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

Optimal. Leaf size=17 \[ 10-\frac {e^x}{5 x}+x+\log \left (\frac {14}{3}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 12, normalized size of antiderivative = 0.71, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 14, 2197} \begin {gather*} x-\frac {e^x}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(1 - x) + 5*x^2)/(5*x^2),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 12, normalized size = 0.71 \begin {gather*} -\frac {e^x}{5 x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*E^x/x + x

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fricas [A]  time = 0.57, size = 15, normalized size = 0.88 \begin {gather*} \frac {5 \, x^{2} - e^{x}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*x^2 - e^x)/x

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giac [A]  time = 0.12, size = 15, normalized size = 0.88 \begin {gather*} \frac {5 \, x^{2} - e^{x}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*x^2 - e^x)/x

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maple [A]  time = 0.07, size = 10, normalized size = 0.59

method	result	size
default	\(x -\frac {{\mathrm e}^{x}}{5 x}\)	\(10\)
risch	\(x -\frac {{\mathrm e}^{x}}{5 x}\)	\(10\)
norman	\(\frac {x^{2}-\frac {{\mathrm e}^{x}}{5}}{x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x-1/5*exp(x)/x

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maxima [C]  time = 0.38, size = 13, normalized size = 0.76 \begin {gather*} 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)+5*x^2)/x^2,x, algorithm="maxima")

[Out]

x - 1/5*Ei(x) + 1/5*gamma(-1, -x)

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mupad [B]  time = 0.05, size = 9, normalized size = 0.53 \begin {gather*} x-\frac {{\mathrm {e}}^x}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - exp(x)/(5*x)

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sympy [A]  time = 0.08, size = 7, normalized size = 0.41 \begin {gather*} x - \frac {e^{x}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - exp(x)/(5*x)

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