Integrand size = 107, antiderivative size = 33 \[ \int \frac {e^{2 x} \left (e^4 (48-96 x)-168 x+168 x^2+e^{\frac {1}{3} (2-2 x)} (-48+128 x)\right )}{48 e^8 x^2+48 e^{\frac {2}{3} (2-2 x)} x^2-168 e^4 x^3+147 x^4+e^{\frac {1}{3} (2-2 x)} \left (-96 e^4 x^2+168 x^3\right )} \, dx=\frac {e^{2 x}}{x \left (-e^4+e^{\frac {2 (1-x)}{3}}+\frac {7 x}{4}\right )} \] Output:
exp(x)^2/(7/4*x+exp(-2/3*x+2/3)-exp(4))/x
Time = 10.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {e^{2 x} \left (e^4 (48-96 x)-168 x+168 x^2+e^{\frac {1}{3} (2-2 x)} (-48+128 x)\right )}{48 e^8 x^2+48 e^{\frac {2}{3} (2-2 x)} x^2-168 e^4 x^3+147 x^4+e^{\frac {1}{3} (2-2 x)} \left (-96 e^4 x^2+168 x^3\right )} \, dx=\frac {8 e^{8 x/3}}{8 e^{2/3} x-8 e^{4+\frac {2 x}{3}} x+14 e^{2 x/3} x^2} \] Input:
Integrate[(E^(2*x)*(E^4*(48 - 96*x) - 168*x + 168*x^2 + E^((2 - 2*x)/3)*(- 48 + 128*x)))/(48*E^8*x^2 + 48*E^((2*(2 - 2*x))/3)*x^2 - 168*E^4*x^3 + 147 *x^4 + E^((2 - 2*x)/3)*(-96*E^4*x^2 + 168*x^3)),x]
Output:
(8*E^((8*x)/3))/(8*E^(2/3)*x - 8*E^(4 + (2*x)/3)*x + 14*E^((2*x)/3)*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.
\(\displaystyle \int \frac {e^{2 x} \left (168 x^2-168 x+e^4 (48-96 x)+e^{\frac {1}{3} (2-2 x)} (128 x-48)\right )}{147 x^4-168 e^4 x^3+48 e^{\frac {2}{3} (2-2 x)} x^2+48 e^8 x^2+e^{\frac {1}{3} (2-2 x)} \left (168 x^3-96 e^4 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{10 x/3} \left (168 x^2-168 x+e^4 (48-96 x)+e^{\frac {1}{3} (2-2 x)} (128 x-48)\right )}{3 x^2 \left (7 e^{2 x/3} x-4 e^{\frac {2 x}{3}+4}+4 e^{2/3}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {8 e^{10 x/3} \left (-21 x^2+21 x+2 e^{\frac {2 (1-x)}{3}} (3-8 x)-6 e^4 (1-2 x)\right )}{x^2 \left (7 e^{2 x/3} x-4 e^{\frac {2 x}{3}+4}+4 e^{2/3}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {8}{3} \int \frac {e^{10 x/3} \left (-21 x^2+21 x+2 e^{\frac {2 (1-x)}{3}} (3-8 x)-6 e^4 (1-2 x)\right )}{x^2 \left (7 e^{2 x/3} x-4 e^{\frac {2 x}{3}+4}+4 e^{2/3}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {8}{3} \int \left (\frac {e^{\frac {10 x}{3}-\frac {2}{3}} \left (4 e^4-7 x\right ) (8 x-3)}{8 x^2 \left (-7 e^{2 x/3} x+4 e^{\frac {2 x}{3}+4}-4 e^{2/3}\right )}-\frac {e^{\frac {8 x}{3}-\frac {2}{3}} (8 x-3)}{8 x^2}+\frac {e^{10 x/3} \left (14 x-8 e^4+21\right )}{2 x \left (7 e^{2 