3.51.59 \(\int \frac {e^{\frac {2 (50 e^2+45 x-5 x^2)}{-4+x}} (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3)}{16-8 x+x^2} \, dx\)

Optimal. Leaf size=25 \[ 100 e^{10 \left (-x+\frac {5 \left (2 e^2+x\right )}{-4+x}\right )} x \]

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Rubi [F]  time = 0.79, 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 {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-36800 x-10000 e^2 x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(50*E^2 + 45*x - 5*x^2))/(-4 + x))*(1600 - 36800*x - 10000*E^2*x + 8100*x^2 - 1000*x^3))/(16 - 8*x
+ x^2),x]

[Out]

100*Defer[Int][E^((2*(50*E^2 + 45*x - 5*x^2))/(-4 + x)), x] - 40000*(2 + E^2)*Defer[Int][E^((2*(50*E^2 + 45*x
- 5*x^2))/(-4 + x))/(-4 + x)^2, x] - 10000*(2 + E^2)*Defer[Int][E^((2*(50*E^2 + 45*x - 5*x^2))/(-4 + x))/(-4 +
x), x] - 1000*Defer[Int][E^((2*(50*E^2 + 45*x - 5*x^2))/(-4 + x))*x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600+\left (-36800-10000 e^2\right ) x+8100 x^2-1000 x^3\right )}{16-8 x+x^2} \, dx\\ &=\int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600+\left (-36800-10000 e^2\right ) x+8100 x^2-1000 x^3\right )}{(-4+x)^2} \, dx\\ &=\int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (1600-400 \left (92+25 e^2\right ) x+8100 x^2-1000 x^3\right )}{(4-x)^2} \, dx\\ &=\int \left (100 e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}}-\frac {40000 e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (2+e^2\right )}{(-4+x)^2}-\frac {10000 e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \left (2+e^2\right )}{-4+x}-1000 e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} x\right ) \, dx\\ &=100 \int e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} \, dx-1000 \int e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}} x \, dx-\left (10000 \left (2+e^2\right )\right ) \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}}}{-4+x} \, dx-\left (40000 \left (2+e^2\right )\right ) \int \frac {e^{\frac {2 \left (50 e^2+45 x-5 x^2\right )}{-4+x}}}{(-4+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 23, normalized size = 0.92 \begin {gather*} 100 e^{\frac {100 e^2-10 (-9+x) x}{-4+x}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(50*E^2 + 45*x - 5*x^2))/(-4 + x))*(1600 - 36800*x - 10000*E^2*x + 8100*x^2 - 1000*x^3))/(16
- 8*x + x^2),x]

[Out]

100*E^((100*E^2 - 10*(-9 + x)*x)/(-4 + x))*x

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fricas [A]  time = 2.04, size = 22, normalized size = 0.88 \begin {gather*} 100 \, x e^{\left (-\frac {10 \, {\left (x^{2} - 9 \, x - 10 \, e^{2}\right )}}{x - 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10000*exp(2)*x-1000*x^3+8100*x^2-36800*x+1600)*exp((50*exp(2)-5*x^2+45*x)/(x-4))^2/(x^2-8*x+16),x,
algorithm="fricas")

[Out]

100*x*e^(-10*(x^2 - 9*x - 10*e^2)/(x - 4))

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giac [A]  time = 0.19, size = 30, normalized size = 1.20 \begin {gather*} 100 \, x e^{\left (-\frac {5 \, {\left (2 \, x^{2} - 5 \, x e^{2} - 18 \, x\right )}}{x - 4} - 25 \, e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10000*exp(2)*x-1000*x^3+8100*x^2-36800*x+1600)*exp((50*exp(2)-5*x^2+45*x)/(x-4))^2/(x^2-8*x+16),x,
algorithm="giac")

[Out]

100*x*e^(-5*(2*x^2 - 5*x*e^2 - 18*x)/(x - 4) - 25*e^2)

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maple [A]  time = 0.24, size = 25, normalized size = 1.00

method	result	size
risch	\(100 x \,{\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}\)	\(25\)
gosper	\(100 x \,{\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}\)	\(27\)
norman	\(\frac {-400 x \,{\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}+100 x^{2} {\mathrm e}^{\frac {-10 x^{2}+100 \,{\mathrm e}^{2}+90 x}{x -4}}}{x -4}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10000*exp(2)*x-1000*x^3+8100*x^2-36800*x+1600)*exp((50*exp(2)-5*x^2+45*x)/(x-4))^2/(x^2-8*x+16),x,method
=_RETURNVERBOSE)

[Out]

100*x*exp(10*(-x^2+10*exp(2)+9*x)/(x-4))

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maxima [A]  time = 0.44, size = 25, normalized size = 1.00 \begin {gather*} 100 \, x e^{\left (-10 \, x + \frac {100 \, e^{2}}{x - 4} + \frac {200}{x - 4} + 50\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10000*exp(2)*x-1000*x^3+8100*x^2-36800*x+1600)*exp((50*exp(2)-5*x^2+45*x)/(x-4))^2/(x^2-8*x+16),x,
algorithm="maxima")

[Out]

100*x*e^(-10*x + 100*e^2/(x - 4) + 200/(x - 4) + 50)

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mupad [B]  time = 3.58, size = 23, normalized size = 0.92 \begin {gather*} 100\,x\,{\mathrm {e}}^{\frac {-10\,x^2+90\,x+100\,{\mathrm {e}}^2}{x-4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*(45*x + 50*exp(2) - 5*x^2))/(x - 4))*(36800*x + 10000*x*exp(2) - 8100*x^2 + 1000*x^3 - 1600))/(x^
2 - 8*x + 16),x)

[Out]

100*x*exp((90*x + 100*exp(2) - 10*x^2)/(x - 4))

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sympy [A]  time = 0.28, size = 20, normalized size = 0.80 \begin {gather*} 100 x e^{\frac {2 \left (- 5 x^{2} + 45 x + 50 e^{2}\right )}{x - 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10000*exp(2)*x-1000*x**3+8100*x**2-36800*x+1600)*exp((50*exp(2)-5*x**2+45*x)/(x-4))**2/(x**2-8*x+1
6),x)

[Out]

100*x*exp(2*(-5*x**2 + 45*x + 50*exp(2))/(x - 4))

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