3.20.12 \(\int \frac {-5 e^{2209}+e^{1+x} (-10 x-5 x^2)+(1-e^{2209} x-e^{1+x} x^2) \log ^2(1-e^{2209} x-e^{1+x} x^2)}{(-1+e^{2209} x+e^{1+x} x^2) \log ^2(1-e^{2209} x-e^{1+x} x^2)} \, dx\)

Optimal. Leaf size=25 \[ -x+\frac {5}{\log \left (1-x \left (e^{2209}+e^{1+x} x\right )\right )} \]

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Rubi [A]  time = 0.41, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 2, integrand size = 101, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6688, 6686} \begin {gather*} \frac {5}{\log \left (-e^{x+1} x^2-e^{2209} x+1\right )}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*E^2209 + E^(1 + x)*(-10*x - 5*x^2) + (1 - E^2209*x - E^(1 + x)*x^2)*Log[1 - E^2209*x - E^(1 + x)*x^2]^
2)/((-1 + E^2209*x + E^(1 + x)*x^2)*Log[1 - E^2209*x - E^(1 + x)*x^2]^2),x]

[Out]

-x + 5/Log[1 - E^2209*x - E^(1 + x)*x^2]

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 \left (-1-\frac {5 e \left (e^{2208}+e^x x (2+x)\right )}{\left (-1+e^{2209} x+e^{1+x} x^2\right ) \log ^2\left (1-e^{2209} x-e^{1+x} x^2\right )}\right ) \, dx\\ &=-x-(5 e) \int \frac {e^{2208}+e^x x (2+x)}{\left (-1+e^{2209} x+e^{1+x} x^2\right ) \log ^2\left (1-e^{2209} x-e^{1+x} x^2\right )} \, dx\\ &=-x+\frac {5}{\log \left (1-e^{2209} x-e^{1+x} x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 1.08 \begin {gather*} -x+\frac {5}{\log \left (1-e^{2209} x-e^{1+x} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*E^2209 + E^(1 + x)*(-10*x - 5*x^2) + (1 - E^2209*x - E^(1 + x)*x^2)*Log[1 - E^2209*x - E^(1 + x)
*x^2]^2)/((-1 + E^2209*x + E^(1 + x)*x^2)*Log[1 - E^2209*x - E^(1 + x)*x^2]^2),x]

[Out]

-x + 5/Log[1 - E^2209*x - E^(1 + x)*x^2]

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fricas [A]  time = 0.70, size = 42, normalized size = 1.68 \begin {gather*} -\frac {x \log \left (-x^{2} e^{\left (x + 1\right )} - x e^{2209} + 1\right ) - 5}{\log \left (-x^{2} e^{\left (x + 1\right )} - x e^{2209} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(1)*exp(x)-x*exp(2209)+1)*log(-x^2*exp(1)*exp(x)-x*exp(2209)+1)^2+(-5*x^2-10*x)*exp(1)*exp
(x)-5*exp(2209))/(x^2*exp(1)*exp(x)+x*exp(2209)-1)/log(-x^2*exp(1)*exp(x)-x*exp(2209)+1)^2,x, algorithm="frica
s")

[Out]

-(x*log(-x^2*e^(x + 1) - x*e^2209 + 1) - 5)/log(-x^2*e^(x + 1) - x*e^2209 + 1)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(1)*exp(x)-x*exp(2209)+1)*log(-x^2*exp(1)*exp(x)-x*exp(2209)+1)^2+(-5*x^2-10*x)*exp(1)*exp
(x)-5*exp(2209))/(x^2*exp(1)*exp(x)+x*exp(2209)-1)/log(-x^2*exp(1)*exp(x)-x*exp(2209)+1)^2,x, algorithm="giac"
)

[Out]

Timed out

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maple [A]  time = 0.09, size = 26, normalized size = 1.04

method	result	size
risch	\(-x +\frac {5}{\ln \left (-x^{2} {\mathrm e}^{x +1}-x \,{\mathrm e}^{2209}+1\right )}\)	\(26\)
norman	\(\frac {5-x \ln \left (-x^{2} {\mathrm e} \,{\mathrm e}^{x}-x \,{\mathrm e}^{2209}+1\right )}{\ln \left (-x^{2} {\mathrm e} \,{\mathrm e}^{x}-x \,{\mathrm e}^{2209}+1\right )}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2*exp(1)*exp(x)-x*exp(2209)+1)*ln(-x^2*exp(1)*exp(x)-x*exp(2209)+1)^2+(-5*x^2-10*x)*exp(1)*exp(x)-5*e
xp(2209))/(x^2*exp(1)*exp(x)+x*exp(2209)-1)/ln(-x^2*exp(1)*exp(x)-x*exp(2209)+1)^2,x,method=_RETURNVERBOSE)

[Out]

-x+5/ln(-x^2*exp(x+1)-x*exp(2209)+1)

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maxima [A]  time = 0.55, size = 42, normalized size = 1.68 \begin {gather*} -\frac {x \log \left (-x^{2} e^{\left (x + 1\right )} - x e^{2209} + 1\right ) - 5}{\log \left (-x^{2} e^{\left (x + 1\right )} - x e^{2209} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(1)*exp(x)-x*exp(2209)+1)*log(-x^2*exp(1)*exp(x)-x*exp(2209)+1)^2+(-5*x^2-10*x)*exp(1)*exp
(x)-5*exp(2209))/(x^2*exp(1)*exp(x)+x*exp(2209)-1)/log(-x^2*exp(1)*exp(x)-x*exp(2209)+1)^2,x, algorithm="maxim
a")

[Out]

-(x*log(-x^2*e^(x + 1) - x*e^2209 + 1) - 5)/log(-x^2*e^(x + 1) - x*e^2209 + 1)

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mupad [B]  time = 13.04, size = 25, normalized size = 1.00 \begin {gather*} \frac {5}{\ln \left (1-x^2\,\mathrm {e}\,{\mathrm {e}}^x-x\,{\mathrm {e}}^{2209}\right )}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(2209) + log(1 - x^2*exp(1)*exp(x) - x*exp(2209))^2*(x*exp(2209) + x^2*exp(1)*exp(x) - 1) + exp(1)*
exp(x)*(10*x + 5*x^2))/(log(1 - x^2*exp(1)*exp(x) - x*exp(2209))^2*(x*exp(2209) + x^2*exp(1)*exp(x) - 1)),x)

[Out]

5/log(1 - x^2*exp(1)*exp(x) - x*exp(2209)) - x

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sympy [A]  time = 0.51, size = 20, normalized size = 0.80 \begin {gather*} - x + \frac {5}{\log {\left (- e x^{2} e^{x} - x e^{2209} + 1 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2*exp(1)*exp(x)-x*exp(2209)+1)*ln(-x**2*exp(1)*exp(x)-x*exp(2209)+1)**2+(-5*x**2-10*x)*exp(1)*
exp(x)-5*exp(2209))/(x**2*exp(1)*exp(x)+x*exp(2209)-1)/ln(-x**2*exp(1)*exp(x)-x*exp(2209)+1)**2,x)

[Out]

-x + 5/log(-E*x**2*exp(x) - x*exp(2209) + 1)

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