3.51.84 \(\int \frac {e^x (-16-16 x)+e^{x+\frac {2}{3} (2 x+2 x^2)} (-1-x)+e^{x+\frac {1}{3} (2 x+2 x^2)} (10+6 x-8 x^2)}{4-4 e^{\frac {1}{3} (2 x+2 x^2)}+e^{\frac {2}{3} (2 x+2 x^2)}} \, dx\)

Optimal. Leaf size=27 \[ e^x \left (-x+\frac {6 x}{-2+e^{\frac {1}{3} x (2+2 x)}}\right ) \]

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Rubi [F]  time = 1.81, 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 {e^x (-16-16 x)+e^{x+\frac {2}{3} \left (2 x+2 x^2\right )} (-1-x)+e^{x+\frac {1}{3} \left (2 x+2 x^2\right )} \left (10+6 x-8 x^2\right )}{4-4 e^{\frac {1}{3} \left (2 x+2 x^2\right )}+e^{\frac {2}{3} \left (2 x+2 x^2\right )}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-16 - 16*x) + E^(x + (2*(2*x + 2*x^2))/3)*(-1 - x) + E^(x + (2*x + 2*x^2)/3)*(10 + 6*x - 8*x^2))/(4
- 4*E^((2*x + 2*x^2)/3) + E^((2*(2*x + 2*x^2))/3)),x]

[Out]

-(E^x*x) + 6*Defer[Int][E^x/(-2 + E^((2*x*(1 + x))/3)), x] - 8*Defer[Int][(E^x*x)/(-2 + E^((2*x*(1 + x))/3))^2
, x] + 2*Defer[Int][(E^x*x)/(-2 + E^((2*x*(1 + x))/3)), x] - 16*Defer[Int][(E^x*x^2)/(-2 + E^((2*x*(1 + x))/3)
)^2, x] - 8*Defer[Int][(E^x*x^2)/(-2 + E^((2*x*(1 + x))/3)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (-16 (1+x)-e^{\frac {4}{3} x (1+x)} (1+x)-2 e^{\frac {2}{3} x (1+x)} \left (-5-3 x+4 x^2\right )\right )}{\left (2-e^{\frac {2}{3} x (1+x)}\right )^2} \, dx\\ &=\int \left (-e^x-e^x x-\frac {8 e^x x (1+2 x)}{\left (-2+e^{\frac {2}{3} x (1+x)}\right )^2}-\frac {2 e^x \left (-3-x+4 x^2\right )}{-2+e^{\frac {2}{3} x (1+x)}}\right ) \, dx\\ &=-\left (2 \int \frac {e^x \left (-3-x+4 x^2\right )}{-2+e^{\frac {2}{3} x (1+x)}} \, dx\right )-8 \int \frac {e^x x (1+2 x)}{\left (-2+e^{\frac {2}{3} x (1+x)}\right )^2} \, dx-\int e^x \, dx-\int e^x x \, dx\\ &=-e^x-e^x x-2 \int \left (-\frac {3 e^x}{-2+e^{\frac {2}{3} x (1+x)}}-\frac {e^x x}{-2+e^{\frac {2}{3} x (1+x)}}+\frac {4 e^x x^2}{-2+e^{\frac {2}{3} x (1+x)}}\right ) \, dx-8 \int \left (\frac {e^x x}{\left (-2+e^{\frac {2}{3} x (1+x)}\right )^2}+\frac {2 e^x x^2}{\left (-2+e^{\frac {2}{3} x (1+x)}\right )^2}\right ) \, dx+\int e^x \, dx\\ &=-e^x x+2 \int \frac {e^x x}{-2+e^{\frac {2}{3} x (1+x)}} \, dx+6 \int \frac {e^x}{-2+e^{\frac {2}{3} x (1+x)}} \, dx-8 \int \frac {e^x x}{\left (-2+e^{\frac {2}{3} x (1+x)}\right )^2} \, dx-8 \int \frac {e^x x^2}{-2+e^{\frac {2}{3} x (1+x)}} \, dx-16 \int \frac {e^x x^2}{\left (-2+e^{\frac {2}{3} x (1+x)}\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.49, size = 24, normalized size = 0.89 \begin {gather*} -e^x \left (1-\frac {6}{-2+e^{\frac {2}{3} x (1+x)}}\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-16 - 16*x) + E^(x + (2*(2*x + 2*x^2))/3)*(-1 - x) + E^(x + (2*x + 2*x^2)/3)*(10 + 6*x - 8*x^2
))/(4 - 4*E^((2*x + 2*x^2)/3) + E^((2*(2*x + 2*x^2))/3)),x]

[Out]

-(E^x*(1 - 6/(-2 + E^((2*x*(1 + x))/3)))*x)

