3.73.50 \(\int \frac {18+9 e^{-4-x} x^2+32 x^3}{9 x^2} \, dx\)

Optimal. Leaf size=22 \[ -e^{-4-x}-\frac {2}{x}+\frac {16 x^2}{9} \]

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 14, 2194} \begin {gather*} \frac {16 x^2}{9}-e^{-x-4}-\frac {2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18 + 9*E^(-4 - x)*x^2 + 32*x^3)/(9*x^2),x]

[Out]

-E^(-4 - x) - 2/x + (16*x^2)/9

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \frac {18+9 e^{-4-x} x^2+32 x^3}{x^2} \, dx\\ &=\frac {1}{9} \int \left (9 e^{-4-x}+\frac {2 \left (9+16 x^3\right )}{x^2}\right ) \, dx\\ &=\frac {2}{9} \int \frac {9+16 x^3}{x^2} \, dx+\int e^{-4-x} \, dx\\ &=-e^{-4-x}+\frac {2}{9} \int \left (\frac {9}{x^2}+16 x\right ) \, dx\\ &=-e^{-4-x}-\frac {2}{x}+\frac {16 x^2}{9}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.00 \begin {gather*} -e^{-4-x}-\frac {2}{x}+\frac {16 x^2}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 + 9*E^(-4 - x)*x^2 + 32*x^3)/(9*x^2),x]

[Out]

-E^(-4 - x) - 2/x + (16*x^2)/9

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fricas [A]  time = 1.08, size = 21, normalized size = 0.95 \begin {gather*} \frac {16 \, x^{3} - 9 \, x e^{\left (-x - 4\right )} - 18}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(9*x^2*exp(-x-4)+32*x^3+18)/x^2,x, algorithm="fricas")

[Out]

1/9*(16*x^3 - 9*x*e^(-x - 4) - 18)/x

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giac [A]  time = 0.15, size = 26, normalized size = 1.18 \begin {gather*} \frac {{\left (16 \, x^{3} e^{4} - 9 \, x e^{\left (-x\right )} - 18 \, e^{4}\right )} e^{\left (-4\right )}}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(9*x^2*exp(-x-4)+32*x^3+18)/x^2,x, algorithm="giac")

[Out]

1/9*(16*x^3*e^4 - 9*x*e^(-x) - 18*e^4)*e^(-4)/x

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maple [A]  time = 0.05, size = 20, normalized size = 0.91

method	result	size
risch	\(\frac {16 x^{2}}{9}-{\mathrm e}^{-x -4}-\frac {2}{x}\)	\(20\)
norman	\(\frac {-2+\frac {16 x^{3}}{9}-x \,{\mathrm e}^{-x -4}}{x}\)	\(21\)
derivativedivides	\(-\frac {2}{x}-\frac {128 x}{9}-\frac {512}{9}+\frac {16 \left (-x -4\right )^{2}}{9}-{\mathrm e}^{-x -4}\)	\(28\)
default	\(-\frac {2}{x}-\frac {128 x}{9}-\frac {512}{9}+\frac {16 \left (-x -4\right )^{2}}{9}-{\mathrm e}^{-x -4}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(9*x^2*exp(-x-4)+32*x^3+18)/x^2,x,method=_RETURNVERBOSE)

[Out]

16/9*x^2-exp(-x-4)-2/x

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maxima [A]  time = 0.35, size = 19, normalized size = 0.86 \begin {gather*} \frac {16}{9} \, x^{2} - \frac {2}{x} - e^{\left (-x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(9*x^2*exp(-x-4)+32*x^3+18)/x^2,x, algorithm="maxima")

[Out]

16/9*x^2 - 2/x - e^(-x - 4)

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mupad [B]  time = 0.07, size = 19, normalized size = 0.86 \begin {gather*} \frac {16\,x^2}{9}-\frac {2}{x}-{\mathrm {e}}^{-x-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(- x - 4) + (32*x^3)/9 + 2)/x^2,x)

[Out]

(16*x^2)/9 - 2/x - exp(- x - 4)

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sympy [A]  time = 0.12, size = 15, normalized size = 0.68 \begin {gather*} \frac {16 x^{2}}{9} - e^{- x - 4} - \frac {2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(9*x**2*exp(-x-4)+32*x**3+18)/x**2,x)

[Out]

16*x**2/9 - exp(-x - 4) - 2/x

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