3.8.69 \(\int \frac {e^{-15-x} (e^{16+x}-2 e^5 x^2+2 x^3-x^4)}{x^2} \, dx\)

Optimal. Leaf size=26 \[ -\frac {e+x}{x}+e^{-10-x} \left (2+\frac {x^2}{e^5}\right ) \]

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Rubi [A]  time = 0.28, antiderivative size = 27, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6742, 2194, 2176} \begin {gather*} e^{-x-15} x^2+2 e^{-x-10}-\frac {e}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-15 - x)*(E^(16 + x) - 2*E^5*x^2 + 2*x^3 - x^4))/x^2,x]

[Out]

2*E^(-10 - x) - E/x + E^(-15 - x)*x^2

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 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2 e^{-10-x}+\frac {e}{x^2}+2 e^{-15-x} x-e^{-15-x} x^2\right ) \, dx\\ &=-\frac {e}{x}-2 \int e^{-10-x} \, dx+2 \int e^{-15-x} x \, dx-\int e^{-15-x} x^2 \, dx\\ &=2 e^{-10-x}-\frac {e}{x}-2 e^{-15-x} x+e^{-15-x} x^2+2 \int e^{-15-x} \, dx-2 \int e^{-15-x} x \, dx\\ &=-2 e^{-15-x}+2 e^{-10-x}-\frac {e}{x}+e^{-15-x} x^2-2 \int e^{-15-x} \, dx\\ &=2 e^{-10-x}-\frac {e}{x}+e^{-15-x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.04 \begin {gather*} 2 e^{-10-x}-\frac {e}{x}+e^{-15-x} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-15 - x)*(E^(16 + x) - 2*E^5*x^2 + 2*x^3 - x^4))/x^2,x]

[Out]

2*E^(-10 - x) - E/x + E^(-15 - x)*x^2

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fricas [A]  time = 0.91, size = 28, normalized size = 1.08 \begin {gather*} \frac {{\left (x^{3} e + 2 \, x e^{6} - e^{\left (x + 17\right )}\right )} e^{\left (-x - 16\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)*exp(5)*exp(x+10)-2*x^2*exp(5)-x^4+2*x^3)/x^2/exp(5)/exp(x+10),x, algorithm="fricas")

[Out]

(x^3*e + 2*x*e^6 - e^(x + 17))*e^(-x - 16)/x

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giac [A]  time = 0.38, size = 30, normalized size = 1.15 \begin {gather*} \frac {{\left (x^{3} e^{\left (-x + 10\right )} + 2 \, x e^{\left (-x + 15\right )} - e^{26}\right )} e^{\left (-25\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)*exp(5)*exp(x+10)-2*x^2*exp(5)-x^4+2*x^3)/x^2/exp(5)/exp(x+10),x, algorithm="giac")

[Out]

(x^3*e^(-x + 10) + 2*x*e^(-x + 15) - e^26)*e^(-25)/x

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maple [A]  time = 0.12, size = 24, normalized size = 0.92

