3.28.86 \(\int \frac {1}{80} (9 e^3 x^2-3 e^3 x^2 \log (9)) \, dx\)

Optimal. Leaf size=16 \[ \frac {1}{80} e^3 x^3 (3-\log (9)) \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6, 12, 30} \begin {gather*} \frac {1}{80} e^3 x^3 (3-\log (9)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9*E^3*x^2 - 3*E^3*x^2*Log[9])/80,x]

[Out]

(E^3*x^3*(3 - Log[9]))/80

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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} &=\int \frac {1}{80} e^3 x^2 (9-3 \log (9)) \, dx\\ &=\frac {1}{80} \left (3 e^3 (3-\log (9))\right ) \int x^2 \, dx\\ &=\frac {1}{80} e^3 x^3 (3-\log (9))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.88 \begin {gather*} -\frac {1}{80} e^3 x^3 (-3+\log (9)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*E^3*x^2 - 3*E^3*x^2*Log[9])/80,x]

[Out]

-1/80*(E^3*x^3*(-3 + Log[9]))

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fricas [A]  time = 0.51, size = 17, normalized size = 1.06 \begin {gather*} -\frac {1}{40} \, x^{3} e^{3} \log \relax (3) + \frac {3}{80} \, x^{3} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3/40*x^2*exp(3)*log(3)+9/80*x^2*exp(3),x, algorithm="fricas")

[Out]

-1/40*x^3*e^3*log(3) + 3/80*x^3*e^3

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giac [A]  time = 0.50, size = 17, normalized size = 1.06 \begin {gather*} -\frac {1}{40} \, x^{3} e^{3} \log \relax (3) + \frac {3}{80} \, x^{3} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3/40*x^2*exp(3)*log(3)+9/80*x^2*exp(3),x, algorithm="giac")

[Out]

-1/40*x^3*e^3*log(3) + 3/80*x^3*e^3

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

method	result	size
gosper	\(-\frac {x^{3} {\mathrm e}^{3} \left (2 \ln \relax (3)-3\right )}{80}\)	\(14\)
norman	\(\left (-\frac {{\mathrm e}^{3} \ln \relax (3)}{40}+\frac {3 \,{\mathrm e}^{3}}{80}\right ) x^{3}\)	\(16\)
default	\(-\frac {x^{3} {\mathrm e}^{3} \ln \relax (3)}{40}+\frac {3 x^{3} {\mathrm e}^{3}}{80}\)	\(18\)
risch	\(-\frac {x^{3} {\mathrm e}^{3} \ln \relax (3)}{40}+\frac {3 x^{3} {\mathrm e}^{3}}{80}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-3/40*x^2*exp(3)*ln(3)+9/80*x^2*exp(3),x,method=_RETURNVERBOSE)

[Out]

-1/80*x^3*exp(3)*(2*ln(3)-3)

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maxima [A]  time = 0.64, size = 17, normalized size = 1.06 \begin {gather*} -\frac {1}{40} \, x^{3} e^{3} \log \relax (3) + \frac {3}{80} \, x^{3} e^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3/40*x^2*exp(3)*log(3)+9/80*x^2*exp(3),x, algorithm="maxima")

[Out]

-1/40*x^3*e^3*log(3) + 3/80*x^3*e^3

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mupad [B]  time = 0.06, size = 13, normalized size = 0.81 \begin {gather*} -\frac {x^3\,{\mathrm {e}}^3\,\left (2\,\ln \relax (3)-3\right )}{80} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x^2*exp(3))/80 - (3*x^2*exp(3)*log(3))/40,x)

[Out]

-(x^3*exp(3)*(2*log(3) - 3))/80

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sympy [A]  time = 0.06, size = 17, normalized size = 1.06 \begin {gather*} x^{3} \left (- \frac {e^{3} \log {\relax (3 )}}{40} + \frac {3 e^{3}}{80}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3/40*x**2*exp(3)*ln(3)+9/80*x**2*exp(3),x)

[Out]

x**3*(-exp(3)*log(3)/40 + 3*exp(3)/80)

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