x/3} x-4 e^{\frac {2 x}{3}+4}+4 e^{2/3}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {8}{3} \left (-\frac {3}{2} \int \frac {e^{\frac {10 x}{3}+\frac {10}{3}}}{x^2 \left (-7 e^{2 x/3} x+4 e^{\frac {2 x}{3}+4}-4 e^{2/3}\right )}dx+7 \int \frac {e^{10 x/3}}{\left (-7 e^{2 x/3} x+4 e^{\frac {2 x}{3}+4}-4 e^{2/3}\right )^2}dx+\frac {1}{2} \left (21-8 e^4\right ) \int \frac {e^{10 x/3}}{x \left (-7 e^{2 x/3} x+4 e^{\frac {2 x}{3}+4}-4 e^{2/3}\right )^2}dx-7 \int \frac {e^{\frac {10 x}{3}-\frac {2}{3}}}{-7 e^{2 x/3} x+4 e^{\frac {2 x}{3}+4}-4 e^{2/3}}dx+\frac {1}{8} \left (21+32 e^4\right ) \int \frac {e^{\frac {10 x}{3}-\frac {2}{3}}}{x \left (-7 e^{2 x/3} x+4 e^{\frac {2 x}{3}+4}-4 e^{2/3}\right )}dx-\frac {3 e^{\frac {8 x}{3}-\frac {2}{3}}}{8 x}\right )\) |
Input:
Int[(E^(2*x)*(E^4*(48 - 96*x) - 168*x + 168*x^2 + E^((2 - 2*x)/3)*(-48 + 1 28*x)))/(48*E^8*x^2 + 48*E^((2*(2 - 2*x))/3)*x^2 - 168*E^4*x^3 + 147*x^4 + E^((2 - 2*x)/3)*(-96*E^4*x^2 + 168*x^3)),x]
Output:
$Aborted
Time = 0.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85
method | result | size |
parallelrisch | \(-\frac {4 \,{\mathrm e}^{2 x}}{x \left (-4 \,{\mathrm e}^{-\frac {2 x}{3}+\frac {2}{3}}-7 x +4 \,{\mathrm e}^{4}\right )}\) | \(28\) |
Input:
int(((128*x-48)*exp(-2/3*x+2/3)+(-96*x+48)*exp(4)+168*x^2-168*x)*exp(x)^2/ (48*x^2*exp(-2/3*x+2/3)^2+(-96*x^2*exp(4)+168*x^3)*exp(-2/3*x+2/3)+48*x^2* exp(4)^2-168*x^3*exp(4)+147*x^4),x,method=_RETURNVERBOSE)
Output:
-4*exp(x)^2/x/(-4*exp(-2/3*x+2/3)-7*x+4*exp(4))
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2 x} \left (e^4 (48-96 x)-168 x+168 x^2+e^{\frac {1}{3} (2-2 x)} (-48+128 x)\right )}{48 e^8 x^2+48 e^{\frac {2}{3} (2-2 x)} x^2-168 e^4 x^3+147 x^4+e^{\frac {1}{3} (2-2 x)} \left (-96 e^4 x^2+168 x^3\right )} \, dx=\frac {4 \, e^{2}}{{\left (7 \, x^{2} - 4 \, x e^{4}\right )} e^{\left (-2 \, x + 2\right )} + 4 \, x e^{\left (-\frac {8}{3} \, x + \frac {8}{3}\right )}} \] Input:
integrate(((128*x-48)*exp(-2/3*x+2/3)+(-96*x+48)*exp(4)+168*x^2-168*x)*exp (x)^2/(48*x^2*exp(-2/3*x+2/3)^2+(-96*x^2*exp(4)+168*x^3)*exp(-2/3*x+2/3)+4 8*x^2*exp(4)^2-168*x^3*exp(4)+147*x^4),x, algorithm="fricas")
Output:
4*e^2/((7*x^2 - 4*x*e^4)*e^(-2*x + 2) + 4*x*e^(-8/3*x + 8/3))
