3.18.70 \(\int \frac {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} (-25-16 x-3 x^2+e^x (25 x+10 x^2+x^3))}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} (25 x+6 x^2+x^3)} \, dx\)

Optimal. Leaf size=26 \[ 6-\log \left (x \left (-2+\frac {(5+x)^2}{-e^{e^x}+x}\right )\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 9.44, 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 {-2 e^{2 e^x}+8 x^2+2 x^3+e^{e^x} \left (-25-16 x-3 x^2+e^x \left (25 x+10 x^2+x^3\right )\right )}{2 e^{2 e^x} x-25 x^2-8 x^3-x^4+e^{e^x} \left (25 x+6 x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*E^(2*E^x) + 8*x^2 + 2*x^3 + E^E^x*(-25 - 16*x - 3*x^2 + E^x*(25*x + 10*x^2 + x^3)))/(2*E^(2*E^x)*x - 2
5*x^2 - 8*x^3 - x^4 + E^E^x*(25*x + 6*x^2 + x^3)),x]

[Out]

2*Defer[Int][(E^E^x - x)^(-1), x] + Defer[Int][E^(E^x + x)/(E^E^x - x), x] - Defer[Int][E^E^x/((E^E^x - x)*x),
x] - (2*Defer[Int][E^(2*E^x)/((E^E^x - x)*x), x])/25 + 10*Defer[Int][1/((E^E^x - x)*(5 + x)^2), x] + 4*Defer[
Int][E^E^x/((E^E^x - x)*(5 + x)^2), x] + (2*Defer[Int][E^(2*E^x)/((E^E^x - x)*(5 + x)^2), x])/5 - 12*Defer[Int
][1/((E^E^x - x)*(5 + x)), x] - 2*Defer[Int][E^E^x/((E^E^x - x)*(5 + x)), x] + (2*Defer[Int][E^(2*E^x)/((E^E^x
- x)*(5 + x)), x])/25 - 4*Defer[Int][(25 + 2*E^E^x + 8*x + x^2)^(-1), x] - 2*Defer[Int][E^(E^x + x)/(25 + 2*E
^E^x + 8*x + x^2), x] + 2*Defer[Int][E^E^x/(x*(25 + 2*E^E^x + 8*x + x^2)), x] + (4*Defer[Int][E^(2*E^x)/(x*(25
+ 2*E^E^x + 8*x + x^2)), x])/25 - 20*Defer[Int][1/((5 + x)^2*(25 + 2*E^E^x + 8*x + x^2)), x] - 8*Defer[Int][E
^E^x/((5 + x)^2*(25 + 2*E^E^x + 8*x + x^2)), x] - (4*Defer[Int][E^(2*E^x)/((5 + x)^2*(25 + 2*E^E^x + 8*x + x^2
)), x])/5 + 24*Defer[Int][1/((5 + x)*(25 + 2*E^E^x + 8*x + x^2)), x] + 4*Defer[Int][E^E^x/((5 + x)*(25 + 2*E^E
^x + 8*x + x^2)), x] - (4*Defer[Int][E^(2*E^x)/((5 + x)*(25 + 2*E^E^x + 8*x + x^2)), x])/25

