3.11.70 \(\int \frac {e^{\frac {(10-5 x) x^2}{-4+x^2 (-8+4 x)}} x^2 (-20+15 x)}{4 x+x^2 (16 x-8 x^2)+x^4 (16 x-16 x^2+4 x^3)} \, dx\)

Optimal. Leaf size=18 \[ e^{\frac {5}{-4+\frac {4}{(-2+x) x^2}}} \]

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Rubi [F]  time = 0.95, 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^{\frac {(10-5 x) x^2}{-4+x^2 (-8+4 x)}} x^2 (-20+15 x)}{4 x+x^2 \left (16 x-8 x^2\right )+x^4 \left (16 x-16 x^2+4 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(((10 - 5*x)*x^2)/(-4 + x^2*(-8 + 4*x)))*x^2*(-20 + 15*x))/(4*x + x^2*(16*x - 8*x^2) + x^4*(16*x - 16*x
^2 + 4*x^3)),x]

[Out]

-5*Defer[Int][(E^(((10 - 5*x)*x^2)/(-4 - 8*x^2 + 4*x^3))*x)/(-1 - 2*x^2 + x^3)^2, x] + (15*Defer[Int][(E^(((10
- 5*x)*x^2)/(-4 - 8*x^2 + 4*x^3))*x^2)/(-1 - 2*x^2 + x^3)^2, x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {(10-5 x) x^2}{-4+x^2 (-8+4 x)}} x (-20+15 x)}{4+16 x^2-8 x^3+16 x^4-16 x^5+4 x^6} \, dx\\ &=\int \frac {e^{\frac {(10-5 x) x^2}{-4-8 x^2+4 x^3}} x (-20+15 x)}{4 \left (1+2 x^2-x^3\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{\frac {(10-5 x) x^2}{-4-8 x^2+4 x^3}} x (-20+15 x)}{\left (1+2 x^2-x^3\right )^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {20 e^{\frac {(10-5 x) x^2}{-4-8 x^2+4 x^3}} x}{\left (-1-2 x^2+x^3\right )^2}+\frac {15 e^{\frac {(10-5 x) x^2}{-4-8 x^2+4 x^3}} x^2}{\left (-1-2 x^2+x^3\right )^2}\right ) \, dx\\ &=\frac {15}{4} \int \frac {e^{\frac {(10-5 x) x^2}{-4-8 x^2+4 x^3}} x^2}{\left (-1-2 x^2+x^3\right )^2} \, dx-5 \int \frac {e^{\frac {(10-5 x) x^2}{-4-8 x^2+4 x^3}} x}{\left (-1-2 x^2+x^3\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 22, normalized size = 1.22 \begin {gather*} e^{-\frac {5}{4}-\frac {5}{4 \left (-1-2 x^2+x^3\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(((10 - 5*x)*x^2)/(-4 + x^2*(-8 + 4*x)))*x^2*(-20 + 15*x))/(4*x + x^2*(16*x - 8*x^2) + x^4*(16*x
- 16*x^2 + 4*x^3)),x]

[Out]

E^(-5/4 - 5/(4*(-1 - 2*x^2 + x^3)))

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fricas [A]  time = 1.17, size = 24, normalized size = 1.33 \begin {gather*} e^{\left (-\frac {5 \, {\left (x^{3} - 2 \, x^{2}\right )}}{4 \, {\left (x^{3} - 2 \, x^{2} - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x-20)*x^2*exp((-5*x+10)*x^2/((4*x-8)*x^2-4))/((4*x^3-16*x^2+16*x)*x^4+(-8*x^2+16*x)*x^2+4*x),x,
algorithm="fricas")

[Out]

e^(-5/4*(x^3 - 2*x^2)/(x^3 - 2*x^2 - 1))

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giac [B]  time = 0.24, size = 36, normalized size = 2.00 \begin {gather*} e^{\left (-\frac {5 \, x^{3}}{4 \, {\left (x^{3} - 2 \, x^{2} - 1\right )}} + \frac {5 \, x^{2}}{2 \, {\left (x^{3} - 2 \, x^{2} - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x-20)*x^2*exp((-5*x+10)*x^2/((4*x-8)*x^2-4))/((4*x^3-16*x^2+16*x)*x^4+(-8*x^2+16*x)*x^2+4*x),x,
algorithm="giac")

[Out]

e^(-5/4*x^3/(x^3 - 2*x^2 - 1) + 5/2*x^2/(x^3 - 2*x^2 - 1))

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maple [A]  time = 0.08, size = 22, normalized size = 1.22

method	result	size
gosper	\({\mathrm e}^{-\frac {5 \left (x -2\right ) x^{2}}{4 \left (x^{3}-2 x^{2}-1\right )}}\)	\(22\)
risch	\({\mathrm e}^{-\frac {5 \left (x -2\right ) x^{2}}{4 \left (x^{3}-2 x^{2}-1\right )}}\)	\(22\)
norman	\(\frac {x^{3} {\mathrm e}^{\frac {\left (-5 x +10\right ) x^{2}}{\left (4 x -8\right ) x^{2}-4}}-2 x^{2} {\mathrm e}^{\frac {\left (-5 x +10\right ) x^{2}}{\left (4 x -8\right ) x^{2}-4}}-{\mathrm e}^{\frac {\left (-5 x +10\right ) x^{2}}{\left (4 x -8\right ) x^{2}-4}}}{x^{3}-2 x^{2}-1}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x-20)*x^2*exp((-5*x+10)*x^2/((4*x-8)*x^2-4))/((4*x^3-16*x^2+16*x)*x^4+(-8*x^2+16*x)*x^2+4*x),x,method=
_RETURNVERBOSE)

[Out]

exp(-5/4*(x-2)*x^2/(x^3-2*x^2-1))

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maxima [A]  time = 0.67, size = 17, normalized size = 0.94 \begin {gather*} e^{\left (-\frac {5}{4 \, {\left (x^{3} - 2 \, x^{2} - 1\right )}} - \frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x-20)*x^2*exp((-5*x+10)*x^2/((4*x-8)*x^2-4))/((4*x^3-16*x^2+16*x)*x^4+(-8*x^2+16*x)*x^2+4*x),x,
algorithm="maxima")

[Out]

e^(-5/4/(x^3 - 2*x^2 - 1) - 5/4)

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mupad [B]  time = 0.28, size = 41, normalized size = 2.28 \begin {gather*} {\mathrm {e}}^{-\frac {5\,x^2}{-2\,x^3+4\,x^2+2}}\,{\mathrm {e}}^{\frac {5\,x^3}{-4\,x^3+8\,x^2+4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(-(x^2*(5*x - 10))/(x^2*(4*x - 8) - 4))*(15*x - 20))/(4*x + x^2*(16*x - 8*x^2) + x^4*(16*x - 16*x^
2 + 4*x^3)),x)

[Out]

exp(-(5*x^2)/(4*x^2 - 2*x^3 + 2))*exp((5*x^3)/(8*x^2 - 4*x^3 + 4))

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sympy [A]  time = 0.22, size = 19, normalized size = 1.06 \begin {gather*} e^{\frac {x^{2} \left (10 - 5 x\right )}{x^{2} \left (4 x - 8\right ) - 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x-20)*x**2*exp((-5*x+10)*x**2/((4*x-8)*x**2-4))/((4*x**3-16*x**2+16*x)*x**4+(-8*x**2+16*x)*x**2+
4*x),x)

[Out]

exp(x**2*(10 - 5*x)/(x**2*(4*x - 8) - 4))

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