3.38.67 \(\int \frac {e^{-x} (4 e^x (-2-56 x)+3 x+3 x^2)}{8 x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac {28-\frac {-1+\frac {3 e^{-x} x}{4}}{2 x}}{x} \]

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Rubi [A]  time = 0.19, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {12, 6742, 2197, 37} \begin {gather*} \frac {(28 x+1)^2}{2 x^2}-\frac {3 e^{-x}}{8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^x*(-2 - 56*x) + 3*x + 3*x^2)/(8*E^x*x^3),x]

[Out]

-3/(8*E^x*x) + (1 + 28*x)^2/(2*x^2)

Rule 12

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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
]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \frac {e^{-x} \left (4 e^x (-2-56 x)+3 x+3 x^2\right )}{x^3} \, dx\\ &=\frac {1}{8} \int \left (\frac {3 e^{-x} (1+x)}{x^2}-\frac {8 (1+28 x)}{x^3}\right ) \, dx\\ &=\frac {3}{8} \int \frac {e^{-x} (1+x)}{x^2} \, dx-\int \frac {1+28 x}{x^3} \, dx\\ &=-\frac {3 e^{-x}}{8 x}+\frac {(1+28 x)^2}{2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 20, normalized size = 0.80 \begin {gather*} \frac {4+224 x-3 e^{-x} x}{8 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^x*(-2 - 56*x) + 3*x + 3*x^2)/(8*E^x*x^3),x]

[Out]

(4 + 224*x - (3*x)/E^x)/(8*x^2)

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fricas [A]  time = 0.72, size = 31, normalized size = 1.24 \begin {gather*} \frac {{\left ({\left (56 \, x + 1\right )} e^{\left (x + 2 \, \log \relax (2)\right )} - 3 \, x\right )} e^{\left (-x - 2 \, \log \relax (2)\right )}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-56*x-2)*exp(x+2*log(2))+3*x^2+3*x)/x^3/exp(x+2*log(2)),x, algorithm="fricas")

[Out]

1/2*((56*x + 1)*e^(x + 2*log(2)) - 3*x)*e^(-x - 2*log(2))/x^2

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giac [A]  time = 0.19, size = 17, normalized size = 0.68 \begin {gather*} -\frac {3 \, x e^{\left (-x\right )} - 224 \, x - 4}{8 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-56*x-2)*exp(x+2*log(2))+3*x^2+3*x)/x^3/exp(x+2*log(2)),x, algorithm="giac")

[Out]

-1/8*(3*x*e^(-x) - 224*x - 4)/x^2

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maple [A]  time = 0.09, size = 21, normalized size = 0.84

method	result	size
risch	\(\frac {56 x +1}{2 x^{2}}-\frac {3 \,{\mathrm e}^{-x}}{8 x}\)	\(21\)
norman	\(\frac {\left (-\frac {3 x}{2}+28 \,{\mathrm e}^{x +2 \ln \relax (2)} x +\frac {{\mathrm e}^{x +2 \ln \relax (2)}}{2}\right ) {\mathrm e}^{-x}}{4 x^{2}}\)	\(37\)
derivativedivides	\(\frac {1}{2 x^{2}}+\frac {28}{x}-\frac {3 \,{\mathrm e}^{-x -2 \ln \relax (2)} \left (2 \ln \relax (2)^{2}-\left (x +2 \ln \relax (2)\right ) \ln \relax (2)+\ln \relax (2)+x \right )}{2 x^{2}}-\frac {3 \,{\mathrm e}^{-x -2 \ln \relax (2)} \ln \relax (2) \left (2 \ln \relax (2)^{2}-\left (x +2 \ln \relax (2)\right ) \ln \relax (2)+\ln \relax (2)+2 x \right )}{x^{2}}+\frac {9 \ln \relax (2) {\mathrm e}^{-x -2 \ln \relax (2)} \left (x +2 \ln \relax (2)\right )}{2 x^{2}}+\frac {3 \,{\mathrm e}^{-x -2 \ln \relax (2)} \ln \relax (2)}{2 x^{2}}-\frac {3 \ln \relax (2)^{2} {\mathrm e}^{-x -2 \ln \relax (2)} \left (x +2 \ln \relax (2)\right )}{x^{2}}+\frac {6 \ln \relax (2)^{3} {\mathrm e}^{-x -2 \ln \relax (2)}}{x^{2}}-\frac {6 \,{\mathrm e}^{-x -2 \ln \relax (2)} \ln \relax (2)^{2}}{x^{2}}\)	\(182\)
default	\(\frac {1}{2 x^{2}}+\frac {28}{x}-\frac {3 \,{\mathrm e}^{-x -2 \ln \relax (2)} \left (2 \ln \relax (2)^{2}-\left (x +2 \ln \relax (2)\right ) \ln \relax (2)+\ln \relax (2)+x \right )}{2 x^{2}}-\frac {3 \,{\mathrm e}^{-x -2 \ln \relax (2)} \ln \relax (2) \left (2 \ln \relax (2)^{2}-\left (x +2 \ln \relax (2)\right ) \ln \relax (2)+\ln \relax (2)+2 x \right )}{x^{2}}+\frac {9 \ln \relax (2) {\mathrm e}^{-x -2 \ln \relax (2)} \left (x +2 \ln \relax (2)\right )}{2 x^{2}}+\frac {3 \,{\mathrm e}^{-x -2 \ln \relax (2)} \ln \relax (2)}{2 x^{2}}-\frac {3 \ln \relax (2)^{2} {\mathrm e}^{-x -2 \ln \relax (2)} \left (x +2 \ln \relax (2)\right )}{x^{2}}+\frac {6 \ln \relax (2)^{3} {\mathrm e}^{-x -2 \ln \relax (2)}}{x^{2}}-\frac {6 \,{\mathrm e}^{-x -2 \ln \relax (2)} \ln \relax (2)^{2}}{x^{2}}\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-56*x-2)*exp(x+2*ln(2))+3*x^2+3*x)/x^3/exp(x+2*ln(2)),x,method=_RETURNVERBOSE)

[Out]

1/2*(56*x+1)/x^2-3/8*exp(-x)/x

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maxima [C]  time = 0.51, size = 22, normalized size = 0.88 \begin {gather*} \frac {28}{x} + \frac {1}{2 \, x^{2}} + \frac {3}{8} \, {\rm Ei}\left (-x\right ) - \frac {3}{8} \, \Gamma \left (-1, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-56*x-2)*exp(x+2*log(2))+3*x^2+3*x)/x^3/exp(x+2*log(2)),x, algorithm="maxima")

[Out]

28/x + 1/2/x^2 + 3/8*Ei(-x) - 3/8*gamma(-1, x)

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mupad [B]  time = 0.08, size = 17, normalized size = 0.68 \begin {gather*} -\frac {x\,\left (\frac {3\,{\mathrm {e}}^{-x}}{8}-28\right )-\frac {1}{2}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(- x - 2*log(2))*((3*x)/2 - (exp(x + 2*log(2))*(56*x + 2))/2 + (3*x^2)/2))/x^3,x)

[Out]

-(x*((3*exp(-x))/8 - 28) - 1/2)/x^2

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sympy [A]  time = 0.12, size = 20, normalized size = 0.80 \begin {gather*} - \frac {3 e^{- x}}{8 x} - \frac {- 56 x - 1}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-56*x-2)*exp(x+2*ln(2))+3*x**2+3*x)/x**3/exp(x+2*ln(2)),x)

[Out]

-3*exp(-x)/(8*x) - (-56*x - 1)/(2*x**2)

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