3.21.89 \(\int \frac {1}{18} e^{\frac {1}{18} (-288+7 x+3 x^2)} (7+6 x) \, dx\)

Optimal. Leaf size=20 \[ e^{-16+\frac {1}{12} \left (2+\frac {14}{3 x}\right ) x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 2244, 2236} \begin {gather*} e^{\frac {x^2}{6}+\frac {7 x}{18}-16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-288 + 7*x + 3*x^2)/18)*(7 + 6*x))/18,x]

[Out]

E^(-16 + (7*x)/18 + x^2/6)

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}{18} \int e^{\frac {1}{18} \left (-288+7 x+3 x^2\right )} (7+6 x) \, dx\\ &=\frac {1}{18} \int e^{-16+\frac {7 x}{18}+\frac {x^2}{6}} (7+6 x) \, dx\\ &=e^{-16+\frac {7 x}{18}+\frac {x^2}{6}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 16, normalized size = 0.80 \begin {gather*} e^{-16+\frac {7 x}{18}+\frac {x^2}{6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-288 + 7*x + 3*x^2)/18)*(7 + 6*x))/18,x]

[Out]

E^(-16 + (7*x)/18 + x^2/6)

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fricas [A]  time = 0.65, size = 11, normalized size = 0.55 \begin {gather*} e^{\left (\frac {1}{6} \, x^{2} + \frac {7}{18} \, x - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(6*x+7)*exp(1/6*x^2+7/18*x-16),x, algorithm="fricas")

[Out]

e^(1/6*x^2 + 7/18*x - 16)

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giac [A]  time = 5.04, size = 11, normalized size = 0.55 \begin {gather*} e^{\left (\frac {1}{6} \, x^{2} + \frac {7}{18} \, x - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(6*x+7)*exp(1/6*x^2+7/18*x-16),x, algorithm="giac")

[Out]

e^(1/6*x^2 + 7/18*x - 16)

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maple [A]  time = 0.03, size = 12, normalized size = 0.60

method	result	size
gosper	\({\mathrm e}^{\frac {1}{6} x^{2}+\frac {7}{18} x -16}\)	\(12\)
default	\({\mathrm e}^{\frac {1}{6} x^{2}+\frac {7}{18} x -16}\)	\(12\)
norman	\({\mathrm e}^{\frac {1}{6} x^{2}+\frac {7}{18} x -16}\)	\(12\)
risch	\({\mathrm e}^{\frac {1}{6} x^{2}+\frac {7}{18} x -16}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/18*(6*x+7)*exp(1/6*x^2+7/18*x-16),x,method=_RETURNVERBOSE)

[Out]

exp(1/6*x^2+7/18*x-16)

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maxima [A]  time = 0.68, size = 11, normalized size = 0.55 \begin {gather*} e^{\left (\frac {1}{6} \, x^{2} + \frac {7}{18} \, x - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(6*x+7)*exp(1/6*x^2+7/18*x-16),x, algorithm="maxima")

[Out]

e^(1/6*x^2 + 7/18*x - 16)

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mupad [B]  time = 1.18, size = 13, normalized size = 0.65 \begin {gather*} {\mathrm {e}}^{\frac {7\,x}{18}}\,{\mathrm {e}}^{-16}\,{\mathrm {e}}^{\frac {x^2}{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((7*x)/18 + x^2/6 - 16)*(6*x + 7))/18,x)

[Out]

exp((7*x)/18)*exp(-16)*exp(x^2/6)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.60 \begin {gather*} e^{\frac {x^{2}}{6} + \frac {7 x}{18} - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(6*x+7)*exp(1/6*x**2+7/18*x-16),x)

[Out]

exp(x**2/6 + 7*x/18 - 16)

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