3.78.81 \(\int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} (-30+e^{2 x-x^2} x)} (-1-2 x+2 x^2)}{20 e^{2 x}+20 e^{\frac {1}{5} (-30+e^{2 x-x^2} x)}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} (-30+e^{2 x-x^2} x)} (-40 e^x+40 \log (2))} \, dx\)

Optimal. Leaf size=32 \[ \frac {1}{2 \left (-e^x+e^{-3+\frac {1}{10} e^{(2-x) x} x}+\log (2)\right )} \]

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Rubi [A]  time = 0.80, antiderivative size = 39, normalized size of antiderivative = 1.22, number of steps used = 3, number of rules used = 3, integrand size = 126, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6688, 12, 6686} \begin {gather*} \frac {e^3}{2 \left (e^{\frac {1}{10} e^{(2-x) x} x}-e^{x+3}+e^3 \log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*E^x + E^(2*x - x^2 + (-30 + E^(2*x - x^2)*x)/10)*(-1 - 2*x + 2*x^2))/(20*E^(2*x) + 20*E^((-30 + E^(2*x
- x^2)*x)/5) - 40*E^x*Log[2] + 20*Log[2]^2 + E^((-30 + E^(2*x - x^2)*x)/10)*(-40*E^x + 40*Log[2])),x]

[Out]

E^3/(2*(E^((E^((2 - x)*x)*x)/10) - E^(3 + x) + E^3*Log[2]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^{6+x}+\exp \left (3+\left (2+\frac {1}{10} e^{-((-2+x) x)}\right ) x-x^2\right ) \left (-1-2 x+2 x^2\right )}{20 \left (e^{\frac {1}{10} e^{-((-2+x) x)} x}-e^{3+x}+e^3 \log (2)\right )^2} \, dx\\ &=\frac {1}{20} \int \frac {10 e^{6+x}+\exp \left (3+\left (2+\frac {1}{10} e^{-((-2+x) x)}\right ) x-x^2\right ) \left (-1-2 x+2 x^2\right )}{\left (e^{\frac {1}{10} e^{-((-2+x) x)} x}-e^{3+x}+e^3 \log (2)\right )^2} \, dx\\ &=\frac {e^3}{2 \left (e^{\frac {1}{10} e^{(2-x) x} x}-e^{3+x}+e^3 \log (2)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 38, normalized size = 1.19 \begin {gather*} \frac {e^3}{2 \left (e^{\frac {1}{10} e^{-((-2+x) x)} x}-e^{3+x}+e^3 \log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*E^x + E^(2*x - x^2 + (-30 + E^(2*x - x^2)*x)/10)*(-1 - 2*x + 2*x^2))/(20*E^(2*x) + 20*E^((-30 +
E^(2*x - x^2)*x)/5) - 40*E^x*Log[2] + 20*Log[2]^2 + E^((-30 + E^(2*x - x^2)*x)/10)*(-40*E^x + 40*Log[2])),x]

[Out]

E^3/(2*(E^(x/(10*E^((-2 + x)*x))) - E^(3 + x) + E^3*Log[2]))

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fricas [A]  time = 0.59, size = 27, normalized size = 0.84 \begin {gather*} \frac {1}{2 \, {\left (e^{\left (\frac {1}{10} \, x e^{\left (-x^{2} + 2 \, x\right )} - 3\right )} - e^{x} + \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-2*x-1)*exp(-x^2+2*x)*exp(1/10*x*exp(-x^2+2*x)-3)+10*exp(x))/(20*exp(1/10*x*exp(-x^2+2*x)-3)^
2+(-40*exp(x)+40*log(2))*exp(1/10*x*exp(-x^2+2*x)-3)+20*exp(x)^2-40*exp(x)*log(2)+20*log(2)^2),x, algorithm="f
ricas")

[Out]

1/2/(e^(1/10*x*e^(-x^2 + 2*x) - 3) - e^x + log(2))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-2*x-1)*exp(-x^2+2*x)*exp(1/10*x*exp(-x^2+2*x)-3)+10*exp(x))/(20*exp(1/10*x*exp(-x^2+2*x)-3)^
2+(-40*exp(x)+40*log(2))*exp(1/10*x*exp(-x^2+2*x)-3)+20*exp(x)^2-40*exp(x)*log(2)+20*log(2)^2),x, algorithm="g
iac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Not invertible Error: Bad Argument Value

