3.3.100 \(\int \frac {-x^3+e^{-2+x+\frac {e^{-2+x} (1+e^{5/x} x^2)}{x^2}} ((2-x) \log (2)+e^{5/x} (5 x-x^3) \log (2))}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} (1+e^{5/x} x^2)}{x^2}} (10 x^3+2 x^4) \log (2)+e^{\frac {2 e^{-2+x} (1+e^{5/x} x^2)}{x^2}} x^3 \log ^2(2)} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)} \]

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Rubi [F]  time = 26.50, 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 {-x^3+e^{-2+x+\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left ((2-x) \log (2)+e^{5/x} \left (5 x-x^3\right ) \log (2)\right )}{25 x^3+10 x^4+x^5+e^{\frac {e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} \left (10 x^3+2 x^4\right ) \log (2)+e^{\frac {2 e^{-2+x} \left (1+e^{5/x} x^2\right )}{x^2}} x^3 \log ^2(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-x^3 + E^(-2 + x + (E^(-2 + x)*(1 + E^(5/x)*x^2))/x^2)*((2 - x)*Log[2] + E^(5/x)*(5*x - x^3)*Log[2]))/(25
*x^3 + 10*x^4 + x^5 + E^((E^(-2 + x)*(1 + E^(5/x)*x^2))/x^2)*(10*x^3 + 2*x^4)*Log[2] + E^((2*E^(-2 + x)*(1 + E
^(5/x)*x^2))/x^2)*x^3*Log[2]^2),x]

[Out]

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

Rubi steps

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

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Mathematica [B]  time = 0.58, size = 137, normalized size = 5.07 \begin {gather*} \frac {-e^4 x^3 \log (2)+e^{2+x} (5+x) (x \log (2)-\log (4))+e^{2+\frac {5}{x}+x} x (5+x) \left (x^2 \log (2)-\log (32)\right )}{e^2 \left (-e^2 x^3+e^x \left (-10+3 x+x^2\right )+e^{\frac {5}{x}+x} x \left (-25-5 x+5 x^2+x^3\right )\right ) \log (2) \left (5+x+e^{e^{-2+x} \left (e^{5/x}+\frac {1}{x^2}\right )} \log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^3 + E^(-2 + x + (E^(-2 + x)*(1 + E^(5/x)*x^2))/x^2)*((2 - x)*Log[2] + E^(5/x)*(5*x - x^3)*Log[2]
))/(25*x^3 + 10*x^4 + x^5 + E^((E^(-2 + x)*(1 + E^(5/x)*x^2))/x^2)*(10*x^3 + 2*x^4)*Log[2] + E^((2*E^(-2 + x)*
(1 + E^(5/x)*x^2))/x^2)*x^3*Log[2]^2),x]

[Out]

(-(E^4*x^3*Log[2]) + E^(2 + x)*(5 + x)*(x*Log[2] - Log[4]) + E^(2 + 5/x + x)*x*(5 + x)*(x^2*Log[2] - Log[32]))
/(E^2*(-(E^2*x^3) + E^x*(-10 + 3*x + x^2) + E^(5/x + x)*x*(-25 - 5*x + 5*x^2 + x^3))*Log[2]*(5 + x + E^(E^(-2
+ x)*(E^(5/x) + x^(-2)))*Log[2]))

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fricas [B]  time = 0.53, size = 50, normalized size = 1.85 \begin {gather*} \frac {e^{\left (x - 2\right )}}{{\left (x + 5\right )} e^{\left (x - 2\right )} + e^{\left (\frac {x^{3} - 2 \, x^{2} + {\left (x^{2} e^{\frac {5}{x}} + 1\right )} e^{\left (x - 2\right )}}{x^{2}}\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+5*x)*log(2)*exp(5/x)+(2-x)*log(2))*exp(x-2)*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)-x^3)/(x^3*log
(2)^2*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)^2+(2*x^4+10*x^3)*log(2)*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)+x^5+10*x^4
+25*x^3),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 7.41, size = 564, normalized size = 20.89 \begin {gather*} \frac {x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - x^{3} e^{2} + 5 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{2} e^{x} - 5 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 3 \, x e^{x} - 25 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - 10 \, e^{x}}{x^{5} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) - x^{4} e^{2} + 10 \, x^{4} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 5 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) - x^{3} e^{\left (\frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) - 5 \, x^{3} e^{2} + x^{3} e^{x} + 20 \, x^{3} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + x^{2} e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \relax (2) - 5 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) + 8 \, x^{2} e^{x} - 50 \, x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} + 3 \, x e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \relax (2) - 25 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}} + 2\right )} \log \relax (2) + 5 \, x e^{x} - 125 \, x e^{\left (\frac {x^{2} - 2 \, x + 5}{x} + 2\right )} - 10 \, e^{\left (x + \frac {x^{2} e^{\left (\frac {x^{2} - 2 \, x + 5}{x}\right )} + e^{\left (x - 2\right )}}{x^{2}}\right )} \log \relax (2) - 50 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+5*x)*log(2)*exp(5/x)+(2-x)*log(2))*exp(x-2)*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)-x^3)/(x^3*log
(2)^2*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)^2+(2*x^4+10*x^3)*log(2)*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)+x^5+10*x^4
+25*x^3),x, algorithm="giac")

[Out]

