \log (-5-x+\log (x))} \, dx\\ &=\frac {3}{x}+2 \left (5+\sqrt [5]{e}\right ) x+x^2+\log ^2(\log (-5-x+\log (x)))+2 \int \left (\frac {3 \left (1+\frac {\sqrt [5]{e}}{3}\right )}{(5+x-\log (x)) \log (-5-x+\log (x))}+\frac {-4-\sqrt [5]{e}}{x (5+x-\log (x)) \log (-5-x+\log (x))}+\frac {x}{(5+x-\log (x)) \log (-5-x+\log (x))}\right ) \, dx+2 \int \log (\log (-5-x+\log (x))) \, dx\\ &=\frac {3}{x}+2 \left (5+\sqrt [5]{e}\right ) x+x^2+\log ^2(\log (-5-x+\log (x)))+2 \int \frac {x}{(5+x-\log (x)) \log (-5-x+\log (x))} \, dx+2 \int \log (\log (-5-x+\log (x))) \, dx+\left (2 \left (3+\sqrt [5]{e}\right )\right ) \int \frac {1}{(5+x-\log (x)) \log (-5-x+\log (x))} \, dx-\left (2 \left (4+\sqrt [5]{e}\right )\right ) \int \frac {1}{x (5+x-\log (x)) \log (-5-x+\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 49, normalized size = 1.75 \begin {gather*} \frac {3}{x}+2 \left (5+\sqrt [5]{e}\right ) x+x^2+2 \left (4+\sqrt [5]{e}+x\right ) \log (\log (-5-x+\log (x)))+\log ^2(\log (-5-x+\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x - 6*x^2 - 2*x^3 + E^(1/5)*(2*x - 2*x^2) + (15 + 3*x - 50*x^2 - 20*x^3 - 2*x^4 + E^(1/5)*(-10*x^
2 - 2*x^3) + (-3 + 10*x^2 + 2*E^(1/5)*x^2 + 2*x^3)*Log[x])*Log[-5 - x + Log[x]] + (2*x - 2*x^2 + (-10*x^2 - 2*
x^3 + 2*x^2*Log[x])*Log[-5 - x + Log[x]])*Log[Log[-5 - x + Log[x]]])/((-5*x^2 - x^3 + x^2*Log[x])*Log[-5 - x +
Log[x]]),x]

[Out]

3/x + 2*(5 + E^(1/5))*x + x^2 + 2*(4 + E^(1/5) + x)*Log[Log[-5 - x + Log[x]]] + Log[Log[-5 - x + Log[x]]]^2

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fricas [B]  time = 0.60, size = 56, normalized size = 2.00 \begin {gather*} \frac {x^{3} + 2 \, x^{2} e^{\frac {1}{5}} + x \log \left (\log \left (-x + \log \relax (x) - 5\right )\right )^{2} + 10 \, x^{2} + 2 \, {\left (x^{2} + x e^{\frac {1}{5}} + 4 \, x\right )} \log \left (\log \left (-x + \log \relax (x) - 5\right )\right ) + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2*log(x)-2*x^3-10*x^2)*log(log(x)-5-x)-2*x^2+2*x)*log(log(log(x)-5-x))+((2*x^2*exp(1/5)+2*x^3
+10*x^2-3)*log(x)+(-2*x^3-10*x^2)*exp(1/5)-2*x^4-20*x^3-50*x^2+3*x+15)*log(log(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2
*x^3-6*x^2+8*x)/(x^2*log(x)-x^3-5*x^2)/log(log(x)-5-x),x, algorithm="fricas")

[Out]

(x^3 + 2*x^2*e^(1/5) + x*log(log(-x + log(x) - 5))^2 + 10*x^2 + 2*(x^2 + x*e^(1/5) + 4*x)*log(log(-x + log(x)
- 5)) + 3)/x

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giac [B]  time = 0.35, size = 74, normalized size = 2.64 \begin {gather*} \frac {x^{3} + 2 \, x^{2} e^{\frac {1}{5}} + 2 \, x^{2} \log \left (\log \left (-x + \log \relax (x) - 5\right )\right ) + 2 \, x e^{\frac {1}{5}} \log \left (\log \left (-x + \log \relax (x) - 5\right )\right ) + x \log \left (\log \left (-x + \log \relax (x) - 5\right )\right )^{2} + 10 \, x^{2} + 8 \, x \log \left (\log \left (-x + \log \relax (x) - 5\right )\right ) + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2*log(x)-2*x^3-10*x^2)*log(log(x)-5-x)-2*x^2+2*x)*log(log(log(x)-5-x))+((2*x^2*exp(1/5)+2*x^3
+10*x^2-3)*log(x)+(-2*x^3-10*x^2)*exp(1/5)-2*x^4-20*x^3-50*x^2+3*x+15)*log(log(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2
*x^3-6*x^2+8*x)/(x^2*log(x)-x^3-5*x^2)/log(log(x)-5-x),x, algorithm="giac")

[Out]

(x^3 + 2*x^2*e^(1/5) + 2*x^2*log(log(-x + log(x) - 5)) + 2*x*e^(1/5)*log(log(-x + log(x) - 5)) + x*log(log(-x
+ log(x) - 5))^2 + 10*x^2 + 8*x*log(log(-x + log(x) - 5)) + 3)/x

