3.68.8 \(\int \frac {e^{-5-\frac {x}{e^5}} (e^{e^2} (-2 e^5-x)-24 x-x^4+e^5 (-48+x^3))}{x^3} \, dx\)

Optimal. Leaf size=25 \[ e^{-\frac {x}{e^5}} \left (-\frac {-24-e^{e^2}}{x^2}+x\right ) \]

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Rubi [A]  time = 0.18, antiderivative size = 30, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 5, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {2199, 2194, 2177, 2178, 2176} \begin {gather*} \frac {\left (24+e^{e^2}\right ) e^{-\frac {x}{e^5}}}{x^2}+e^{-\frac {x}{e^5}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-5 - x/E^5)*(E^E^2*(-2*E^5 - x) - 24*x - x^4 + E^5*(-48 + x^3)))/x^3,x]

[Out]

(24 + E^E^2)/(E^(x/E^5)*x^2) + x/E^(x/E^5)

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 2177

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{-\frac {x}{e^5}}-\frac {2 e^{-\frac {x}{e^5}} \left (24+e^{e^2}\right )}{x^3}-\frac {e^{-5-\frac {x}{e^5}} \left (24+e^{e^2}\right )}{x^2}-e^{-5-\frac {x}{e^5}} x\right ) \, dx\\ &=\left (-24-e^{e^2}\right ) \int \frac {e^{-5-\frac {x}{e^5}}}{x^2} \, dx-\left (2 \left (24+e^{e^2}\right )\right ) \int \frac {e^{-\frac {x}{e^5}}}{x^3} \, dx+\int e^{-\frac {x}{e^5}} \, dx-\int e^{-5-\frac {x}{e^5}} x \, dx\\ &=-e^{5-\frac {x}{e^5}}+\frac {e^{-\frac {x}{e^5}} \left (24+e^{e^2}\right )}{x^2}+\frac {e^{-5-\frac {x}{e^5}} \left (24+e^{e^2}\right )}{x}+e^{-\frac {x}{e^5}} x-e^5 \int e^{-5-\frac {x}{e^5}} \, dx+\frac {\left (24+e^{e^2}\right ) \int \frac {e^{-\frac {x}{e^5}}}{x^2} \, dx}{e^5}+\frac {\left (24+e^{e^2}\right ) \int \frac {e^{-5-\frac {x}{e^5}}}{x} \, dx}{e^5}\\ &=\frac {e^{-\frac {x}{e^5}} \left (24+e^{e^2}\right )}{x^2}+e^{-\frac {x}{e^5}} x+\frac {\left (24+e^{e^2}\right ) \text {Ei}\left (-\frac {x}{e^5}\right )}{e^{10}}-\frac {\left (24+e^{e^2}\right ) \int \frac {e^{-\frac {x}{e^5}}}{x} \, dx}{e^{10}}\\ &=\frac {e^{-\frac {x}{e^5}} \left (24+e^{e^2}\right )}{x^2}+e^{-\frac {x}{e^5}} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 22, normalized size = 0.88 \begin {gather*} \frac {e^{-\frac {x}{e^5}} \left (24+e^{e^2}+x^3\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-5 - x/E^5)*(E^E^2*(-2*E^5 - x) - 24*x - x^4 + E^5*(-48 + x^3)))/x^3,x]

[Out]

(24 + E^E^2 + x^3)/(E^(x/E^5)*x^2)

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fricas [B]  time = 0.58, size = 39, normalized size = 1.56 \begin {gather*} \frac {{\left (x^{3} + 24\right )} e^{\left (-{\left (x + 5 \, e^{5}\right )} e^{\left (-5\right )} + 5\right )} + e^{\left (-{\left (x + 5 \, e^{5}\right )} e^{\left (-5\right )} + e^{2} + 5\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(5)-x)*exp(exp(2))+(x^3-48)*exp(5)-x^4-24*x)/x^3/exp(5)/exp(x/exp(5)),x, algorithm="fricas")

[Out]