(x^4*e^((x^2 - 2*x + 5)/x + 2) - x^3*e^2 + 5*x^3*e^((x^2 - 2*x + 5)/x + 2) + x^2*e^x - 5*x^2*e^((x^2 - 2*x + 5
)/x + 2) + 3*x*e^x - 25*x*e^((x^2 - 2*x + 5)/x + 2) - 10*e^x)/(x^5*e^((x^2 - 2*x + 5)/x + 2) + x^4*e^((x^2 - 2
*x + 5)/x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2 + 2)*log(2) - x^4*e^2 + 10*x^4*e^((x^2 - 2*x + 5)/x +
2) + 5*x^3*e^((x^2 - 2*x + 5)/x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2 + 2)*log(2) - x^3*e^((x^2*e^((x^
2 - 2*x + 5)/x) + e^(x - 2))/x^2 + 2)*log(2) - 5*x^3*e^2 + x^3*e^x + 20*x^3*e^((x^2 - 2*x + 5)/x + 2) + x^2*e^
(x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2)*log(2) - 5*x^2*e^((x^2 - 2*x + 5)/x + (x^2*e^((x^2 - 2*x + 5
)/x) + e^(x - 2))/x^2 + 2)*log(2) + 8*x^2*e^x - 50*x^2*e^((x^2 - 2*x + 5)/x + 2) + 3*x*e^(x + (x^2*e^((x^2 - 2
*x + 5)/x) + e^(x - 2))/x^2)*log(2) - 25*x*e^((x^2 - 2*x + 5)/x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^2
+ 2)*log(2) + 5*x*e^x - 125*x*e^((x^2 - 2*x + 5)/x + 2) - 10*e^(x + (x^2*e^((x^2 - 2*x + 5)/x) + e^(x - 2))/x^
2)*log(2) - 50*e^x)

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maple [A]  time = 0.21, size = 30, normalized size = 1.11

method	result	size
risch	\(\frac {1}{\ln \relax (2) {\mathrm e}^{\frac {\left (x^{2} {\mathrm e}^{\frac {5}{x}}+1\right ) {\mathrm e}^{x -2}}{x^{2}}}+x +5}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^3+5*x)*ln(2)*exp(5/x)+(2-x)*ln(2))*exp(x-2)*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)-x^3)/(x^3*ln(2)^2*exp
((x^2*exp(5/x)+1)*exp(x-2)/x^2)^2+(2*x^4+10*x^3)*ln(2)*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)+x^5+10*x^4+25*x^3),x
,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.94, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{e^{\left (\frac {e^{\left (x - 2\right )}}{x^{2}} + e^{\left (x + \frac {5}{x} - 2\right )}\right )} \log \relax (2) + x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3+5*x)*log(2)*exp(5/x)+(2-x)*log(2))*exp(x-2)*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)-x^3)/(x^3*log
(2)^2*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)^2+(2*x^4+10*x^3)*log(2)*exp((x^2*exp(5/x)+1)*exp(x-2)/x^2)+x^5+10*x^4
+25*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.71, size = 93, normalized size = 3.44 \begin {gather*} \frac {x\,\left (\ln \relax (2)-{\mathrm {e}}^{5/x}\,\ln \left (32\right )\right )-\ln \relax (4)+x^3\,{\mathrm {e}}^{5/x}\,\ln \relax (2)}{{\ln \relax (2)}^2\,\left ({\mathrm {e}}^{{\mathrm {e}}^{-2}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^x+\frac {{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{x^2}}+\frac {x+5}{\ln \relax (2)}\right )\,\left (x-5\,x\,{\mathrm {e}}^{5/x}+x^3\,{\mathrm {e}}^{5/x}-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3 + exp((exp(x - 2)*(x^2*exp(5/x) + 1))/x^2)*exp(x - 2)*(log(2)*(x - 2) - exp(5/x)*log(2)*(5*x - x^3))
)/(25*x^3 + 10*x^4 + x^5 + x^3*exp((2*exp(x - 2)*(x^2*exp(5/x) + 1))/x^2)*log(2)^2 + exp((exp(x - 2)*(x^2*exp(
5/x) + 1))/x^2)*log(2)*(10*x^3 + 2*x^4)),x)

[Out]

(x*(log(2) - exp(5/x)*log(32)) - log(4) + x^3*exp(5/x)*log(2))/(log(2)^2*(exp(exp(-2)*exp(5/x)*exp(x) + (exp(-
2)*exp(x))/x^2) + (x + 5)/log(2))*(x - 5*x*exp(5/x) + x^3*exp(5/x) - 2))

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sympy [A]  time = 0.72, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{x + e^{\frac {\left (x^{2} e^{\frac {5}{x}} + 1\right ) e^{x - 2}}{x^{2}}} \log {\relax (2 )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**3+5*x)*ln(2)*exp(5/x)+(2-x)*ln(2))*exp(x-2)*exp((x**2*exp(5/x)+1)*exp(x-2)/x**2)-x**3)/(x**3*
ln(2)**2*exp((x**2*exp(5/x)+1)*exp(x-2)/x**2)**2+(2*x**4+10*x**3)*ln(2)*exp((x**2*exp(5/x)+1)*exp(x-2)/x**2)+x
**5+10*x**4+25*x**3),x)

[Out]

1/(x + exp((x**2*exp(5/x) + 1)*exp(x - 2)/x**2)*log(2) + 5)

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