3.21.50 \(\int \frac {e^2 (2+e)}{-e^2+6 e^{2 x+e x}} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{3}+\log \left (-3+\frac {1}{2} e^{2-(2+e) x}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {12, 2282, 36, 29, 31} \begin {gather*} -e x-2 x+\log \left (e^2-6 e^{(2+e) x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(2 + E))/(-E^2 + 6*E^(2*x + E*x)),x]

[Out]

-2*x - E*x + Log[E^2 - 6*E^((2 + 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 (e^2 (2+e)\right ) \int \frac {1}{-e^2+6 e^{2 x+e x}} \, dx\\ &=e^2 \operatorname {Subst}\left (\int \frac {1}{x \left (-e^2+6 x\right )} \, dx,x,e^{2 x+e x}\right )\\ &=6 \operatorname {Subst}\left (\int \frac {1}{-e^2+6 x} \, dx,x,e^{2 x+e x}\right )-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x+e x}\right )\\ &=-2 x-e x+\log \left (e^2-6 e^{(2+e) x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.00 \begin {gather*} -((2+e) x)+\log \left (e^2-6 e^{(2+e) x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(2 + E))/(-E^2 + 6*E^(2*x + E*x)),x]

[Out]

-((2 + E)*x) + Log[E^2 - 6*E^((2 + E)*x)]

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fricas [A]  time = 0.52, size = 26, normalized size = 1.24 \begin {gather*} -x e - 2 \, x + \log \left (-e^{2} + 6 \, e^{\left (x e + 2 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)+2)*exp(2)/(6*exp(x*exp(1)+2*x)-exp(2)),x, algorithm="fricas")

[Out]

-x*e - 2*x + log(-e^2 + 6*e^(x*e + 2*x))

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giac [A]  time = 0.24, size = 33, normalized size = 1.57 \begin {gather*} -{\left (x {\left (e + 2\right )} e^{\left (-2\right )} - e^{\left (-2\right )} \log \left ({\left | -e^{2} + 6 \, e^{\left (x {\left (e + 2\right )}\right )} \right |}\right )\right )} e^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)+2)*exp(2)/(6*exp(x*exp(1)+2*x)-exp(2)),x, algorithm="giac")

[Out]

-(x*(e + 2)*e^(-2) - e^(-2)*log(abs(-e^2 + 6*e^(x*(e + 2)))))*e^2

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maple [A]  time = 0.05, size = 23, normalized size = 1.10

method	result	size
risch	\(-x \,{\mathrm e}-2 x +\ln \left (-\frac {{\mathrm e}^{2}}{6}+{\mathrm e}^{x \left ({\mathrm e}+2\right )}\right )\)	\(23\)
norman	\(\left (-{\mathrm e}-2\right ) x +\ln \left (-6 \,{\mathrm e}^{x \,{\mathrm e}+2 x}+{\mathrm e}^{2}\right )\)	\(25\)
derivativedivides	\({\mathrm e}^{2} \left ({\mathrm e}^{-2} \ln \left (6 \,{\mathrm e}^{x \,{\mathrm e}+2 x}-{\mathrm e}^{2}\right )-{\mathrm e}^{-2} \ln \left ({\mathrm e}^{x \,{\mathrm e}+2 x}\right )\right )\)	\(43\)
default	\({\mathrm e}^{2} \left ({\mathrm e}^{-2} \ln \left (6 \,{\mathrm e}^{x \,{\mathrm e}+2 x}-{\mathrm e}^{2}\right )-{\mathrm e}^{-2} \ln \left ({\mathrm e}^{x \,{\mathrm e}+2 x}\right )\right )\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1)+2)*exp(2)/(6*exp(x*exp(1)+2*x)-exp(2)),x,method=_RETURNVERBOSE)

[Out]

-x*exp(1)-2*x+ln(-1/6*exp(2)+exp(x*(exp(1)+2)))

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maxima [B]  time = 0.39, size = 38, normalized size = 1.81 \begin {gather*} -{\left (x e^{\left (-2\right )} - \frac {e^{\left (-2\right )} \log \left (-e^{2} + 6 \, e^{\left (x {\left (e + 2\right )}\right )}\right )}{e + 2}\right )} {\left (e + 2\right )} e^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)+2)*exp(2)/(6*exp(x*exp(1)+2*x)-exp(2)),x, algorithm="maxima")

[Out]

-(x*e^(-2) - e^(-2)*log(-e^2 + 6*e^(x*(e + 2)))/(e + 2))*(e + 2)*e^2

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mupad [B]  time = 0.16, size = 25, normalized size = 1.19 \begin {gather*} \ln \left (6\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{x\,\mathrm {e}}-{\mathrm {e}}^2\right )-x\,\left (\mathrm {e}+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)*(exp(1) + 2))/(exp(2) - 6*exp(2*x + x*exp(1))),x)

[Out]

log(6*exp(2*x)*exp(x*exp(1)) - exp(2)) - x*(exp(1) + 2)

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sympy [A]  time = 0.16, size = 24, normalized size = 1.14 \begin {gather*} x \left (- e - 2\right ) + \log {\left (e^{2 x + e x} - \frac {e^{2}}{6} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)+2)*exp(2)/(6*exp(x*exp(1)+2*x)-exp(2)),x)

[Out]

x*(-E - 2) + log(exp(2*x + E*x) - exp(2)/6)

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