((x^3 + 24)*e^(-(x + 5*e^5)*e^(-5) + 5) + e^(-(x + 5*e^5)*e^(-5) + e^2 + 5))/x^2

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giac [B]  time = 0.13, size = 39, normalized size = 1.56 \begin {gather*} \frac {{\left (x^{3} e^{\left (-x e^{\left (-5\right )} + 10\right )} + e^{\left (-x e^{\left (-5\right )} + e^{2} + 10\right )} + 24 \, e^{\left (-x e^{\left (-5\right )} + 10\right )}\right )} e^{\left (-10\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(5)-x)*exp(exp(2))+(x^3-48)*exp(5)-x^4-24*x)/x^3/exp(5)/exp(x/exp(5)),x, algorithm="giac")

[Out]

(x^3*e^(-x*e^(-5) + 10) + e^(-x*e^(-5) + e^2 + 10) + 24*e^(-x*e^(-5) + 10))*e^(-10)/x^2

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maple [A]  time = 0.10, size = 19, normalized size = 0.76

method	result	size
risch	\(\frac {\left (x^{3}+{\mathrm e}^{{\mathrm e}^{2}}+24\right ) {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{x^{2}}\)	\(19\)
gosper	\(\frac {\left (x^{3}+{\mathrm e}^{{\mathrm e}^{2}}+24\right ) {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{x^{2}}\)	\(22\)
norman	\(\frac {\left (x^{3}+{\mathrm e}^{{\mathrm e}^{2}}+24\right ) {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{x^{2}}\)	\(22\)
derivativedivides	\({\mathrm e}^{-15} \left (-{\mathrm e}^{-x \,{\mathrm e}^{-5}} {\mathrm e}^{20}-48 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{10} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2 x^{2}}+\frac {{\mathrm e}^{5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2 x}-\frac {\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )}{2}\right )-24 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{x}+\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )\right )-{\mathrm e}^{20} \left (-x \,{\mathrm e}^{-5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}-{\mathrm e}^{-x \,{\mathrm e}^{-5}}\right )-2 \,{\mathrm e}^{{\mathrm e}^{2}} {\mathrm e}^{5} \left (-\frac {{\mathrm e}^{10} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2 x^{2}}+\frac {{\mathrm e}^{5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2 x}-\frac {\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )}{2}\right )-{\mathrm e}^{{\mathrm e}^{2}} {\mathrm e}^{5} \left (-\frac {{\mathrm e}^{5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{x}+\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )\right )\right )\)	\(218\)
default	\({\mathrm e}^{-15} \left (-{\mathrm e}^{-x \,{\mathrm e}^{-5}} {\mathrm e}^{20}-48 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{10} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2 x^{2}}+\frac {{\mathrm e}^{5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2 x}-\frac {\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )}{2}\right )-24 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{x}+\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )\right )-{\mathrm e}^{20} \left (-x \,{\mathrm e}^{-5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}-{\mathrm e}^{-x \,{\mathrm e}^{-5}}\right )-2 \,{\mathrm e}^{{\mathrm e}^{2}} {\mathrm e}^{5} \left (-\frac {{\mathrm e}^{10} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2 x^{2}}+\frac {{\mathrm e}^{5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2 x}-\frac {\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )}{2}\right )-{\mathrm e}^{{\mathrm e}^{2}} {\mathrm e}^{5} \left (-\frac {{\mathrm e}^{5} {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{x}+\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )\right )\right )\)	\(218\)
meijerg	\(-{\mathrm e}^{5} \left (1-\frac {\left (2 x \,{\mathrm e}^{-5}+2\right ) {\mathrm e}^{-x \,{\mathrm e}^{-5}}}{2}\right )+{\mathrm e}^{5} \left (1-{\mathrm e}^{-x \,{\mathrm e}^{-5}}\right )+\left (-24-{\mathrm e}^{{\mathrm e}^{2}}\right ) {\mathrm e}^{-10} \left (-\frac {{\mathrm e}^{5}}{x}+6-\ln \relax (x )+\frac {{\mathrm e}^{5} \left (2-2 x \,{\mathrm e}^{-5}\right )}{2 x}-\frac {{\mathrm e}^{5-x \,{\mathrm e}^{-5}}}{x}+\ln \left (x \,{\mathrm e}^{-5}\right )+\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )\right )-2 \,{\mathrm e}^{-10+{\mathrm e}^{2}} \left (-\frac {{\mathrm e}^{10}}{2 x^{2}}+\frac {{\mathrm e}^{5}}{x}-\frac {13}{4}+\frac {\ln \relax (x )}{2}+\frac {{\mathrm e}^{10} \left (9 x^{2} {\mathrm e}^{-10}-12 x \,{\mathrm e}^{-5}+6\right )}{12 x^{2}}-\frac {{\mathrm e}^{10-x \,{\mathrm e}^{-5}} \left (3-3 x \,{\mathrm e}^{-5}\right )}{6 x^{2}}-\frac {\ln \left (x \,{\mathrm e}^{-5}\right )}{2}-\frac {\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )}{2}\right )-48 \,{\mathrm e}^{-10} \left (-\frac {{\mathrm e}^{10}}{2 x^{2}}+\frac {{\mathrm e}^{5}}{x}-\frac {13}{4}+\frac {\ln \relax (x )}{2}+\frac {{\mathrm e}^{10} \left (9 x^{2} {\mathrm e}^{-10}-12 x \,{\mathrm e}^{-5}+6\right )}{12 x^{2}}-\frac {{\mathrm e}^{10-x \,{\mathrm e}^{-5}} \left (3-3 x \,{\mathrm e}^{-5}\right )}{6 x^{2}}-\frac {\ln \left (x \,{\mathrm e}^{-5}\right )}{2}-\frac {\expIntegralEi \left (1, x \,{\mathrm e}^{-5}\right )}{2}\right )\)	\(258\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(5)-x)*exp(exp(2))+(x^3-48)*exp(5)-x^4-24*x)/x^3/exp(5)/exp(x/exp(5)),x,method=_RETURNVERBOSE)

