3.2.78 \(\int \frac {1}{2} e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} (-16 x^2+e^x (-12 x^2-4 x^3)-12 x^2 \log (x))) \, dx\)

Optimal. Leaf size=28 \[ e^{e^{1-\frac {x}{2}}-e^{2 x^3 \left (1+e^x+\log (x)\right )}} \]

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Rubi [A]  time = 1.03, antiderivative size = 39, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 2, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {12, 6706} \begin {gather*} e^{e^{\frac {2-x}{2}}-e^{2 e^x x^3+2 x^3} x^{2 x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^((2 - x)/2) - E^(2*x^3 + 2*E^x*x^3 + 2*x^3*Log[x]))*(-E^((2 - x)/2) + E^(2*x^3 + 2*E^x*x^3 + 2*x^3*L
og[x])*(-16*x^2 + E^x*(-12*x^2 - 4*x^3) - 12*x^2*Log[x])))/2,x]

[Out]

E^(E^((2 - x)/2) - E^(2*x^3 + 2*E^x*x^3)*x^(2*x^3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} &=\frac {1}{2} \int e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3+2 x^3 \log (x)}} \left (-e^{\frac {2-x}{2}}+e^{2 x^3+2 e^x x^3+2 x^3 \log (x)} \left (-16 x^2+e^x \left (-12 x^2-4 x^3\right )-12 x^2 \log (x)\right )\right ) \, dx\\ &=e^{e^{\frac {2-x}{2}}-e^{2 x^3+2 e^x x^3} x^{2 x^3}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 33, normalized size = 1.18 \begin {gather*} e^{e^{1-\frac {x}{2}}-e^{2 \left (1+e^x\right ) x^3} x^{2 x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^((2 - x)/2) - E^(2*x^3 + 2*E^x*x^3 + 2*x^3*Log[x]))*(-E^((2 - x)/2) + E^(2*x^3 + 2*E^x*x^3 + 2
*x^3*Log[x])*(-16*x^2 + E^x*(-12*x^2 - 4*x^3) - 12*x^2*Log[x])))/2,x]

[Out]

E^(E^(1 - x/2) - E^(2*(1 + E^x)*x^3)*x^(2*x^3))

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fricas [B]  time = 0.71, size = 46, normalized size = 1.64 \begin {gather*} e^{\left (-e^{\left (2 \, {\left (x^{3} e^{\left (-x + 2\right )} \log \relax (x) + x^{3} e^{2} + x^{3} e^{\left (-x + 2\right )}\right )} e^{\left (x - 2\right )}\right )} + e^{\left (-\frac {1}{2} \, x + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-12*x^2*log(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*log(x)+2*exp(x)*x^3+2*x^3)-exp(1-1/2*x
))*exp(-exp(2*x^3*log(x)+2*exp(x)*x^3+2*x^3)+exp(1-1/2*x)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-12*x^2*log(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*log(x)+2*exp(x)*x^3+2*x^3)-exp(1-1/2*x
))*exp(-exp(2*x^3*log(x)+2*exp(x)*x^3+2*x^3)+exp(1-1/2*x)),x, algorithm="giac")

[Out]

integrate(-1/2*(4*(3*x^2*log(x) + 4*x^2 + (x^3 + 3*x^2)*e^x)*e^(2*x^3*e^x + 2*x^3*log(x) + 2*x^3) + e^(-1/2*x
+ 1))*e^(-e^(2*x^3*e^x + 2*x^3*log(x) + 2*x^3) + e^(-1/2*x + 1)), x)

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maple [A]  time = 0.14, size = 28, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{-x^{2 x^{3}} {\mathrm e}^{2 x^{3} \left ({\mathrm e}^{x}+1\right )}+{\mathrm e}^{1-\frac {x}{2}}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-12*x^2*ln(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*ln(x)+2*exp(x)*x^3+2*x^3)-exp(1-1/2*x))*exp(-
exp(2*x^3*ln(x)+2*exp(x)*x^3+2*x^3)+exp(1-1/2*x)),x,method=_RETURNVERBOSE)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-12*x^2*log(x)+(-4*x^3-12*x^2)*exp(x)-16*x^2)*exp(2*x^3*log(x)+2*exp(x)*x^3+2*x^3)-exp(1-1/2*x
))*exp(-exp(2*x^3*log(x)+2*exp(x)*x^3+2*x^3)+exp(1-1/2*x)),x, algorithm="maxima")

[Out]

-1/2*integrate((4*(3*x^2*log(x) + 4*x^2 + (x^3 + 3*x^2)*e^x)*e^(2*x^3*e^x + 2*x^3*log(x) + 2*x^3) + e^(-1/2*x
+ 1))*e^(-e^(2*x^3*e^x + 2*x^3*log(x) + 2*x^3) + e^(-1/2*x + 1)), x)

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mupad [B]  time = 0.47, size = 33, normalized size = 1.18 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{-\frac {x}{2}}\,\mathrm {e}}\,{\mathrm {e}}^{-x^{2\,x^3}\,{\mathrm {e}}^{2\,x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(1 - x/2) - exp(2*x^3*exp(x) + 2*x^3*log(x) + 2*x^3))*(exp(1 - x/2) + exp(2*x^3*exp(x) + 2*x^3*lo
g(x) + 2*x^3)*(exp(x)*(12*x^2 + 4*x^3) + 12*x^2*log(x) + 16*x^2)))/2,x)

[Out]

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

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sympy [A]  time = 7.64, size = 34, normalized size = 1.21 \begin {gather*} e^{- e^{2 x^{3} e^{x} + 2 x^{3} \log {\relax (x )} + 2 x^{3}} + \frac {e}{\sqrt {e^{x}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-12*x**2*ln(x)+(-4*x**3-12*x**2)*exp(x)-16*x**2)*exp(2*x**3*ln(x)+2*exp(x)*x**3+2*x**3)-exp(1-
1/2*x))*exp(-exp(2*x**3*ln(x)+2*exp(x)*x**3+2*x**3)+exp(1-1/2*x)),x)

[Out]

exp(-exp(2*x**3*exp(x) + 2*x**3*log(x) + 2*x**3) + E/sqrt(exp(x)))

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