3.98.100 \(\int e^{-e^{16}+e^{-e^{16}} (-4 x^2+e^{e^{16}} (-4 x+e^x x)+(x^2-x^3) \log (5))} (-8 x+e^{e^{16}} (-4+e^x (1+x))+(2 x-3 x^2) \log (5)) \, dx\)

Optimal. Leaf size=28 \[ e^{x \left (-4+e^x+e^{-e^{16}} x (-4+(1-x) \log (5))\right )} \]

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Rubi [A]  time = 0.50, antiderivative size = 53, normalized size of antiderivative = 1.89, number of steps used = 1, number of rules used = 1, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6706} \begin {gather*} 5^{e^{-e^{16}} \left (x^2-x^3\right )} \exp \left (-e^{-e^{16}} \left (4 x^2+e^{e^{16}} \left (4 x-e^x x\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-E^16 + (-4*x^2 + E^E^16*(-4*x + E^x*x) + (x^2 - x^3)*Log[5])/E^E^16)*(-8*x + E^E^16*(-4 + E^x*(1 + x))
+ (2*x - 3*x^2)*Log[5]),x]

[Out]

5^((x^2 - x^3)/E^E^16)/E^((4*x^2 + E^E^16*(4*x - E^x*x))/E^E^16)

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=5^{e^{-e^{16}} \left (x^2-x^3\right )} \exp \left (-e^{-e^{16}} \left (4 x^2+e^{e^{16}} \left (4 x-e^x x\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.52, size = 0, normalized size = 0.00 \begin {gather*} \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[E^(-E^16 + (-4*x^2 + E^E^16*(-4*x + E^x*x) + (x^2 - x^3)*Log[5])/E^E^16)*(-8*x + E^E^16*(-4 + E^x*(1
+ x)) + (2*x - 3*x^2)*Log[5]),x]

[Out]

Integrate[E^(-E^16 + (-4*x^2 + E^E^16*(-4*x + E^x*x) + (x^2 - x^3)*Log[5])/E^E^16)*(-8*x + E^E^16*(-4 + E^x*(1
+ x)) + (2*x - 3*x^2)*Log[5]), x]

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fricas [A]  time = 0.71, size = 46, normalized size = 1.64 \begin {gather*} e^{\left (-{\left (4 \, x^{2} - {\left (x e^{x} - 4 \, x - e^{16}\right )} e^{\left (e^{16}\right )} + {\left (x^{3} - x^{2}\right )} \log \relax (5)\right )} e^{\left (-e^{16}\right )} + e^{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*log(5)-8*x)*exp(((exp(x)*x-4*x)*exp(exp(16))+(-x^3+x^2)*
log(5)-4*x^2)/exp(exp(16)))/exp(exp(16)),x, algorithm="fricas")

[Out]

e^(-(4*x^2 - (x*e^x - 4*x - e^16)*e^(e^16) + (x^3 - x^2)*log(5))*e^(-e^16) + e^16)

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giac [A]  time = 0.17, size = 42, normalized size = 1.50 \begin {gather*} e^{\left (-x^{3} e^{\left (-e^{16}\right )} \log \relax (5) + x^{2} e^{\left (-e^{16}\right )} \log \relax (5) - 4 \, x^{2} e^{\left (-e^{16}\right )} + x e^{x} - 4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*log(5)-8*x)*exp(((exp(x)*x-4*x)*exp(exp(16))+(-x^3+x^2)*
log(5)-4*x^2)/exp(exp(16)))/exp(exp(16)),x, algorithm="giac")

[Out]

e^(-x^3*e^(-e^16)*log(5) + x^2*e^(-e^16)*log(5) - 4*x^2*e^(-e^16) + x*e^x - 4*x)

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maple [A]  time = 0.11, size = 37, normalized size = 1.32

method	result	size
risch	\({\mathrm e}^{-x \left (x^{2} \ln \relax (5)-x \ln \relax (5)-{\mathrm e}^{x +{\mathrm e}^{16}}+4 \,{\mathrm e}^{{\mathrm e}^{16}}+4 x \right ) {\mathrm e}^{-{\mathrm e}^{16}}}\)	\(37\)
norman	\({\mathrm e}^{\left (\left ({\mathrm e}^{x} x -4 x \right ) {\mathrm e}^{{\mathrm e}^{16}}+\left (-x^{3}+x^{2}\right ) \ln \relax (5)-4 x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{16}}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*ln(5)-8*x)*exp(((exp(x)*x-4*x)*exp(exp(16))+(-x^3+x^2)*ln(5)-4
*x^2)/exp(exp(16)))/exp(exp(16)),x,method=_RETURNVERBOSE)

[Out]

exp(-x*(x^2*ln(5)-x*ln(5)-exp(x+exp(16))+4*exp(exp(16))+4*x)*exp(-exp(16)))

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maxima [A]  time = 0.65, size = 42, normalized size = 1.50 \begin {gather*} e^{\left (-x^{3} e^{\left (-e^{16}\right )} \log \relax (5) + x^{2} e^{\left (-e^{16}\right )} \log \relax (5) - 4 \, x^{2} e^{\left (-e^{16}\right )} + x e^{x} - 4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*log(5)-8*x)*exp(((exp(x)*x-4*x)*exp(exp(16))+(-x^3+x^2)*
log(5)-4*x^2)/exp(exp(16)))/exp(exp(16)),x, algorithm="maxima")

[Out]

e^(-x^3*e^(-e^16)*log(5) + x^2*e^(-e^16)*log(5) - 4*x^2*e^(-e^16) + x*e^x - 4*x)

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mupad [B]  time = 5.87, size = 38, normalized size = 1.36 \begin {gather*} 5^{{\mathrm {e}}^{-{\mathrm {e}}^{16}}\,\left (x^2-x^3\right )}\,{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-4\,x^2\,{\mathrm {e}}^{-{\mathrm {e}}^{16}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-exp(16))*exp(-exp(-exp(16))*(exp(exp(16))*(4*x - x*exp(x)) - log(5)*(x^2 - x^3) + 4*x^2))*(log(5)*(2*
x - 3*x^2) - 8*x + exp(exp(16))*(exp(x)*(x + 1) - 4)),x)

[Out]

5^(exp(-exp(16))*(x^2 - x^3))*exp(x*exp(x))*exp(-4*x)*exp(-4*x^2*exp(-exp(16)))

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sympy [A]  time = 0.27, size = 34, normalized size = 1.21 \begin {gather*} e^{\frac {- 4 x^{2} + \left (- x^{3} + x^{2}\right ) \log {\relax (5 )} + \left (x e^{x} - 4 x\right ) e^{e^{16}}}{e^{e^{16}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)-4)*exp(exp(16))+(-3*x**2+2*x)*ln(5)-8*x)*exp(((exp(x)*x-4*x)*exp(exp(16))+(-x**3+x**2
)*ln(5)-4*x**2)/exp(exp(16)))/exp(exp(16)),x)

[Out]

exp((-4*x**2 + (-x**3 + x**2)*log(5) + (x*exp(x) - 4*x)*exp(exp(16)))*exp(-exp(16)))

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