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {16 e^{e^x}}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )}-\frac {25 e^{e^x}}{\left (e^{e^x}-x\right ) x \left (25+2 e^{e^x}+8 x+x^2\right )}-\frac {2 e^{2 e^x}}{\left (e^{e^x}-x\right ) x \left (25+2 e^{e^x}+8 x+x^2\right )}+\frac {8 x}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )}-\frac {3 e^{e^x} x}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )}+\frac {2 x^2}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )}+\frac {e^{e^x+x} (5+x)^2}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) x \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx\right )+2 \int \frac {x^2}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-3 \int \frac {e^{e^x} x}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+8 \int \frac {x}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-16 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-25 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) x \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+\int \frac {e^{e^x+x} (5+x)^2}{\left (e^{e^x}-x\right ) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx\\ &=-\left (2 \int \left (\frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) x (5+x)^2}-\frac {2 e^{2 e^x}}{x (5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx\right )+2 \int \left (\frac {x^2}{\left (e^{e^x}-x\right ) (5+x)^2}-\frac {2 x^2}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx-3 \int \left (\frac {e^{e^x} x}{\left (e^{e^x}-x\right ) (5+x)^2}-\frac {2 e^{e^x} x}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx+8 \int \left (\frac {x}{\left (e^{e^x}-x\right ) (5+x)^2}-\frac {2 x}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx-16 \int \left (\frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)^2}-\frac {2 e^{e^x}}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx-25 \int \left (\frac {e^{e^x}}{\left (e^{e^x}-x\right ) x (5+x)^2}-\frac {2 e^{e^x}}{x (5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx+\int \left (\frac {e^{e^x+x}}{e^{e^x}-x}-\frac {2 e^{e^x+x}}{25+2 e^{e^x}+8 x+x^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) x (5+x)^2} \, dx\right )+2 \int \frac {x^2}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx-2 \int \frac {e^{e^x+x}}{25+2 e^{e^x}+8 x+x^2} \, dx-3 \int \frac {e^{e^x} x}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx+4 \int \frac {e^{2 e^x}}{x (5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-4 \int \frac {x^2}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+6 \int \frac {e^{e^x} x}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+8 \int \frac {x}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx-16 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx-16 \int \frac {x}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-25 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) x (5+x)^2} \, dx+32 \int \frac {e^{e^x}}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+50 \int \frac {e^{e^x}}{x (5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+\int \frac {e^{e^x+x}}{e^{e^x}-x} \, dx\\ &=-\left (2 \int \frac {e^{e^x+x}}{25+2 e^{e^x}+8 x+x^2} \, dx\right )+2 \int \left (\frac {1}{e^{e^x}-x}+\frac {25}{\left (e^{e^x}-x\right ) (5+x)^2}-\frac {10}{\left (e^{e^x}-x\right ) (5+x)}\right ) \, dx-2 \int \left (\frac {e^{2 e^x}}{25 \left (e^{e^x}-x\right ) x}-\frac {e^{2 e^x}}{5 \left (e^{e^x}-x\right ) (5+x)^2}-\frac {e^{2 e^x}}{25 \left (e^{e^x}-x\right ) (5+x)}\right ) \, dx-3 \int \left (-\frac {5 e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)^2}+\frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)}\right ) \, dx-4 \int \left (\frac {1}{25+2 e^{e^x}+8 x+x^2}+\frac {25}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}-\frac {10}{(5+x) \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx+4 \int \left (\frac {e^{2 e^x}}{25 x \left (25+2 e^{e^x}+8 x+x^2\right )}-\frac {e^{2 e^x}}{5 (5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}-\frac {e^{2 e^x}}{25 (5+x) \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx+6 \int \left (-\frac {5 e^{e^x}}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}+\frac {e^{e^x}}{(5+x) \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx+8 \int \left (-\frac {5}{\left (e^{e^x}-x\right ) (5+x)^2}+\frac {1}{\left (e^{e^x}-x\right ) (5+x)}\right ) \, dx-16 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx-16 \int \left (-\frac {5}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}+\frac {1}{(5+x) \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx-25 \int \left (\frac {e^{e^x}}{25 \left (e^{e^x}-x\right ) x}-\frac {e^{e^x}}{5 \left (e^{e^x}-x\right ) (5+x)^2}-\frac {e^{e^x}}{25 \left (e^{e^x}-x\right ) (5+x)}\right ) \, dx+32 \int \frac {e^{e^x}}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+50 \int \left (\frac {e^{e^x}}{25 x \left (25+2 e^{e^x}+8 x+x^2\right )}-\frac {e^{e^x}}{5 (5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )}-\frac {e^{e^x}}{25 (5+x) \left (25+2 e^{e^x}+8 x+x^2\right )}\right ) \, dx+\int \frac {e^{e^x+x}}{e^{e^x}-x} \, dx\\ &=-\left (\frac {2}{25} \int \frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) x} \, dx\right )+\frac {2}{25} \int \frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) (5+x)} \, dx+\frac {4}{25} \int \frac {e^{2 e^x}}{x \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-\frac {4}{25} \int \frac {e^{2 e^x}}{(5+x) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+\frac {2}{5} \int \frac {e^{2 e^x}}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx-\frac {4}{5} \int \frac {e^{2 e^x}}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+2 \int \frac {1}{e^{e^x}-x} \, dx-2 \int \frac {e^{e^x+x}}{25+2 e^{e^x}+8 x+x^2} \, dx+2 \int \frac {e^{e^x}}{x \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-2 \int \frac {e^{e^x}}{(5+x) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-3 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)} \, dx-4 \int \frac {1}{25+2 e^{e^x}+8 x+x^2} \, dx+5 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx+6 \int \frac {e^{e^x}}{(5+x) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+8 \int \frac {1}{\left (e^{e^x}-x\right ) (5+x)} \, dx-10 \int \frac {e^{e^x}}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+15 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx-16 \int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx-16 \int \frac {1}{(5+x) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-20 \int \frac {1}{\left (e^{e^x}-x\right ) (5+x)} \, dx-30 \int \frac {e^{e^x}}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+32 \int \frac {e^{e^x}}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-40 \int \frac {1}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx+40 \int \frac {1}{(5+x) \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+50 \int \frac {1}{\left (e^{e^x}-x\right ) (5+x)^2} \, dx+80 \int \frac {1}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx-100 \int \frac {1}{(5+x)^2 \left (25+2 e^{e^x}+8 x+x^2\right )} \, dx+\int \frac {e^{e^x+x}}{e^{e^x}-x} \, dx-\int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) x} \, dx+\int \frac {e^{e^x}}{\left (e^{e^x}-x\right ) (5+x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 33, normalized size = 1.27 \begin {gather*} \log \left (e^{e^x}-x\right )-\log (x)-\log \left (25+2 e^{e^x}+8 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(2*E^x) + 8*x^2 + 2*x^3 + E^E^x*(-25 - 16*x - 3*x^2 + E^x*(25*x + 10*x^2 + x^3)))/(2*E^(2*E^x)
*x - 25*x^2 - 8*x^3 - x^4 + E^E^x*(25*x + 6*x^2 + x^3)),x]

