3.63.19 \(\int \frac {e^{-12+x} (-25+25 x)}{x^2} \, dx\)

Optimal. Leaf size=15 \[ \frac {25 \left (e^x-x\right )}{e^{12} x} \]

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Rubi [A]  time = 0.03, antiderivative size = 10, normalized size of antiderivative = 0.67, number of steps used = 1, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2197} \begin {gather*} \frac {25 e^{x-12}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-12 + x)*(-25 + 25*x))/x^2,x]

[Out]

(25*E^(-12 + x))/x

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} &=\frac {25 e^{-12+x}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 0.67 \begin {gather*} \frac {25 e^{-12+x}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-12 + x)*(-25 + 25*x))/x^2,x]

[Out]

(25*E^(-12 + x))/x

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fricas [A]  time = 0.51, size = 9, normalized size = 0.60 \begin {gather*} \frac {25 \, e^{\left (x - 12\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x-25)*exp(x)/x^2/exp(6)^2,x, algorithm="fricas")

[Out]

25*e^(x - 12)/x

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giac [A]  time = 0.20, size = 9, normalized size = 0.60 \begin {gather*} \frac {25 \, e^{\left (x - 12\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x-25)*exp(x)/x^2/exp(6)^2,x, algorithm="giac")

[Out]

25*e^(x - 12)/x

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maple [A]  time = 0.11, size = 10, normalized size = 0.67

method	result	size
risch	\(\frac {25 \,{\mathrm e}^{x -12}}{x}\)	\(10\)
gosper	\(\frac {25 \,{\mathrm e}^{x} {\mathrm e}^{-12}}{x}\)	\(12\)
default	\(\frac {25 \,{\mathrm e}^{x} {\mathrm e}^{-12}}{x}\)	\(12\)
norman	\(\frac {25 \,{\mathrm e}^{x} {\mathrm e}^{-12}}{x}\)	\(12\)
meijerg	\(25 \,{\mathrm e}^{-12} \left (-\ln \left (-x \right )-\expIntegralEi \left (1, -x \right )+\ln \relax (x )+i \pi \right )+25 \,{\mathrm e}^{-12} \left (-\frac {2 x +2}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\expIntegralEi \left (1, -x \right )+1-\ln \relax (x )-i \pi +\frac {1}{x}\right )\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x-25)*exp(x)/x^2/exp(6)^2,x,method=_RETURNVERBOSE)

[Out]

25/x*exp(x-12)

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maxima [C]  time = 0.38, size = 16, normalized size = 1.07 \begin {gather*} 25 \, {\rm Ei}\relax (x) e^{\left (-12\right )} - 25 \, e^{\left (-12\right )} \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x-25)*exp(x)/x^2/exp(6)^2,x, algorithm="maxima")

[Out]

25*Ei(x)*e^(-12) - 25*e^(-12)*gamma(-1, -x)

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mupad [B]  time = 4.17, size = 9, normalized size = 0.60 \begin {gather*} \frac {25\,{\mathrm {e}}^{-12}\,{\mathrm {e}}^x}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-12)*exp(x)*(25*x - 25))/x^2,x)

[Out]

(25*exp(-12)*exp(x))/x

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sympy [A]  time = 0.10, size = 8, normalized size = 0.53 \begin {gather*} \frac {25 e^{x}}{x e^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*x-25)*exp(x)/x**2/exp(6)**2,x)

[Out]

25*exp(-12)*exp(x)/x

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