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

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

________________________________________________________________________________________

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 {gather*} 2 \log (x)-\log \left (5-e^x\right ) \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {5}{-5+e^x}+\frac {2-x}{x}\right ) \, dx\\ &=-\left (5 \int \frac {1}{-5+e^x} \, dx\right )+\int \frac {2-x}{x} \, dx\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {1}{(-5+x) x} \, dx,x,e^x\right )\right )+\int \left (-1+\frac {2}{x}\right ) \, dx\\ &=-x+2 \log (x)-\operatorname {Subst}\left (\int \frac {1}{-5+x} \, dx,x,e^x\right )+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )\\ &=-\log \left (5-e^x\right )+2 \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 15, normalized size = 0.94 \begin {gather*} -\log \left (5-e^x\right )+2 \log (x) \end {gather*}

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]

________________________________________________________________________________________

fricas [A]  time = 0.88, size = 12, normalized size = 0.75 \begin {gather*} 2 \, \log \relax (x) - \log \left (e^{x} - 5\right ) \end {gather*}

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)

________________________________________________________________________________________

giac [A]  time = 0.26, size = 12, normalized size = 0.75 \begin {gather*} 2 \, \log \relax (x) - \log \left (e^{x} - 5\right ) \end {gather*}

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)

________________________________________________________________________________________

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

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

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)

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 12, normalized size = 0.75 \begin {gather*} 2 \, \log \relax (x) - \log \left (e^{x} - 5\right ) \end {gather*}

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)

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 12, normalized size = 0.75 \begin {gather*} 2\,\ln \relax (x)-\ln \left ({\mathrm {e}}^x-5\right ) \end {gather*}

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)

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 10, normalized size = 0.62 \begin {gather*} 2 \log {\relax (x )} - \log {\left (e^{x} - 5 \right )} \end {gather*}

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)

________________________________________________________________________________________