3.50.53 \(\int \frac {2-e^{25}+e^x (-1+x)}{5 x^2} \, dx\)

Optimal. Leaf size=18 \[ \frac {-2+e^{25}+e^x-x}{5 x} \]

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Rubi [A]  time = 0.04, antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 2197} \begin {gather*} \frac {e^x}{5 x}-\frac {2-e^{25}}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - E^25 + E^x*(-1 + x))/(5*x^2),x]

[Out]

E^x/(5*x) - (2 - E^25)/(5*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 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} &=\frac {1}{5} \int \frac {2-e^{25}+e^x (-1+x)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {2-e^{25}}{x^2}+\frac {e^x (-1+x)}{x^2}\right ) \, dx\\ &=-\frac {2-e^{25}}{5 x}+\frac {1}{5} \int \frac {e^x (-1+x)}{x^2} \, dx\\ &=\frac {e^x}{5 x}-\frac {2-e^{25}}{5 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.83 \begin {gather*} \frac {-2+e^{25}+e^x}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - E^25 + E^x*(-1 + x))/(5*x^2),x]

[Out]

(-2 + E^25 + E^x)/(5*x)

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fricas [A]  time = 0.81, size = 11, normalized size = 0.61 \begin {gather*} \frac {e^{25} + e^{x} - 2}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x-1)*exp(x)-exp(25)+2)/x^2,x, algorithm="fricas")

[Out]

1/5*(e^25 + e^x - 2)/x

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giac [A]  time = 0.14, size = 11, normalized size = 0.61 \begin {gather*} \frac {e^{25} + e^{x} - 2}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x-1)*exp(x)-exp(25)+2)/x^2,x, algorithm="giac")

[Out]

1/5*(e^25 + e^x - 2)/x

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maple [A]  time = 0.03, size = 15, normalized size = 0.83

method	result	size
norman	\(\frac {\frac {{\mathrm e}^{x}}{5}+\frac {{\mathrm e}^{25}}{5}-\frac {2}{5}}{x}\)	\(15\)
default	\(-\frac {2}{5 x}+\frac {{\mathrm e}^{25}}{5 x}+\frac {{\mathrm e}^{x}}{5 x}\)	\(21\)
risch	\(-\frac {2}{5 x}+\frac {{\mathrm e}^{25}}{5 x}+\frac {{\mathrm e}^{x}}{5 x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((x-1)*exp(x)-exp(25)+2)/x^2,x,method=_RETURNVERBOSE)

[Out]

(1/5*exp(x)+1/5*exp(25)-2/5)/x

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maxima [C]  time = 0.37, size = 24, normalized size = 1.33 \begin {gather*} \frac {e^{25}}{5 \, x} - \frac {2}{5 \, x} + \frac {1}{5} \, {\rm Ei}\relax (x) - \frac {1}{5} \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x-1)*exp(x)-exp(25)+2)/x^2,x, algorithm="maxima")

[Out]

1/5*e^25/x - 2/5/x + 1/5*Ei(x) - 1/5*gamma(-1, -x)

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mupad [B]  time = 3.22, size = 11, normalized size = 0.61 \begin {gather*} \frac {{\mathrm {e}}^{25}+{\mathrm {e}}^x-2}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*(x - 1))/5 - exp(25)/5 + 2/5)/x^2,x)

[Out]

(exp(25) + exp(x) - 2)/(5*x)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.83 \begin {gather*} \frac {e^{x}}{5 x} - \frac {\frac {2}{5} - \frac {e^{25}}{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((x-1)*exp(x)-exp(25)+2)/x**2,x)

[Out]

exp(x)/(5*x) - (2/5 - exp(25)/5)/x

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