Integrand size = 112, antiderivative size = 28 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=3-2 x+\frac {x}{4 \left (1+e^{x^3}-x-\log ^2(2)\right )} \]
Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=\frac {1}{4} \left (-8 x+\frac {x}{1+e^{x^3}-x-\log ^2(2)}\right ) \]
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 ]
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 |
\(\displaystyle \frac {1}{4} \left (\int \frac {1}{-x+e^{x^3}-\log ^2(2)+1}dx+\int \frac {x}{\left (x-e^{x^3}+\log ^2(2)-1\right )^2}dx+3 \left (1-\log ^2(2)\right ) \int \frac {x^3}{\left (x-e^{x^3}+\log ^2(2)-1\right )^2}dx+3 \int \frac {x^3}{x-e^{x^3}+\log ^2(2)-1}dx-3 \int \frac {x^4}{\left (x-e^{x^3}+\log ^2(2)-1\right )^2}dx-8 x\right )\) |
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]
3.27.59.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-2 x -\frac {x}{4 \left (\ln \left (2\right )^{2}+x -{\mathrm e}^{x^{3}}-1\right )}\) | \(23\) |
parallelrisch | \(-\frac {8 x \ln \left (2\right )^{2}+8 x^{2}-8 \,{\mathrm e}^{x^{3}} x -7 x}{4 \left (\ln \left (2\right )^{2}+x -{\mathrm e}^{x^{3}}-1\right )}\) | \(41\) |
norman | \(\frac {\left (\frac {7}{4}-2 \ln \left (2\right )^{2}\right ) {\mathrm e}^{x^{3}}-2 x^{2}+2 \,{\mathrm e}^{x^{3}} x +2 \ln \left (2\right )^{4}-\frac {15 \ln \left (2\right )^{2}}{4}+\frac {7}{4}}{\ln \left (2\right )^{2}+x -{\mathrm e}^{x^{3}}-1}\) | \(56\) |
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)
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {8 \, x \log \left (2\right )^{2} + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )} - 7 \, x}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \]
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=\
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=- 2 x + \frac {x}{- 4 x + 4 e^{x^{3}} - 4 \log {\left (2 \right )}^{2} + 4} \]
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)
Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {{\left (8 \, \log \left (2\right )^{2} - 7\right )} x + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )}}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \]
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=\
Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {8 \, x \log \left (2\right )^{2} + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )} - 7 \, x}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \]
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=\
Time = 11.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {x\,\left (8\,x-8\,{\mathrm {e}}^{x^3}+8\,{\ln \left (2\right )}^2-7\right )}{4\,\left (x-{\mathrm {e}}^{x^3}+{\ln \left (2\right )}^2-1\right )} \]