3.39.52 \(\int \frac {e^{-\frac {16-x^3}{x}} (240+3 e^{\frac {16-x^3}{x}} x^2+30 x^3+e^{x+\frac {16-x^3}{x}} (5 x^2+3 x^3))}{3 x^2} \, dx\)

Optimal. Leaf size=26 \[ 4+5 e^{-\frac {16}{x}+x^2}+\left (1+e^x\right ) \left (\frac {2}{3}+x\right ) \]

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Rubi [A]  time = 0.60, antiderivative size = 32, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 5, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {12, 6688, 2176, 2194, 6706} \begin {gather*} 5 e^{x^2-\frac {16}{x}}+x-e^x+\frac {1}{3} e^x (3 x+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(240 + 3*E^((16 - x^3)/x)*x^2 + 30*x^3 + E^(x + (16 - x^3)/x)*(5*x^2 + 3*x^3))/(3*E^((16 - x^3)/x)*x^2),x]

[Out]

-E^x + 5*E^(-16/x + x^2) + x + (E^x*(5 + 3*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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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}{3} \int \frac {e^{-\frac {16-x^3}{x}} \left (240+3 e^{\frac {16-x^3}{x}} x^2+30 x^3+e^{x+\frac {16-x^3}{x}} \left (5 x^2+3 x^3\right )\right )}{x^2} \, dx\\ &=\frac {1}{3} \int \left (3+e^x (5+3 x)+\frac {30 e^{-\frac {16}{x}+x^2} \left (8+x^3\right )}{x^2}\right ) \, dx\\ &=x+\frac {1}{3} \int e^x (5+3 x) \, dx+10 \int \frac {e^{-\frac {16}{x}+x^2} \left (8+x^3\right )}{x^2} \, dx\\ &=5 e^{-\frac {16}{x}+x^2}+x+\frac {1}{3} e^x (5+3 x)-\int e^x \, dx\\ &=-e^x+5 e^{-\frac {16}{x}+x^2}+x+\frac {1}{3} e^x (5+3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 24, normalized size = 0.92 \begin {gather*} 5 e^{-\frac {16}{x}+x^2}+x+e^x \left (\frac {2}{3}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(240 + 3*E^((16 - x^3)/x)*x^2 + 30*x^3 + E^(x + (16 - x^3)/x)*(5*x^2 + 3*x^3))/(3*E^((16 - x^3)/x)*x
^2),x]

[Out]

5*E^(-16/x + x^2) + x + E^x*(2/3 + x)

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fricas [A]  time = 0.53, size = 38, normalized size = 1.46 \begin {gather*} \frac {1}{3} \, {\left ({\left (3 \, x + 2\right )} e^{\left (-\frac {x^{3} - x^{2} - 16}{x}\right )} + 15\right )} e^{\left (\frac {x^{3} - 16}{x}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^3+5*x^2)*exp((-x^3+16)/x)*exp(x)+3*x^2*exp((-x^3+16)/x)+30*x^3+240)/x^2/exp((-x^3+16)/x),x
, algorithm="fricas")

[Out]

1/3*((3*x + 2)*e^(-(x^3 - x^2 - 16)/x) + 15)*e^((x^3 - 16)/x) + x

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giac [A]  time = 0.20, size = 22, normalized size = 0.85 \begin {gather*} x e^{x} + x + \frac {2}{3} \, e^{x} + 5 \, e^{\left (\frac {x^{3} - 16}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^3+5*x^2)*exp((-x^3+16)/x)*exp(x)+3*x^2*exp((-x^3+16)/x)+30*x^3+240)/x^2/exp((-x^3+16)/x),x
, algorithm="giac")

[Out]

x*e^x + x + 2/3*e^x + 5*e^((x^3 - 16)/x)

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maple [A]  time = 0.07, size = 24, normalized size = 0.92

method	result	size
risch	\(x +\frac {\left (3 x +2\right ) {\mathrm e}^{x}}{3}+5 \,{\mathrm e}^{\frac {x^{3}-16}{x}}\)	\(24\)
norman	\(\frac {\left (x^{2} {\mathrm e}^{\frac {-x^{3}+16}{x}}+{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {-x^{3}+16}{x}}+5 x +\frac {2 \,{\mathrm e}^{x} x \,{\mathrm e}^{\frac {-x^{3}+16}{x}}}{3}\right ) {\mathrm e}^{-\frac {-x^{3}+16}{x}}}{x}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((3*x^3+5*x^2)*exp((-x^3+16)/x)*exp(x)+3*x^2*exp((-x^3+16)/x)+30*x^3+240)/x^2/exp((-x^3+16)/x),x,metho
d=_RETURNVERBOSE)

[Out]

x+1/3*(3*x+2)*exp(x)+5*exp((x^3-16)/x)

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maxima [A]  time = 0.41, size = 24, normalized size = 0.92 \begin {gather*} {\left (x - 1\right )} e^{x} + x + 5 \, e^{\left (x^{2} - \frac {16}{x}\right )} + \frac {5}{3} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x^3+5*x^2)*exp((-x^3+16)/x)*exp(x)+3*x^2*exp((-x^3+16)/x)+30*x^3+240)/x^2/exp((-x^3+16)/x),x
, algorithm="maxima")

[Out]

(x - 1)*e^x + x + 5*e^(x^2 - 16/x) + 5/3*e^x

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mupad [B]  time = 2.42, size = 22, normalized size = 0.85 \begin {gather*} x+5\,{\mathrm {e}}^{x^2-\frac {16}{x}}+\frac {2\,{\mathrm {e}}^x}{3}+x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x^3 - 16)/x)*(x^2*exp(-(x^3 - 16)/x) + 10*x^3 + (exp(-(x^3 - 16)/x)*exp(x)*(5*x^2 + 3*x^3))/3 + 80))
/x^2,x)

[Out]

x + 5*exp(x^2 - 16/x) + (2*exp(x))/3 + x*exp(x)

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sympy [A]  time = 0.21, size = 20, normalized size = 0.77 \begin {gather*} x + \frac {\left (3 x + 2\right ) e^{x}}{3} + 5 e^{- \frac {16 - x^{3}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((3*x**3+5*x**2)*exp((-x**3+16)/x)*exp(x)+3*x**2*exp((-x**3+16)/x)+30*x**3+240)/x**2/exp((-x**3+
16)/x),x)

[Out]

x + (3*x + 2)*exp(x)/3 + 5*exp(-(16 - x**3)/x)

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