Integrand size = 92, antiderivative size = 34 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=x^2 \left (7-x-e^x x \left (\frac {e^{x \left (-3+\frac {2}{4+x}\right )}}{x}+x\right )\right ) \] Output:
(7-exp(x)*(x+exp((2/(4+x)-3)*x)/x)*x-x)*x^2
Time = 5.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=x^2 \left (7-e^{-\frac {2 x (3+x)}{4+x}}-x-e^x x^2\right ) \] Input:
Integrate[(224*x + 64*x^2 - 10*x^3 - 3*x^4 + E^x*(-64*x^3 - 48*x^4 - 12*x^ 5 - x^6 + E^((-10*x - 3*x^2)/(4 + x))*(-32*x + 8*x^2 + 14*x^3 + 2*x^4)))/( 16 + 8*x + x^2),x]
Output:
x^2*(7 - E^((-2*x*(3 + x))/(4 + x)) - x - 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.
\(\displaystyle \int \frac {-3 x^4-10 x^3+64 x^2+e^x \left (-x^6-12 x^5-48 x^4-64 x^3+e^{\frac {-3 x^2-10 x}{x+4}} \left (2 x^4+14 x^3+8 x^2-32 x\right )\right )+224 x}{x^2+8 x+16} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-3 x^4-10 x^3+64 x^2+e^x \left (-x^6-12 x^5-48 x^4-64 x^3+e^{\frac {-3 x^2-10 x}{x+4}} \left (2 x^4+14 x^3+8 x^2-32 x\right )\right )+224 x}{(x+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {e^x x^6}{(x+4)^2}-\frac {12 e^x x^5}{(x+4)^2}-\frac {48 e^x x^4}{(x+4)^2}-\frac {3 x^4}{(x+4)^2}-\frac {64 e^x x^3}{(x+4)^2}-\frac {10 x^3}{(x+4)^2}+\frac {64 x^2}{(x+4)^2}+\frac {2 e^{-\frac {2 x (x+3)}{x+4}} \left (x^3+7 x^2+4 x-16\right ) x}{(x+4)^2}+\frac {224 x}{(x+4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int e^{-\frac {2 x (x+3)}{x+4}} x^2dx-8 \int e^{-\frac {2 x (x+3)}{x+4}}dx-2 \int e^{-\frac {2 x (x+3)}{x+4}} xdx-128 \int \frac {e^{-\frac {2 x (x+3)}{x+4}}}{(x+4)^2}dx+64 \int \frac {e^{-\frac {2 x (x+3)}{x+4}}}{x+4}dx-e^x x^4-x^3+7 x^2\) |
Input:
Int[(224*x + 64*x^2 - 10*x^3 - 3*x^4 + E^x*(-64*x^3 - 48*x^4 - 12*x^5 - x^ 6 + E^((-10*x - 3*x^2)/(4 + x))*(-32*x + 8*x^2 + 14*x^3 + 2*x^4)))/(16 + 8 *x + x^2),x]
Output:
$Aborted
Time = 2.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06
method | result | size |
risch | \(-{\mathrm e}^{x} x^{4}-x^{2} {\mathrm e}^{-\frac {2 \left (3+x \right ) x}{4+x}}-x^{3}+7 x^{2}\) | \(36\) |
parallelrisch | \(-{\mathrm e}^{x} x^{4}-{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}-x^{3}+7 x^{2}-96\) | \(43\) |
parts | \(\frac {-4 \,{\mathrm e}^{x} x^{4}-x^{5} {\mathrm e}^{x}-4 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}-{\mathrm e}^{x} x^{3} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}}{4+x}-x^{3}+7 x^{2}\) | \(79\) |
norman | \(\frac {28 x^{2}+3 x^{3}-x^{4}-x^{5} {\mathrm e}^{x}-4 \,{\mathrm e}^{x} x^{4}-4 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}-{\mathrm e}^{x} x^{3} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}}{4+x}\) | \(83\) |
orering | \(\text {Expression too large to display}\) | \(1614\) |
Input:
int((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5-48*x^4 -64*x^3)*exp(x)-3*x^4-10*x^3+64*x^2+224*x)/(x^2+8*x+16),x,method=_RETURNVE RBOSE)
