Integrand size = 90, antiderivative size = 22 \[ \int \frac {100 x^2+100 x^3+25 x^4+e^{e^{\frac {3+200 x+100 x^2}{50 x+25 x^2}}+\frac {3+200 x+100 x^2}{50 x+25 x^2}} (6+6 x)}{100 x^2+100 x^3+25 x^4} \, dx=-e^{e^{4+\frac {3}{25 x (2+x)}}}+x \] Output:
x-exp(exp(4+3/25/x/(2+x)))
Time = 2.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {100 x^2+100 x^3+25 x^4+e^{e^{\frac {3+200 x+100 x^2}{50 x+25 x^2}}+\frac {3+200 x+100 x^2}{50 x+25 x^2}} (6+6 x)}{100 x^2+100 x^3+25 x^4} \, dx=\frac {1}{25} \left (-25 e^{e^{4+\frac {3}{50 x}-\frac {3}{50 (2+x)}}}+25 x\right ) \] Input:
Integrate[(100*x^2 + 100*x^3 + 25*x^4 + E^(E^((3 + 200*x + 100*x^2)/(50*x + 25*x^2)) + (3 + 200*x + 100*x^2)/(50*x + 25*x^2))*(6 + 6*x))/(100*x^2 + 100*x^3 + 25*x^4),x]
Output:
(-25*E^E^(4 + 3/(50*x) - 3/(50*(2 + x))) + 25*x)/25
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(6 x+6) \exp \left (\frac {100 x^2+200 x+3}{25 x^2+50 x}+e^{\frac {100 x^2+200 x+3}{25 x^2+50 x}}\right )+25 x^4+100 x^3+100 x^2}{25 x^4+100 x^3+100 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {(6 x+6) \exp \left (\frac {100 x^2+200 x+3}{25 x^2+50 x}+e^{\frac {100 x^2+200 x+3}{25 x^2+50 x}}\right )+25 x^4+100 x^3+100 x^2}{x^2 \left (25 x^2+100 x+100\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {(6 x+6) \exp \left (\frac {100 x^2+200 x+3}{25 x^2+50 x}+e^{\frac {100 x^2+200 x+3}{25 x^2+50 x}}\right )+25 x^4+100 x^3+100 x^2}{x^2 (5 x+10)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {6 (x+1) \exp \left (\frac {25 x^2 \exp \left (\frac {100 x^2}{25 x^2+50 x}+\frac {200 x}{25 x^2+50 x}+\frac {3}{25 x^2+50 x}\right )+50 x \exp \left (\frac {100 x^2}{25 x^2+50 x}+\frac {200 x}{25 x^2+50 x}+\frac {3}{25 x^2+50 x}\right )+100 x^2+200 x+3}{25 x (x+2)}\right )}{25 x^2 (x+2)^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{50} \int \frac {\exp \left (\frac {25 \exp \left (\frac {100 x^2}{25 x^2+50 x}+\frac {200 x}{25 x^2+50 x}+\frac {3}{25 x^2+50 x}\right ) x^2+100 x^2+50 \exp \left (\frac {100 x^2}{25 x^2+50 x}+\frac {200 x}{25 x^2+50 x}+\frac {3}{25 x^2+50 x}\right ) x+200 x+3}{25 x (x+2)}\right )}{x^2}dx-\frac {3}{50} \int \frac {\exp \left (\frac {25 \exp \left (\frac {100 x^2}{25 x^2+50 x}+\frac {200 x}{25 x^2+50 x}+\frac {3}{25 x^2+50 x}\right ) x^2+100 x^2+50 \exp \left (\frac {100 x^2}{25 x^2+50 x}+\frac {200 x}{25 x^2+50 x}+\frac {3}{25 x^2+50 x}\right ) x+200 x+3}{25 x (x+2)}\right )}{(x+2)^2}dx+x\) |
Input:
Int[(100*x^2 + 100*x^3 + 25*x^4 + E^(E^((3 + 200*x + 100*x^2)/(50*x + 25*x ^2)) + (3 + 200*x + 100*x^2)/(50*x + 25*x^2))*(6 + 6*x))/(100*x^2 + 100*x^ 3 + 25*x^4),x]
Output:
$Aborted
Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23
method | result | size |
risch | \(x -{\mathrm e}^{{\mathrm e}^{\frac {100 x^{2}+200 x +3}{25 x \left (2+x \right )}}}\) | \(27\) |
parallelrisch | \(x -{\mathrm e}^{{\mathrm e}^{\frac {100 x^{2}+200 x +3}{25 x \left (2+x \right )}}}-3\) | \(28\) |
