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

Optimal. Leaf size=28 \[ 3-2 x+\frac {x}{4 \left (1+e^{x^3}-x-\log ^2(2)\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 1.40, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 28, normalized size = 1.00 \begin {gather*} \frac {1}{4} \left (-8 x+\frac {x}{1+e^{x^3}-x-\log ^2(2)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 40, normalized size = 1.43 \begin {gather*} -\frac {8 \, x \log \relax (2)^{2} + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )} - 7 \, x}{4 \, {\left (\log \relax (2)^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.70, size = 40, normalized size = 1.43 \begin {gather*} -\frac {8 \, x \log \relax (2)^{2} + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )} - 7 \, x}{4 \, {\left (\log \relax (2)^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.18, size = 23, normalized size = 0.82

method	result	size
risch	\(-2 x -\frac {x}{4 \left (\ln \relax (2)^{2}+x -{\mathrm e}^{x^{3}}-1\right )}\)	\(23\)
norman	\(\frac {\left (\frac {7}{4}-2 \ln \relax (2)^{2}\right ) {\mathrm e}^{x^{3}}-2 x^{2}+2 \,{\mathrm e}^{x^{3}} x +2 \ln \relax (2)^{4}-\frac {15 \ln \relax (2)^{2}}{4}+\frac {7}{4}}{\ln \relax (2)^{2}+x -{\mathrm e}^{x^{3}}-1}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 40, normalized size = 1.43 \begin {gather*} -\frac {{\left (8 \, \log \relax (2)^{2} - 7\right )} x + 8 \, x^{2} - 8 \, x e^{\left (x^{3}\right )}}{4 \, {\left (\log \relax (2)^{2} + x - e^{\left (x^{3}\right )} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 3.61, size = 39, normalized size = 1.39 \begin {gather*} -\frac {x\,\left (8\,x-8\,{\mathrm {e}}^{x^3}+8\,{\ln \relax (2)}^2-7\right )}{4\,\left (x-{\mathrm {e}}^{x^3}+{\ln \relax (2)}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 22, normalized size = 0.79 \begin {gather*} - 2 x + \frac {x}{- 4 x + 4 e^{x^{3}} - 4 \log {\relax (2 )}^{2} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________