3.72.73 \(\int \frac {e^{-5-x} (-1024-1024 x-21 x^2-15 x^3)}{4 x^2} \, dx\)

Optimal. Leaf size=20 \[ e^{-5-x} \left (9+\frac {256}{x}+\frac {15 x}{4}\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 34, normalized size of antiderivative = 1.70, number of steps used = 9, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {12, 2199, 2194, 2177, 2178, 2176} \begin {gather*} \frac {15}{4} e^{-x-5} x+9 e^{-x-5}+\frac {256 e^{-x-5}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-5 - x)*(-1024 - 1024*x - 21*x^2 - 15*x^3))/(4*x^2),x]

[Out]

9*E^(-5 - x) + (256*E^(-5 - x))/x + (15*E^(-5 - x)*x)/4

Rule 12

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

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} &=\frac {1}{4} \int \frac {e^{-5-x} \left (-1024-1024 x-21 x^2-15 x^3\right )}{x^2} \, dx\\ &=\frac {1}{4} \int \left (-21 e^{-5-x}-\frac {1024 e^{-5-x}}{x^2}-\frac {1024 e^{-5-x}}{x}-15 e^{-5-x} x\right ) \, dx\\ &=-\left (\frac {15}{4} \int e^{-5-x} x \, dx\right )-\frac {21}{4} \int e^{-5-x} \, dx-256 \int \frac {e^{-5-x}}{x^2} \, dx-256 \int \frac {e^{-5-x}}{x} \, dx\\ &=\frac {21 e^{-5-x}}{4}+\frac {256 e^{-5-x}}{x}+\frac {15}{4} e^{-5-x} x-\frac {256 \text {Ei}(-x)}{e^5}-\frac {15}{4} \int e^{-5-x} \, dx+256 \int \frac {e^{-5-x}}{x} \, dx\\ &=9 e^{-5-x}+\frac {256 e^{-5-x}}{x}+\frac {15}{4} e^{-5-x} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 24, normalized size = 1.20 \begin {gather*} \frac {e^{-5-x} \left (1024+36 x+15 x^2\right )}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-5 - x)*(-1024 - 1024*x - 21*x^2 - 15*x^3))/(4*x^2),x]

[Out]

(E^(-5 - x)*(1024 + 36*x + 15*x^2))/(4*x)

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fricas [A]  time = 0.75, size = 21, normalized size = 1.05 \begin {gather*} \frac {{\left (15 \, x^{2} + 36 \, x + 1024\right )} e^{\left (-x - 5\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-15*x^3-21*x^2-1024*x-1024)/x^2/exp(5)/exp(x),x, algorithm="fricas")

[Out]

1/4*(15*x^2 + 36*x + 1024)*e^(-x - 5)/x

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giac [A]  time = 0.20, size = 30, normalized size = 1.50 \begin {gather*} \frac {{\left (15 \, x^{2} e^{\left (-x\right )} + 36 \, x e^{\left (-x\right )} + 1024 \, e^{\left (-x\right )}\right )} e^{\left (-5\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-15*x^3-21*x^2-1024*x-1024)/x^2/exp(5)/exp(x),x, algorithm="giac")

[Out]

1/4*(15*x^2*e^(-x) + 36*x*e^(-x) + 1024*e^(-x))*e^(-5)/x

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maple [A]  time = 0.05, size = 22, normalized size = 1.10

method	result	size
risch	\(\frac {\left (15 x^{2}+36 x +1024\right ) {\mathrm e}^{-x -5}}{4 x}\)	\(22\)
gosper	\(\frac {\left (15 x^{2}+36 x +1024\right ) {\mathrm e}^{-x} {\mathrm e}^{-5}}{4 x}\)	\(24\)
default	\(\frac {{\mathrm e}^{-5} \left (36 \,{\mathrm e}^{-x}+15 x \,{\mathrm e}^{-x}+\frac {1024 \,{\mathrm e}^{-x}}{x}\right )}{4}\)	\(30\)
norman	\(\frac {\left (256 \,{\mathrm e}^{-5}+9 x \,{\mathrm e}^{-5}+\frac {15 x^{2} {\mathrm e}^{-5}}{4}\right ) {\mathrm e}^{-x}}{x}\)	\(32\)
meijerg	\(-256 \,{\mathrm e}^{-5} \left (-\frac {1}{x}+1+\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )\right )-\frac {15 \,{\mathrm e}^{-5} \left (1-\frac {\left (2 x +2\right ) {\mathrm e}^{-x}}{2}\right )}{4}-\frac {21 \,{\mathrm e}^{-5} \left (1-{\mathrm e}^{-x}\right )}{4}+256 \,{\mathrm e}^{-5} \expIntegralEi \left (1, x\right )\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-15*x^3-21*x^2-1024*x-1024)/x^2/exp(5)/exp(x),x,method=_RETURNVERBOSE)

[Out]

1/4*(15*x^2+36*x+1024)/x*exp(-x-5)

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maxima [C]  time = 0.38, size = 35, normalized size = 1.75 \begin {gather*} -256 \, {\rm Ei}\left (-x\right ) e^{\left (-5\right )} + \frac {15}{4} \, {\left (x + 1\right )} e^{\left (-x - 5\right )} + 256 \, e^{\left (-5\right )} \Gamma \left (-1, x\right ) + \frac {21}{4} \, e^{\left (-x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-15*x^3-21*x^2-1024*x-1024)/x^2/exp(5)/exp(x),x, algorithm="maxima")

[Out]

-256*Ei(-x)*e^(-5) + 15/4*(x + 1)*e^(-x - 5) + 256*e^(-5)*gamma(-1, x) + 21/4*e^(-x - 5)

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mupad [B]  time = 0.07, size = 21, normalized size = 1.05 \begin {gather*} \frac {{\mathrm {e}}^{-x-5}\,\left (15\,x^2+36\,x+1024\right )}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*exp(-5)*(256*x + (21*x^2)/4 + (15*x^3)/4 + 256))/x^2,x)

[Out]

(exp(- x - 5)*(36*x + 15*x^2 + 1024))/(4*x)

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sympy [A]  time = 0.13, size = 19, normalized size = 0.95 \begin {gather*} \frac {\left (15 x^{2} + 36 x + 1024\right ) e^{- x}}{4 x e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-15*x**3-21*x**2-1024*x-1024)/x**2/exp(5)/exp(x),x)

[Out]

(15*x**2 + 36*x + 1024)*exp(-5)*exp(-x)/(4*x)

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