parts | \(x +\frac {-2 x \,{\mathrm e}^{{\mathrm e}^{\frac {100 x^{2}+200 x +3}{25 x^{2}+50 x}}}-x^{2} {\mathrm e}^{{\mathrm e}^{\frac {100 x^{2}+200 x +3}{25 x^{2}+50 x}}}}{x \left (2+x \right )}\) | \(69\) |
norman | \(\frac {x^{3}-4 x -2 x \,{\mathrm e}^{{\mathrm e}^{\frac {100 x^{2}+200 x +3}{25 x^{2}+50 x}}}-x^{2} {\mathrm e}^{{\mathrm e}^{\frac {100 x^{2}+200 x +3}{25 x^{2}+50 x}}}}{x \left (2+x \right )}\) | \(73\) |
Input:
int(((6+6*x)*exp((100*x^2+200*x+3)/(25*x^2+50*x))*exp(exp((100*x^2+200*x+3 )/(25*x^2+50*x)))+25*x^4+100*x^3+100*x^2)/(25*x^4+100*x^3+100*x^2),x,metho d=_RETURNVERBOSE)
Output:
x-exp(exp(1/25*(100*x^2+200*x+3)/x/(2+x)))
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (18) = 36\).
Time = 0.10 (sec) , antiderivative size = 103, normalized size of antiderivative = 4.68 \[ \int \frac {100 x^2+100 x^3+25 x^4+e^{e^{\frac {3+200 x+100 x^2}{50 x+25 x^2}}+\frac {3+200 x+100 x^2}{50 x+25 x^2}} (6+6 x)}{100 x^2+100 x^3+25 x^4} \, dx={\left (x e^{\left (\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )} - e^{\left (\frac {100 \, x^{2} + 25 \, {\left (x^{2} + 2 \, x\right )} e^{\left (\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )}\right )} e^{\left (-\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )} \] Input:
integrate(((6+6*x)*exp((100*x^2+200*x+3)/(25*x^2+50*x))*exp(exp((100*x^2+2 00*x+3)/(25*x^2+50*x)))+25*x^4+100*x^3+100*x^2)/(25*x^4+100*x^3+100*x^2),x , algorithm="fricas")
Output:
(x*e^(1/25*(100*x^2 + 200*x + 3)/(x^2 + 2*x)) - e^(1/25*(100*x^2 + 25*(x^2 + 2*x)*e^(1/25*(100*x^2 + 200*x + 3)/(x^2 + 2*x)) + 200*x + 3)/(x^2 + 2*x )))*e^(-1/25*(100*x^2 + 200*x + 3)/(x^2 + 2*x))
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {100 x^2+100 x^3+25 x^4+e^{e^{\frac {3+200 x+100 x^2}{50 x+25 x^2}}+\frac {3+200 x+100 x^2}{50 x+25 x^2}} (6+6 x)}{100 x^2+100 x^3+25 x^4} \, dx=x - e^{e^{\frac {100 x^{2} + 200 x + 3}{25 x^{2} + 50 x}}} \] Input:
integrate(((6+6*x)*exp((100*x**2+200*x+3)/(25*x**2+50*x))*exp(exp((100*x** 2+200*x+3)/(25*x**2+50*x)))+25*x**4+100*x**3+100*x**2)/(25*x**4+100*x**3+1 00*x**2),x)
Output:
x - exp(exp((100*x**2 + 200*x + 3)/(25*x**2 + 50*x)))
\[ \int \frac {100 x^2+100 x^3+25 x^4+e^{e^{\frac {3+200 x+100 x^2}{50 x+25 x^2}}+\frac {3+200 x+100 x^2}{50 x+25 x^2}} (6+6 x)}{100 x^2+100 x^3+25 x^4} \, dx=\int { \frac {25 \, x^{4} + 100 \, x^{3} + 100 \, x^{2} + 6 \, {\left (x + 1\right )} e^{\left (\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}} + e^{\left (\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )}\right )}}{25 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} \,d x } \] Input:
integrate(((6+6*x)*exp((100*x^2+200*x+3)/(25*x^2+50*x))*exp(exp((100*x^2+2 00*x+3)/(25*x^2+50*x)))+25*x^4+100*x^3+100*x^2)/(25*x^4+100*x^3+100*x^2),x , algorithm="maxima")
Output:
x + 1/25*integrate(6*(x*e^4 + e^4)*e^(-3/50/(x + 2) + 3/50/x + e^(-3/50/(x + 2) + 3/50/x + 4))/(x^4 + 4*x^3 + 4*x^2), x)
Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (18) = 36\).
