3.84.63 \(\int \frac {1}{3} e^{e^x} (e^{1+2 x} (1-x)+e^{10} (-1+2 x)+e^x (-e x+e^{10} (-x+x^2))) \, dx\)

Optimal. Leaf size=26 \[ \frac {1}{3} e^{1+e^x} (1-x) \left (e^x-e^9 x\right ) \]

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Rubi [F]  time = 0.53, 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^{e^x} \left (e^{1+2 x} (1-x)+e^{10} (-1+2 x)+e^x \left (-e x+e^{10} \left (-x+x^2\right )\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-1/3*E^(1 + E^x) + E^(1 + E^x + x)/3 - (E^10*ExpIntegralEi[E^x])/3 + (2*Defer[Int][E^(10 + E^x)*x, x])/3 - ((1
+ E^9)*Defer[Int][E^(1 + E^x + x)*x, x])/3 - Defer[Int][E^(1 + E^x + 2*x)*x, x]/3 + Defer[Int][E^(10 + E^x +
x)*x^2, x]/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int e^{e^x} \left (e^{1+2 x} (1-x)+e^{10} (-1+2 x)+e^x \left (-e x+e^{10} \left (-x+x^2\right )\right )\right ) \, dx\\ &=\frac {1}{3} \int \left (-e^{1+e^x+2 x} (-1+x)+e^{10+e^x} (-1+2 x)+e^{1+e^x+x} x \left (-1-e^9+e^9 x\right )\right ) \, dx\\ &=-\left (\frac {1}{3} \int e^{1+e^x+2 x} (-1+x) \, dx\right )+\frac {1}{3} \int e^{10+e^x} (-1+2 x) \, dx+\frac {1}{3} \int e^{1+e^x+x} x \left (-1-e^9+e^9 x\right ) \, dx\\ &=\frac {1}{3} \int \left (-e^{10+e^x}+2 e^{10+e^x} x\right ) \, dx-\frac {1}{3} \int \left (-e^{1+e^x+2 x}+e^{1+e^x+2 x} x\right ) \, dx+\frac {1}{3} \int \left (-e^{1+e^x+x} \left (1+e^9\right ) x+e^{10+e^x+x} x^2\right ) \, dx\\ &=-\left (\frac {1}{3} \int e^{10+e^x} \, dx\right )+\frac {1}{3} \int e^{1+e^x+2 x} \, dx-\frac {1}{3} \int e^{1+e^x+2 x} x \, dx+\frac {1}{3} \int e^{10+e^x+x} x^2 \, dx+\frac {2}{3} \int e^{10+e^x} x \, dx+\frac {1}{3} \left (-1-e^9\right ) \int e^{1+e^x+x} x \, dx\\ &=-\left (\frac {1}{3} \int e^{1+e^x+2 x} x \, dx\right )+\frac {1}{3} \int e^{10+e^x+x} x^2 \, dx-\frac {1}{3} \operatorname {Subst}\left (\int \frac {e^{10+x}}{x} \, dx,x,e^x\right )+\frac {1}{3} \operatorname {Subst}\left (\int e^{1+x} x \, dx,x,e^x\right )+\frac {2}{3} \int e^{10+e^x} x \, dx+\frac {1}{3} \left (-1-e^9\right ) \int e^{1+e^x+x} x \, dx\\ &=\frac {1}{3} e^{1+e^x+x}-\frac {e^{10} \text {Ei}\left (e^x\right )}{3}-\frac {1}{3} \int e^{1+e^x+2 x} x \, dx+\frac {1}{3} \int e^{10+e^x+x} x^2 \, dx-\frac {1}{3} \operatorname {Subst}\left (\int e^{1+x} \, dx,x,e^x\right )+\frac {2}{3} \int e^{10+e^x} x \, dx+\frac {1}{3} \left (-1-e^9\right ) \int e^{1+e^x+x} x \, dx\\ &=-\frac {1}{3} e^{1+e^x}+\frac {1}{3} e^{1+e^x+x}-\frac {e^{10} \text {Ei}\left (e^x\right )}{3}-\frac {1}{3} \int e^{1+e^x+2 x} x \, dx+\frac {1}{3} \int e^{10+e^x+x} x^2 \, dx+\frac {2}{3} \int e^{10+e^x} x \, dx+\frac {1}{3} \left (-1-e^9\right ) \int e^{1+e^x+x} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 25, normalized size = 0.96 \begin {gather*} \frac {1}{3} e^{1+e^x} (-1+x) \left (-e^x+e^9 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(1 + E^x)*(-1 + x)*(-E^x + E^9*x))/3

