3.82.72 \(\int \frac {e^{5-x} (80+52 x+12 x^2)}{x^3} \, dx\)

Optimal. Leaf size=19 \[ \frac {4 e^{5-x} \left (-3-\frac {10}{x}\right )}{x} \]

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Rubi [A]  time = 0.13, antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2199, 2177, 2178} \begin {gather*} -\frac {40 e^{5-x}}{x^2}-\frac {12 e^{5-x}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5 - x)*(80 + 52*x + 12*x^2))/x^3,x]

[Out]

(-40*E^(5 - x))/x^2 - (12*E^(5 - x))/x

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 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 (\frac {80 e^{5-x}}{x^3}+\frac {52 e^{5-x}}{x^2}+\frac {12 e^{5-x}}{x}\right ) \, dx\\ &=12 \int \frac {e^{5-x}}{x} \, dx+52 \int \frac {e^{5-x}}{x^2} \, dx+80 \int \frac {e^{5-x}}{x^3} \, dx\\ &=-\frac {40 e^{5-x}}{x^2}-\frac {52 e^{5-x}}{x}+12 e^5 \text {Ei}(-x)-40 \int \frac {e^{5-x}}{x^2} \, dx-52 \int \frac {e^{5-x}}{x} \, dx\\ &=-\frac {40 e^{5-x}}{x^2}-\frac {12 e^{5-x}}{x}-40 e^5 \text {Ei}(-x)+40 \int \frac {e^{5-x}}{x} \, dx\\ &=-\frac {40 e^{5-x}}{x^2}-\frac {12 e^{5-x}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 17, normalized size = 0.89 \begin {gather*} -\frac {4 e^{5-x} (10+3 x)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 - x)*(80 + 52*x + 12*x^2))/x^3,x]

[Out]

(-4*E^(5 - x)*(10 + 3*x))/x^2

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fricas [A]  time = 0.55, size = 16, normalized size = 0.84 \begin {gather*} -\frac {4 \, {\left (3 \, x + 10\right )} e^{\left (-x + 5\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2+52*x+80)*exp(5)/exp(x)/x^3,x, algorithm="fricas")

[Out]

-4*(3*x + 10)*e^(-x + 5)/x^2

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giac [A]  time = 0.21, size = 23, normalized size = 1.21 \begin {gather*} -\frac {4 \, {\left (3 \, x e^{\left (-x + 5\right )} + 10 \, e^{\left (-x + 5\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2+52*x+80)*exp(5)/exp(x)/x^3,x, algorithm="giac")

[Out]

-4*(3*x*e^(-x + 5) + 10*e^(-x + 5))/x^2

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maple [A]  time = 0.06, size = 17, normalized size = 0.89

method	result	size
gosper	\(-\frac {4 \,{\mathrm e}^{5} \left (3 x +10\right ) {\mathrm e}^{-x}}{x^{2}}\)	\(17\)
risch	\(-\frac {4 \left (3 x +10\right ) {\mathrm e}^{5-x}}{x^{2}}\)	\(17\)
norman	\(\frac {\left (-12 x \,{\mathrm e}^{5}-40 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{x^{2}}\)	\(19\)
default	\(4 \,{\mathrm e}^{5} \left (-\frac {10 \,{\mathrm e}^{-x}}{x^{2}}-\frac {3 \,{\mathrm e}^{-x}}{x}\right )\)	\(24\)
meijerg	\(-12 \,{\mathrm e}^{5} \expIntegralEi \left (1, x\right )+52 \,{\mathrm e}^{5} \left (-\frac {1}{x}+1+\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )\right )+80 \,{\mathrm e}^{5} \left (-\frac {1}{2 x^{2}}+\frac {1}{x}-\frac {3}{4}+\frac {9 x^{2}-12 x +6}{12 x^{2}}-\frac {\left (-3 x +3\right ) {\mathrm e}^{-x}}{6 x^{2}}-\frac {\expIntegralEi \left (1, x\right )}{2}\right )\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x^2+52*x+80)*exp(5)/exp(x)/x^3,x,method=_RETURNVERBOSE)

[Out]

-4*exp(5)*(3*x+10)/exp(x)/x^2

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maxima [C]  time = 0.41, size = 23, normalized size = 1.21 \begin {gather*} 12 \, {\rm Ei}\left (-x\right ) e^{5} - 52 \, e^{5} \Gamma \left (-1, x\right ) - 80 \, e^{5} \Gamma \left (-2, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x^2+52*x+80)*exp(5)/exp(x)/x^3,x, algorithm="maxima")

[Out]

12*Ei(-x)*e^5 - 52*e^5*gamma(-1, x) - 80*e^5*gamma(-2, x)

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mupad [B]  time = 5.38, size = 16, normalized size = 0.84 \begin {gather*} -\frac {4\,{\mathrm {e}}^{5-x}\,\left (3\,x+10\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*exp(5)*(52*x + 12*x^2 + 80))/x^3,x)

[Out]

-(4*exp(5 - x)*(3*x + 10))/x^2

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sympy [A]  time = 0.12, size = 19, normalized size = 1.00 \begin {gather*} \frac {\left (- 12 x e^{5} - 40 e^{5}\right ) e^{- x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x**2+52*x+80)*exp(5)/exp(x)/x**3,x)

[Out]

(-12*x*exp(5) - 40*exp(5))*exp(-x)/x**2

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