3.2.18 \(\int \frac {e^{-x} (-45-45 x+e^{-\frac {e^5 x}{5}} (5+5 x+e^5 x))}{5 x^2} \, dx\) [118]

Optimal. Leaf size=29 \[ \frac {e^{-x} \left (9-e^{\frac {1}{5} \left (x-\left (1+e^5\right ) x\right )}\right )}{x} \]

[Out]

(9-exp(1/5*x-1/5*(exp(5)+1)*x))/x/exp(x)

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Rubi [A]
time = 0.23, antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {12, 6874, 2228} \begin {gather*} \frac {9 e^{-x}}{x}-\frac {e^{-\frac {1}{5} \left (5+e^5\right ) x}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45 - 45*x + (5 + 5*x + E^5*x)/E^((E^5*x)/5))/(5*E^x*x^2),x]

[Out]

9/(E^x*x) - 1/(E^(((5 + E^5)*x)/5)*x)

Rule 12

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

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{-x} \left (-45-45 x+e^{-\frac {e^5 x}{5}} \left (5+5 x+e^5 x\right )\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {45 e^{-x} (1+x)}{x^2}+\frac {e^{-x-\frac {e^5 x}{5}} \left (5+\left (5+e^5\right ) x\right )}{x^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{-x-\frac {e^5 x}{5}} \left (5+\left (5+e^5\right ) x\right )}{x^2} \, dx-9 \int \frac {e^{-x} (1+x)}{x^2} \, dx\\ &=\frac {9 e^{-x}}{x}-\frac {e^{-\frac {1}{5} \left (5+e^5\right ) x}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 1.30, size = 23, normalized size = 0.79 \begin {gather*} \frac {e^{-x} \left (9-e^{-\frac {e^5 x}{5}}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 - 45*x + (5 + 5*x + E^5*x)/E^((E^5*x)/5))/(5*E^x*x^2),x]

[Out]

(9 - E^(-1/5*(E^5*x)))/(E^x*x)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.09, size = 82, normalized size = 2.83

method	result	size
norman	\(\frac {\left (9-{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{5}}\right ) {\mathrm e}^{-x}}{x}\)	\(19\)
risch	\(\frac {9 \,{\mathrm e}^{-x}}{x}-\frac {{\mathrm e}^{-\frac {x \left ({\mathrm e}^{5}+5\right )}{5}}}{x}\)	\(24\)
default	\(-\frac {{\mathrm e}^{5} \expIntegral \left (1, -\left (-1-\frac {{\mathrm e}^{5}}{5}\right ) x \right )}{5}+\frac {9 \,{\mathrm e}^{-x}}{x}+\left (-1-\frac {{\mathrm e}^{5}}{5}\right ) \left (-\frac {{\mathrm e}^{\left (-1-\frac {{\mathrm e}^{5}}{5}\right ) x}}{\left (-1-\frac {{\mathrm e}^{5}}{5}\right ) x}-\expIntegral \left (1, -\left (-1-\frac {{\mathrm e}^{5}}{5}\right ) x \right )\right )-\expIntegral \left (1, -\left (-1-\frac {{\mathrm e}^{5}}{5}\right ) x \right )\)	\(82\)
meijerg	\(-\frac {9 \left (-2 x +2\right )}{2 x}+\frac {9 \,{\mathrm e}^{-x}}{x}-4+\frac {9}{x}+\frac {{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right ) \left (\frac {5 \,{\mathrm e}^{-5} \left (2-\frac {2 x \,{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right )}{5}\right )}{2 x \left (1+5 \,{\mathrm e}^{-5}\right )}-\frac {5 \,{\mathrm e}^{-5-\frac {x \,{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right )}{5}}}{x \left (1+5 \,{\mathrm e}^{-5}\right )}+\ln \left (\frac {x \,{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right )}{5}\right )+\expIntegral \left (1, \frac {x \,{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right )}{5}\right )-4-\ln \left (x \right )+\ln \left (5\right )-\ln \left (1+5 \,{\mathrm e}^{-5}\right )-\frac {5 \,{\mathrm e}^{-5}}{x \left (1+5 \,{\mathrm e}^{-5}\right )}\right )}{5}-\ln \left (\frac {x \,{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right )}{5}\right )-\expIntegral \left (1, \frac {x \,{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right )}{5}\right )+\ln \left (x \right )-\ln \left (5\right )+\ln \left (1+5 \,{\mathrm e}^{-5}\right )+\frac {{\mathrm e}^{5} \left (-\ln \left (\frac {x \,{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right )}{5}\right )-\expIntegral \left (1, \frac {x \,{\mathrm e}^{5} \left (1+5 \,{\mathrm e}^{-5}\right )}{5}\right )+\ln \left (x \right )-\ln \left (5\right )+5+\ln \left (1+5 \,{\mathrm e}^{-5}\right )\right )}{5}\)	\(239\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((x*exp(5)+5*x+5)*exp(-1/5*x*exp(5))-45*x-45)/exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/5*exp(5)*Ei(1,-(-1-1/5*exp(5))*x)+9/exp(x)/x+(-1-1/5*exp(5))*(-1/(-1-1/5*exp(5))/x*exp((-1-1/5*exp(5))*x)-E
i(1,-(-1-1/5*exp(5))*x))-Ei(1,-(-1-1/5*exp(5))*x)