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maple [B]  time = 0.07, size = 72, normalized size = 2.57

method	result	size
risch	\(\ln \left (\ln \left (\ln \relax (x )-5-x \right )\right )^{2}+2 \ln \left (\ln \left (\ln \relax (x )-5-x \right )\right ) x +\frac {2 \ln \left (\ln \left (\ln \relax (x )-5-x \right )\right ) x \,{\mathrm e}^{\frac {1}{5}}+2 x^{2} {\mathrm e}^{\frac {1}{5}}+x^{3}+8 \ln \left (\ln \left (\ln \relax (x )-5-x \right )\right ) x +10 x^{2}+3}{x}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2*ln(x)-2*x^3-10*x^2)*ln(ln(x)-5-x)-2*x^2+2*x)*ln(ln(ln(x)-5-x))+((2*x^2*exp(1/5)+2*x^3+10*x^2-3)*l
n(x)+(-2*x^3-10*x^2)*exp(1/5)-2*x^4-20*x^3-50*x^2+3*x+15)*ln(ln(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2*x^3-6*x^2+8*x)
/(x^2*ln(x)-x^3-5*x^2)/ln(ln(x)-5-x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(ln(x)-5-x))^2+2*ln(ln(ln(x)-5-x))*x+(2*ln(ln(ln(x)-5-x))*x*exp(1/5)+2*x^2*exp(1/5)+x^3+8*ln(ln(ln(x)-5-x
))*x+10*x^2+3)/x

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maxima [B]  time = 0.81, size = 52, normalized size = 1.86 \begin {gather*} \frac {x^{3} + 2 \, x^{2} {\left (e^{\frac {1}{5}} + 5\right )} + x \log \left (\log \left (-x + \log \relax (x) - 5\right )\right )^{2} + 2 \, {\left (x^{2} + x {\left (e^{\frac {1}{5}} + 4\right )}\right )} \log \left (\log \left (-x + \log \relax (x) - 5\right )\right ) + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2*log(x)-2*x^3-10*x^2)*log(log(x)-5-x)-2*x^2+2*x)*log(log(log(x)-5-x))+((2*x^2*exp(1/5)+2*x^3
+10*x^2-3)*log(x)+(-2*x^3-10*x^2)*exp(1/5)-2*x^4-20*x^3-50*x^2+3*x+15)*log(log(x)-5-x)+(-2*x^2+2*x)*exp(1/5)-2
*x^3-6*x^2+8*x)/(x^2*log(x)-x^3-5*x^2)/log(log(x)-5-x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.67, size = 56, normalized size = 2.00 \begin {gather*} {\ln \left (\ln \left (\ln \relax (x)-x-5\right )\right )}^2+\ln \left (\ln \left (\ln \relax (x)-x-5\right )\right )\,\left (2\,{\mathrm {e}}^{1/5}+8\right )+2\,x\,\ln \left (\ln \left (\ln \relax (x)-x-5\right )\right )+\frac {3}{x}+x^2+x\,\left (2\,{\mathrm {e}}^{1/5}+10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x) - x - 5)*(exp(1/5)*(10*x^2 + 2*x^3) - 3*x - log(x)*(2*x^2*exp(1/5) + 10*x^2 + 2*x^3 - 3) + 50*
x^2 + 20*x^3 + 2*x^4 - 15) - 8*x - exp(1/5)*(2*x - 2*x^2) + log(log(log(x) - x - 5))*(log(log(x) - x - 5)*(10*
x^2 - 2*x^2*log(x) + 2*x^3) - 2*x + 2*x^2) + 6*x^2 + 2*x^3)/(log(log(x) - x - 5)*(5*x^2 - x^2*log(x) + x^3)),x
)

[Out]

log(log(log(x) - x - 5))^2 + log(log(log(x) - x - 5))*(2*exp(1/5) + 8) + 2*x*log(log(log(x) - x - 5)) + 3/x +
x^2 + x*(2*exp(1/5) + 10)

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sympy [B]  time = 1.40, size = 60, normalized size = 2.14 \begin {gather*} x^{2} + 2 x \log {\left (\log {\left (- x + \log {\relax (x )} - 5 \right )} \right )} + x \left (2 e^{\frac {1}{5}} + 10\right ) + \log {\left (\log {\left (- x + \log {\relax (x )} - 5 \right )} \right )}^{2} + 2 \left (e^{\frac {1}{5}} + 4\right ) \log {\left (\log {\left (- x + \log {\relax (x )} - 5 \right )} \right )} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2*ln(x)-2*x**3-10*x**2)*ln(ln(x)-5-x)-2*x**2+2*x)*ln(ln(ln(x)-5-x))+((2*x**2*exp(1/5)+2*x**3
+10*x**2-3)*ln(x)+(-2*x**3-10*x**2)*exp(1/5)-2*x**4-20*x**3-50*x**2+3*x+15)*ln(ln(x)-5-x)+(-2*x**2+2*x)*exp(1/
5)-2*x**3-6*x**2+8*x)/(x**2*ln(x)-x**3-5*x**2)/ln(ln(x)-5-x),x)

[Out]

x**2 + 2*x*log(log(-x + log(x) - 5)) + x*(2*exp(1/5) + 10) + log(log(-x + log(x) - 5))**2 + 2*(exp(1/5) + 4)*l
og(log(-x + log(x) - 5)) + 3/x

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