3.98.34 \(\int \frac {1}{9} (9-32 e^{\frac {16 x^2}{9}} x) \, dx\)

Optimal. Leaf size=14 \[ 2-e^{\frac {16 x^2}{9}}+x \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2209} \begin {gather*} x-e^{\frac {16 x^2}{9}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9 - 32*E^((16*x^2)/9)*x)/9,x]

[Out]

-E^((16*x^2)/9) + 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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \left (9-32 e^{\frac {16 x^2}{9}} x\right ) \, dx\\ &=x-\frac {32}{9} \int e^{\frac {16 x^2}{9}} x \, dx\\ &=-e^{\frac {16 x^2}{9}}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.93 \begin {gather*} -e^{\frac {16 x^2}{9}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 - 32*E^((16*x^2)/9)*x)/9,x]

[Out]

-E^((16*x^2)/9) + x

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fricas [A]  time = 1.03, size = 10, normalized size = 0.71 \begin {gather*} x - e^{\left (\frac {16}{9} \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-32/9*x*exp(16/9*x^2)+1,x, algorithm="fricas")

[Out]

x - e^(16/9*x^2)

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giac [A]  time = 0.17, size = 10, normalized size = 0.71 \begin {gather*} x - e^{\left (\frac {16}{9} \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-32/9*x*exp(16/9*x^2)+1,x, algorithm="giac")

[Out]

x - e^(16/9*x^2)

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maple [A]  time = 0.02, size = 11, normalized size = 0.79

method	result	size
default	\(x -{\mathrm e}^{\frac {16 x^{2}}{9}}\)	\(11\)
norman	\(x -{\mathrm e}^{\frac {16 x^{2}}{9}}\)	\(11\)
risch	\(x -{\mathrm e}^{\frac {16 x^{2}}{9}}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-32/9*x*exp(16/9*x^2)+1,x,method=_RETURNVERBOSE)

[Out]

x-exp(16/9*x^2)

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maxima [A]  time = 0.37, size = 10, normalized size = 0.71 \begin {gather*} x - e^{\left (\frac {16}{9} \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-32/9*x*exp(16/9*x^2)+1,x, algorithm="maxima")

[Out]

x - e^(16/9*x^2)

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mupad [B]  time = 0.07, size = 10, normalized size = 0.71 \begin {gather*} x-{\mathrm {e}}^{\frac {16\,x^2}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1 - (32*x*exp((16*x^2)/9))/9,x)

[Out]

x - exp((16*x^2)/9)

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sympy [A]  time = 0.09, size = 8, normalized size = 0.57 \begin {gather*} x - e^{\frac {16 x^{2}}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-32/9*x*exp(16/9*x**2)+1,x)

[Out]

x - exp(16*x**2/9)

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