[Out]

Log[E^E^x - x] - Log[x] - Log[25 + 2*E^E^x + 8*x + x^2]

________________________________________________________________________________________

fricas [A]  time = 0.97, size = 29, normalized size = 1.12 \begin {gather*} -\log \left (x^{2} + 8 \, x + 2 \, e^{\left (e^{x}\right )} + 25\right ) - \log \relax (x) + \log \left (-x + e^{\left (e^{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(exp(x))^2+((x^3+10*x^2+25*x)*exp(x)-3*x^2-16*x-25)*exp(exp(x))+2*x^3+8*x^2)/(2*x*exp(exp(x))
^2+(x^3+6*x^2+25*x)*exp(exp(x))-x^4-8*x^3-25*x^2),x, algorithm="fricas")

[Out]

-log(x^2 + 8*x + 2*e^(e^x) + 25) - log(x) + log(-x + e^(e^x))

________________________________________________________________________________________

giac [A]  time = 1.02, size = 29, normalized size = 1.12 \begin {gather*} -\log \left (x^{2} + 8 \, x + 2 \, e^{\left (e^{x}\right )} + 25\right ) + \log \left (x - e^{\left (e^{x}\right )}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(exp(x))^2+((x^3+10*x^2+25*x)*exp(x)-3*x^2-16*x-25)*exp(exp(x))+2*x^3+8*x^2)/(2*x*exp(exp(x))
^2+(x^3+6*x^2+25*x)*exp(exp(x))-x^4-8*x^3-25*x^2),x, algorithm="giac")

