3.69.87 $\int \frac{-10+e^x(2-x)}{-5x+e^{x}x}dx$

Optimal. Leaf size=16 $$\log \left(\frac{4e^4{x}^2}{-5+e^x}\right)$$

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Rubi [A]  time = 0.14, antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 6, integrand size = 23, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.261, Rules used = {6742, 2282, 36, 31, 29, 43} $$\begin{array}{c}2\log (x)-\log (5-e^x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-10 + E^x*(2 - x))/(-5*x + E^x*x),x]

[Out]

-Log[5 - E^x] + 2*Log[x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{5}{-5+e^x}+\frac{2-x}{x})dx\\ & =-(5\int \frac{1}{-5+e^x}dx)+\int \frac{2-x}{x}dx\\ & =-\left(5\mathrm{Subst}(\int \frac{1}{(-5+x)x}dx,x,e^x)\right)+\int (-1+\frac{2}{x})dx\\ & =-x+2\log (x)-\mathrm{Subst}(\int \frac{1}{-5+x}dx,x,e^x)+\mathrm{Subst}(\int \frac{1}{x}dx,x,e^x)\\ & =-\log (5-e^x)+2\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 15, normalized size = 0.94 $$\begin{array}{c}-\log (5-e^x)+2\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-10 + E^x*(2 - x))/(-5*x + E^x*x),x]

[Out]

-Log[5 - E^x] + 2*Log[x]

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fricas [A]  time = 0.88, size = 12, normalized size = 0.75 $$\begin{array}{c}2\log (x)-\log (e^x-5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2-x)*exp(x)-10)/(exp(x)*x-5*x),x, algorithm="fricas")

[Out]

2*log(x) - log(e^x - 5)

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giac [A]  time = 0.26, size = 12, normalized size = 0.75 $$\begin{array}{c}2\log (x)-\log (e^x-5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2-x)*exp(x)-10)/(exp(x)*x-5*x),x, algorithm="giac")

[Out]

2*log(x) - log(e^x - 5)

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maple [A]  time = 0.03, size = 13, normalized size = 0.81

method	result	size
norman	$2\ln (x)-\ln ({\mathrm{e}}^x-5)$	$13$
risch	$2\ln (x)-\ln ({\mathrm{e}}^x-5)$	$13$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2-x)*exp(x)-10)/(exp(x)*x-5*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)-ln(exp(x)-5)

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maxima [A]  time = 0.38, size = 12, normalized size = 0.75 $$\begin{array}{c}2\log (x)-\log (e^x-5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2-x)*exp(x)-10)/(exp(x)*x-5*x),x, algorithm="maxima")

[Out]

2*log(x) - log(e^x - 5)

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mupad [B]  time = 0.06, size = 12, normalized size = 0.75 $$\begin{array}{c}2\ln (x)-\ln ({\mathrm{e}}^x-5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x - 2) + 10)/(5*x - x*exp(x)),x)

[Out]

2*log(x) - log(exp(x) - 5)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.62 $$\begin{array}{c}2\log (x)-\log (e^x-5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2-x)*exp(x)-10)/(exp(x)*x-5*x),x)

[Out]

2*log(x) - log(exp(x) - 5)

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