Integrand size = 157, antiderivative size = 22 \[ \int \frac {-3+e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )} \left (-1+10 x+2 x^2+e^x \left (40 x+56 x^2+8 x^3\right )+e^{2 x} \left (32 x^2+32 x^3\right )\right )}{9+6 e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )}+e^{50+20 x+2 x^2+32 e^{2 x} x^2+2 e^x \left (40 x+8 x^2\right )}} \, dx=9-\frac {x}{3+e^{\left (5+x+4 e^x x\right )^2}} \] Output:
9-x/(exp((x+5+4*exp(x)*x)^2)+3)
Time = 0.54 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-3+e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )} \left (-1+10 x+2 x^2+e^x \left (40 x+56 x^2+8 x^3\right )+e^{2 x} \left (32 x^2+32 x^3\right )\right )}{9+6 e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )}+e^{50+20 x+2 x^2+32 e^{2 x} x^2+2 e^x \left (40 x+8 x^2\right )}} \, dx=-\frac {x}{3+e^{\left (5+x+4 e^x x\right )^2}} \] Input:
Integrate[(-3 + E^(25 + 10*x + x^2 + 16*E^(2*x)*x^2 + E^x*(40*x + 8*x^2))* (-1 + 10*x + 2*x^2 + E^x*(40*x + 56*x^2 + 8*x^3) + E^(2*x)*(32*x^2 + 32*x^ 3)))/(9 + 6*E^(25 + 10*x + x^2 + 16*E^(2*x)*x^2 + E^x*(40*x + 8*x^2)) + E^ (50 + 20*x + 2*x^2 + 32*E^(2*x)*x^2 + 2*E^x*(40*x + 8*x^2))),x]
Output:
-(x/(3 + E^(5 + x + 4*E^x*x)^2))
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.
\(\displaystyle \int \frac {\left (2 x^2+e^x \left (8 x^3+56 x^2+40 x\right )+e^{2 x} \left (32 x^3+32 x^2\right )+10 x-1\right ) \exp \left (16 e^{2 x} x^2+x^2+e^x \left (8 x^2+40 x\right )+10 x+25\right )-3}{6 \exp \left (16 e^{2 x} x^2+x^2+e^x \left (8 x^2+40 x\right )+10 x+25\right )+\exp \left (32 e^{2 x} x^2+2 x^2+2 e^x \left (8 x^2+40 x\right )+20 x+50\right )+9} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\left (4 e^x x+x+5\right )^2} \left (32 e^{2 x} (x+1) x^2+2 x^2+8 e^x \left (x^2+7 x+5\right ) x+10 x-1\right )-3}{\left (e^{\left (4 e^x x+x+5\right )^2}+3\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {8 e^x x^3+32 e^{2 x} x^3+56 e^x x^2+32 e^{2 x} x^2+2 x^2+40 e^x x+10 x-1}{e^{\left (4 e^x x+x+5\right )^2}+3}-\frac {6 x \left (4 e^x x^2+16 e^{2 x} x^2+28 e^x x+16 e^{2 x} x+x+20 e^x+5\right )}{\left (e^{\left (4 e^x x+x+5\right )^2}+3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -24 \int \frac {e^x x^3}{\left (3+e^{\left (4 e^x x+x+5\right )^2}\right )^2}dx-96 \int \frac {e^{2 x} x^3}{\left (3+e^{\left (4 e^x x+x+5\right )^2}\right )^2}dx+8 \int \frac {e^x x^3}{3+e^{\left (4 e^x x+x+5\right )^2}}dx+32 \int \frac {e^{2 x} x^3}{3+e^{\left (4 e^x x+x+5\right )^2}}dx-6 \int \frac {x^2}{\left (3+e^{\left (4 e^x x+x+5\right )^2}\right )^2}dx-168 \int \frac {e^x x^2}{\left (3+e^{\left (4 e^x x+x+5\right )^2}\right )^2}dx-96 \int \frac {e^{2 x} x^2}{\left (3+e^{\left (4 e^x x+x+5\right )^2}\right )^2}dx+2 \int \frac {x^2}{3+e^{\left (4 e^x x+x+5\right )^2}}dx+56 \int \frac {e^x x^2}{3+e^{\left (4 e^x x+x+5\right )^2}}dx+32 \int \frac {e^{2 x} x^2}{3+e^{\left (4 e^x x+x+5\right )^2}}dx-\int \frac {1}{3+e^{\left (4 e^x x+x+5\right )^2}}dx-30 \int \frac {x}{\left (3+e^{\left (4 e^x x+x+5\right )^2}\right )^2}dx-120 \int \frac {e^x x}{\left (3+e^{\left (4 e^x x+x+5\right )^2}\right )^2}dx+10 \int \frac {x}{3+e^{\left (4 e^x x+x+5\right )^2}}dx+40 \int \frac {e^x x}{3+e^{\left (4 e^x x+x+5\right )^2}}dx\) |
