Integrand size = 99, antiderivative size = 23 \[ \int \frac {3+e^{5+x} (-18+6 e+6 x)}{32+2 e^2-16 x+2 x^2+e (-16+4 x)+e^{5+x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )+e^{10+2 x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )} \, dx=\frac {3}{\left (2+4 e^{5+x}\right ) (4-e-x)} \] Output:
3/(4-x-exp(1))/(2+4*exp(5+x))
Time = 1.57 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {3+e^{5+x} (-18+6 e+6 x)}{32+2 e^2-16 x+2 x^2+e (-16+4 x)+e^{5+x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )+e^{10+2 x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )} \, dx=-\frac {3}{2 \left (1+2 e^{5+x}\right ) (-4+e+x)} \] Input:
Integrate[(3 + E^(5 + x)*(-18 + 6*E + 6*x))/(32 + 2*E^2 - 16*x + 2*x^2 + E *(-16 + 4*x) + E^(5 + x)*(128 + 8*E^2 - 64*x + 8*x^2 + E*(-64 + 16*x)) + E ^(10 + 2*x)*(128 + 8*E^2 - 64*x + 8*x^2 + E*(-64 + 16*x))),x]
Output:
-3/(2*(1 + 2*E^(5 + x))*(-4 + E + x))
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.
\(\displaystyle \int \frac {e^{x+5} (6 x+6 e-18)+3}{2 x^2+e^{x+5} \left (8 x^2-64 x+e (16 x-64)+8 e^2+128\right )+e^{2 x+10} \left (8 x^2-64 x+e (16 x-64)+8 e^2+128\right )-16 x+e (4 x-16)+2 e^2+32} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {6 e^{x+5} (x-3)+6 e^{x+6}+3}{2 \left (2 e^{x+5}+1\right )^2 (-x-e+4)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {3 \left (-2 e^{x+5} (3-x)+2 e^{x+6}+1\right )}{\left (1+2 e^{x+5}\right )^2 (-x-e+4)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{2} \int \frac {-2 e^{x+5} (3-x)+2 e^{x+6}+1}{\left (1+2 e^{x+5}\right )^2 (-x-e+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {3}{2} \int \left (\frac {x+e-3}{\left (1+2 e^{x+5}\right ) (x+e-4)^2}-\frac {1}{\left (1+2 e^{x+5}\right )^2 (x+e-4)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{2} \left (\int \frac {1}{\left (1+2 e^{x+5}\right ) (x+e-4)^2}dx-\int \frac {1}{\left (1+2 e^{x+5}\right )^2 (x+e-4)}dx+\int \frac {1}{\left (1+2 e^{x+5}\right ) (x+e-4)}dx\right )\) |
Input:
Int[(3 + E^(5 + x)*(-18 + 6*E + 6*x))/(32 + 2*E^2 - 16*x + 2*x^2 + E*(-16 + 4*x) + E^(5 + x)*(128 + 8*E^2 - 64*x + 8*x^2 + E*(-64 + 16*x)) + E^(10 + 2*x)*(128 + 8*E^2 - 64*x + 8*x^2 + E*(-64 + 16*x))),x]
Output:
$Aborted
Time = 0.55 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
norman | \(-\frac {3}{2 \left (2 \,{\mathrm e}^{5+x}+1\right ) \left ({\mathrm e}+x -4\right )}\) | \(20\) |
risch | \(-\frac {3}{2 \left (2 \,{\mathrm e}^{5+x}+1\right ) \left ({\mathrm e}+x -4\right )}\) | \(20\) |
parallelrisch | \(-\frac {3}{2 \left (2 \,{\mathrm e}^{5+x}+1\right ) \left ({\mathrm e}+x -4\right )}\) | \(20\) |
Input:
int(((6*exp(1)+6*x-18)*exp(5+x)+3)/((8*exp(1)^2+(16*x-64)*exp(1)+8*x^2-64* x+128)*exp(5+x)^2+(8*exp(1)^2+(16*x-64)*exp(1)+8*x^2-64*x+128)*exp(5+x)+2* exp(1)^2+(4*x-16)*exp(1)+2*x^2-16*x+32),x,method=_RETURNVERBOSE)
Output:
-3/2/(2*exp(5+x)+1)/(exp(1)+x-4)
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {3+e^{5+x} (-18+6 e+6 x)}{32+2 e^2-16 x+2 x^2+e (-16+4 x)+e^{5+x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )+e^{10+2 x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )} \, dx=-\frac {3}{2 \, {\left (2 \, {\left (x + e - 4\right )} e^{\left (x + 5\right )} + x + e - 4\right )}} \] Input:
integrate(((6*exp(1)+6*x-18)*exp(5+x)+3)/((8*exp(1)^2+(16*x-64)*exp(1)+8*x ^2-64*x+128)*exp(5+x)^2+(8*exp(1)^2+(16*x-64)*exp(1)+8*x^2-64*x+128)*exp(5 +x)+2*exp(1)^2+(4*x-16)*exp(1)+2*x^2-16*x+32),x, algorithm="fricas")
