3.85.61 \(\int \frac {e^x (25-25 x)}{16 x^2} \, dx\)

Optimal. Leaf size=13 \[ -\frac {25}{16} \left (-1+\frac {e^x}{x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 10, normalized size of antiderivative = 0.77, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 2197} \begin {gather*} -\frac {25 e^x}{16 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(25 - 25*x))/(16*x^2),x]

[Out]

(-25*E^x)/(16*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 {1}{16} \int \frac {e^x (25-25 x)}{x^2} \, dx\\ &=-\frac {25 e^x}{16 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 0.77 \begin {gather*} -\frac {25 e^x}{16 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(25 - 25*x))/(16*x^2),x]

[Out]

(-25*E^x)/(16*x)

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fricas [A]  time = 0.49, size = 7, normalized size = 0.54 \begin {gather*} -\frac {25 \, e^{x}}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-25*x+25)*exp(x)/x^2,x, algorithm="fricas")

[Out]

-25/16*e^x/x

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giac [A]  time = 0.15, size = 7, normalized size = 0.54 \begin {gather*} -\frac {25 \, e^{x}}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-25*x+25)*exp(x)/x^2,x, algorithm="giac")

[Out]

-25/16*e^x/x

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maple [A]  time = 0.06, size = 8, normalized size = 0.62

method	result	size
gosper	\(-\frac {25 \,{\mathrm e}^{x}}{16 x}\)	\(8\)
default	\(-\frac {25 \,{\mathrm e}^{x}}{16 x}\)	\(8\)
norman	\(-\frac {25 \,{\mathrm e}^{x}}{16 x}\)	\(8\)
risch	\(-\frac {25 \,{\mathrm e}^{x}}{16 x}\)	\(8\)
meijerg	\(-\frac {25}{16 x}-\frac {25}{16}+\frac {\frac {25 x}{16}+\frac {25}{16}}{x}-\frac {25 \,{\mathrm e}^{x}}{16 x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(-25*x+25)*exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-25/16*exp(x)/x

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maxima [C]  time = 0.37, size = 12, normalized size = 0.92 \begin {gather*} -\frac {25}{16} \, {\rm Ei}\relax (x) + \frac {25}{16} \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-25*x+25)*exp(x)/x^2,x, algorithm="maxima")

[Out]

-25/16*Ei(x) + 25/16*gamma(-1, -x)

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mupad [B]  time = 5.23, size = 7, normalized size = 0.54 \begin {gather*} -\frac {25\,{\mathrm {e}}^x}{16\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(25*x - 25))/(16*x^2),x)

[Out]

-(25*exp(x))/(16*x)

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sympy [A]  time = 0.08, size = 8, normalized size = 0.62 \begin {gather*} - \frac {25 e^{x}}{16 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-25*x+25)*exp(x)/x**2,x)

[Out]

-25*exp(x)/(16*x)

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