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fricas [B]  time = 0.66, size = 53, normalized size = 1.96 \begin {gather*} -\frac {x e^{\left (\frac {8}{3} \, x^{2} + \frac {17}{3} \, x\right )} - 8 \, x e^{\left (2 \, x^{2} + 5 \, x\right )}}{e^{\left (\frac {8}{3} \, x^{2} + \frac {14}{3} \, x\right )} - 2 \, e^{\left (2 \, x^{2} + 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-1)*exp(x)*exp(2/3*x^2+2/3*x)^2+(-8*x^2+6*x+10)*exp(x)*exp(2/3*x^2+2/3*x)+(-16*x-16)*exp(x))/(ex
p(2/3*x^2+2/3*x)^2-4*exp(2/3*x^2+2/3*x)+4),x, algorithm="fricas")

[Out]

-(x*e^(8/3*x^2 + 17/3*x) - 8*x*e^(2*x^2 + 5*x))/(e^(8/3*x^2 + 14/3*x) - 2*e^(2*x^2 + 4*x))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-1)*exp(x)*exp(2/3*x^2+2/3*x)^2+(-8*x^2+6*x+10)*exp(x)*exp(2/3*x^2+2/3*x)+(-16*x-16)*exp(x))/(ex
p(2/3*x^2+2/3*x)^2-4*exp(2/3*x^2+2/3*x)+4),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.08, size = 23, normalized size = 0.85

method	result	size
risch	\(-{\mathrm e}^{x} x +\frac {6 x \,{\mathrm e}^{x}}{{\mathrm e}^{\frac {2 \left (x +1\right ) x}{3}}-2}\)	\(23\)
norman	\(\frac {8 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x \,{\mathrm e}^{\frac {2}{3} x^{2}+\frac {2}{3} x}}{{\mathrm e}^{\frac {2}{3} x^{2}+\frac {2}{3} x}-2}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x-1)*exp(x)*exp(2/3*x^2+2/3*x)^2+(-8*x^2+6*x+10)*exp(x)*exp(2/3*x^2+2/3*x)+(-16*x-16)*exp(x))/(exp(2/3*
x^2+2/3*x)^2-4*exp(2/3*x^2+2/3*x)+4),x,method=_RETURNVERBOSE)

[Out]

-exp(x)*x+6*x*exp(x)/(exp(2/3*(x+1)*x)-2)

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maxima [A]  time = 0.38, size = 34, normalized size = 1.26 \begin {gather*} -\frac {x e^{\left (\frac {2}{3} \, x^{2} + \frac {5}{3} \, x\right )} - 8 \, x e^{x}}{e^{\left (\frac {2}{3} \, x^{2} + \frac {2}{3} \, x\right )} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-1)*exp(x)*exp(2/3*x^2+2/3*x)^2+(-8*x^2+6*x+10)*exp(x)*exp(2/3*x^2+2/3*x)+(-16*x-16)*exp(x))/(ex
p(2/3*x^2+2/3*x)^2-4*exp(2/3*x^2+2/3*x)+4),x, algorithm="maxima")

[Out]

-(x*e^(2/3*x^2 + 5/3*x) - 8*x*e^x)/(e^(2/3*x^2 + 2/3*x) - 2)

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mupad [B]  time = 3.66, size = 25, normalized size = 0.93 \begin {gather*} \frac {6\,x\,{\mathrm {e}}^x}{{\mathrm {e}}^{\frac {2\,x^2}{3}+\frac {2\,x}{3}}-2}-x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(16*x + 16) - exp((2*x)/3 + (2*x^2)/3)*exp(x)*(6*x - 8*x^2 + 10) + exp((4*x)/3 + (4*x^2)/3)*exp(x
)*(x + 1))/(exp((4*x)/3 + (4*x^2)/3) - 4*exp((2*x)/3 + (2*x^2)/3) + 4),x)

[Out]

(6*x*exp(x))/(exp((2*x)/3 + (2*x^2)/3) - 2) - x*exp(x)

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sympy [A]  time = 0.17, size = 26, normalized size = 0.96 \begin {gather*} - x e^{x} + \frac {6 x e^{x}}{e^{\frac {2 x^{2}}{3} + \frac {2 x}{3}} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-1)*exp(x)*exp(2/3*x**2+2/3*x)**2+(-8*x**2+6*x+10)*exp(x)*exp(2/3*x**2+2/3*x)+(-16*x-16)*exp(x))
/(exp(2/3*x**2+2/3*x)**2-4*exp(2/3*x**2+2/3*x)+4),x)

[Out]

-x*exp(x) + 6*x*exp(x)/(exp(2*x**2/3 + 2*x/3) - 2)

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