\(\int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx\) [2]

Optimal result

Integrand size = 18, antiderivative size = 24 \[ \int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx=-\frac {\text {arctanh}\left (\sqrt {\frac {2}{5}} e^{m x}\right )}{\sqrt {10} m} \]

[Out]

-1/10*arctanh(1/5*exp(m*x)*10^(1/2))/m*10^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2320, 213} \[ \int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx=-\frac {\text {arctanh}\left (\sqrt {\frac {2}{5}} e^{m x}\right )}{\sqrt {10} m} \]

[In]

Int[(-5/E^(m*x) + 2*E^(m*x))^(-1),x]

[Out]

-(ArcTanh[Sqrt[2/5]*E^(m*x)]/(Sqrt[10]*m))

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 2320

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{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{-5+2 x^2} \, dx,x,e^{m x}\right )}{m} \\ & = -\frac {\text {arctanh}\left (\sqrt {\frac {2}{5}} e^{m x}\right )}{\sqrt {10} m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx=-\frac {\text {arctanh}\left (\sqrt {\frac {2}{5}} e^{m x}\right )}{\sqrt {10} m} \]

[In]

Integrate[(-5/E^(m*x) + 2*E^(m*x))^(-1),x]

[Out]

-(ArcTanh[Sqrt[2/5]*E^(m*x)]/(Sqrt[10]*m))

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {\operatorname {arctanh}\left (\frac {{\mathrm e}^{m x} \sqrt {10}}{5}\right ) \sqrt {10}}{10 m}\)	\(19\)
default	\(-\frac {\operatorname {arctanh}\left (\frac {{\mathrm e}^{m x} \sqrt {10}}{5}\right ) \sqrt {10}}{10 m}\)	\(19\)
risch	\(\frac {\sqrt {10}\, \ln \left ({\mathrm e}^{m x}-\frac {\sqrt {10}}{2}\right )}{20 m}-\frac {\sqrt {10}\, \ln \left ({\mathrm e}^{m x}+\frac {\sqrt {10}}{2}\right )}{20 m}\)	\(40\)

[In]

int(1/(-5/exp(m*x)+2*exp(m*x)),x,method=_RETURNVERBOSE)

[Out]

-1/10*arctanh(1/5*exp(m*x)*10^(1/2))/m*10^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (18) = 36\).

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx=\frac {\sqrt {10} \log \left (-\frac {2 \, \sqrt {10} e^{\left (m x\right )} - 2 \, e^{\left (2 \, m x\right )} - 5}{2 \, e^{\left (2 \, m x\right )} - 5}\right )}{20 \, m} \]

[In]

integrate(1/(-5/exp(m*x)+2*exp(m*x)),x, algorithm="fricas")

[Out]

1/20*sqrt(10)*log(-(2*sqrt(10)*e^(m*x) - 2*e^(2*m*x) - 5)/(2*e^(2*m*x) - 5))/m

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx=\frac {\operatorname {RootSum} {\left (40 z^{2} - 1, \left ( i \mapsto i \log {\left (- 4 i + e^{- m x} \right )} \right )\right )}}{m} \]

[In]

integrate(1/(-5/exp(m*x)+2*exp(m*x)),x)

[Out]

RootSum(40*_z**2 - 1, Lambda(_i, _i*log(-4*_i + exp(-m*x))))/m

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx=\frac {\sqrt {10} \log \left (-\frac {\sqrt {10} - 5 \, e^{\left (-m x\right )}}{\sqrt {10} + 5 \, e^{\left (-m x\right )}}\right )}{20 \, m} \]

[In]

integrate(1/(-5/exp(m*x)+2*exp(m*x)),x, algorithm="maxima")

[Out]

1/20*sqrt(10)*log(-(sqrt(10) - 5*e^(-m*x))/(sqrt(10) + 5*e^(-m*x)))/m

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (18) = 36\).

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx=-\frac {\sqrt {10} \log \left (\frac {1}{2} \, \sqrt {10} + e^{\left (m x\right )}\right ) - \sqrt {10} \log \left ({\left | -\frac {1}{2} \, \sqrt {10} + e^{\left (m x\right )} \right |}\right )}{20 \, m} \]

[In]

integrate(1/(-5/exp(m*x)+2*exp(m*x)),x, algorithm="giac")

[Out]

-1/20*(sqrt(10)*log(1/2*sqrt(10) + e^(m*x)) - sqrt(10)*log(abs(-1/2*sqrt(10) + e^(m*x))))/m

Mupad [B] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {1}{-5 e^{-m x}+2 e^{m x}} \, dx=-\frac {\sqrt {10}\,\mathrm {atanh}\left (\frac {\sqrt {10}\,{\mathrm {e}}^{m\,x}}{5}\right )}{10\,m} \]

[In]

int(1/(2*exp(m*x) - 5*exp(-m*x)),x)

[Out]

-(10^(1/2)*atanh((10^(1/2)*exp(m*x))/5))/(10*m)