Timed out. \[ \int \frac {e^{2 x} \left (e^4 (48-96 x)-168 x+168 x^2+e^{\frac {1}{3} (2-2 x)} (-48+128 x)\right )}{48 e^8 x^2+48 e^{\frac {2}{3} (2-2 x)} x^2-168 e^4 x^3+147 x^4+e^{\frac {1}{3} (2-2 x)} \left (-96 e^4 x^2+168 x^3\right )} \, dx=\text {Timed out} \] Input:
integrate(((128*x-48)*exp(-2/3*x+2/3)+(-96*x+48)*exp(4)+168*x**2-168*x)*ex p(x)**2/(48*x**2*exp(-2/3*x+2/3)**2+(-96*x**2*exp(4)+168*x**3)*exp(-2/3*x+ 2/3)+48*x**2*exp(4)**2-168*x**3*exp(4)+147*x**4),x)
Output:
Timed out
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2 x} \left (e^4 (48-96 x)-168 x+168 x^2+e^{\frac {1}{3} (2-2 x)} (-48+128 x)\right )}{48 e^8 x^2+48 e^{\frac {2}{3} (2-2 x)} x^2-168 e^4 x^3+147 x^4+e^{\frac {1}{3} (2-2 x)} \left (-96 e^4 x^2+168 x^3\right )} \, dx=\frac {4 \, e^{\left (\frac {8}{3} \, x + \frac {1}{3}\right )}}{4 \, x e + {\left (7 \, x^{2} e^{\frac {1}{3}} - 4 \, x e^{\frac {13}{3}}\right )} e^{\left (\frac {2}{3} \, x\right )}} \] Input:
integrate(((128*x-48)*exp(-2/3*x+2/3)+(-96*x+48)*exp(4)+168*x^2-168*x)*exp (x)^2/(48*x^2*exp(-2/3*x+2/3)^2+(-96*x^2*exp(4)+168*x^3)*exp(-2/3*x+2/3)+4 8*x^2*exp(4)^2-168*x^3*exp(4)+147*x^4),x, algorithm="maxima")
Output:
4*e^(8/3*x + 1/3)/(4*x*e + (7*x^2*e^(1/3) - 4*x*e^(13/3))*e^(2/3*x))
Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (27) = 54\).
Time = 0.23 (sec) , antiderivative size = 386, normalized size of antiderivative = 11.70 \[ \int \frac {e^{2 x} \left (e^4 (48-96 x)-168 x+168 x^2+e^{\frac {1}{3} (2-2 x)} (-48+128 x)\right )}{48 e^8 x^2+48 e^{\frac {2}{3} (2-2 x)} x^2-168 e^4 x^3+147 x^4+e^{\frac {1}{3} (2-2 x)} \left (-96 e^4 x^2+168 x^3\right )} \, dx =\text {Too large to display} \] Input:
integrate(((128*x-48)*exp(-2/3*x+2/3)+(-96*x+48)*exp(4)+168*x^2-168*x)*exp (x)^2/(48*x^2*exp(-2/3*x+2/3)^2+(-96*x^2*exp(4)+168*x^3)*exp(-2/3*x+2/3)+4 8*x^2*exp(4)^2-168*x^3*exp(4)+147*x^4),x, algorithm="giac")
Output:
4*(343*(x - 1)^3*e^(2*x) + 1029*(x - 1)^2*e^(2*x) - 588*(x - 1)^2*e^(2*x + 4) + 1029*(x - 1)*e^(2*x) + 336*(x - 1)*e^(2*x + 8) - 1176*(x - 1)*e^(2*x + 4) + 343*e^(2*x) - 64*e^(2*x + 12) + 336*e^(2*x + 8) - 588*e^(2*x + 4)) /(2401*(x - 1)^5 - 5488*(x - 1)^4*e^4 + 1372*(x - 1)^4*e^(-2/3*x + 2/3) + 12005*(x - 1)^4 + 4704*(x - 1)^3*e^8 - 21952*(x - 1)^3*e^4 - 2352*(x - 1)^ 3*e^(-2/3*x + 14/3) + 5488*(x - 1)^3*e^(-2/3*x + 2/3) + 24010*(x - 1)^3 - 1792*(x - 1)^2*e^12 + 