method	result	size
risch	\(-\frac {{\mathrm e}}{x}+\left (2 \,{\mathrm e}^{5}+x^{2}\right ) {\mathrm e}^{-x -15}\)	\(24\)
norman	\(\frac {\left ({\mathrm e}^{-5} x^{3}+2 x -{\mathrm e} \,{\mathrm e}^{x +10}\right ) {\mathrm e}^{-x -10}}{x}\)	\(31\)
derivativedivides	\({\mathrm e}^{-5} \left (-\frac {{\mathrm e} \,{\mathrm e}^{5}}{x}+660 \,{\mathrm e}^{-x -10}-42 \left (x +31\right ) {\mathrm e}^{-x -10}+\left (\left (x +10\right )^{2}+22 x +542\right ) {\mathrm e}^{-x -10}-200 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{-x -10}}{x}+{\mathrm e}^{-10} \expIntegralEi \left (1, x\right )\right )+40 \,{\mathrm e}^{5} \left (-\frac {10 \,{\mathrm e}^{-x -10}}{x}+9 \,{\mathrm e}^{-10} \expIntegralEi \left (1, x\right )\right )-2 \,{\mathrm e}^{5} \left (-{\mathrm e}^{-x -10}-\frac {100 \,{\mathrm e}^{-x -10}}{x}+80 \,{\mathrm e}^{-10} \expIntegralEi \left (1, x\right )\right )\right )\)	\(128\)
default	\({\mathrm e}^{-5} \left (-\frac {{\mathrm e} \,{\mathrm e}^{5}}{x}+660 \,{\mathrm e}^{-x -10}-42 \left (x +31\right ) {\mathrm e}^{-x -10}+\left (\left (x +10\right )^{2}+22 x +542\right ) {\mathrm e}^{-x -10}-200 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{-x -10}}{x}+{\mathrm e}^{-10} \expIntegralEi \left (1, x\right )\right )+40 \,{\mathrm e}^{5} \left (-\frac {10 \,{\mathrm e}^{-x -10}}{x}+9 \,{\mathrm e}^{-10} \expIntegralEi \left (1, x\right )\right )-2 \,{\mathrm e}^{5} \left (-{\mathrm e}^{-x -10}-\frac {100 \,{\mathrm e}^{-x -10}}{x}+80 \,{\mathrm e}^{-10} \expIntegralEi \left (1, x\right )\right )\right )\)	\(128\)
meijerg	\({\mathrm e}^{-x -9+x \,{\mathrm e}^{-10}} \left (-{\mathrm e}^{10}+1\right ) \left (\frac {{\mathrm e}^{10} \left (2-2 x \,{\mathrm e}^{-10} \left (-{\mathrm e}^{10}+1\right )\right )}{2 x \left (-{\mathrm e}^{10}+1\right )}-\frac {{\mathrm e}^{10-x \,{\mathrm e}^{-10} \left (-{\mathrm e}^{10}+1\right )}}{x \left (-{\mathrm e}^{10}+1\right )}+\ln \left (x \,{\mathrm e}^{-10} \left (-{\mathrm e}^{10}+1\right )\right )+\expIntegralEi \left (1, x \,{\mathrm e}^{-10} \left (-{\mathrm e}^{10}+1\right )\right )+11-\ln \relax (x )-\ln \left (-{\mathrm e}^{10}+1\right )-\frac {{\mathrm e}^{10}}{x \left (-{\mathrm e}^{10}+1\right )}\right )-2 \,{\mathrm e}^{-x +x \,{\mathrm e}^{-10}} \left (1-{\mathrm e}^{-x \,{\mathrm e}^{-10}}\right )-{\mathrm e}^{15-x +x \,{\mathrm e}^{-10}} \left (2-\frac {\left (3 x^{2} {\mathrm e}^{-20}+6 x \,{\mathrm e}^{-10}+6\right ) {\mathrm e}^{-x \,{\mathrm e}^{-10}}}{3}\right )+2 \,{\mathrm e}^{-x +5+x \,{\mathrm e}^{-10}} \left (1-\frac {\left (2+2 x \,{\mathrm e}^{-10}\right ) {\mathrm e}^{-x \,{\mathrm e}^{-10}}}{2}\right )\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1)*exp(5)*exp(x+10)-2*x^2*exp(5)-x^4+2*x^3)/x^2/exp(5)/exp(x+10),x,method=_RETURNVERBOSE)

[Out]

-exp(1)/x+(2*exp(5)+x^2)*exp(-x-15)

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maxima [A]  time = 0.58, size = 42, normalized size = 1.62 \begin {gather*} {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x - 15\right )} - 2 \, {\left (x + 1\right )} e^{\left (-x - 15\right )} - \frac {e}{x} + 2 \, e^{\left (-x - 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)*exp(5)*exp(x+10)-2*x^2*exp(5)-x^4+2*x^3)/x^2/exp(5)/exp(x+10),x, algorithm="maxima")

[Out]

(x^2 + 2*x + 2)*e^(-x - 15) - 2*(x + 1)*e^(-x - 15) - e/x + 2*e^(-x - 10)

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mupad [B]  time = 0.10, size = 26, normalized size = 1.00 \begin {gather*} 2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-10}-\frac {\mathrm {e}}{x}+x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-5)*exp(- x - 10)*(exp(x + 10)*exp(6) - 2*x^2*exp(5) + 2*x^3 - x^4))/x^2,x)

[Out]

2*exp(-x)*exp(-10) - exp(1)/x + x^2*exp(-x)*exp(-15)

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sympy [A]  time = 0.12, size = 22, normalized size = 0.85 \begin {gather*} \frac {\left (x^{2} + 2 e^{5}\right ) e^{- x - 10}}{e^{5}} - \frac {e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)*exp(5)*exp(x+10)-2*x**2*exp(5)-x**4+2*x**3)/x**2/exp(5)/exp(x+10),x)

[Out]

(x**2 + 2*exp(5))*exp(-5)*exp(-x - 10) - E/x

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