3.46.45 \(\int \frac {1}{9} (9+e^{\frac {1}{9} (160 x-32 x^2)} (160-64 x)) \, dx\)

Optimal. Leaf size=14 \[ e^{\frac {32}{9} (5-x) x}+x \]

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2244, 2236} \begin {gather*} e^{\frac {160 x}{9}-\frac {32 x^2}{9}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9 + E^((160*x - 32*x^2)/9)*(160 - 64*x))/9,x]

[Out]

E^((160*x)/9 - (32*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 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \left (9+e^{\frac {1}{9} \left (160 x-32 x^2\right )} (160-64 x)\right ) \, dx\\ &=x+\frac {1}{9} \int e^{\frac {1}{9} \left (160 x-32 x^2\right )} (160-64 x) \, dx\\ &=x+\frac {1}{9} \int e^{\frac {160 x}{9}-\frac {32 x^2}{9}} (160-64 x) \, dx\\ &=e^{\frac {160 x}{9}-\frac {32 x^2}{9}}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 1.21 \begin {gather*} e^{\frac {160 x}{9}-\frac {32 x^2}{9}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 + E^((160*x - 32*x^2)/9)*(160 - 64*x))/9,x]

[Out]

E^((160*x)/9 - (32*x^2)/9) + x

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fricas [A]  time = 0.60, size = 12, normalized size = 0.86 \begin {gather*} x + e^{\left (-\frac {32}{9} \, x^{2} + \frac {160}{9} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-64*x+160)*exp(-32/9*x^2+160/9*x)+1,x, algorithm="fricas")

[Out]

x + e^(-32/9*x^2 + 160/9*x)

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giac [A]  time = 0.18, size = 12, normalized size = 0.86 \begin {gather*} x + e^{\left (-\frac {32}{9} \, x^{2} + \frac {160}{9} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-64*x+160)*exp(-32/9*x^2+160/9*x)+1,x, algorithm="giac")

[Out]

x + e^(-32/9*x^2 + 160/9*x)

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maple [A]  time = 0.02, size = 10, normalized size = 0.71

method	result	size
risch	\(x +{\mathrm e}^{-\frac {32 \left (x -5\right ) x}{9}}\)	\(10\)
default	\(x +{\mathrm e}^{-\frac {32}{9} x^{2}+\frac {160}{9} x}\)	\(13\)
norman	\(x +{\mathrm e}^{-\frac {32}{9} x^{2}+\frac {160}{9} x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(-64*x+160)*exp(-32/9*x^2+160/9*x)+1,x,method=_RETURNVERBOSE)

[Out]

x+exp(-32/9*(x-5)*x)

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maxima [A]  time = 0.43, size = 12, normalized size = 0.86 \begin {gather*} x + e^{\left (-\frac {32}{9} \, x^{2} + \frac {160}{9} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-64*x+160)*exp(-32/9*x^2+160/9*x)+1,x, algorithm="maxima")

[Out]

x + e^(-32/9*x^2 + 160/9*x)

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mupad [B]  time = 0.14, size = 12, normalized size = 0.86 \begin {gather*} x+{\mathrm {e}}^{\frac {160\,x}{9}-\frac {32\,x^2}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1 - (exp((160*x)/9 - (32*x^2)/9)*(64*x - 160))/9,x)

[Out]

x + exp((160*x)/9 - (32*x^2)/9)

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sympy [A]  time = 0.10, size = 14, normalized size = 1.00 \begin {gather*} x + e^{- \frac {32 x^{2}}{9} + \frac {160 x}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-64*x+160)*exp(-32/9*x**2+160/9*x)+1,x)

[Out]

x + exp(-32*x**2/9 + 160*x/9)

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