3.85.64 \(\int \frac {8 e^{e^{e^{25}}} x^3}{\log (2)} \, dx\)

Optimal. Leaf size=16 \[ \frac {2 e^{e^{e^{25}}} x^4}{\log (2)} \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 30} \begin {gather*} \frac {2 e^{e^{e^{25}}} x^4}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*E^E^E^25*x^3)/Log[2],x]

[Out]

(2*E^E^E^25*x^4)/Log[2]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\left (8 e^{e^{e^{25}}}\right ) \int x^3 \, dx}{\log (2)}\\ &=\frac {2 e^{e^{e^{25}}} x^4}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.00 \begin {gather*} \frac {2 e^{e^{e^{25}}} x^4}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^E^E^25*x^3)/Log[2],x]

[Out]

(2*E^E^E^25*x^4)/Log[2]

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fricas [A]  time = 0.94, size = 13, normalized size = 0.81 \begin {gather*} \frac {2 \, x^{4} e^{\left (e^{\left (e^{25}\right )}\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x^3*exp(exp(exp(25)))/log(2),x, algorithm="fricas")

[Out]

2*x^4*e^(e^(e^25))/log(2)

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giac [A]  time = 0.14, size = 13, normalized size = 0.81 \begin {gather*} \frac {2 \, x^{4} e^{\left (e^{\left (e^{25}\right )}\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x^3*exp(exp(exp(25)))/log(2),x, algorithm="giac")

[Out]

2*x^4*e^(e^(e^25))/log(2)

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maple [A]  time = 0.03, size = 14, normalized size = 0.88

method	result	size
gosper	\(\frac {2 x^{4} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{25}}}}{\ln \relax (2)}\)	\(14\)
default	\(\frac {2 x^{4} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{25}}}}{\ln \relax (2)}\)	\(14\)
norman	\(\frac {2 x^{4} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{25}}}}{\ln \relax (2)}\)	\(14\)
risch	\(\frac {2 x^{4} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{25}}}}{\ln \relax (2)}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x^3*exp(exp(exp(25)))/ln(2),x,method=_RETURNVERBOSE)

[Out]

2*x^4*exp(exp(exp(25)))/ln(2)

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maxima [A]  time = 0.35, size = 13, normalized size = 0.81 \begin {gather*} \frac {2 \, x^{4} e^{\left (e^{\left (e^{25}\right )}\right )}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x^3*exp(exp(exp(25)))/log(2),x, algorithm="maxima")

[Out]

2*x^4*e^(e^(e^25))/log(2)

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mupad [B]  time = 5.10, size = 13, normalized size = 0.81 \begin {gather*} \frac {2\,x^4\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{25}}}}{\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^3*exp(exp(exp(25))))/log(2),x)

[Out]

(2*x^4*exp(exp(exp(25))))/log(2)

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sympy [A]  time = 0.06, size = 14, normalized size = 0.88 \begin {gather*} \frac {2 x^{4} e^{e^{e^{25}}}}{\log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x**3*exp(exp(exp(25)))/ln(2),x)

[Out]

2*x**4*exp(exp(exp(25)))/log(2)

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