3.15.50 \(\int \frac {e^{\frac {1-e^8 \log (e^2 (x+e x^2))}{e^8}} (-1-2 e x)}{x+e x^2} \, dx\)

Optimal. Leaf size=17 \[ \frac {e^{-2+\frac {1}{e^8}}}{x+e x^2} \]

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Rubi [A]  time = 0.37, antiderivative size = 18, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {1593, 2274, 12, 1584, 74} \begin {gather*} \frac {e^{\frac {1}{e^8}-2}}{x (e x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((1 - E^8*Log[E^2*(x + E*x^2)])/E^8)*(-1 - 2*E*x))/(x + E*x^2),x]

[Out]

E^(-2 + E^(-8))/(x*(1 + 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 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1-e^8 \log \left (e^2 \left (x+e x^2\right )\right )}{e^8}} (-1-2 e x)}{x (1+e x)} \, dx\\ &=\int \frac {e^{-2+\frac {1}{e^8}} (-1-2 e x)}{x (1+e x) \left (x+e x^2\right )} \, dx\\ &=e^{-2+\frac {1}{e^8}} \int \frac {-1-2 e x}{x (1+e x) \left (x+e x^2\right )} \, dx\\ &=e^{-2+\frac {1}{e^8}} \int \frac {-1-2 e x}{x^2 (1+e x)^2} \, dx\\ &=\frac {e^{-2+\frac {1}{e^8}}}{x (1+e x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{-2+\frac {1}{e^8}}}{x+e x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((1 - E^8*Log[E^2*(x + E*x^2)])/E^8)*(-1 - 2*E*x))/(x + E*x^2),x]

[Out]

E^(-2 + E^(-8))/(x + E*x^2)

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fricas [A]  time = 1.91, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{\left (e^{\left (-8\right )}\right )}}{x^{2} e^{3} + x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(1)-1)*exp((-exp(4)^2*log((x^2*exp(1)+x)*exp(2))+1)/exp(4)^2)/(x^2*exp(1)+x),x, algorithm="
fricas")

[Out]

e^(e^(-8))/(x^2*e^3 + x*e^2)

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giac [A]  time = 0.31, size = 16, normalized size = 0.94 \begin {gather*} \frac {e^{\left (e^{\left (-8\right )} - 2\right )}}{x^{2} e + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(1)-1)*exp((-exp(4)^2*log((x^2*exp(1)+x)*exp(2))+1)/exp(4)^2)/(x^2*exp(1)+x),x, algorithm="
giac")

[Out]

e^(e^(-8) - 2)/(x^2*e + x)

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maple [A]  time = 0.25, size = 18, normalized size = 1.06

method	result	size
risch	\(\frac {{\mathrm e}^{-2+{\mathrm e}^{-8}}}{x \left (x \,{\mathrm e}+1\right )}\)	\(18\)
norman	\(\frac {{\mathrm e}^{{\mathrm e}^{-8}} {\mathrm e}^{-2}}{x \left (x \,{\mathrm e}+1\right )}\)	\(22\)
default	\({\mathrm e}^{-2+{\mathrm e}^{-8}} \left (\frac {1}{x}-\frac {{\mathrm e}}{x \,{\mathrm e}+1}\right )\)	\(23\)
gosper	\({\mathrm e}^{-\left ({\mathrm e}^{8} \ln \left (x \left (x \,{\mathrm e}+1\right ) {\mathrm e}^{2}\right )-1\right ) {\mathrm e}^{-8}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*exp(1)-1)*exp((-exp(4)^2*ln((x^2*exp(1)+x)*exp(2))+1)/exp(4)^2)/(x^2*exp(1)+x),x,method=_RETURNVERBO
SE)

[Out]

1/x/(x*exp(1)+1)*exp(-2+exp(-8))

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maxima [A]  time = 0.72, size = 22, normalized size = 1.29 \begin {gather*} e^{\left (-{\left (e^{8} \log \left ({\left (x^{2} e + x\right )} e^{2}\right ) - 1\right )} e^{\left (-8\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(1)-1)*exp((-exp(4)^2*log((x^2*exp(1)+x)*exp(2))+1)/exp(4)^2)/(x^2*exp(1)+x),x, algorithm="
maxima")

[Out]

e^(-(e^8*log((x^2*e + x)*e^2) - 1)*e^(-8))

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mupad [B]  time = 0.15, size = 17, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{-8}-2}}{x\,\left (x\,\mathrm {e}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-exp(-8)*(exp(8)*log(exp(2)*(x + x^2*exp(1))) - 1))*(2*x*exp(1) + 1))/(x + x^2*exp(1)),x)

[Out]

exp(exp(-8) - 2)/(x*(x*exp(1) + 1))

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sympy [A]  time = 0.39, size = 17, normalized size = 1.00 \begin {gather*} \frac {e^{e^{-8}}}{x^{2} e^{3} + x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*exp(1)-1)*exp((-exp(4)**2*ln((x**2*exp(1)+x)*exp(2))+1)/exp(4)**2)/(x**2*exp(1)+x),x)

[Out]

exp(exp(-8))/(x**2*exp(3) + x*exp(2))

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