3.71.93 \(\int \frac {1}{5} (-2 x+30 x^2+e^x (-10 x-5 x^2)+e^{e^x x^2} (-1+20 x+e^{2 x} (-10 x^2-5 x^3)+e^x (-5-5 x-2 x^2+19 x^3+10 x^4))) \, dx\)

Optimal. Leaf size=25 \[ x \left (e^{e^x x^2}+x\right ) \left (-\frac {1}{5}-e^x+2 x\right ) \]

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Rubi [F]  time = 1.22, 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}{5} \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x + 30*x^2 + E^x*(-10*x - 5*x^2) + E^(E^x*x^2)*(-1 + 20*x + E^(2*x)*(-10*x^2 - 5*x^3) + E^x*(-5 - 5*x
- 2*x^2 + 19*x^3 + 10*x^4)))/5,x]

[Out]

-1/5*x^2 - E^x*x^2 + 2*x^3 - Defer[Int][E^(E^x*x^2), x]/5 - Defer[Int][E^(x + E^x*x^2), x] + 4*Defer[Int][E^(E
^x*x^2)*x, x] - Defer[Int][E^(x + E^x*x^2)*x, x] - (2*Defer[Int][E^(x + E^x*x^2)*x^2, x])/5 - 2*Defer[Int][E^(
2*x + E^x*x^2)*x^2, x] + (19*Defer[Int][E^(x + E^x*x^2)*x^3, x])/5 - Defer[Int][E^(2*x + E^x*x^2)*x^3, x] + 2*
Defer[Int][E^(x + E^x*x^2)*x^4, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx\\ &=-\frac {x^2}{5}+2 x^3+\frac {1}{5} \int e^x \left (-10 x-5 x^2\right ) \, dx+\frac {1}{5} \int e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right ) \, dx\\ &=-\frac {x^2}{5}+2 x^3+\frac {1}{5} \int e^x (-10-5 x) x \, dx+\frac {1}{5} \int \left (-e^{e^x x^2}+20 e^{e^x x^2} x-5 e^{2 x+e^x x^2} x^2 (2+x)+e^{x+e^x x^2} \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right ) \, dx\\ &=-\frac {x^2}{5}+2 x^3-\frac {1}{5} \int e^{e^x x^2} \, dx+\frac {1}{5} \int \left (-10 e^x x-5 e^x x^2\right ) \, dx+\frac {1}{5} \int e^{x+e^x x^2} \left (-5-5 x-2 x^2+19 x^3+10 x^4\right ) \, dx+4 \int e^{e^x x^2} x \, dx-\int e^{2 x+e^x x^2} x^2 (2+x) \, dx\\ &=-\frac {x^2}{5}+2 x^3-\frac {1}{5} \int e^{e^x x^2} \, dx+\frac {1}{5} \int \left (-5 e^{x+e^x x^2}-5 e^{x+e^x x^2} x-2 e^{x+e^x x^2} x^2+19 e^{x+e^x x^2} x^3+10 e^{x+e^x x^2} x^4\right ) \, dx-2 \int e^x x \, dx+4 \int e^{e^x x^2} x \, dx-\int e^x x^2 \, dx-\int \left (2 e^{2 x+e^x x^2} x^2+e^{2 x+e^x x^2} x^3\right ) \, dx\\ &=-2 e^x x-\frac {x^2}{5}-e^x x^2+2 x^3-\frac {1}{5} \int e^{e^x x^2} \, dx-\frac {2}{5} \int e^{x+e^x x^2} x^2 \, dx+2 \int e^x \, dx+2 \int e^x x \, dx-2 \int e^{2 x+e^x x^2} x^2 \, dx+2 \int e^{x+e^x x^2} x^4 \, dx+\frac {19}{5} \int e^{x+e^x x^2} x^3 \, dx+4 \int e^{e^x x^2} x \, dx-\int e^{x+e^x x^2} \, dx-\int e^{x+e^x x^2} x \, dx-\int e^{2 x+e^x x^2} x^3 \, dx\\ &=2 e^x-\frac {x^2}{5}-e^x x^2+2 x^3-\frac {1}{5} \int e^{e^x x^2} \, dx-\frac {2}{5} \int e^{x+e^x x^2} x^2 \, dx-2 \int e^x \, dx-2 \int e^{2 x+e^x x^2} x^2 \, dx+2 \int e^{x+e^x x^2} x^4 \, dx+\frac {19}{5} \int e^{x+e^x x^2} x^3 \, dx+4 \int e^{e^x x^2} x \, dx-\int e^{x+e^x x^2} \, dx-\int e^{x+e^x x^2} x \, dx-\int e^{2 x+e^x x^2} x^3 \, dx\\ &=-\frac {x^2}{5}-e^x x^2+2 x^3-\frac {1}{5} \int e^{e^x x^2} \, dx-\frac {2}{5} \int e^{x+e^x x^2} x^2 \, dx-2 \int e^{2 x+e^x x^2} x^2 \, dx+2 \int e^{x+e^x x^2} x^4 \, dx+\frac {19}{5} \int e^{x+e^x x^2} x^3 \, dx+4 \int e^{e^x x^2} x \, dx-\int e^{x+e^x x^2} \, dx-\int e^{x+e^x x^2} x \, dx-\int e^{2 x+e^x x^2} x^3 \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.72, size = 26, normalized size = 1.04 \begin {gather*} \frac {1}{5} x \left (e^{e^x x^2}+x\right ) \left (-1-5 e^x+10 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + 30*x^2 + E^x*(-10*x - 5*x^2) + E^(E^x*x^2)*(-1 + 20*x + E^(2*x)*(-10*x^2 - 5*x^3) + E^x*(-5
- 5*x - 2*x^2 + 19*x^3 + 10*x^4)))/5,x]

