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

Optimal. Leaf size=26 \[ -1-e^x+\frac {4}{5} \left (\frac {e^3}{3}-x\right )+\log (2 x) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.50, 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*} -\frac {4 x}{5}-e^x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 4*x - 5*E^x*x)/(5*x),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.50 \begin {gather*} -e^x-\frac {4 x}{5}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^x - (4*x)/5 + Log[x]

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fricas [A]  time = 0.51, size = 10, normalized size = 0.38 \begin {gather*} -\frac {4}{5} \, x - e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5*x - e^x + log(x)

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giac [A]  time = 0.13, size = 10, normalized size = 0.38 \begin {gather*} -\frac {4}{5} \, x - e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5*x - e^x + log(x)

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maple [A]  time = 0.02, size = 11, normalized size = 0.42

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 10, normalized size = 0.38 \begin {gather*} -\frac {4}{5} \, x - e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/5*x - e^x + log(x)

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mupad [B]  time = 0.21, size = 10, normalized size = 0.38 \begin {gather*} \ln \relax (x)-{\mathrm {e}}^x-\frac {4\,x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((4*x)/5 + x*exp(x) - 1)/x,x)

[Out]

log(x) - exp(x) - (4*x)/5

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sympy [A]  time = 0.09, size = 10, normalized size = 0.38 \begin {gather*} - \frac {4 x}{5} - e^{x} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x/5 - exp(x) + log(x)

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