3.3.29 \(\int -\frac {5}{5+\frac {e^x}{5}} \, dx\)

Optimal. Leaf size=12 \[ 3+\log \left (1+25 e^{-x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {12, 2282, 36, 29, 31} \begin {gather*} \log \left (e^x+25\right )-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-5/(5 + E^x/5),x]

[Out]

-x + Log[25 + E^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 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 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]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (5 \int \frac {1}{5+\frac {e^x}{5}} \, dx\right )\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {5}{x (25+x)} \, dx,x,e^x\right )\right )\\ &=-\left (25 \operatorname {Subst}\left (\int \frac {1}{x (25+x)} \, dx,x,e^x\right )\right )\\ &=-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\operatorname {Subst}\left (\int \frac {1}{25+x} \, dx,x,e^x\right )\\ &=-x+\log \left (25+e^x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.50 \begin {gather*} -25 \left (\frac {x}{25}-\frac {1}{25} \log \left (25+e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-5/(5 + E^x/5),x]

[Out]

-25*(x/25 - Log[25 + E^x]/25)

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fricas [A]  time = 0.64, size = 14, normalized size = 1.17 \begin {gather*} -x + \log \left (e^{\left (x - \log \relax (5)\right )} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/(exp(-log(5)+x)+5),x, algorithm="fricas")

[Out]

-x + log(e^(x - log(5)) + 5)

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giac [A]  time = 0.38, size = 16, normalized size = 1.33 \begin {gather*} -x + \log \relax (5) + \log \left (e^{\left (x - \log \relax (5)\right )} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/(exp(-log(5)+x)+5),x, algorithm="giac")

[Out]

-x + log(5) + log(e^(x - log(5)) + 5)

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maple [A]  time = 0.02, size = 14, normalized size = 1.17

method	result	size
risch	\(\ln \relax (5)-x +\ln \left (\frac {{\mathrm e}^{x}}{5}+5\right )\)	\(14\)
norman	\(-x +\ln \left ({\mathrm e}^{-\ln \relax (5)+x}+5\right )\)	\(15\)
derivativedivides	\(\ln \left ({\mathrm e}^{-\ln \relax (5)+x}+5\right )-\ln \left ({\mathrm e}^{-\ln \relax (5)+x}\right )\)	\(22\)
default	\(\ln \left ({\mathrm e}^{-\ln \relax (5)+x}+5\right )-\ln \left ({\mathrm e}^{-\ln \relax (5)+x}\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/(exp(-ln(5)+x)+5),x,method=_RETURNVERBOSE)

[Out]

ln(5)-x+ln(1/5*exp(x)+5)

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maxima [A]  time = 0.41, size = 9, normalized size = 0.75 \begin {gather*} -x + \log \left (e^{x} + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/(exp(-log(5)+x)+5),x, algorithm="maxima")

[Out]

-x + log(e^x + 25)

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mupad [B]  time = 0.30, size = 11, normalized size = 0.92 \begin {gather*} \ln \left (\frac {{\mathrm {e}}^x}{5}+5\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/(exp(x - log(5)) + 5),x)

[Out]

log(exp(x)/5 + 5) - x

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sympy [A]  time = 0.08, size = 7, normalized size = 0.58 \begin {gather*} - x + \log {\left (e^{x} + 25 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/(exp(-ln(5)+x)+5),x)

[Out]

-x + log(exp(x) + 25)

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