Input:
Int[(-3 + E^(25 + 10*x + x^2 + 16*E^(2*x)*x^2 + E^x*(40*x + 8*x^2))*(-1 + 10*x + 2*x^2 + E^x*(40*x + 56*x^2 + 8*x^3) + E^(2*x)*(32*x^2 + 32*x^3)))/( 9 + 6*E^(25 + 10*x + x^2 + 16*E^(2*x)*x^2 + E^x*(40*x + 8*x^2)) + E^(50 + 20*x + 2*x^2 + 32*E^(2*x)*x^2 + 2*E^x*(40*x + 8*x^2))),x]
Output:
$Aborted
Time = 0.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73
method | result | size |
risch | \(-\frac {x}{{\mathrm e}^{8 \,{\mathrm e}^{x} x^{2}+16 \,{\mathrm e}^{2 x} x^{2}+40 \,{\mathrm e}^{x} x +x^{2}+10 x +25}+3}\) | \(38\) |
parallelrisch | \(-\frac {x}{{\mathrm e}^{16 \,{\mathrm e}^{2 x} x^{2}+\left (8 x^{2}+40 x \right ) {\mathrm e}^{x}+x^{2}+10 x +25}+3}\) | \(38\) |
Input:
int((((32*x^3+32*x^2)*exp(x)^2+(8*x^3+56*x^2+40*x)*exp(x)+2*x^2+10*x-1)*ex p(16*exp(x)^2*x^2+(8*x^2+40*x)*exp(x)+x^2+10*x+25)-3)/(exp(16*exp(x)^2*x^2 +(8*x^2+40*x)*exp(x)+x^2+10*x+25)^2+6*exp(16*exp(x)^2*x^2+(8*x^2+40*x)*exp (x)+x^2+10*x+25)+9),x,method=_RETURNVERBOSE)
Output:
-x/(exp(8*exp(x)*x^2+16*exp(2*x)*x^2+40*exp(x)*x+x^2+10*x+25)+3)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {-3+e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )} \left (-1+10 x+2 x^2+e^x \left (40 x+56 x^2+8 x^3\right )+e^{2 x} \left (32 x^2+32 x^3\right )\right )}{9+6 e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )}+e^{50+20 x+2 x^2+32 e^{2 x} x^2+2 e^x \left (40 x+8 x^2\right )}} \, dx=-\frac {x}{e^{\left (16 \, x^{2} e^{\left (2 \, x\right )} + x^{2} + 8 \, {\left (x^{2} + 5 \, x\right )} e^{x} + 10 \, x + 25\right )} + 3} \] Input:
integrate((((32*x^3+32*x^2)*exp(x)^2+(8*x^3+56*x^2+40*x)*exp(x)+2*x^2+10*x -1)*exp(16*exp(x)^2*x^2+(8*x^2+40*x)*exp(x)+x^2+10*x+25)-3)/(exp(16*exp(x) ^2*x^2+(8*x^2+40*x)*exp(x)+x^2+10*x+25)^2+6*exp(16*exp(x)^2*x^2+(8*x^2+40* x)*exp(x)+x^2+10*x+25)+9),x, algorithm="fricas")
Output:
-x/(e^(16*x^2*e^(2*x) + x^2 + 8*(x^2 + 5*x)*e^x + 10*x + 25) + 3)
Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (17) = 34\).
Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {-3+e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )} \left (-1+10 x+2 x^2+e^x \left (40 x+56 x^2+8 x^3\right )+e^{2 x} \left (32 x^2+32 x^3\right )\right )}{9+6 e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )}+e^{50+20 x+2 x^2+32 e^{2 x} x^2+2 e^x \left (40 x+8 x^2\right )}} \, dx=- \frac {x}{e^{16 x^{2} e^{2 x} + x^{2} + 10 x + \left (8 x^{2} + 40 x\right ) e^{x} + 25} + 3} \] Input:
integrate((((32*x**3+32*x**2)*exp(x)**2+(8*x**3+56*x**2+40*x)*exp(x)+2*x** 2+10*x-1)*exp(16*exp(x)**2*x**2+(8*x**2+40*x)*exp(x)+x**2+10*x+25)-3)/(exp (16*exp(x)**2*x**2+(8*x**2+40*x)*exp(x)+x**2+10*x+25)**2+6*exp(16*exp(x)** 2*x**2+(8*x**2+40*x)*exp(x)+x**2+10*x+25)+9),x)
Output:
-x/(exp(16*x**2*exp(2*x) + x**2 + 10*x + (8*x**2 + 40*x)*exp(x) + 25) + 3)
Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-3+e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )} \left (-1+10 x+2 x^2+e^x \left (40 x+56 x^2+8 x^3\right )+e^{2 x} \left (32 x^2+32 x^3\right )\right )}{9+6 e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )}+e^{50+20 x+2 x^2+32 e^{2 x} x^2+2 e^x \left (40 x+8 x^2\right )}} \, dx=-\frac {x}{e^{\left (16 \, x^{2} e^{\left (2 \, x\right )} + 8 \, x^{2} e^{x} + x^{2} + 40 \, x e^{x} + 10 \, x + 25\right )} + 3} \] Input:
integrate((((32*x^3+32*x^2)*exp(x)^2+(8*x^3+56*x^2+40*x)*exp(x)+2*x^2+10*x -1)*exp(16*exp(x)^2*x^2+(8*x^2+40*x)*exp(x)+x^2+10*x+25)-3)/(exp(16*exp(x) ^2*x^2+(8*x^2+40*x)*exp(x)+x^2+10*x+25)^2+6*exp(16*exp(x)^2*x^2+(8*x^2+40* x)*exp(x)+x^2+10*x+25)+9),x, algorithm="maxima")
Output:
-x/(e^(16*x^2*e^(2*x) + 8*x^2*e^x + x^2 + 40*x*e^x + 10*x + 25) + 3)
Time = 0.38 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-3+e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )} \left (-1+10 x+2 x^2+e^x \left (40 x+56 x^2+8 x^3\right )+e^{2 x} \left (32 x^2+32 x^3\right )\right )}{9+6 e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )}+e^{50+20 x+2 x^2+32 e^{2 x} x^2+2 e^x \left (40 x+8 x^2\right )}} \, dx=-\frac {x}{e^{\left (16 \, x^{2} e^{\left (2 \, x\right )} + 8 \, x^{2} e^{x} + x^{2} + 40 \, x e^{x} + 10 \, x + 25\right )} + 3} \] Input:
integrate((((32*x^3+32*x^2)*exp(x)^2+(8*x^3+56*x^2+40*x)*exp(x)+2*x^2+10*x -1)*exp(16*exp(x)^2*x^2+(8*x^2+40*x)*exp(x)+x^2+10*x+25)-3)/(exp(16*exp(x) ^2*x^2+(8*x^2+40*x)*exp(x)+x^2+10*x+25)^2+6*exp(16*exp(x)^2*x^2+(8*x^2+40* x)*exp(x)+x^2+10*x+25)+9),x, algorithm="giac")
Output:
-x/(e^(16*x^2*e^(2*x) + 8*x^2*e^x + x^2 + 40*x*e^x + 10*x + 25) + 3)
Time = 3.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {-3+e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )} \left (-1+10 x+2 x^2+e^x \left (40 x+56 x^2+8 x^3\right )+e^{2 x} \left (32 x^2+32 x^3\right )\right )}{9+6 e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )}+e^{50+20 x+2 x^2+32 e^{2 x} x^2+2 e^x \left (40 x+8 x^2\right )}} \, dx=-\frac {x}{{\mathrm {e}}^{40\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{25}\,{\mathrm {e}}^{8\,x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{16\,x^2\,{\mathrm {e}}^{2\,x}}+3} \] Input:
int((exp(10*x + 16*x^2*exp(2*x) + exp(x)*(40*x + 8*x^2) + x^2 + 25)*(10*x + exp(2*x)*(32*x^2 + 32*x^3) + 2*x^2 + exp(x)*(40*x + 56*x^2 + 8*x^3) - 1) - 3)/(6*exp(10*x + 16*x^2*exp(2*x) + exp(x)*(40*x + 8*x^2) + x^2 + 25) + exp(20*x + 32*x^2*exp(2*x) + 2*exp(x)*(40*x + 8*x^2) + 2*x^2 + 50) + 9),x)
Output:
-x/(exp(40*x*exp(x))*exp(10*x)*exp(x^2)*exp(25)*exp(8*x^2*exp(x))*exp(16*x ^2*exp(2*x)) + 3)
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {-3+e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )} \left (-1+10 x+2 x^2+e^x \left (40 x+56 x^2+8 x^3\right )+e^{2 x} \left (32 x^2+32 x^3\right )\right )}{9+6 e^{25+10 x+x^2+16 e^{2 x} x^2+e^x \left (40 x+8 x^2\right )}+e^{50+20 x+2 x^2+32 e^{2 x} x^2+2 e^x \left (40 x+8 x^2\right )}} \, dx=-\frac {x}{e^{16 e^{2 x} x^{2}+8 e^{x} x^{2}+40 e^{x} x +x^{2}+10 x} e^{25}+3} \] Input:
int((((32*x^3+32*x^2)*exp(x)^2+(8*x^3+56*x^2+40*x)*exp(x)+2*x^2+10*x-1)*ex p(16*exp(x)^2*x^2+(8*x^2+40*x)*exp(x)+x^2+10*x+25)-3)/(exp(16*exp(x)^2*x^2 +(8*x^2+40*x)*exp(x)+x^2+10*x+25)^2+6*exp(16*exp(x)^2*x^2+(8*x^2+40*x)*exp (x)+x^2+10*x+25)+9),x)
Output:
( - x)/(e**(16*e**(2*x)*x**2 + 8*e**x*x**2 + 40*e**x*x + x**2 + 10*x)*e**2 5 + 3)