3.50.74 \(\int \frac {14+25 e^{-5+e^5-x} (-2-x)}{x^3} \, dx\)

Optimal. Leaf size=27 \[ 5-\frac {7-25 e^{-5+e^5-x}}{x^2}-\log ^2(3) \]

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Rubi [A]  time = 0.05, antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {14, 2197} \begin {gather*} \frac {25 e^{-x+e^5-5}}{x^2}-\frac {7}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(14 + 25*E^(-5 + E^5 - x)*(-2 - x))/x^3,x]

[Out]

-7/x^2 + (25*E^(-5 + E^5 - x))/x^2

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 2197

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
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {14}{x^3}-\frac {25 e^{-5+e^5-x} (2+x)}{x^3}\right ) \, dx\\ &=-\frac {7}{x^2}-25 \int \frac {e^{-5+e^5-x} (2+x)}{x^3} \, dx\\ &=-\frac {7}{x^2}+\frac {25 e^{-5+e^5-x}}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 18, normalized size = 0.67 \begin {gather*} \frac {-7+25 e^{-5+e^5-x}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(14 + 25*E^(-5 + E^5 - x)*(-2 - x))/x^3,x]

[Out]

(-7 + 25*E^(-5 + E^5 - x))/x^2

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fricas [A]  time = 0.98, size = 18, normalized size = 0.67 \begin {gather*} \frac {e^{\left (-x + e^{5} + 2 \, \log \relax (5) - 5\right )} - 7}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(2*log(5)+exp(5)-x-5)+14)/x^3,x, algorithm="fricas")

[Out]

(e^(-x + e^5 + 2*log(5) - 5) - 7)/x^2

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giac [A]  time = 0.15, size = 16, normalized size = 0.59 \begin {gather*} \frac {25 \, e^{\left (-x + e^{5} - 5\right )} - 7}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(2*log(5)+exp(5)-x-5)+14)/x^3,x, algorithm="giac")

[Out]

(25*e^(-x + e^5 - 5) - 7)/x^2

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maple [A]  time = 0.09, size = 19, normalized size = 0.70

method	result	size
norman	\(\frac {-7+{\mathrm e}^{2 \ln \relax (5)+{\mathrm e}^{5}-x -5}}{x^{2}}\)	\(19\)
risch	\(\frac {25 \,{\mathrm e}^{{\mathrm e}^{5}-x -5}}{x^{2}}-\frac {7}{x^{2}}\)	\(20\)
derivativedivides	\(\frac {{\mathrm e}^{2 \ln \relax (5)+{\mathrm e}^{5}-x -5}}{x^{2}}-\frac {7}{x^{2}}\)	\(23\)
default	\(\frac {{\mathrm e}^{2 \ln \relax (5)+{\mathrm e}^{5}-x -5}}{x^{2}}-\frac {7}{x^{2}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x-2)*exp(2*ln(5)+exp(5)-x-5)+14)/x^3,x,method=_RETURNVERBOSE)

[Out]

(-7+exp(2*ln(5)+exp(5)-x-5))/x^2

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maxima [C]  time = 0.47, size = 26, normalized size = 0.96 \begin {gather*} 25 \, e^{\left (e^{5} - 5\right )} \Gamma \left (-1, x\right ) + 50 \, e^{\left (e^{5} - 5\right )} \Gamma \left (-2, x\right ) - \frac {7}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(2*log(5)+exp(5)-x-5)+14)/x^3,x, algorithm="maxima")

[Out]

25*e^(e^5 - 5)*gamma(-1, x) + 50*e^(e^5 - 5)*gamma(-2, x) - 7/x^2

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mupad [B]  time = 3.19, size = 20, normalized size = 0.74 \begin {gather*} \frac {25\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^5}}{x^2}-\frac {7}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(5) - x + 2*log(5) - 5)*(x + 2) - 14)/x^3,x)

[Out]

(25*exp(-x)*exp(-5)*exp(exp(5)))/x^2 - 7/x^2

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sympy [A]  time = 0.10, size = 17, normalized size = 0.63 \begin {gather*} \frac {25 e^{- x - 5 + e^{5}}}{x^{2}} - \frac {7}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x-2)*exp(2*ln(5)+exp(5)-x-5)+14)/x**3,x)

[Out]

25*exp(-x - 5 + exp(5))/x**2 - 7/x**2

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