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maple [A]  time = 0.45, size = 25, normalized size = 0.78

method	result	size
risch	\(\frac {1}{2 \ln \relax (2)-2 \,{\mathrm e}^{x}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-\left (x -2\right ) x}}{10}-3}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-2*x-1)*exp(-x^2+2*x)*exp(1/10*x*exp(-x^2+2*x)-3)+10*exp(x))/(20*exp(1/10*x*exp(-x^2+2*x)-3)^2+(-40
*exp(x)+40*ln(2))*exp(1/10*x*exp(-x^2+2*x)-3)+20*exp(x)^2-40*exp(x)*ln(2)+20*ln(2)^2),x,method=_RETURNVERBOSE)

[Out]

1/2/(ln(2)-exp(x)+exp(1/10*x*exp(-(x-2)*x)-3))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{20} \, \int \frac {{\left (2 \, x^{2} - 2 \, x - 1\right )} e^{\left (-x^{2} + \frac {1}{10} \, x e^{\left (-x^{2} + 2 \, x\right )} + 2 \, x - 3\right )} + 10 \, e^{x}}{2 \, {\left (e^{x} - \log \relax (2)\right )} e^{\left (\frac {1}{10} \, x e^{\left (-x^{2} + 2 \, x\right )} - 3\right )} + 2 \, e^{x} \log \relax (2) - \log \relax (2)^{2} - e^{\left (\frac {1}{5} \, x e^{\left (-x^{2} + 2 \, x\right )} - 6\right )} - e^{\left (2 \, x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-2*x-1)*exp(-x^2+2*x)*exp(1/10*x*exp(-x^2+2*x)-3)+10*exp(x))/(20*exp(1/10*x*exp(-x^2+2*x)-3)^
2+(-40*exp(x)+40*log(2))*exp(1/10*x*exp(-x^2+2*x)-3)+20*exp(x)^2-40*exp(x)*log(2)+20*log(2)^2),x, algorithm="m
axima")

[Out]

-1/20*integrate(((2*x^2 - 2*x - 1)*e^(-x^2 + 1/10*x*e^(-x^2 + 2*x) + 2*x - 3) + 10*e^x)/(2*(e^x - log(2))*e^(1
/10*x*e^(-x^2 + 2*x) - 3) + 2*e^x*log(2) - log(2)^2 - e^(1/5*x*e^(-x^2 + 2*x) - 6) - e^(2*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {10\,{\mathrm {e}}^x-{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x-x^2}}{10}-3}\,{\mathrm {e}}^{2\,x-x^2}\,\left (-2\,x^2+2\,x+1\right )}{20\,{\mathrm {e}}^{2\,x}+20\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x-x^2}}{5}-6}+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x-x^2}}{10}-3}\,\left (40\,\ln \relax (2)-40\,{\mathrm {e}}^x\right )-40\,{\mathrm {e}}^x\,\ln \relax (2)+20\,{\ln \relax (2)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(x) - exp((x*exp(2*x - x^2))/10 - 3)*exp(2*x - x^2)*(2*x - 2*x^2 + 1))/(20*exp(2*x) + 20*exp((x*exp
(2*x - x^2))/5 - 6) + exp((x*exp(2*x - x^2))/10 - 3)*(40*log(2) - 40*exp(x)) - 40*exp(x)*log(2) + 20*log(2)^2)
,x)

[Out]

int((10*exp(x) - exp((x*exp(2*x - x^2))/10 - 3)*exp(2*x - x^2)*(2*x - 2*x^2 + 1))/(20*exp(2*x) + 20*exp((x*exp
(2*x - x^2))/5 - 6) + exp((x*exp(2*x - x^2))/10 - 3)*(40*log(2) - 40*exp(x)) - 40*exp(x)*log(2) + 20*log(2)^2)
, x)

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sympy [A]  time = 0.32, size = 27, normalized size = 0.84 \begin {gather*} \frac {1}{- 2 e^{x} + 2 e^{\frac {x e^{- x^{2} + 2 x}}{10} - 3} + 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-2*x-1)*exp(-x**2+2*x)*exp(1/10*x*exp(-x**2+2*x)-3)+10*exp(x))/(20*exp(1/10*x*exp(-x**2+2*x)
-3)**2+(-40*exp(x)+40*ln(2))*exp(1/10*x*exp(-x**2+2*x)-3)+20*exp(x)**2-40*exp(x)*ln(2)+20*ln(2)**2),x)

[Out]

1/(-2*exp(x) + 2*exp(x*exp(-x**2 + 2*x)/10 - 3) + 2*log(2))

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