Output:
-3/2/(2*(x + e - 4)*e^(x + 5) + x + e - 4)
Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {3+e^{5+x} (-18+6 e+6 x)}{32+2 e^2-16 x+2 x^2+e (-16+4 x)+e^{5+x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )+e^{10+2 x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )} \, dx=- \frac {3}{2 x + \left (4 x - 16 + 4 e\right ) e^{x + 5} - 8 + 2 e} \] Input:
integrate(((6*exp(1)+6*x-18)*exp(5+x)+3)/((8*exp(1)**2+(16*x-64)*exp(1)+8* x**2-64*x+128)*exp(5+x)**2+(8*exp(1)**2+(16*x-64)*exp(1)+8*x**2-64*x+128)* exp(5+x)+2*exp(1)**2+(4*x-16)*exp(1)+2*x**2-16*x+32),x)
Output:
-3/(2*x + (4*x - 16 + 4*E)*exp(x + 5) - 8 + 2*E)
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {3+e^{5+x} (-18+6 e+6 x)}{32+2 e^2-16 x+2 x^2+e (-16+4 x)+e^{5+x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )+e^{10+2 x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )} \, dx=-\frac {3}{2 \, {\left (2 \, {\left (x e^{5} + e^{6} - 4 \, e^{5}\right )} e^{x} + x + e - 4\right )}} \] Input:
integrate(((6*exp(1)+6*x-18)*exp(5+x)+3)/((8*exp(1)^2+(16*x-64)*exp(1)+8*x ^2-64*x+128)*exp(5+x)^2+(8*exp(1)^2+(16*x-64)*exp(1)+8*x^2-64*x+128)*exp(5 +x)+2*exp(1)^2+(4*x-16)*exp(1)+2*x^2-16*x+32),x, algorithm="maxima")
Output:
-3/2/(2*(x*e^5 + e^6 - 4*e^5)*e^x + x + e - 4)
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {3+e^{5+x} (-18+6 e+6 x)}{32+2 e^2-16 x+2 x^2+e (-16+4 x)+e^{5+x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )+e^{10+2 x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )} \, dx=-\frac {3}{2 \, {\left (2 \, {\left (x + 5\right )} e^{\left (x + 5\right )} + x + e + 2 \, e^{\left (x + 6\right )} - 18 \, e^{\left (x + 5\right )} - 4\right )}} \] Input:
integrate(((6*exp(1)+6*x-18)*exp(5+x)+3)/((8*exp(1)^2+(16*x-64)*exp(1)+8*x ^2-64*x+128)*exp(5+x)^2+(8*exp(1)^2+(16*x-64)*exp(1)+8*x^2-64*x+128)*exp(5 +x)+2*exp(1)^2+(4*x-16)*exp(1)+2*x^2-16*x+32),x, algorithm="giac")
Output:
-3/2/(2*(x + 5)*e^(x + 5) + x + e + 2*e^(x + 6) - 18*e^(x + 5) - 4)
Time = 1.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.61 \[ \int \frac {3+e^{5+x} (-18+6 e+6 x)}{32+2 e^2-16 x+2 x^2+e (-16+4 x)+e^{5+x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )+e^{10+2 x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )} \, dx=\frac {6\,{\mathrm {e}}^{x+5}+\frac {3\,x}{\mathrm {e}-4}+\frac {6\,x\,{\mathrm {e}}^{x+5}}{\mathrm {e}-4}}{2\,x-16\,{\mathrm {e}}^{x+5}+4\,{\mathrm {e}}^{x+6}+2\,\mathrm {e}+4\,x\,{\mathrm {e}}^{x+5}-8} \] Input:
int((exp(x + 5)*(6*x + 6*exp(1) - 18) + 3)/(2*exp(2) - 16*x + exp(2*x + 10 )*(8*exp(2) - 64*x + 8*x^2 + exp(1)*(16*x - 64) + 128) + exp(x + 5)*(8*exp (2) - 64*x + 8*x^2 + exp(1)*(16*x - 64) + 128) + 2*x^2 + exp(1)*(4*x - 16) + 32),x)
Output:
(6*exp(x + 5) + (3*x)/(exp(1) - 4) + (6*x*exp(x + 5))/(exp(1) - 4))/(2*x - 16*exp(x + 5) + 4*exp(x + 6) + 2*exp(1) + 4*x*exp(x + 5) - 8)
Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {3+e^{5+x} (-18+6 e+6 x)}{32+2 e^2-16 x+2 x^2+e (-16+4 x)+e^{5+x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )+e^{10+2 x} \left (128+8 e^2-64 x+8 x^2+e (-64+16 x)\right )} \, dx=-\frac {3}{4 e^{x} e^{6}+4 e^{x} e^{5} x -16 e^{x} e^{5}+2 e +2 x -8} \] Input:
int(((6*exp(1)+6*x-18)*exp(5+x)+3)/((8*exp(1)^2+(16*x-64)*exp(1)+8*x^2-64* x+128)*exp(5+x)^2+(8*exp(1)^2+(16*x-64)*exp(1)+8*x^2-64*x+128)*exp(5+x)+2* exp(1)^2+(4*x-16)*exp(1)+2*x^2-16*x+32),x)
Output:
( - 3)/(2*(2*e**x*e**6 + 2*e**x*e**5*x - 8*e**x*e**5 + e + x - 4))