3.38.9 \(\int \frac {e^{\frac {2 x}{-60+5 e^x+5 x+5 x^2+5 \log (x)}} (-26+e^x (2-2 x)-2 x^2+2 \log (x))}{720+5 e^{2 x}-120 x-115 x^2+10 x^3+5 x^4+e^x (-120+10 x+10 x^2)+(-120+10 e^x+10 x+10 x^2) \log (x)+5 \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ e^{\frac {2 x}{5 \left (-12+e^x+x+x^2+\log (x)\right )}} \]

________________________________________________________________________________________

Rubi [A]  time = 1.80, antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 3, number of rules used = 3, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6688, 12, 6706} \begin {gather*} e^{-\frac {2 x}{5 \left (-x^2-x-e^x-\log (x)+12\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*x)/(-60 + 5*E^x + 5*x + 5*x^2 + 5*Log[x]))*(-26 + E^x*(2 - 2*x) - 2*x^2 + 2*Log[x]))/(720 + 5*E^(2*
x) - 120*x - 115*x^2 + 10*x^3 + 5*x^4 + E^x*(-120 + 10*x + 10*x^2) + (-120 + 10*E^x + 10*x + 10*x^2)*Log[x] +
5*Log[x]^2),x]

[Out]

E^((-2*x)/(5*(12 - E^x - x - x^2 - Log[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 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 {2 e^{\frac {2 x}{5 \left (-12+e^x+x+x^2+\log (x)\right )}} \left (-13-e^x (-1+x)-x^2+\log (x)\right )}{5 \left (12-e^x-x-x^2-\log (x)\right )^2} \, dx\\ &=\frac {2}{5} \int \frac {e^{\frac {2 x}{5 \left (-12+e^x+x+x^2+\log (x)\right )}} \left (-13-e^x (-1+x)-x^2+\log (x)\right )}{\left (12-e^x-x-x^2-\log (x)\right )^2} \, dx\\ &=e^{-\frac {2 x}{5 \left (12-e^x-x-x^2-\log (x)\right )}}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 2.12, size = 20, normalized size = 1.00 \begin {gather*} e^{\frac {2 x}{5 \left (-12+e^x+x+x^2+\log (x)\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*x)/(-60 + 5*E^x + 5*x + 5*x^2 + 5*Log[x]))*(-26 + E^x*(2 - 2*x) - 2*x^2 + 2*Log[x]))/(720 + 5
*E^(2*x) - 120*x - 115*x^2 + 10*x^3 + 5*x^4 + E^x*(-120 + 10*x + 10*x^2) + (-120 + 10*E^x + 10*x + 10*x^2)*Log
[x] + 5*Log[x]^2),x]

[Out]

E^((2*x)/(5*(-12 + E^x + x + x^2 + Log[x])))

________________________________________________________________________________________

fricas [A]  time = 0.73, size = 16, normalized size = 0.80 \begin {gather*} e^{\left (\frac {2 \, x}{5 \, {\left (x^{2} + x + e^{x} + \log \relax (x) - 12\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+(-2*x+2)*exp(x)-2*x^2-26)*exp(2*x/(5*log(x)+5*exp(x)+5*x^2+5*x-60))/(5*log(x)^2+(10*exp(x)
+10*x^2+10*x-120)*log(x)+5*exp(x)^2+(10*x^2+10*x-120)*exp(x)+5*x^4+10*x^3-115*x^2-120*x+720),x, algorithm="fri
cas")

[Out]

e^(2/5*x/(x^2 + x + e^x + log(x) - 12))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (x^{2} + {\left (x - 1\right )} e^{x} - \log \relax (x) + 13\right )} e^{\left (\frac {2 \, x}{5 \, {\left (x^{2} + x + e^{x} + \log \relax (x) - 12\right )}}\right )}}{5 \, {\left (x^{4} + 2 \, x^{3} - 23 \, x^{2} + 2 \, {\left (x^{2} + x - 12\right )} e^{x} + 2 \, {\left (x^{2} + x + e^{x} - 12\right )} \log \relax (x) + \log \relax (x)^{2} - 24 \, x + e^{\left (2 \, x\right )} + 144\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+(-2*x+2)*exp(x)-2*x^2-26)*exp(2*x/(5*log(x)+5*exp(x)+5*x^2+5*x-60))/(5*log(x)^2+(10*exp(x)
+10*x^2+10*x-120)*log(x)+5*exp(x)^2+(10*x^2+10*x-120)*exp(x)+5*x^4+10*x^3-115*x^2-120*x+720),x, algorithm="gia
c")