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.52, size = 47, normalized size = 1.62 \begin {gather*} \frac {1}{5} \, {\rm Ei}\left (-\frac {1}{5} \, x {\left (e^{5} + 5\right )}\right ) e^{5} - \frac {1}{5} \, {\left (e^{5} + 5\right )} \Gamma \left (-1, \frac {1}{5} \, x {\left (e^{5} + 5\right )}\right ) + {\rm Ei}\left (-\frac {1}{5} \, x {\left (e^{5} + 5\right )}\right ) - 9 \, {\rm Ei}\left (-x\right ) + 9 \, \Gamma \left (-1, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x*exp(5)+5*x+5)*exp(-1/5*x*exp(5))-45*x-45)/exp(x)/x^2,x, algorithm="maxima")

[Out]

1/5*Ei(-1/5*x*(e^5 + 5))*e^5 - 1/5*(e^5 + 5)*gamma(-1, 1/5*x*(e^5 + 5)) + Ei(-1/5*x*(e^5 + 5)) - 9*Ei(-x) + 9*
gamma(-1, x)

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Fricas [A]
time = 0.30, size = 17, normalized size = 0.59 \begin {gather*} -\frac {{\left (e^{\left (-\frac {1}{5} \, x e^{5}\right )} - 9\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x*exp(5)+5*x+5)*exp(-1/5*x*exp(5))-45*x-45)/exp(x)/x^2,x, algorithm="fricas")

[Out]

-(e^(-1/5*x*e^5) - 9)*e^(-x)/x

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Sympy [A]
time = 0.25, size = 20, normalized size = 0.69 \begin {gather*} \frac {9 e^{- x}}{x} - \frac {e^{- x} e^{- \frac {x e^{5}}{5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x*exp(5)+5*x+5)*exp(-1/5*x*exp(5))-45*x-45)/exp(x)/x**2,x)

[Out]

9*exp(-x)/x - exp(-x)*exp(-x*exp(5)/5)/x

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Giac [A]
time = 0.40, size = 22, normalized size = 0.76 \begin {gather*} -\frac {e^{\left (-\frac {1}{5} \, x e^{5} - x\right )} - 9 \, e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x*exp(5)+5*x+5)*exp(-1/5*x*exp(5))-45*x-45)/exp(x)/x^2,x, algorithm="giac")

[Out]

-(e^(-1/5*x*e^5 - x) - 9*e^(-x))/x

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Mupad [B]
time = 0.11, size = 23, normalized size = 0.79 \begin {gather*} \frac {9\,{\mathrm {e}}^{-x}-{\mathrm {e}}^{-x-\frac {x\,{\mathrm {e}}^5}{5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(9*x - (exp(-(x*exp(5))/5)*(5*x + x*exp(5) + 5))/5 + 9))/x^2,x)

[Out]

(9*exp(-x) - exp(- x - (x*exp(5))/5))/x

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