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

Optimal. Leaf size=24 \[ e^{1+x} \log \left (x^2+e^{4+x} \left (x-x^2\right )\right ) \]

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Rubi [F]  time = 2.99, 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 {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 21, normalized size = 0.88 \begin {gather*} e^{1+x} \log \left (x \left (-e^{4+x} (-1+x)+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(1 + x)*Log[x*(-(E^(4 + x)*(-1 + x)) + x)]

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fricas [A]  time = 0.77, size = 23, normalized size = 0.96 \begin {gather*} e^{\left (x + 1\right )} \log \left (x^{2} - {\left (x^{2} - x\right )} e^{\left (x + 4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 25, normalized size = 1.04 \begin {gather*} e^{\left (x + 1\right )} \log \left (-x^{2} e^{\left (x + 4\right )} + x^{2} + x e^{\left (x + 4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 27, normalized size = 1.12

method	result	size
norman	\({\mathrm e}^{-3} {\mathrm e}^{4+x} \ln \left (\left (-x^{2}+x \right ) {\mathrm e}^{4+x}+x^{2}\right )\)	\(27\)
risch	\(\ln \left (\left ({\mathrm e}^{4+x}-1\right ) x -{\mathrm e}^{4+x}\right ) {\mathrm e}^{x +1}+\ln \relax (x ) {\mathrm e}^{x +1}-i \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{4+x}-1\right ) x -{\mathrm e}^{4+x}\right )\right )^{2} \pi \,{\mathrm e}^{x +1}-\frac {i \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{4+x}-1\right ) x -{\mathrm e}^{4+x}\right )\right ) \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{4+x}-1\right ) x -{\mathrm e}^{4+x}\right )\right ) \mathrm {csgn}\left (i x \right ) \pi \,{\mathrm e}^{x +1}}{2}+\frac {i \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{4+x}-1\right ) x -{\mathrm e}^{4+x}\right )\right )^{2} \mathrm {csgn}\left (i x \right ) \pi \,{\mathrm e}^{x +1}}{2}+\frac {i \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{4+x}-1\right ) x -{\mathrm e}^{4+x}\right )\right )^{2} \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{4+x}-1\right ) x -{\mathrm e}^{4+x}\right )\right ) \pi \,{\mathrm e}^{x +1}}{2}+\frac {i \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{4+x}-1\right ) x -{\mathrm e}^{4+x}\right )\right )^{3} \pi \,{\mathrm e}^{x +1}}{2}+i \pi \,{\mathrm e}^{x +1}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 29, normalized size = 1.21 \begin {gather*} e^{\left (x + 1\right )} \log \left (-{\left (x e^{4} - e^{4}\right )} e^{x} + x\right ) + e^{\left (x + 1\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x + 1)*log(-(x*e^4 - e^4)*e^x + x) + e^(x + 1)*log(x)

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mupad [B]  time = 7.02, size = 22, normalized size = 0.92 \begin {gather*} \ln \left (x^2+{\mathrm {e}}^4\,{\mathrm {e}}^x\,\left (x-x^2\right )\right )\,{\mathrm {e}}^{x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^2 + exp(4)*exp(x)*(x - x^2))*exp(x + 1)

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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((((x**2-x)*exp(x+1)*exp(4+x)-x**2*exp(x+1))*ln((-x**2+x)*exp(4+x)+x**2)+(x**2+x-1)*exp(x+1)*exp(4+x)
-2*x*exp(x+1))/((x**2-x)*exp(4+x)-x**2),x)

[Out]

Timed out

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