3.45.93 \(\int \frac {5-5 x-4 e^x x}{5 x} \, dx\)

Optimal. Leaf size=18 \[ 2-\frac {4 e^x}{5}-x+\log \left (\frac {x}{9}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 14, 2194, 43} \begin {gather*} -x-\frac {4 e^x}{5}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 5*x - 4*E^x*x)/(5*x),x]

[Out]

(-4*E^x)/5 - x + Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {5-5 x-4 e^x x}{x} \, dx\\ &=\frac {1}{5} \int \left (-4 e^x-\frac {5 (-1+x)}{x}\right ) \, dx\\ &=-\frac {4 \int e^x \, dx}{5}-\int \frac {-1+x}{x} \, dx\\ &=-\frac {4 e^x}{5}-\int \left (1-\frac {1}{x}\right ) \, dx\\ &=-\frac {4 e^x}{5}-x+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.72 \begin {gather*} -\frac {4 e^x}{5}-x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 5*x - 4*E^x*x)/(5*x),x]

[Out]

(-4*E^x)/5 - x + Log[x]

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fricas [A]  time = 0.46, size = 10, normalized size = 0.56 \begin {gather*} -x - \frac {4}{5} \, e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*exp(x)*x-5*x+5)/x,x, algorithm="fricas")

[Out]

-x - 4/5*e^x + log(x)

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giac [A]  time = 0.19, size = 10, normalized size = 0.56 \begin {gather*} -x - \frac {4}{5} \, e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*exp(x)*x-5*x+5)/x,x, algorithm="giac")

[Out]

-x - 4/5*e^x + log(x)

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maple [A]  time = 0.02, size = 11, normalized size = 0.61

method	result	size
default	\(\ln \relax (x )-x -\frac {4 \,{\mathrm e}^{x}}{5}\)	\(11\)
norman	\(\ln \relax (x )-x -\frac {4 \,{\mathrm e}^{x}}{5}\)	\(11\)
risch	\(\ln \relax (x )-x -\frac {4 \,{\mathrm e}^{x}}{5}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-4*exp(x)*x-5*x+5)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)-x-4/5*exp(x)

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maxima [A]  time = 0.37, size = 10, normalized size = 0.56 \begin {gather*} -x - \frac {4}{5} \, e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*exp(x)*x-5*x+5)/x,x, algorithm="maxima")

[Out]

-x - 4/5*e^x + log(x)

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mupad [B]  time = 3.33, size = 10, normalized size = 0.56 \begin {gather*} \ln \relax (x)-\frac {4\,{\mathrm {e}}^x}{5}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + (4*x*exp(x))/5 - 1)/x,x)

[Out]

log(x) - (4*exp(x))/5 - x

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sympy [A]  time = 0.09, size = 10, normalized size = 0.56 \begin {gather*} - x - \frac {4 e^{x}}{5} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-4*exp(x)*x-5*x+5)/x,x)

[Out]

-x - 4*exp(x)/5 + log(x)

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