[Out]

(x*(E^(E^x*x^2) + x)*(-1 - 5*E^x + 10*x))/5

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fricas [A]  time = 0.59, size = 41, normalized size = 1.64 \begin {gather*} 2 \, x^{3} - x^{2} e^{x} - \frac {1}{5} \, x^{2} + \frac {1}{5} \, {\left (10 \, x^{2} - 5 \, x e^{x} - x\right )} e^{\left (x^{2} e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^3-10*x^2)*exp(x)^2+(10*x^4+19*x^3-2*x^2-5*x-5)*exp(x)+20*x-1)*exp(exp(x)*x^2)+1/5*(-5*x^2
-10*x)*exp(x)+6*x^2-2/5*x,x, algorithm="fricas")

[Out]

2*x^3 - x^2*e^x - 1/5*x^2 + 1/5*(10*x^2 - 5*x*e^x - x)*e^(x^2*e^x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int 6 \, x^{2} - \frac {1}{5} \, {\left (5 \, {\left (x^{3} + 2 \, x^{2}\right )} e^{\left (2 \, x\right )} - {\left (10 \, x^{4} + 19 \, x^{3} - 2 \, x^{2} - 5 \, x - 5\right )} e^{x} - 20 \, x + 1\right )} e^{\left (x^{2} e^{x}\right )} - {\left (x^{2} + 2 \, x\right )} e^{x} - \frac {2}{5} \, x\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^3-10*x^2)*exp(x)^2+(10*x^4+19*x^3-2*x^2-5*x-5)*exp(x)+20*x-1)*exp(exp(x)*x^2)+1/5*(-5*x^2
-10*x)*exp(x)+6*x^2-2/5*x,x, algorithm="giac")

[Out]

integrate(6*x^2 - 1/5*(5*(x^3 + 2*x^2)*e^(2*x) - (10*x^4 + 19*x^3 - 2*x^2 - 5*x - 5)*e^x - 20*x + 1)*e^(x^2*e^
x) - (x^2 + 2*x)*e^x - 2/5*x, x)

