3.21.31 \(\int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{5 x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{5} \left (3-25 e^{2 e^x}+e^{4/x}-x\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {12, 14, 2282, 2194, 2209} \begin {gather*} -\frac {x}{5}-5 e^{2 e^x}+\frac {e^{4/x}}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*E^(4/x) - x^2 - 50*E^(2*E^x + x)*x^2)/(5*x^2),x]

[Out]

-5*E^(2*E^x) + E^(4/x)/5 - x/5

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-4 e^{4/x}-x^2-50 e^{2 e^x+x} x^2}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-50 e^{2 e^x+x}-\frac {4 e^{4/x}+x^2}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {4 e^{4/x}+x^2}{x^2} \, dx\right )-10 \int e^{2 e^x+x} \, dx\\ &=-\left (\frac {1}{5} \int \left (1+\frac {4 e^{4/x}}{x^2}\right ) \, dx\right )-10 \operatorname {Subst}\left (\int e^{2 x} \, dx,x,e^x\right )\\ &=-5 e^{2 e^x}-\frac {x}{5}-\frac {4}{5} \int \frac {e^{4/x}}{x^2} \, dx\\ &=-5 e^{2 e^x}+\frac {e^{4/x}}{5}-\frac {x}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.04 \begin {gather*} -5 e^{2 e^x}+\frac {e^{4/x}}{5}-\frac {x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^(4/x) - x^2 - 50*E^(2*E^x + x)*x^2)/(5*x^2),x]

[Out]

-5*E^(2*E^x) + E^(4/x)/5 - x/5

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fricas [A]  time = 0.70, size = 29, normalized size = 1.16 \begin {gather*} -\frac {1}{5} \, {\left ({\left (x - e^{\frac {4}{x}}\right )} e^{x} + 25 \, e^{\left (x + 2 \, e^{x}\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-50*x^2*exp(x)*exp(exp(x))^2-4*exp(2/x)^2-x^2)/x^2,x, algorithm="fricas")

[Out]

-1/5*((x - e^(4/x))*e^x + 25*e^(x + 2*e^x))*e^(-x)

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giac [A]  time = 0.34, size = 29, normalized size = 1.16 \begin {gather*} -\frac {1}{5} \, {\left (x e^{x} + 25 \, e^{\left (x + 2 \, e^{x}\right )}\right )} e^{\left (-x\right )} + \frac {1}{5} \, e^{\frac {4}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-50*x^2*exp(x)*exp(exp(x))^2-4*exp(2/x)^2-x^2)/x^2,x, algorithm="giac")

[Out]

-1/5*(x*e^x + 25*e^(x + 2*e^x))*e^(-x) + 1/5*e^(4/x)

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maple [A]  time = 0.03, size = 20, normalized size = 0.80

method	result	size
default	\(\frac {{\mathrm e}^{\frac {4}{x}}}{5}-5 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}-\frac {x}{5}\)	\(20\)
risch	\(\frac {{\mathrm e}^{\frac {4}{x}}}{5}-5 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}-\frac {x}{5}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-50*x^2*exp(x)*exp(exp(x))^2-4*exp(2/x)^2-x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/5*x+1/5*exp(1/x)^4-5*exp(exp(x))^2

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maxima [A]  time = 0.36, size = 19, normalized size = 0.76 \begin {gather*} -\frac {1}{5} \, x - 5 \, e^{\left (2 \, e^{x}\right )} + \frac {1}{5} \, e^{\frac {4}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-50*x^2*exp(x)*exp(exp(x))^2-4*exp(2/x)^2-x^2)/x^2,x, algorithm="maxima")

[Out]

-1/5*x - 5*e^(2*e^x) + 1/5*e^(4/x)

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mupad [B]  time = 1.36, size = 19, normalized size = 0.76 \begin {gather*} \frac {{\mathrm {e}}^{4/x}}{5}-\frac {x}{5}-5\,{\mathrm {e}}^{2\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((4*exp(4/x))/5 + x^2/5 + 10*x^2*exp(2*exp(x))*exp(x))/x^2,x)

[Out]

exp(4/x)/5 - x/5 - 5*exp(2*exp(x))

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sympy [A]  time = 0.36, size = 17, normalized size = 0.68 \begin {gather*} - \frac {x}{5} + \frac {e^{\frac {4}{x}}}{5} - 5 e^{2 e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-50*x**2*exp(x)*exp(exp(x))**2-4*exp(2/x)**2-x**2)/x**2,x)

[Out]

-x/5 + exp(4/x)/5 - 5*exp(2*exp(x))

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