3.52.67 \(\int \frac {1}{3} e^{-x-e^{-x} (75 x-15 x^2+2 e^x x^2)} (-75 x+105 x^2-15 x^3+e^x (1-4 x^2)) \, dx\)

Optimal. Leaf size=26 \[ \frac {1}{3} e^{-15 e^{-x} (5-x) x-2 x^2} x \]

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Rubi [F]  time = 1.26, 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 {1}{3} e^{-x-e^{-x} \left (75 x-15 x^2+2 e^x x^2\right )} \left (-75 x+105 x^2-15 x^3+e^x \left (1-4 x^2\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x - (75*x - 15*x^2 + 2*E^x*x^2)/E^x)*(-75*x + 105*x^2 - 15*x^3 + E^x*(1 - 4*x^2)))/3,x]

[Out]

Defer[Int][E^(-((x*(75 - 15*x + 2*E^x*x))/E^x)), x]/3 - 25*Defer[Int][x/E^((x*(75 + E^x - 15*x + 2*E^x*x))/E^x
), x] - (4*Defer[Int][x^2/E^((x*(75 - 15*x + 2*E^x*x))/E^x), x])/3 + 35*Defer[Int][x^2/E^((x*(75 + E^x - 15*x
+ 2*E^x*x))/E^x), x] - 5*Defer[Int][x^3/E^((x*(75 + E^x - 15*x + 2*E^x*x))/E^x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int e^{-x-e^{-x} \left (75 x-15 x^2+2 e^x x^2\right )} \left (-75 x+105 x^2-15 x^3+e^x \left (1-4 x^2\right )\right ) \, dx\\ &=\frac {1}{3} \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} \left (-75 x+105 x^2-15 x^3+e^x \left (1-4 x^2\right )\right ) \, dx\\ &=\frac {1}{3} \int \left (-75 e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x+105 e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^2-15 e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^3-e^{x-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} \left (-1+4 x^2\right )\right ) \, dx\\ &=-\left (\frac {1}{3} \int e^{x-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} \left (-1+4 x^2\right ) \, dx\right )-5 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^3 \, dx-25 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x \, dx+35 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^2 \, dx\\ &=-\left (\frac {1}{3} \int e^{-e^{-x} x \left (75-15 x+2 e^x x\right )} \left (-1+4 x^2\right ) \, dx\right )-5 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^3 \, dx-25 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x \, dx+35 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^2 \, dx\\ &=-\left (\frac {1}{3} \int \left (-e^{-e^{-x} x \left (75-15 x+2 e^x x\right )}+4 e^{-e^{-x} x \left (75-15 x+2 e^x x\right )} x^2\right ) \, dx\right )-5 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^3 \, dx-25 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x \, dx+35 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^2 \, dx\\ &=\frac {1}{3} \int e^{-e^{-x} x \left (75-15 x+2 e^x x\right )} \, dx-\frac {4}{3} \int e^{-e^{-x} x \left (75-15 x+2 e^x x\right )} x^2 \, dx-5 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^3 \, dx-25 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x \, dx+35 \int e^{-e^{-x} x \left (75+e^x-15 x+2 e^x x\right )} x^2 \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 24, normalized size = 0.92 \begin {gather*} \frac {1}{3} e^{15 e^{-x} (-5+x) x-2 x^2} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x - (75*x - 15*x^2 + 2*E^x*x^2)/E^x)*(-75*x + 105*x^2 - 15*x^3 + E^x*(1 - 4*x^2)))/3,x]

[Out]

(E^((15*(-5 + x)*x)/E^x - 2*x^2)*x)/3

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fricas [A]  time = 0.46, size = 31, normalized size = 1.19 \begin {gather*} \frac {1}{3} \, x e^{\left ({\left (15 \, x^{2} - {\left (2 \, x^{2} + x\right )} e^{x} - 75 \, x\right )} e^{\left (-x\right )} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*x^2+1)*exp(x)-15*x^3+105*x^2-75*x)/exp(x)/exp((2*exp(x)*x^2-15*x^2+75*x)/exp(x)),x, algorit
hm="fricas")

[Out]

1/3*x*e^((15*x^2 - (2*x^2 + x)*e^x - 75*x)*e^(-x) + x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {1}{3} \, {\left (15 \, x^{3} - 105 \, x^{2} + {\left (4 \, x^{2} - 1\right )} e^{x} + 75 \, x\right )} e^{\left (-{\left (2 \, x^{2} e^{x} - 15 \, x^{2} + 75 \, x\right )} e^{\left (-x\right )} - x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*x^2+1)*exp(x)-15*x^3+105*x^2-75*x)/exp(x)/exp((2*exp(x)*x^2-15*x^2+75*x)/exp(x)),x, algorit
hm="giac")

[Out]

integrate(-1/3*(15*x^3 - 105*x^2 + (4*x^2 - 1)*e^x + 75*x)*e^(-(2*x^2*e^x - 15*x^2 + 75*x)*e^(-x) - x), x)

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maple [A]  time = 0.06, size = 22, normalized size = 0.85

method	result	size
risch	\(\frac {x \,{\mathrm e}^{-x \left (2 \,{\mathrm e}^{x} x -15 x +75\right ) {\mathrm e}^{-x}}}{3}\)	\(22\)
norman	\(\frac {x \,{\mathrm e}^{-\left (2 \,{\mathrm e}^{x} x^{2}-15 x^{2}+75 x \right ) {\mathrm e}^{-x}}}{3}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-4*x^2+1)*exp(x)-15*x^3+105*x^2-75*x)/exp(x)/exp((2*exp(x)*x^2-15*x^2+75*x)/exp(x)),x,method=_RETURN
VERBOSE)

[Out]

1/3*x*exp(-x*(2*exp(x)*x-15*x+75)*exp(-x))

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maxima [A]  time = 0.48, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, x e^{\left (15 \, x^{2} e^{\left (-x\right )} - 2 \, x^{2} - 75 \, x e^{\left (-x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*x^2+1)*exp(x)-15*x^3+105*x^2-75*x)/exp(x)/exp((2*exp(x)*x^2-15*x^2+75*x)/exp(x)),x, algorit
hm="maxima")

[Out]

1/3*x*e^(15*x^2*e^(-x) - 2*x^2 - 75*x*e^(-x))

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mupad [B]  time = 3.30, size = 27, normalized size = 1.04 \begin {gather*} \frac {x\,{\mathrm {e}}^{-75\,x\,{\mathrm {e}}^{-x}}\,{\mathrm {e}}^{-2\,x^2}\,{\mathrm {e}}^{15\,x^2\,{\mathrm {e}}^{-x}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-x)*exp(-exp(-x)*(75*x + 2*x^2*exp(x) - 15*x^2))*(25*x + (exp(x)*(4*x^2 - 1))/3 - 35*x^2 + 5*x^3),x)

[Out]

(x*exp(-75*x*exp(-x))*exp(-2*x^2)*exp(15*x^2*exp(-x)))/3

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sympy [A]  time = 0.24, size = 24, normalized size = 0.92 \begin {gather*} \frac {x e^{- \left (2 x^{2} e^{x} - 15 x^{2} + 75 x\right ) e^{- x}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-4*x**2+1)*exp(x)-15*x**3+105*x**2-75*x)/exp(x)/exp((2*exp(x)*x**2-15*x**2+75*x)/exp(x)),x)

[Out]

x*exp(-(2*x**2*exp(x) - 15*x**2 + 75*x)*exp(-x))/3

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