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maple [A]  time = 0.06, size = 38, normalized size = 1.52

method	result	size
risch	\(\frac {\left (10 x -5 \,{\mathrm e}^{x}-1\right ) x \,{\mathrm e}^{{\mathrm e}^{x} x^{2}}}{5}-{\mathrm e}^{x} x^{2}+2 x^{3}-\frac {x^{2}}{5}\)	\(38\)
default	\(-\frac {x^{2}}{5}+2 x^{3}+2 x^{2} {\mathrm e}^{{\mathrm e}^{x} x^{2}}-\frac {{\mathrm e}^{{\mathrm e}^{x} x^{2}} x}{5}-{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x} x^{2}} x -{\mathrm e}^{x} x^{2}\)	\(53\)
norman	\(-\frac {x^{2}}{5}+2 x^{3}+2 x^{2} {\mathrm e}^{{\mathrm e}^{x} x^{2}}-\frac {{\mathrm e}^{{\mathrm e}^{x} x^{2}} x}{5}-{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x} x^{2}} x -{\mathrm e}^{x} x^{2}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-5*x^3-10*x^2)*exp(x)^2+(10*x^4+19*x^3-2*x^2-5*x-5)*exp(x)+20*x-1)*exp(exp(x)*x^2)+1/5*(-5*x^2-10*x)
*exp(x)+6*x^2-2/5*x,x,method=_RETURNVERBOSE)

[Out]

1/5*(10*x-5*exp(x)-1)*x*exp(exp(x)*x^2)-exp(x)*x^2+2*x^3-1/5*x^2

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maxima [A]  time = 0.38, size = 41, normalized size = 1.64 \begin {gather*} 2 \, x^{3} - x^{2} e^{x} - \frac {1}{5} \, x^{2} + \frac {1}{5} \, {\left (10 \, x^{2} - 5 \, x e^{x} - x\right )} e^{\left (x^{2} e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^3-10*x^2)*exp(x)^2+(10*x^4+19*x^3-2*x^2-5*x-5)*exp(x)+20*x-1)*exp(exp(x)*x^2)+1/5*(-5*x^2
-10*x)*exp(x)+6*x^2-2/5*x,x, algorithm="maxima")

[Out]

2*x^3 - x^2*e^x - 1/5*x^2 + 1/5*(10*x^2 - 5*x*e^x - x)*e^(x^2*e^x)

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mupad [B]  time = 4.15, size = 21, normalized size = 0.84 \begin {gather*} -\frac {x\,\left (x+{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}\right )\,\left (5\,{\mathrm {e}}^x-10\,x+1\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(6*x^2 - (exp(x^2*exp(x))*(exp(x)*(5*x + 2*x^2 - 19*x^3 - 10*x^4 + 5) - 20*x + exp(2*x)*(10*x^2 + 5*x^3) +
1))/5 - (exp(x)*(10*x + 5*x^2))/5 - (2*x)/5,x)

[Out]

-(x*(x + exp(x^2*exp(x)))*(5*exp(x) - 10*x + 1))/5

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sympy [A]  time = 0.40, size = 39, normalized size = 1.56 \begin {gather*} 2 x^{3} - x^{2} e^{x} - \frac {x^{2}}{5} + \frac {\left (10 x^{2} - 5 x e^{x} - x\right ) e^{x^{2} e^{x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x**3-10*x**2)*exp(x)**2+(10*x**4+19*x**3-2*x**2-5*x-5)*exp(x)+20*x-1)*exp(exp(x)*x**2)+1/5*
(-5*x**2-10*x)*exp(x)+6*x**2-2/5*x,x)

[Out]

2*x**3 - x**2*exp(x) - x**2/5 + (10*x**2 - 5*x*exp(x) - x)*exp(x**2*exp(x))/5

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