3.35.38 \(\int \frac {4 e^x x^3+e^{\frac {-4-3 x-x^2}{4 x^2}} (24+9 x)}{12 x^3} \, dx\)

Optimal. Leaf size=27 \[ 18+\frac {e^x}{3}+e^{\frac {x-(2+x)^2}{4 x^2}} \]

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Rubi [A]  time = 0.20, antiderivative size = 26, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {12, 14, 2194, 6706} \begin {gather*} e^{-\frac {1}{x^2}-\frac {3}{4 x}-\frac {1}{4}}+\frac {e^x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^x*x^3 + E^((-4 - 3*x - x^2)/(4*x^2))*(24 + 9*x))/(12*x^3),x]

[Out]

E^(-1/4 - x^(-2) - 3/(4*x)) + E^x/3

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]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{12} \int \frac {4 e^x x^3+e^{\frac {-4-3 x-x^2}{4 x^2}} (24+9 x)}{x^3} \, dx\\ &=\frac {1}{12} \int \left (4 e^x+\frac {3 e^{-\frac {1}{4}-\frac {1}{x^2}-\frac {3}{4 x}} (8+3 x)}{x^3}\right ) \, dx\\ &=\frac {1}{4} \int \frac {e^{-\frac {1}{4}-\frac {1}{x^2}-\frac {3}{4 x}} (8+3 x)}{x^3} \, dx+\frac {\int e^x \, dx}{3}\\ &=e^{-\frac {1}{4}-\frac {1}{x^2}-\frac {3}{4 x}}+\frac {e^x}{3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 26, normalized size = 0.96 \begin {gather*} e^{-\frac {1}{4}-\frac {1}{x^2}-\frac {3}{4 x}}+\frac {e^x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^x*x^3 + E^((-4 - 3*x - x^2)/(4*x^2))*(24 + 9*x))/(12*x^3),x]

[Out]

E^(-1/4 - x^(-2) - 3/(4*x)) + E^x/3

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fricas [A]  time = 0.69, size = 19, normalized size = 0.70 \begin {gather*} \frac {1}{3} \, e^{x} + e^{\left (-\frac {x^{2} + 3 \, x + 4}{4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(4*exp(x)*x^3+(9*x+24)*exp(1/4*(-x^2-3*x-4)/x^2))/x^3,x, algorithm="fricas")

[Out]

1/3*e^x + e^(-1/4*(x^2 + 3*x + 4)/x^2)

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giac [A]  time = 0.18, size = 18, normalized size = 0.67 \begin {gather*} \frac {1}{3} \, e^{x} + e^{\left (-\frac {3}{4 \, x} - \frac {1}{x^{2}} - \frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(4*exp(x)*x^3+(9*x+24)*exp(1/4*(-x^2-3*x-4)/x^2))/x^3,x, algorithm="giac")

[Out]

1/3*e^x + e^(-3/4/x - 1/x^2 - 1/4)

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maple [A]  time = 0.04, size = 19, normalized size = 0.70

method	result	size
default	\({\mathrm e}^{-\frac {1}{4}-\frac {3}{4 x}-\frac {1}{x^{2}}}+\frac {{\mathrm e}^{x}}{3}\)	\(19\)
risch	\(\frac {{\mathrm e}^{x}}{3}+{\mathrm e}^{-\frac {x^{2}+3 x +4}{4 x^{2}}}\)	\(20\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {-x^{2}-3 x -4}{4 x^{2}}}+\frac {{\mathrm e}^{x} x^{2}}{3}}{x^{2}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/12*(4*exp(x)*x^3+(9*x+24)*exp(1/4*(-x^2-3*x-4)/x^2))/x^3,x,method=_RETURNVERBOSE)

[Out]

exp(-1/4-3/4/x-1/x^2)+1/3*exp(x)

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maxima [A]  time = 0.48, size = 26, normalized size = 0.96 \begin {gather*} \frac {1}{3} \, {\left (e^{\left (x + \frac {1}{x^{2}} + \frac {1}{4}\right )} + 3 \, e^{\left (-\frac {3}{4 \, x}\right )}\right )} e^{\left (-\frac {1}{x^{2}} - \frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(4*exp(x)*x^3+(9*x+24)*exp(1/4*(-x^2-3*x-4)/x^2))/x^3,x, algorithm="maxima")

[Out]

1/3*(e^(x + 1/x^2 + 1/4) + 3*e^(-3/4/x))*e^(-1/x^2 - 1/4)

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mupad [B]  time = 2.24, size = 18, normalized size = 0.67 \begin {gather*} {\mathrm {e}}^{-\frac {3}{4\,x}-\frac {1}{x^2}-\frac {1}{4}}+\frac {{\mathrm {e}}^x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3*exp(x))/3 + (exp(-((3*x)/4 + x^2/4 + 1)/x^2)*(9*x + 24))/12)/x^3,x)

[Out]

exp(- 3/(4*x) - 1/x^2 - 1/4) + exp(x)/3

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sympy [A]  time = 0.20, size = 22, normalized size = 0.81 \begin {gather*} \frac {e^{x}}{3} + e^{\frac {- \frac {x^{2}}{4} - \frac {3 x}{4} - 1}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(4*exp(x)*x**3+(9*x+24)*exp(1/4*(-x**2-3*x-4)/x**2))/x**3,x)

[Out]

exp(x)/3 + exp((-x**2/4 - 3*x/4 - 1)/x**2)

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