3.13.79 \(\int \frac {1}{8} e^{2 e^3} (-5+16 x^3) \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{4} e^{2 e^3} x^3 \left (-\frac {5}{2 x^2}+2 x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12} \begin {gather*} \frac {1}{2} e^{2 e^3} x^4-\frac {5}{8} e^{2 e^3} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*E^3)*(-5 + 16*x^3))/8,x]

[Out]

(-5*E^(2*E^3)*x)/8 + (E^(2*E^3)*x^4)/2

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} e^{2 e^3} \int \left (-5+16 x^3\right ) \, dx\\ &=-\frac {5}{8} e^{2 e^3} x+\frac {1}{2} e^{2 e^3} x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 0.80 \begin {gather*} \frac {1}{8} e^{2 e^3} \left (-5 x+4 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(2*E^3)*(-5*x + 4*x^4))/8

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fricas [A]  time = 0.82, size = 16, normalized size = 0.64 \begin {gather*} \frac {1}{8} \, {\left (4 \, x^{4} - 5 \, x\right )} e^{\left (2 \, e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*x^4 - 5*x)*e^(2*e^3)

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giac [A]  time = 0.26, size = 16, normalized size = 0.64 \begin {gather*} \frac {1}{8} \, {\left (4 \, x^{4} - 5 \, x\right )} e^{\left (2 \, e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*x^4 - 5*x)*e^(2*e^3)

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maple [A]  time = 0.02, size = 16, normalized size = 0.64

method	result	size
gosper	\(\frac {x \left (4 x^{3}-5\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}}}{8}\)	\(16\)
default	\(\frac {\left (4 x^{4}-5 x \right ) {\mathrm e}^{2 \,{\mathrm e}^{3}}}{8}\)	\(17\)
norman	\(-\frac {5 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{8}+\frac {x^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}}{2}\)	\(20\)
risch	\(-\frac {5 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{8}+\frac {x^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}}{2}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*(4*x^3-5)*exp(exp(3))^2

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maxima [A]  time = 0.45, size = 16, normalized size = 0.64 \begin {gather*} \frac {1}{8} \, {\left (4 \, x^{4} - 5 \, x\right )} e^{\left (2 \, e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*x^4 - 5*x)*e^(2*e^3)

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mupad [B]  time = 0.87, size = 15, normalized size = 0.60 \begin {gather*} \frac {x\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (4\,x^3-5\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp(3))*(16*x^3 - 5))/8,x)

[Out]

(x*exp(2*exp(3))*(4*x^3 - 5))/8

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sympy [A]  time = 0.06, size = 22, normalized size = 0.88 \begin {gather*} \frac {x^{4} e^{2 e^{3}}}{2} - \frac {5 x e^{2 e^{3}}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*exp(2*exp(3))/2 - 5*x*exp(2*exp(3))/8

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