3.80.10 \(\int \frac {(20+x^2-x^3) \log (6)+(20-x^2+2 x^3) \log (6) \log (x)+e^{\frac {e^x}{\log (6)}} (-20 \log (6)+(20 e^x x-20 \log (6)) \log (x))}{5 x^2 \log (6) \log ^2(x)} \, dx\)

Optimal. Leaf size=34 \[ \frac {\frac {4 \left (-1+e^{\frac {e^x}{\log (6)}}\right )}{x}+\frac {1}{5} \left (-x+x^2\right )}{\log (x)} \]

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Rubi [F]  time = 1.33, 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 {\left (20+x^2-x^3\right ) \log (6)+\left (20-x^2+2 x^3\right ) \log (6) \log (x)+e^{\frac {e^x}{\log (6)}} \left (-20 \log (6)+\left (20 e^x x-20 \log (6)\right ) \log (x)\right )}{5 x^2 \log (6) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((20 + x^2 - x^3)*Log[6] + (20 - x^2 + 2*x^3)*Log[6]*Log[x] + E^(E^x/Log[6])*(-20*Log[6] + (20*E^x*x - 20*
Log[6])*Log[x]))/(5*x^2*Log[6]*Log[x]^2),x]

[Out]

(4*E^(E^x/Log[6]))/(x*Log[x]) + Defer[Int][(20 + x^2 - x^3)/(x^2*Log[x]^2), x]/5 + Defer[Int][(20 - x^2 + 2*x^
3)/(x^2*Log[x]), x]/5

Rubi steps

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

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Mathematica [A]  time = 0.40, size = 33, normalized size = 0.97 \begin {gather*} \frac {-20+20 e^{\frac {e^x}{\log (6)}}-x^2+x^3}{5 x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((20 + x^2 - x^3)*Log[6] + (20 - x^2 + 2*x^3)*Log[6]*Log[x] + E^(E^x/Log[6])*(-20*Log[6] + (20*E^x*x
- 20*Log[6])*Log[x]))/(5*x^2*Log[6]*Log[x]^2),x]

[Out]

(-20 + 20*E^(E^x/Log[6]) - x^2 + x^3)/(5*x*Log[x])

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fricas [A]  time = 0.58, size = 29, normalized size = 0.85 \begin {gather*} \frac {x^{3} - x^{2} + 20 \, e^{\left (\frac {e^{x}}{\log \relax (6)}\right )} - 20}{5 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^3 - x^2 + 20*e^(e^x/log(6)) - 20)/(x*log(x))

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giac [B]  time = 0.32, size = 58, normalized size = 1.71 \begin {gather*} \frac {{\left (x^{3} e^{x} \log \relax (6) - x^{2} e^{x} \log \relax (6) - 20 \, e^{x} \log \relax (6) + 20 \, e^{\left (\frac {x \log \relax (6) + e^{x}}{\log \relax (6)}\right )} \log \relax (6)\right )} e^{\left (-x\right )}}{5 \, x \log \relax (6) \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^3*e^x*log(6) - x^2*e^x*log(6) - 20*e^x*log(6) + 20*e^((x*log(6) + e^x)/log(6))*log(6))*e^(-x)/(x*log(6)
*log(x))

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maple [B]  time = 0.07, size = 73, normalized size = 2.15

method	result	size
risch	\(\frac {x^{3} \ln \relax (2)+x^{3} \ln \relax (3)-x^{2} \ln \relax (2)-x^{2} \ln \relax (3)-20 \ln \relax (2)-20 \ln \relax (3)}{5 \left (\ln \relax (2)+\ln \relax (3)\right ) x \ln \relax (x )}+\frac {4 \,{\mathrm e}^{\frac {{\mathrm e}^{x}}{\ln \relax (2)+\ln \relax (3)}}}{\ln \relax (x ) x}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(((20*exp(x)*x-20*ln(6))*ln(x)-20*ln(6))*exp(exp(x)/ln(6))+(2*x^3-x^2+20)*ln(6)*ln(x)+(-x^3+x^2+20)*ln
(6))/x^2/ln(6)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/5/(ln(2)+ln(3))/x*(x^3*ln(2)+x^3*ln(3)-x^2*ln(2)-x^2*ln(3)-20*ln(2)-20*ln(3))/ln(x)+4/ln(x)/x*exp(exp(x)/(ln
(2)+ln(3)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\Gamma \left (-1, -\log \relax (x)\right ) \log \relax (6) - 2 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) \log \relax (6) - 20 \, \Gamma \left (-1, \log \relax (x)\right ) \log \relax (6) + 2 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) \log \relax (3) + 20 \, {\rm Ei}\left (-\log \relax (x)\right ) \log \relax (3) - {\rm Ei}\left (\log \relax (x)\right ) \log \relax (3) + 2 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) \log \relax (2) + 20 \, {\rm Ei}\left (-\log \relax (x)\right ) \log \relax (2) - {\rm Ei}\left (\log \relax (x)\right ) \log \relax (2) + \frac {20 \, {\left (\log \relax (3) + \log \relax (2)\right )} e^{\left (\frac {e^{x}}{\log \relax (3) + \log \relax (2)}\right )}}{x \log \relax (x)}}{5 \, \log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(gamma(-1, -log(x))*log(6) - 2*gamma(-1, -2*log(x))*log(6) - 20*gamma(-1, log(x))*log(6) + 20*(log(3) + lo
g(2))*e^(e^x/(log(3) + log(2)))/(x*log(x)) + integrate((2*x^3*(log(3) + log(2)) - x^2*(log(3) + log(2)) + 20*l
og(3) + 20*log(2))/(x^2*log(x)), x))/log(6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\frac {\ln \relax (6)\,\left (-x^3+x^2+20\right )}{5}-\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{\ln \relax (6)}}\,\left (20\,\ln \relax (6)+\ln \relax (x)\,\left (20\,\ln \relax (6)-20\,x\,{\mathrm {e}}^x\right )\right )}{5}+\frac {\ln \relax (6)\,\ln \relax (x)\,\left (2\,x^3-x^2+20\right )}{5}}{x^2\,\ln \relax (6)\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(6)*(x^2 - x^3 + 20))/5 - (exp(exp(x)/log(6))*(20*log(6) + log(x)*(20*log(6) - 20*x*exp(x))))/5 + (lo
g(6)*log(x)*(2*x^3 - x^2 + 20))/5)/(x^2*log(6)*log(x)^2),x)

[Out]

int(((log(6)*(x^2 - x^3 + 20))/5 - (exp(exp(x)/log(6))*(20*log(6) + log(x)*(20*log(6) - 20*x*exp(x))))/5 + (lo
g(6)*log(x)*(2*x^3 - x^2 + 20))/5)/(x^2*log(6)*log(x)^2), x)

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sympy [A]  time = 0.39, size = 29, normalized size = 0.85 \begin {gather*} \frac {x^{3} - x^{2} - 20}{5 x \log {\relax (x )}} + \frac {4 e^{\frac {e^{x}}{\log {\relax (6 )}}}}{x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((20*exp(x)*x-20*ln(6))*ln(x)-20*ln(6))*exp(exp(x)/ln(6))+(2*x**3-x**2+20)*ln(6)*ln(x)+(-x**3+x
**2+20)*ln(6))/x**2/ln(6)/ln(x)**2,x)

[Out]

(x**3 - x**2 - 20)/(5*x*log(x)) + 4*exp(exp(x)/log(6))/(x*log(x))

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