[Out]

integrate(-2/5*(x^2 + (x - 1)*e^x - log(x) + 13)*e^(2/5*x/(x^2 + x + e^x + log(x) - 12))/(x^4 + 2*x^3 - 23*x^2
+ 2*(x^2 + x - 12)*e^x + 2*(x^2 + x + e^x - 12)*log(x) + log(x)^2 - 24*x + e^(2*x) + 144), x)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 17, normalized size = 0.85

method	result	size
risch	\({\mathrm e}^{\frac {2 x}{5 \left (x^{2}+\ln \relax (x )+{\mathrm e}^{x}+x -12\right )}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(x)+(-2*x+2)*exp(x)-2*x^2-26)*exp(2*x/(5*ln(x)+5*exp(x)+5*x^2+5*x-60))/(5*ln(x)^2+(10*exp(x)+10*x^2+1
0*x-120)*ln(x)+5*exp(x)^2+(10*x^2+10*x-120)*exp(x)+5*x^4+10*x^3-115*x^2-120*x+720),x,method=_RETURNVERBOSE)

[Out]

exp(2/5*x/(x^2+ln(x)+exp(x)+x-12))

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 16, normalized size = 0.80 \begin {gather*} e^{\left (\frac {2 \, x}{5 \, {\left (x^{2} + x + e^{x} + \log \relax (x) - 12\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)+(-2*x+2)*exp(x)-2*x^2-26)*exp(2*x/(5*log(x)+5*exp(x)+5*x^2+5*x-60))/(5*log(x)^2+(10*exp(x)
+10*x^2+10*x-120)*log(x)+5*exp(x)^2+(10*x^2+10*x-120)*exp(x)+5*x^4+10*x^3-115*x^2-120*x+720),x, algorithm="max
ima")

[Out]

e^(2/5*x/(x^2 + x + e^x + log(x) - 12))

________________________________________________________________________________________

mupad [B]  time = 2.38, size = 24, normalized size = 1.20 \begin {gather*} {\mathrm {e}}^{\frac {2\,x}{5\,\left (x+{\mathrm {e}}^x+\ln \relax (x)+x^2-12\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*x)/(5*x + 5*exp(x) + 5*log(x) + 5*x^2 - 60))*(exp(x)*(2*x - 2) - 2*log(x) + 2*x^2 + 26))/(5*exp(2
*x) - 120*x + 5*log(x)^2 + exp(x)*(10*x + 10*x^2 - 120) - 115*x^2 + 10*x^3 + 5*x^4 + log(x)*(10*x + 10*exp(x)
+ 10*x^2 - 120) + 720),x)

[Out]

exp((2*x)/(5*(x + exp(x) + log(x) + x^2 - 12)))

________________________________________________________________________________________

sympy [A]  time = 3.47, size = 24, normalized size = 1.20 \begin {gather*} e^{\frac {2 x}{5 x^{2} + 5 x + 5 e^{x} + 5 \log {\relax (x )} - 60}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(x)+(-2*x+2)*exp(x)-2*x**2-26)*exp(2*x/(5*ln(x)+5*exp(x)+5*x**2+5*x-60))/(5*ln(x)**2+(10*exp(x)
+10*x**2+10*x-120)*ln(x)+5*exp(x)**2+(10*x**2+10*x-120)*exp(x)+5*x**4+10*x**3-115*x**2-120*x+720),x)

[Out]

exp(2*x/(5*x**2 + 5*x + 5*exp(x) + 5*log(x) - 60))

________________________________________________________________________________________