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fricas [A]  time = 0.53, size = 25, normalized size = 0.96 \begin {gather*} \frac {1}{3} \, {\left ({\left (x^{2} - x\right )} e^{10} - {\left (x - 1\right )} e^{\left (x + 1\right )}\right )} e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-x+1)*exp(1)*exp(x)^2+((x^2-x)*exp(1)*exp(9)-x*exp(1))*exp(x)+(2*x-1)*exp(1)*exp(9))*exp(exp(x
)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.15, size = 45, normalized size = 1.73 \begin {gather*} \frac {1}{3} \, {\left (x^{2} e^{\left (x + e^{x} + 10\right )} - x e^{\left (2 \, x + e^{x} + 1\right )} - x e^{\left (x + e^{x} + 10\right )} + e^{\left (2 \, x + e^{x} + 1\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-x+1)*exp(1)*exp(x)^2+((x^2-x)*exp(1)*exp(9)-x*exp(1))*exp(x)+(2*x-1)*exp(1)*exp(9))*exp(exp(x
)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 29, normalized size = 1.12

method	result	size
risch	\(\frac {\left (x^{2} {\mathrm e}^{10}-x \,{\mathrm e}^{10}-x \,{\mathrm e}^{x +1}+{\mathrm e}^{x +1}\right ) {\mathrm e}^{{\mathrm e}^{x}}}{3}\)	\(29\)
norman	\(\frac {{\mathrm e} \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}}{3}+\frac {{\mathrm e} \,{\mathrm e}^{9} x^{2} {\mathrm e}^{{\mathrm e}^{x}}}{3}-\frac {{\mathrm e}^{{\mathrm e}^{x}} x \,{\mathrm e} \,{\mathrm e}^{x}}{3}-\frac {{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e} \,{\mathrm e}^{9} x}{3}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((1-x)*exp(1)*exp(x)^2+((x^2-x)*exp(1)*exp(9)-x*exp(1))*exp(x)+(2*x-1)*exp(1)*exp(9))*exp(exp(x)),x,me
thod=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{3} \, {\rm Ei}\left (e^{x}\right ) e^{10} + \frac {1}{3} \, {\left (x^{2} e^{10} - x e^{10} - {\left (x e - e\right )} e^{x}\right )} e^{\left (e^{x}\right )} + \frac {1}{3} \, \int e^{\left (e^{x} + 10\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-x+1)*exp(1)*exp(x)^2+((x^2-x)*exp(1)*exp(9)-x*exp(1))*exp(x)+(2*x-1)*exp(1)*exp(9))*exp(exp(x
)),x, algorithm="maxima")

[Out]

-1/3*Ei(e^x)*e^10 + 1/3*(x^2*e^10 - x*e^10 - (x*e - e)*e^x)*e^(e^x) + 1/3*integrate(e^(e^x + 10), x)

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mupad [B]  time = 0.08, size = 18, normalized size = 0.69 \begin {gather*} -\frac {{\mathrm {e}}^{{\mathrm {e}}^x+1}\,\left ({\mathrm {e}}^x-x\,{\mathrm {e}}^9\right )\,\left (x-1\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(exp(x) + 1)*(exp(x) - x*exp(9))*(x - 1))/3

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sympy [A]  time = 0.25, size = 32, normalized size = 1.23 \begin {gather*} \frac {\left (x^{2} e^{10} - e x e^{x} - x e^{10} + e e^{x}\right ) e^{e^{x}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-x+1)*exp(1)*exp(x)**2+((x**2-x)*exp(1)*exp(9)-x*exp(1))*exp(x)+(2*x-1)*exp(1)*exp(9))*exp(exp
(x)),x)

[Out]

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

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