Integrand size = 112, antiderivative size = 28 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=3-2 x+\frac {x}{4 \left (1+e^{x^3}-x-\log ^2(2)\right )} \] Output:
3+1/4*x/(exp(x^3)-x-ln(2)^2+1)-2*x
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=\frac {1}{4} \left (-8 x+\frac {x}{1+e^{x^3}-x-\log ^2(2)}\right ) \] Input:
Integrate[(-7 - 8*E^(2*x^3) + 16*x - 8*x^2 + (15 - 16*x)*Log[2]^2 - 8*Log[ 2]^4 + E^x^3*(-15 + 16*x - 3*x^3 + 16*Log[2]^2))/(4 + 4*E^(2*x^3) - 8*x + 4*x^2 + (-8 + 8*x)*Log[2]^2 + 4*Log[2]^4 + E^x^3*(8 - 8*x - 8*Log[2]^2)),x ]
Output:
(-8*x + x/(1 + E^x^3 - x - Log[2]^2))/4
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.
\(\displaystyle \int \frac {-8 e^{2 x^3}+e^{x^3} \left (-3 x^3+16 x-15+16 \log ^2(2)\right )-8 x^2+16 x+(15-16 x) \log ^2(2)-7-8 \log ^4(2)}{4 e^{2 x^3}+e^{x^3} \left (-8 x+8-8 \log ^2(2)\right )+4 x^2-8 x+(8 x-8) \log ^2(2)+4+4 \log ^4(2)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-8 e^{2 x^3}+e^{x^3} \left (-3 x^3+16 x-15+16 \log ^2(2)\right )-8 x^2+16 x+(15-16 x) \log ^2(2)-7 \left (1+\frac {8 \log ^4(2)}{7}\right )}{4 \left (e^{x^3}-x+1-\log ^2(2)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {8 x^2-16 x+8 e^{2 x^3}+e^{x^3} \left (3 x^3-16 x-16 \log ^2(2)+15\right )+8 \log ^4(2)-(15-16 x) \log ^2(2)+7}{\left (-x+e^{x^3}-\log ^2(2)+1\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {8 x^2-16 x+8 e^{2 x^3}+e^{x^3} \left (3 x^3-16 x-16 \log ^2(2)+15\right )+8 \log ^4(2)-(15-16 x) \log ^2(2)+7}{\left (-x+e^{x^3}-\log ^2(2)+1\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {1}{4} \int \frac {8 x^2-16 x+8 e^{2 x^3}+e^{x^3} \left (3 x^3-16 x-16 \log ^2(2)+15\right )+7 \left (1+\frac {8 \log ^4(2)}{7}\right )-(15-16 x) \log ^2(2)}{\left (-x+e^{x^3}-\log ^2(2)+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (-\frac {3 x^3-1}{x-e^{x^3}+\log ^2(2)-1}+\frac {x \left (3 x^3-3 \left (1-\log ^2(2)\right ) x^2-1\right )}{\left (-x+e^{x^3}-\log ^2(2)+1\right )^2}+8\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\int \frac {1}{-x+e^{x^3}-\log ^2(2)+1}dx+\int \frac {x}{\left (x-e^{x^3}+\log ^2(2)-1\right )^2}dx+3 \left (1-\log ^2(2)\right ) \int \frac {x^3}{\left (x-e^{x^3}+\log ^2(2)-1\right )^2}dx+3 \int \frac {x^3}{x-e^{x^3}+\log ^2(2)-1}dx-3 \int \frac {x^4}{\left (x-e^{x^3}+\log ^2(2)-1\right )^2}dx-8 x\right )\) |
Input:
Int[(-7 - 8*E^(2*x^3) + 16*x - 8*x^2 + (15 - 16*x)*Log[2]^2 - 8*Log[2]^4 + E^x^3*(-15 + 16*x - 3*x^3 + 16*Log[2]^2))/(4 + 4*E^(2*x^3) - 8*x + 4*x^2 + (-8 + 8*x)*Log[2]^2 + 4*Log[2]^4 + E^x^3*(8 - 8*x - 8*Log[2]^2)),x]
Output:
$Aborted
Time = 0.43 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-2 x -\frac {x}{4 \left (\ln \left (2\right )^{2}-{\mathrm e}^{x^{3}}+x -1\right )}\) | \(23\) |
parallelrisch | \(-\frac {8 x \ln \left (2\right )^{2}+8 x^{2}-8 x \,{\mathrm e}^{x^{3}}-7 x}{4 \left (\ln \left (2\right )^{2}-{\mathrm e}^{x^{3}}+x -1\right )}\) | \(41\) |
norman | \(\frac {\left (\frac {7}{4}-2 \ln \left (2\right )^{2}\right ) {\mathrm e}^{x^{3}}-2 x^{2}+2 x \,{\mathrm e}^{x^{3}}+2 \ln \left (2\right )^{4}-\frac {15 \ln \left (2\right )^{2}}{4}+\frac {7}{4}}{\ln \left (2\right )^{2}-{\mathrm e}^{x^{3}}+x -1}\) | \(56\) |
Input:
int((-8*exp(x^3)^2+(16*ln(2)^2-3*x^3+16*x-15)*exp(x^3)-8*ln(2)^4+(-16*x+15 )*ln(2)^2-8*x^2+16*x-7)/(4*exp(x^3)^2+(-8*ln(2)^2-8*x+8)*exp(x^3)+4*ln(2)^ 4+(8*x-8)*ln(2)^2+4*x^2-8*x+4),x,method=_RETURNVERBOSE)
Output:
-2*x-1/4*x/(ln(2)^2-exp(x^3)+x-1)
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {8 \, x \log \left (2\right )^{2} + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )} - 7 \, x}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \] Input:
integrate((-8*exp(x^3)^2+(16*log(2)^2-3*x^3+16*x-15)*exp(x^3)-8*log(2)^4+( -16*x+15)*log(2)^2-8*x^2+16*x-7)/(4*exp(x^3)^2+(-8*log(2)^2-8*x+8)*exp(x^3 )+4*log(2)^4+(8*x-8)*log(2)^2+4*x^2-8*x+4),x, algorithm="fricas")
Output:
-1/4*(8*x*log(2)^2 + 8*x^2 - 8*x*e^(x^3) - 7*x)/(log(2)^2 + x - e^(x^3) - 1)
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=- 2 x + \frac {x}{- 4 x + 4 e^{x^{3}} - 4 \log {\left (2 \right )}^{2} + 4} \] Input:
integrate((-8*exp(x**3)**2+(16*ln(2)**2-3*x**3+16*x-15)*exp(x**3)-8*ln(2)* *4+(-16*x+15)*ln(2)**2-8*x**2+16*x-7)/(4*exp(x**3)**2+(-8*ln(2)**2-8*x+8)* exp(x**3)+4*ln(2)**4+(8*x-8)*ln(2)**2+4*x**2-8*x+4),x)
Output:
-2*x + x/(-4*x + 4*exp(x**3) - 4*log(2)**2 + 4)
Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {{\left (8 \, \log \left (2\right )^{2} - 7\right )} x + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )}}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \] Input:
integrate((-8*exp(x^3)^2+(16*log(2)^2-3*x^3+16*x-15)*exp(x^3)-8*log(2)^4+( -16*x+15)*log(2)^2-8*x^2+16*x-7)/(4*exp(x^3)^2+(-8*log(2)^2-8*x+8)*exp(x^3 )+4*log(2)^4+(8*x-8)*log(2)^2+4*x^2-8*x+4),x, algorithm="maxima")
Output:
-1/4*((8*log(2)^2 - 7)*x + 8*x^2 - 8*x*e^(x^3))/(log(2)^2 + x - e^(x^3) - 1)
Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {8 \, x \log \left (2\right )^{2} + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )} - 7 \, x}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \] Input:
integrate((-8*exp(x^3)^2+(16*log(2)^2-3*x^3+16*x-15)*exp(x^3)-8*log(2)^4+( -16*x+15)*log(2)^2-8*x^2+16*x-7)/(4*exp(x^3)^2+(-8*log(2)^2-8*x+8)*exp(x^3 )+4*log(2)^4+(8*x-8)*log(2)^2+4*x^2-8*x+4),x, algorithm="giac")
Output:
-1/4*(8*x*log(2)^2 + 8*x^2 - 8*x*e^(x^3) - 7*x)/(log(2)^2 + x - e^(x^3) - 1)
Time = 2.56 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {x\,\left (8\,x-8\,{\mathrm {e}}^{x^3}+8\,{\ln \left (2\right )}^2-7\right )}{4\,\left (x-{\mathrm {e}}^{x^3}+{\ln \left (2\right )}^2-1\right )} \] Input:
int(-(8*exp(2*x^3) - 16*x - exp(x^3)*(16*x + 16*log(2)^2 - 3*x^3 - 15) + l og(2)^2*(16*x - 15) + 8*log(2)^4 + 8*x^2 + 7)/(4*exp(2*x^3) - 8*x + log(2) ^2*(8*x - 8) + 4*log(2)^4 + 4*x^2 - exp(x^3)*(8*x + 8*log(2)^2 - 8) + 4),x )
Output:
-(x*(8*x - 8*exp(x^3) + 8*log(2)^2 - 7))/(4*(x - exp(x^3) + log(2)^2 - 1))
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=\frac {8 e^{x^{3}} \mathrm {log}\left (2\right )^{2}-8 e^{x^{3}} x -7 e^{x^{3}}-8 \mathrm {log}\left (2\right )^{4}+15 \mathrm {log}\left (2\right )^{2}+8 x^{2}-7}{4 e^{x^{3}}-4 \mathrm {log}\left (2\right )^{2}-4 x +4} \] Input:
int((-8*exp(x^3)^2+(16*log(2)^2-3*x^3+16*x-15)*exp(x^3)-8*log(2)^4+(-16*x+ 15)*log(2)^2-8*x^2+16*x-7)/(4*exp(x^3)^2+(-8*log(2)^2-8*x+8)*exp(x^3)+4*lo g(2)^4+(8*x-8)*log(2)^2+4*x^2-8*x+4),x)
Output:
(8*e**(x**3)*log(2)**2 - 8*e**(x**3)*x - 7*e**(x**3) - 8*log(2)**4 + 15*lo g(2)**2 + 8*x**2 - 7)/(4*(e**(x**3) - log(2)**2 - x + 1))