[Out]

(x^3+exp(exp(2))+24)/x^2*exp(-x*exp(-5))

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maxima [C]  time = 0.42, size = 67, normalized size = 2.68 \begin {gather*} {\left (x + e^{5}\right )} e^{\left (-x e^{\left (-5\right )}\right )} + 24 \, e^{\left (-10\right )} \Gamma \left (-1, x e^{\left (-5\right )}\right ) + e^{\left (e^{2} - 10\right )} \Gamma \left (-1, x e^{\left (-5\right )}\right ) + 48 \, e^{\left (-10\right )} \Gamma \left (-2, x e^{\left (-5\right )}\right ) + 2 \, e^{\left (e^{2} - 10\right )} \Gamma \left (-2, x e^{\left (-5\right )}\right ) - e^{\left (-x e^{\left (-5\right )} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(5)-x)*exp(exp(2))+(x^3-48)*exp(5)-x^4-24*x)/x^3/exp(5)/exp(x/exp(5)),x, algorithm="maxima")

[Out]

(x + e^5)*e^(-x*e^(-5)) + 24*e^(-10)*gamma(-1, x*e^(-5)) + e^(e^2 - 10)*gamma(-1, x*e^(-5)) + 48*e^(-10)*gamma
(-2, x*e^(-5)) + 2*e^(e^2 - 10)*gamma(-2, x*e^(-5)) - e^(-x*e^(-5) + 5)

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mupad [B]  time = 0.12, size = 28, normalized size = 1.12 \begin {gather*} \frac {{\mathrm {e}}^{-x\,{\mathrm {e}}^{-5}-5}\,\left ({\mathrm {e}}^5\,x^3+{\mathrm {e}}^{{\mathrm {e}}^2+5}+24\,{\mathrm {e}}^5\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-5)*exp(-x*exp(-5))*(24*x + exp(exp(2))*(x + 2*exp(5)) + x^4 - exp(5)*(x^3 - 48)))/x^3,x)

[Out]

(exp(- x*exp(-5) - 5)*(exp(exp(2) + 5) + 24*exp(5) + x^3*exp(5)))/x^2

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sympy [A]  time = 0.15, size = 19, normalized size = 0.76 \begin {gather*} \frac {\left (x^{3} + 24 + e^{e^{2}}\right ) e^{- \frac {x}{e^{5}}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(5)-x)*exp(exp(2))+(x**3-48)*exp(5)-x**4-24*x)/x**3/exp(5)/exp(x/exp(5)),x)

[Out]

(x**3 + 24 + exp(exp(2)))*exp(-x*exp(-5))/x**2

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