3.22.33 \(\int \frac {4 e^{-2+i \pi +2 x} (2-2 x)-x}{x^3} \, dx\)

Optimal. Leaf size=20 \[ \frac {-4 e^{-2+i \pi +2 x}+x}{x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {14, 2197} \begin {gather*} \frac {4 e^{2 x-2}}{x^2}+\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*E^(-2 + 2*x))/x^2 + x^(-1)

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

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Mathematica [A]  time = 0.03, size = 16, normalized size = 0.80 \begin {gather*} \frac {4 e^{-2+2 x}}{x^2}+\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*E^(-2 + 2*x))/x^2 + x^(-1)

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fricas [A]  time = 1.17, size = 16, normalized size = 0.80 \begin {gather*} \frac {x + e^{\left (2 \, x + 2 \, \log \relax (2) - 2\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + e^(2*x + 2*log(2) - 2))/x^2

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giac [A]  time = 0.18, size = 17, normalized size = 0.85 \begin {gather*} \frac {{\left (x e^{2} + 4 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*e^2 + 4*e^(2*x))*e^(-2)/x^2

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maple [A]  time = 0.12, size = 16, normalized size = 0.80

method	result	size
risch	\(\frac {1}{x}+\frac {4 \,{\mathrm e}^{2 x -2}}{x^{2}}\)	\(16\)
norman	\(\frac {x +{\mathrm e}^{2 \ln \relax (2)-2+2 x}}{x^{2}}\)	\(17\)
derivativedivides	\(\frac {1}{x}+\frac {{\mathrm e}^{2 \ln \relax (2)-2+2 x}}{x^{2}}\)	\(19\)
default	\(\frac {1}{x}+\frac {{\mathrm e}^{2 \ln \relax (2)-2+2 x}}{x^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x+4*exp(2*x-2)/x^2

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maxima [C]  time = 0.45, size = 22, normalized size = 1.10 \begin {gather*} 16 \, e^{\left (-2\right )} \Gamma \left (-1, -2 \, x\right ) + 32 \, e^{\left (-2\right )} \Gamma \left (-2, -2 \, x\right ) + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*e^(-2)*gamma(-1, -2*x) + 32*e^(-2)*gamma(-2, -2*x) + 1/x

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mupad [B]  time = 0.06, size = 14, normalized size = 0.70 \begin {gather*} \frac {x+4\,{\mathrm {e}}^{2\,x-2}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + 4*exp(2*x - 2))/x^2

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sympy [A]  time = 0.09, size = 14, normalized size = 0.70 \begin {gather*} \frac {1}{x} + \frac {4 e^{2 x - 2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x + 4*exp(2*x - 2)/x**2

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