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

Optimal. Leaf size=31 \[ e^{-4-2 e^{e^9-(x+\log (1-x))^2}+2 x} x^3 \]

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Rubi [F]  time = 16.15, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) x^2 (3+2 x)-4 \exp \left (-4+e^9-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x-x^2-\log ^2(1-x)\right ) (1-x)^{-1-2 x} x^4 (x+\log (1-x))\right ) \, dx\\ &=-\left (4 \int \exp \left (-4+e^9-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x-x^2-\log ^2(1-x)\right ) (1-x)^{-1-2 x} x^4 (x+\log (1-x)) \, dx\right )+\int \exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) x^2 (3+2 x) \, dx\\ &=-\left (4 \int \exp \left (-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}-4 \left (1-\frac {e^9}{4}\right )+2 x-x^2-\log ^2(1-x)\right ) (1-x)^{-1-2 x} x^4 (x+\log (1-x)) \, dx\right )+\int \left (3 \exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) x^2+2 \exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) x^3\right ) \, dx\\ &=2 \int \exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) x^3 \, dx+3 \int \exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) x^2 \, dx-4 \int \left (\exp \left (-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}-4 \left (1-\frac {e^9}{4}\right )+2 x-x^2-\log ^2(1-x)\right ) (1-x)^{-1-2 x} x^5+\exp \left (-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}-4 \left (1-\frac {e^9}{4}\right )+2 x-x^2-\log ^2(1-x)\right ) (1-x)^{-1-2 x} x^4 \log (1-x)\right ) \, dx\\ &=2 \int \exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) x^3 \, dx+3 \int \exp \left (-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x\right ) x^2 \, dx-4 \int \exp \left (-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}-4 \left (1-\frac {e^9}{4}\right )+2 x-x^2-\log ^2(1-x)\right ) (1-x)^{-1-2 x} x^5 \, dx-4 \int \exp \left (-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}-4 \left (1-\frac {e^9}{4}\right )+2 x-x^2-\log ^2(1-x)\right ) (1-x)^{-1-2 x} x^4 \log (1-x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 43, normalized size = 1.39 \begin {gather*} e^{-4-2 e^{e^9-x^2-\log ^2(1-x)} (1-x)^{-2 x}+2 x} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-4 - (2*E^(E^9 - x^2 - Log[1 - x]^2))/(1 - x)^(2*x) + 2*x)*x^3

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fricas [A]  time = 0.87, size = 40, normalized size = 1.29 \begin {gather*} x^{3} e^{\left (2 \, x - 2 \, e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*e^(2*x - 2*e^(-x^2 - 2*x*log(-x + 1) - log(-x + 1)^2 + e^9) - 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{4} + x^{3} - 3 \, x^{2} + 4 \, {\left (x^{5} + x^{4} \log \left (-x + 1\right )\right )} e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )}\right )} e^{\left (2 \, x - 2 \, e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )} - 4\right )}}{x - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*x^4 + x^3 - 3*x^2 + 4*(x^5 + x^4*log(-x + 1))*e^(-x^2 - 2*x*log(-x + 1) - log(-x + 1)^2 + e^9))*e
^(2*x - 2*e^(-x^2 - 2*x*log(-x + 1) - log(-x + 1)^2 + e^9) - 4)/(x - 1), x)

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maple [A]  time = 0.31, size = 41, normalized size = 1.32

method	result	size
risch	\(x^{3} {\mathrm e}^{-2 \left (1-x \right )^{-2 x} {\mathrm e}^{-\ln \left (1-x \right )^{2}+{\mathrm e}^{9}-x^{2}}-4+2 x}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*exp(-2*(1-x)^(-2*x)*exp(-ln(1-x)^2+exp(9)-x^2)-4+2*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{4} + x^{3} - 3 \, x^{2} + 4 \, {\left (x^{5} + x^{4} \log \left (-x + 1\right )\right )} e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )}\right )} e^{\left (2 \, x - 2 \, e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )} - 4\right )}}{x - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*x^4 + x^3 - 3*x^2 + 4*(x^5 + x^4*log(-x + 1))*e^(-x^2 - 2*x*log(-x + 1) - log(-x + 1)^2 + e^9))*e
^(2*x - 2*e^(-x^2 - 2*x*log(-x + 1) - log(-x + 1)^2 + e^9) - 4)/(x - 1), x)

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mupad [B]  time = 1.16, size = 44, normalized size = 1.42 \begin {gather*} x^3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{-{\ln \left (1-x\right )}^2}\,{\mathrm {e}}^{{\mathrm {e}}^9}}{{\left (1-x\right )}^{2\,x}}}\,{\mathrm {e}}^{-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*exp(2*x)*exp(-(2*exp(-x^2)*exp(-log(1 - x)^2)*exp(exp(9)))/(1 - x)^(2*x))*exp(-4)

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sympy [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(((4*x**4*ln(-x+1)+4*x**5)*exp(-ln(-x+1)**2-2*x*ln(-x+1)+exp(9)-x**2)+2*x**4+x**3-3*x**2)/(x-1)/exp(e
xp(-ln(-x+1)**2-2*x*ln(-x+1)+exp(9)-x**2)+2-x)**2,x)

[Out]

Timed out

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