Time = 0.45 (sec) , antiderivative size = 124, normalized size of antiderivative = 5.64 \[ \int \frac {100 x^2+100 x^3+25 x^4+e^{e^{\frac {3+200 x+100 x^2}{50 x+25 x^2}}+\frac {3+200 x+100 x^2}{50 x+25 x^2}} (6+6 x)}{100 x^2+100 x^3+25 x^4} \, dx={\left (x e^{\left (\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )} - e^{\left (\frac {25 \, x^{2} e^{\left (\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )} + 100 \, x^{2} + 50 \, x e^{\left (\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )}\right )} e^{\left (-\frac {100 \, x^{2} + 200 \, x + 3}{25 \, {\left (x^{2} + 2 \, x\right )}}\right )} \] Input:
integrate(((6+6*x)*exp((100*x^2+200*x+3)/(25*x^2+50*x))*exp(exp((100*x^2+2 00*x+3)/(25*x^2+50*x)))+25*x^4+100*x^3+100*x^2)/(25*x^4+100*x^3+100*x^2),x , algorithm="giac")
Output:
(x*e^(1/25*(100*x^2 + 200*x + 3)/(x^2 + 2*x)) - e^(1/25*(25*x^2*e^(1/25*(1 00*x^2 + 200*x + 3)/(x^2 + 2*x)) + 100*x^2 + 50*x*e^(1/25*(100*x^2 + 200*x + 3)/(x^2 + 2*x)) + 200*x + 3)/(x^2 + 2*x)))*e^(-1/25*(100*x^2 + 200*x + 3)/(x^2 + 2*x))
Time = 4.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {100 x^2+100 x^3+25 x^4+e^{e^{\frac {3+200 x+100 x^2}{50 x+25 x^2}}+\frac {3+200 x+100 x^2}{50 x+25 x^2}} (6+6 x)}{100 x^2+100 x^3+25 x^4} \, dx=x-{\mathrm {e}}^{{\mathrm {e}}^{\frac {4\,x}{x+2}}\,{\mathrm {e}}^{\frac {3}{25\,x^2+50\,x}}\,{\mathrm {e}}^{\frac {8}{x+2}}} \] Input:
int((100*x^2 + 100*x^3 + 25*x^4 + exp(exp((200*x + 100*x^2 + 3)/(50*x + 25 *x^2)))*exp((200*x + 100*x^2 + 3)/(50*x + 25*x^2))*(6*x + 6))/(100*x^2 + 1 00*x^3 + 25*x^4),x)
Output:
x - exp(exp((4*x)/(x + 2))*exp(3/(50*x + 25*x^2))*exp(8/(x + 2)))
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {100 x^2+100 x^3+25 x^4+e^{e^{\frac {3+200 x+100 x^2}{50 x+25 x^2}}+\frac {3+200 x+100 x^2}{50 x+25 x^2}} (6+6 x)}{100 x^2+100 x^3+25 x^4} \, dx=-e^{e^{\frac {3}{25 x^{2}+50 x}} e^{4}}+x \] Input:
int(((6+6*x)*exp((100*x^2+200*x+3)/(25*x^2+50*x))*exp(exp((100*x^2+200*x+3 )/(25*x^2+50*x)))+25*x^4+100*x^3+100*x^2)/(25*x^4+100*x^3+100*x^2),x)
Output:
- e**(e**(3/(25*x**2 + 50*x))*e**4) + x