3.35.84 \(\int \frac {e^{-3+e^x} (e^{2+4 x} (x^2-4 x^3+e^x x^3)-2 x \log (2)-e^x x^2 \log (2))}{2 e^{4+8 x} x^2-4 e^{2+4 x} x \log (2)+2 \log ^2(2)} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^{-3+e^x} x}{2 \left (e^{2+4 x}-\frac {\log (2)}{x}\right )} \]

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Rubi [F]  time = 3.72, 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 {e^{-3+e^x} \left (e^{2+4 x} \left (x^2-4 x^3+e^x x^3\right )-2 x \log (2)-e^x x^2 \log (2)\right )}{2 e^{4+8 x} x^2-4 e^{2+4 x} x \log (2)+2 \log ^2(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-3 + E^x)*(E^(2 + 4*x)*(x^2 - 4*x^3 + E^x*x^3) - 2*x*Log[2] - E^x*x^2*Log[2]))/(2*E^(4 + 8*x)*x^2 - 4*
E^(2 + 4*x)*x*Log[2] + 2*Log[2]^2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 30, normalized size = 1.03 \begin {gather*} \frac {e^{-3+e^x} x^2}{2 \left (e^{2+4 x} x-\log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 + E^x)*(E^(2 + 4*x)*(x^2 - 4*x^3 + E^x*x^3) - 2*x*Log[2] - E^x*x^2*Log[2]))/(2*E^(4 + 8*x)*x^
2 - 4*E^(2 + 4*x)*x*Log[2] + 2*Log[2]^2),x]

[Out]

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

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fricas [A]  time = 0.62, size = 25, normalized size = 0.86 \begin {gather*} \frac {x^{2} e^{\left (e^{x} - 3\right )}}{2 \, {\left (x e^{\left (4 \, x + 2\right )} - \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.43, size = 29, normalized size = 1.00 \begin {gather*} \frac {x^{2} e^{\left (x + e^{x}\right )}}{2 \, {\left (x e^{\left (5 \, x + 5\right )} - e^{\left (x + 3\right )} \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*e^(x + e^x)/(x*e^(5*x + 5) - e^(x + 3)*log(2))

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maple [A]  time = 0.06, size = 26, normalized size = 0.90

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{x}-3}}{2 x \,{\mathrm e}^{4 x +2}-2 \ln \relax (2)}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 25, normalized size = 0.86 \begin {gather*} \frac {x^{2} e^{\left (e^{x}\right )}}{2 \, {\left (x e^{\left (4 \, x + 5\right )} - e^{3} \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.50, size = 25, normalized size = 0.86 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-5}}{2\,\left (x\,{\mathrm {e}}^{4\,x}-{\mathrm {e}}^{-2}\,\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x) - 3)*(2*x*log(2) - exp(4*x + 2)*(x^3*exp(x) + x^2 - 4*x^3) + x^2*exp(x)*log(2)))/(2*log(2)^2
+ 2*x^2*exp(8*x + 4) - 4*x*exp(4*x + 2)*log(2)),x)

[Out]

(x^2*exp(exp(x))*exp(-5))/(2*(x*exp(4*x) - exp(-2)*log(2)))

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sympy [A]  time = 0.24, size = 26, normalized size = 0.90 \begin {gather*} \frac {x^{2} e^{e^{x} - 3}}{2 x e^{2} e^{4 x} - 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x**3-4*x**3+x**2)*exp(4*x+2)-x**2*ln(2)*exp(x)-2*x*ln(2))/(2*x**2*exp(4*x+2)**2-4*x*ln(2)*e
xp(4*x+2)+2*ln(2)**2)/exp(-exp(x)+3),x)

[Out]

x**2*exp(exp(x) - 3)/(2*x*exp(2)*exp(4*x) - 2*log(2))

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