\(\int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} (-15+16 x-3 x^3+16 \log ^2(2))}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} (8-8 x-8 \log ^2(2))} \, dx\) [5659]

Optimal result

Integrand size = 112, antiderivative size = 28 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=3-2 x+\frac {x}{4 \left (1+e^{x^3}-x-\log ^2(2)\right )} \]

[Out]

3+1/4*x/(exp(x^3)-x-ln(2)^2+1)-2*x

Rubi [F]

\[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=\int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx \]

[In]

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]

[Out]

-2*x + Defer[Int][(1 + E^x^3 - x - Log[2]^2)^(-1), x]/4 + Defer[Int][x/(-1 - E^x^3 + x + Log[2]^2)^2, x]/4 + (
3*(1 - Log[2]^2)*Defer[Int][x^3/(-1 - E^x^3 + x + Log[2]^2)^2, x])/4 - (3*Defer[Int][x^4/(-1 - E^x^3 + x + Log
[2]^2)^2, x])/4 + (3*Defer[Int][x^3/(-1 - E^x^3 + x + Log[2]^2), x])/4

Rubi steps \begin{align*} \text {integral}& = \int \frac {-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )-7 \left (1+\frac {8 \log ^4(2)}{7}\right )}{4 \left (1+e^{x^3}-x-\log ^2(2)\right )^2} \, dx \\ & = \frac {1}{4} \int \frac {-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )-7 \left (1+\frac {8 \log ^4(2)}{7}\right )}{\left (1+e^{x^3}-x-\log ^2(2)\right )^2} \, dx \\ & = \frac {1}{4} \int \left (-8+\frac {-1+3 x^3}{-1-e^{x^3}+x+\log ^2(2)}+\frac {x \left (1-3 x^3+3 x^2 \left (1-\log ^2(2)\right )\right )}{\left (1+e^{x^3}-x-\log ^2(2)\right )^2}\right ) \, dx \\ & = -2 x+\frac {1}{4} \int \frac {-1+3 x^3}{-1-e^{x^3}+x+\log ^2(2)} \, dx+\frac {1}{4} \int \frac {x \left (1-3 x^3+3 x^2 \left (1-\log ^2(2)\right )\right )}{\left (1+e^{x^3}-x-\log ^2(2)\right )^2} \, dx \\ & = -2 x+\frac {1}{4} \int \left (\frac {x}{\left (-1-e^{x^3}+x+\log ^2(2)\right )^2}-\frac {3 x^4}{\left (-1-e^{x^3}+x+\log ^2(2)\right )^2}-\frac {3 x^3 (-1+\log (2)) (1+\log (2))}{\left (-1-e^{x^3}+x+\log ^2(2)\right )^2}\right ) \, dx+\frac {1}{4} \int \left (\frac {1}{1+e^{x^3}-x-\log ^2(2)}+\frac {3 x^3}{-1-e^{x^3}+x+\log ^2(2)}\right ) \, dx \\ & = -2 x+\frac {1}{4} \int \frac {1}{1+e^{x^3}-x-\log ^2(2)} \, dx+\frac {1}{4} \int \frac {x}{\left (-1-e^{x^3}+x+\log ^2(2)\right )^2} \, dx-\frac {3}{4} \int \frac {x^4}{\left (-1-e^{x^3}+x+\log ^2(2)\right )^2} \, dx+\frac {3}{4} \int \frac {x^3}{-1-e^{x^3}+x+\log ^2(2)} \, dx-\frac {1}{4} (3 (-1+\log (2)) (1+\log (2))) \int \frac {x^3}{\left (-1-e^{x^3}+x+\log ^2(2)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=\frac {1}{4} \left (-8 x+\frac {x}{1+e^{x^3}-x-\log ^2(2)}\right ) \]

[In]

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 + 1
6*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]

[Out]

(-8*x + x/(1 + E^x^3 - x - Log[2]^2))/4

Maple [A] (verified)

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	\(\frac {\left (\frac {7}{4}-2 \ln \left (2\right )^{2}\right ) {\mathrm e}^{x^{3}}-2 x^{2}+2 \,{\mathrm e}^{x^{3}} x +2 \ln \left (2\right )^{4}-\frac {15 \ln \left (2\right )^{2}}{4}+\frac {7}{4}}{\ln \left (2\right )^{2}+x -{\mathrm e}^{x^{3}}-1}\)	\(56\)

[In]

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)

[Out]

-2*x-1/4*x/(ln(2)^2+x-exp(x^3)-1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {8 \, x \log \left (2\right )^{2} + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )} - 7 \, x}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \]

[In]

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")

[Out]

-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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=- 2 x + \frac {x}{- 4 x + 4 e^{x^{3}} - 4 \log {\left (2 \right )}^{2} + 4} \]

[In]

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)

[Out]

-2*x + x/(-4*x + 4*exp(x**3) - 4*log(2)**2 + 4)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {{\left (8 \, \log \left (2\right )^{2} - 7\right )} x + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )}}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \]

[In]

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")

[Out]

-1/4*((8*log(2)^2 - 7)*x + 8*x^2 - 8*x*e^(x^3))/(log(2)^2 + x - e^(x^3) - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {8 \, x \log \left (2\right )^{2} + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )} - 7 \, x}{4 \, {\left (\log \left (2\right )^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \]

[In]

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")

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 11.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-7-8 e^{2 x^3}+16 x-8 x^2+(15-16 x) \log ^2(2)-8 \log ^4(2)+e^{x^3} \left (-15+16 x-3 x^3+16 \log ^2(2)\right )}{4+4 e^{2 x^3}-8 x+4 x^2+(-8+8 x) \log ^2(2)+4 \log ^4(2)+e^{x^3} \left (8-8 x-8 \log ^2(2)\right )} \, dx=-\frac {x\,\left (8\,x-8\,{\mathrm {e}}^{x^3}+8\,{\ln \left (2\right )}^2-7\right )}{4\,\left (x-{\mathrm {e}}^{x^3}+{\ln \left (2\right )}^2-1\right )} \]

[In]

int(-(8*exp(2*x^3) - 16*x - exp(x^3)*(16*x + 16*log(2)^2 - 3*x^3 - 15) + log(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)

[Out]

-(x*(8*x - 8*exp(x^3) + 8*log(2)^2 - 7))/(4*(x - exp(x^3) + log(2)^2 - 1))