Output:
-exp(x)*x^4-x^2*exp(-2*(3+x)*x/(4+x))-x^3+7*x^2
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=-x^{3} + 7 \, x^{2} - {\left (x^{4} + x^{2} e^{\left (-\frac {3 \, x^{2} + 10 \, x}{x + 4}\right )}\right )} e^{x} \] Input:
integrate((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5- 48*x^4-64*x^3)*exp(x)-3*x^4-10*x^3+64*x^2+224*x)/(x^2+8*x+16),x, algorithm ="fricas")
Output:
-x^3 + 7*x^2 - (x^4 + x^2*e^(-(3*x^2 + 10*x)/(x + 4)))*e^x
Time = 3.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=- x^{4} e^{x} - x^{3} - x^{2} e^{x} e^{\frac {- 3 x^{2} - 10 x}{x + 4}} + 7 x^{2} \] Input:
integrate((((2*x**4+14*x**3+8*x**2-32*x)*exp((-3*x**2-10*x)/(4+x))-x**6-12 *x**5-48*x**4-64*x**3)*exp(x)-3*x**4-10*x**3+64*x**2+224*x)/(x**2+8*x+16), x)
Output:
-x**4*exp(x) - x**3 - x**2*exp(x)*exp((-3*x**2 - 10*x)/(x + 4)) + 7*x**2
Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=-x^{3} + 7 \, x^{2} - {\left (x^{4} e^{\left (3 \, x\right )} + x^{2} e^{\left (-\frac {8}{x + 4} + 2\right )}\right )} e^{\left (-2 \, x\right )} \] Input:
integrate((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5- 48*x^4-64*x^3)*exp(x)-3*x^4-10*x^3+64*x^2+224*x)/(x^2+8*x+16),x, algorithm ="maxima")
Output:
-x^3 + 7*x^2 - (x^4*e^(3*x) + x^2*e^(-8/(x + 4) + 2))*e^(-2*x)
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=-x^{4} e^{x} - x^{3} - x^{2} e^{\left (-\frac {2 \, {\left (x^{2} + 3 \, x\right )}}{x + 4}\right )} + 7 \, x^{2} \] Input:
integrate((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5- 48*x^4-64*x^3)*exp(x)-3*x^4-10*x^3+64*x^2+224*x)/(x^2+8*x+16),x, algorithm ="giac")
Output:
-x^4*e^x - x^3 - x^2*e^(-2*(x^2 + 3*x)/(x + 4)) + 7*x^2
Time = 4.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.29 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=7\,x^2-x^4\,{\mathrm {e}}^x-x^3-x^2\,{\mathrm {e}}^{x-\frac {10\,x}{x+4}-\frac {3\,x^2}{x+4}} \] Input:
int(-(exp(x)*(64*x^3 - exp(-(10*x + 3*x^2)/(x + 4))*(8*x^2 - 32*x + 14*x^3 + 2*x^4) + 48*x^4 + 12*x^5 + x^6) - 224*x - 64*x^2 + 10*x^3 + 3*x^4)/(8*x + x^2 + 16),x)
Output:
7*x^2 - x^4*exp(x) - x^3 - x^2*exp(x - (10*x)/(x + 4) - (3*x^2)/(x + 4))
Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.76 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=\frac {x^{2} \left (-e^{\frac {3 x^{2}+12 x +8}{x +4}} x^{2}-e^{\frac {2 x^{2}+8 x +8}{x +4}} x +7 e^{\frac {2 x^{2}+8 x +8}{x +4}}-e^{2}\right )}{e^{\frac {2 x^{2}+8 x +8}{x +4}}} \] Input:
int((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5-48*x^4 -64*x^3)*exp(x)-3*x^4-10*x^3+64*x^2+224*x)/(x^2+8*x+16),x)
Output:
(x**2*( - e**((3*x**2 + 12*x + 8)/(x + 4))*x**2 - e**((2*x**2 + 8*x + 8)/( x + 4))*x + 7*e**((2*x**2 + 8*x + 8)/(x + 4)) - e**2))/e**((2*x**2 + 8*x + 8)/(x + 4))