[Out]

-log(x^2 + 8*x + 2*e^(e^x) + 25) + log(x - e^(e^x)) - log(x)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 30, normalized size = 1.15

method	result	size
norman	\(-\ln \relax (x )-\ln \left (x^{2}+8 x +2 \,{\mathrm e}^{{\mathrm e}^{x}}+25\right )+\ln \left (x -{\mathrm e}^{{\mathrm e}^{x}}\right )\)	\(30\)
risch	\(-\ln \relax (x )+\ln \left ({\mathrm e}^{{\mathrm e}^{x}}-x \right )-\ln \left (\frac {x^{2}}{2}+4 x +{\mathrm e}^{{\mathrm e}^{x}}+\frac {25}{2}\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(exp(x))^2+((x^3+10*x^2+25*x)*exp(x)-3*x^2-16*x-25)*exp(exp(x))+2*x^3+8*x^2)/(2*x*exp(exp(x))^2+(x^
3+6*x^2+25*x)*exp(exp(x))-x^4-8*x^3-25*x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(x)-ln(x^2+8*x+2*exp(exp(x))+25)+ln(x-exp(exp(x)))

________________________________________________________________________________________

maxima [A]  time = 0.46, size = 29, normalized size = 1.12 \begin {gather*} -\log \left (\frac {1}{2} \, x^{2} + 4 \, x + e^{\left (e^{x}\right )} + \frac {25}{2}\right ) - \log \relax (x) + \log \left (-x + e^{\left (e^{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(exp(x))^2+((x^3+10*x^2+25*x)*exp(x)-3*x^2-16*x-25)*exp(exp(x))+2*x^3+8*x^2)/(2*x*exp(exp(x))
^2+(x^3+6*x^2+25*x)*exp(exp(x))-x^4-8*x^3-25*x^2),x, algorithm="maxima")

[Out]

-log(1/2*x^2 + 4*x + e^(e^x) + 25/2) - log(x) + log(-x + e^(e^x))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {2\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (16\,x-{\mathrm {e}}^x\,\left (x^3+10\,x^2+25\,x\right )+3\,x^2+25\right )-8\,x^2-2\,x^3}{25\,x^2-2\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+8\,x^3+x^4-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x^3+6\,x^2+25\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(2*exp(x)) + exp(exp(x))*(16*x - exp(x)*(25*x + 10*x^2 + x^3) + 3*x^2 + 25) - 8*x^2 - 2*x^3)/(25*x^2
- 2*x*exp(2*exp(x)) + 8*x^3 + x^4 - exp(exp(x))*(25*x + 6*x^2 + x^3)),x)

[Out]

int((2*exp(2*exp(x)) + exp(exp(x))*(16*x - exp(x)*(25*x + 10*x^2 + x^3) + 3*x^2 + 25) - 8*x^2 - 2*x^3)/(25*x^2
- 2*x*exp(2*exp(x)) + 8*x^3 + x^4 - exp(exp(x))*(25*x + 6*x^2 + x^3)), x)

________________________________________________________________________________________

sympy [A]  time = 0.51, size = 29, normalized size = 1.12 \begin {gather*} - \log {\relax (x )} + \log {\left (- x + e^{e^{x}} \right )} - \log {\left (\frac {x^{2}}{2} + 4 x + e^{e^{x}} + \frac {25}{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(exp(x))**2+((x**3+10*x**2+25*x)*exp(x)-3*x**2-16*x-25)*exp(exp(x))+2*x**3+8*x**2)/(2*x*exp(e
xp(x))**2+(x**3+6*x**2+25*x)*exp(exp(x))-x**4-8*x**3-25*x**2),x)

[Out]

-log(x) + log(-x + exp(exp(x))) - log(x**2/2 + 4*x + exp(exp(x)) + 25/2)

________________________________________________________________________________________