14112*(x - 1)^2*e^8 - 32928*(x - 1)^2*e^4 + 1344*(x - 1)^2*e^(-2/3*x + 26/3) - 7056*(x - 1)^2*e^(-2/3*x + 14/3) + 8232*(x - 1) ^2*e^(-2/3*x + 2/3) + 24010*(x - 1)^2 + 256*(x - 1)*e^16 - 3584*(x - 1)*e^ 12 + 14112*(x - 1)*e^8 - 21952*(x - 1)*e^4 - 256*(x - 1)*e^(-2/3*x + 38/3) + 2688*(x - 1)*e^(-2/3*x + 26/3) - 7056*(x - 1)*e^(-2/3*x + 14/3) + 5488* (x - 1)*e^(-2/3*x + 2/3) + 12005*x + 256*e^16 - 1792*e^12 + 4704*e^8 - 548 8*e^4 - 256*e^(-2/3*x + 38/3) + 1344*e^(-2/3*x + 26/3) - 2352*e^(-2/3*x + 14/3) + 1372*e^(-2/3*x + 2/3) - 9604)
Timed out. \[ \int \frac {e^{2 x} \left (e^4 (48-96 x)-168 x+168 x^2+e^{\frac {1}{3} (2-2 x)} (-48+128 x)\right )}{48 e^8 x^2+48 e^{\frac {2}{3} (2-2 x)} x^2-168 e^4 x^3+147 x^4+e^{\frac {1}{3} (2-2 x)} \left (-96 e^4 x^2+168 x^3\right )} \, dx=-\int \frac {{\mathrm {e}}^{2\,x}\,\left (168\,x-{\mathrm {e}}^{\frac {2}{3}-\frac {2\,x}{3}}\,\left (128\,x-48\right )-168\,x^2+{\mathrm {e}}^4\,\left (96\,x-48\right )\right )}{48\,x^2\,{\mathrm {e}}^8-168\,x^3\,{\mathrm {e}}^4-{\mathrm {e}}^{\frac {2}{3}-\frac {2\,x}{3}}\,\left (96\,x^2\,{\mathrm {e}}^4-168\,x^3\right )+48\,x^2\,{\mathrm {e}}^{\frac {4}{3}-\frac {4\,x}{3}}+147\,x^4} \,d x \] Input:
int(-(exp(2*x)*(168*x - exp(2/3 - (2*x)/3)*(128*x - 48) - 168*x^2 + exp(4) *(96*x - 48)))/(48*x^2*exp(8) - 168*x^3*exp(4) - exp(2/3 - (2*x)/3)*(96*x^ 2*exp(4) - 168*x^3) + 48*x^2*exp(4/3 - (4*x)/3) + 147*x^4),x)
Output:
-int((exp(2*x)*(168*x - exp(2/3 - (2*x)/3)*(128*x - 48) - 168*x^2 + exp(4) *(96*x - 48)))/(48*x^2*exp(8) - 168*x^3*exp(4) - exp(2/3 - (2*x)/3)*(96*x^ 2*exp(4) - 168*x^3) + 48*x^2*exp(4/3 - (4*x)/3) + 147*x^4), x)
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^{2 x} \left (e^4 (48-96 x)-168 x+168 x^2+e^{\frac {1}{3} (2-2 x)} (-48+128 x)\right )}{48 e^8 x^2+48 e^{\frac {2}{3} (2-2 x)} x^2-168 e^4 x^3+147 x^4+e^{\frac {1}{3} (2-2 x)} \left (-96 e^4 x^2+168 x^3\right )} \, dx=-\frac {4 e^{\frac {8 x}{3}}}{x \left (4 e^{\frac {2 x}{3}} e^{4}-7 e^{\frac {2 x}{3}} x -4 e^{\frac {2}{3}}\right )} \] Input:
int(((128*x-48)*exp(-2/3*x+2/3)+(-96*x+48)*exp(4)+168*x^2-168*x)*exp(x)^2/ (48*x^2*exp(-2/3*x+2/3)^2+(-96*x^2*exp(4)+168*x^3)*exp(-2/3*x+2/3)+48*x^2* exp(4)^2-168*x^3*exp(4)+147*x^4),x)
Output:
( - 4*e**((8*x)/3))/(x*(4*e**((2*x)/3)*e**4 - 7*e**((2*x)/3)*x - 4*e**(2/3 )))