3.95.3 \(\int \frac {e^{5+\frac {e^5}{-3-x-x^2+\log (5)}} (1+2 x)}{9+6 x+7 x^2+2 x^3+x^4+(-6-2 x-2 x^2) \log (5)+\log ^2(5)} \, dx\)

Optimal. Leaf size=20 \[ e^{\frac {e^5}{-3-x-x^2+\log (5)}} \]

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Rubi [A]  time = 0.44, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6688, 6706} \begin {gather*} e^{-\frac {e^5}{x^2+x+3-\log (5)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5 + E^5/(-3 - x - x^2 + Log[5]))*(1 + 2*x))/(9 + 6*x + 7*x^2 + 2*x^3 + x^4 + (-6 - 2*x - 2*x^2)*Log[5]
+ Log[5]^2),x]

[Out]

E^(-(E^5/(3 + x + x^2 - Log[5])))

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{5-\frac {e^5}{3+x+x^2-\log (5)}} (1+2 x)}{\left (3+x+x^2-\log (5)\right )^2} \, dx\\ &=e^{-\frac {e^5}{3+x+x^2-\log (5)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.58, size = 19, normalized size = 0.95 \begin {gather*} e^{-\frac {e^5}{3+x+x^2-\log (5)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + E^5/(-3 - x - x^2 + Log[5]))*(1 + 2*x))/(9 + 6*x + 7*x^2 + 2*x^3 + x^4 + (-6 - 2*x - 2*x^2)*
Log[5] + Log[5]^2),x]

[Out]

E^(-(E^5/(3 + x + x^2 - Log[5])))

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fricas [A]  time = 0.61, size = 34, normalized size = 1.70 \begin {gather*} e^{\left (\frac {5 \, x^{2} + 5 \, x - e^{5} - 5 \, \log \relax (5) + 15}{x^{2} + x - \log \relax (5) + 3} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(5)*exp(exp(5)/(log(5)-x^2-x-3))/(log(5)^2+(-2*x^2-2*x-6)*log(5)+x^4+2*x^3+7*x^2+6*x+9),x
, algorithm="fricas")

[Out]

e^((5*x^2 + 5*x - e^5 - 5*log(5) + 15)/(x^2 + x - log(5) + 3) - 5)

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giac [B]  time = 0.19, size = 81, normalized size = 4.05 \begin {gather*} e^{\left (\frac {5 \, x^{2}}{x^{2} + x - \log \relax (5) + 3} + \frac {5 \, x}{x^{2} + x - \log \relax (5) + 3} - \frac {e^{5}}{x^{2} + x - \log \relax (5) + 3} - \frac {5 \, \log \relax (5)}{x^{2} + x - \log \relax (5) + 3} + \frac {15}{x^{2} + x - \log \relax (5) + 3} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(5)*exp(exp(5)/(log(5)-x^2-x-3))/(log(5)^2+(-2*x^2-2*x-6)*log(5)+x^4+2*x^3+7*x^2+6*x+9),x
, algorithm="giac")

[Out]

e^(5*x^2/(x^2 + x - log(5) + 3) + 5*x/(x^2 + x - log(5) + 3) - e^5/(x^2 + x - log(5) + 3) - 5*log(5)/(x^2 + x
- log(5) + 3) + 15/(x^2 + x - log(5) + 3) - 5)

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maple [A]  time = 0.27, size = 18, normalized size = 0.90

method	result	size
default	\({\mathrm e}^{-\frac {{\mathrm e}^{5}}{-\ln \relax (5)+x^{2}+x +3}}\)	\(18\)
gosper	\({\mathrm e}^{\frac {{\mathrm e}^{5}}{\ln \relax (5)-x^{2}-x -3}}\)	\(19\)
risch	\({\mathrm e}^{\frac {{\mathrm e}^{5}}{\ln \relax (5)-x^{2}-x -3}}\)	\(19\)
norman	\(\frac {\left (\ln \relax (5)-3\right ) {\mathrm e}^{\frac {{\mathrm e}^{5}}{\ln \relax (5)-x^{2}-x -3}}-x \,{\mathrm e}^{\frac {{\mathrm e}^{5}}{\ln \relax (5)-x^{2}-x -3}}-x^{2} {\mathrm e}^{\frac {{\mathrm e}^{5}}{\ln \relax (5)-x^{2}-x -3}}}{\ln \relax (5)-x^{2}-x -3}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x+1)*exp(5)*exp(exp(5)/(ln(5)-x^2-x-3))/(ln(5)^2+(-2*x^2-2*x-6)*ln(5)+x^4+2*x^3+7*x^2+6*x+9),x,method=_
RETURNVERBOSE)

[Out]

exp(-exp(5)/(-ln(5)+x^2+x+3))

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maxima [A]  time = 0.57, size = 17, normalized size = 0.85 \begin {gather*} e^{\left (-\frac {e^{5}}{x^{2} + x - \log \relax (5) + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(5)*exp(exp(5)/(log(5)-x^2-x-3))/(log(5)^2+(-2*x^2-2*x-6)*log(5)+x^4+2*x^3+7*x^2+6*x+9),x
, algorithm="maxima")

[Out]

e^(-e^5/(x^2 + x - log(5) + 3))

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mupad [B]  time = 9.56, size = 17, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{-\frac {{\mathrm {e}}^5}{x^2+x-\ln \relax (5)+3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(5)/(x - log(5) + x^2 + 3))*exp(5)*(2*x + 1))/(6*x - log(5)*(2*x + 2*x^2 + 6) + log(5)^2 + 7*x^2
+ 2*x^3 + x^4 + 9),x)

[Out]

exp(-exp(5)/(x - log(5) + x^2 + 3))

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sympy [A]  time = 0.50, size = 14, normalized size = 0.70 \begin {gather*} e^{\frac {e^{5}}{- x^{2} - x - 3 + \log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x+1)*exp(5)*exp(exp(5)/(ln(5)-x**2-x-3))/(ln(5)**2+(-2*x**2-2*x-6)*ln(5)+x**4+2*x**3+7*x**2+6*x+9
),x)

[Out]

exp(exp(5)/(-x**2 - x - 3 + log(5)))

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