3.6.20 \(\int \frac {-2+x-2 e^{10-2 x} x}{x} \, dx\)

Optimal. Leaf size=14 \[ 1+e^{10-2 x}+x-2 \log (x) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {14, 2194, 43} \begin {gather*} x+e^{10-2 x}-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + x - 2*E^(10 - 2*x)*x)/x,x]

[Out]

E^(10 - 2*x) + x - 2*Log[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} &=\int \left (-2 e^{10-2 x}+\frac {-2+x}{x}\right ) \, dx\\ &=-\left (2 \int e^{10-2 x} \, dx\right )+\int \frac {-2+x}{x} \, dx\\ &=e^{10-2 x}+\int \left (1-\frac {2}{x}\right ) \, dx\\ &=e^{10-2 x}+x-2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.93 \begin {gather*} e^{10-2 x}+x-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + x - 2*E^(10 - 2*x)*x)/x,x]

[Out]

E^(10 - 2*x) + x - 2*Log[x]

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fricas [A]  time = 0.66, size = 12, normalized size = 0.86 \begin {gather*} x + e^{\left (-2 \, x + 10\right )} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + e^(-2*x + 10) - 2*log(x)

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giac [A]  time = 0.27, size = 12, normalized size = 0.86 \begin {gather*} x + e^{\left (-2 \, x + 10\right )} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + e^(-2*x + 10) - 2*log(x)

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maple [A]  time = 0.09, size = 13, normalized size = 0.93

method	result	size
risch	\(x +{\mathrm e}^{-2 x +10}-2 \ln \relax (x )\)	\(13\)
norman	\(x +{\mathrm e}^{-2 x +10}-2 \ln \relax (x )\)	\(15\)
derivativedivides	\(3 \ln \relax (x )-5+x -5 \ln \left (-x \right )+{\mathrm e}^{-2 x +10}\)	\(22\)
default	\(3 \ln \relax (x )-5+x -5 \ln \left (-x \right )+{\mathrm e}^{-2 x +10}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+exp(-2*x+10)-2*ln(x)

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maxima [A]  time = 0.47, size = 12, normalized size = 0.86 \begin {gather*} x + e^{\left (-2 \, x + 10\right )} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + e^(-2*x + 10) - 2*log(x)

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mupad [B]  time = 0.05, size = 12, normalized size = 0.86 \begin {gather*} x+{\mathrm {e}}^{10-2\,x}-2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*exp(10 - 2*x) - x + 2)/x,x)

[Out]

x + exp(10 - 2*x) - 2*log(x)

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sympy [A]  time = 0.12, size = 12, normalized size = 0.86 \begin {gather*} x + e^{10 - 2 x} - 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(10 - 2*x) - 2*log(x)

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