3.4.74 \(\int \frac {e^{-x} (e^x (2 x^2+4 x^3+2 x^4)+e^{\frac {2 e^{-x} (e^{16}+\log (4))}{x}} (e^x x^2+e^{16} (2+4 x+2 x^2)+(2+4 x+2 x^2) \log (4)))}{2 x^2+4 x^3+2 x^4} \, dx\)

Optimal. Leaf size=29 \[ x-\frac {e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}}}{2+2 x} \]

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Rubi [F]  time = 3.38, 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^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2+4 x^3+2 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(2*x^2 + 4*x^3 + 2*x^4) + E^((2*(E^16 + Log[4]))/(E^x*x))*(E^x*x^2 + E^16*(2 + 4*x + 2*x^2) + (2 + 4*
x + 2*x^2)*Log[4]))/(E^x*(2*x^2 + 4*x^3 + 2*x^4)),x]

[Out]

x + (E^16 + Log[4])*Defer[Int][E^(-x + (2*(E^16 + Log[4]))/(E^x*x))/x^2, x] + Defer[Int][E^((2*(E^16 + Log[4])
)/(E^x*x))/(1 + x)^2, x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{x^2 \left (2+4 x+2 x^2\right )} \, dx\\ &=\int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2 (1+x)^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{x^2 (1+x)^2} \, dx\\ &=\frac {1}{2} \int \left (2+\frac {e^{-x+\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+2 e^{16} (1+x)^2 \left (1+\frac {\log (4)}{e^{16}}\right )\right )}{x^2 (1+x)^2}\right ) \, dx\\ &=x+\frac {1}{2} \int \frac {e^{-x+\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+2 e^{16} (1+x)^2 \left (1+\frac {\log (4)}{e^{16}}\right )\right )}{x^2 (1+x)^2} \, dx\\ &=x+\frac {1}{2} \int \left (\frac {e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}}}{(1+x)^2}+\frac {2 e^{-x+\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^{16}+\log (4)\right )}{x^2}\right ) \, dx\\ &=x+\frac {1}{2} \int \frac {e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}}}{(1+x)^2} \, dx+\left (e^{16}+\log (4)\right ) \int \frac {e^{-x+\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}}}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 29, normalized size = 1.00 \begin {gather*} x-\frac {e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}}}{2 (1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(2*x^2 + 4*x^3 + 2*x^4) + E^((2*(E^16 + Log[4]))/(E^x*x))*(E^x*x^2 + E^16*(2 + 4*x + 2*x^2) + (
2 + 4*x + 2*x^2)*Log[4]))/(E^x*(2*x^2 + 4*x^3 + 2*x^4)),x]

[Out]

x - E^((2*(E^16 + Log[4]))/(E^x*x))/(2*(1 + x))

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fricas [A]  time = 1.00, size = 35, normalized size = 1.21 \begin {gather*} \frac {2 \, x^{2} + 2 \, x - e^{\left (\frac {2 \, {\left (e^{16} + 2 \, \log \relax (2)\right )} e^{\left (-x\right )}}{x}\right )}}{2 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^2+2*(2*x^2+4*x+2)*log(2)+(2*x^2+4*x+2)*exp(16))*exp((2*log(2)+exp(16))/exp(x)/x)^2+(2*x^4
+4*x^3+2*x^2)*exp(x))/(2*x^4+4*x^3+2*x^2)/exp(x),x, algorithm="fricas")

[Out]

1/2*(2*x^2 + 2*x - e^(2*(e^16 + 2*log(2))*e^(-x)/x))/(x + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} e^{x} + {\left (x^{2} e^{x} + 2 \, {\left (x^{2} + 2 \, x + 1\right )} e^{16} + 4 \, {\left (x^{2} + 2 \, x + 1\right )} \log \relax (2)\right )} e^{\left (\frac {2 \, {\left (e^{16} + 2 \, \log \relax (2)\right )} e^{\left (-x\right )}}{x}\right )}\right )} e^{\left (-x\right )}}{2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^2+2*(2*x^2+4*x+2)*log(2)+(2*x^2+4*x+2)*exp(16))*exp((2*log(2)+exp(16))/exp(x)/x)^2+(2*x^4
+4*x^3+2*x^2)*exp(x))/(2*x^4+4*x^3+2*x^2)/exp(x),x, algorithm="giac")

[Out]

integrate(1/2*(2*(x^4 + 2*x^3 + x^2)*e^x + (x^2*e^x + 2*(x^2 + 2*x + 1)*e^16 + 4*(x^2 + 2*x + 1)*log(2))*e^(2*
(e^16 + 2*log(2))*e^(-x)/x))*e^(-x)/(x^4 + 2*x^3 + x^2), x)

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maple [A]  time = 0.24, size = 27, normalized size = 0.93

method	result	size
risch	\(x -\frac {{\mathrm e}^{\frac {2 \left (2 \ln \relax (2)+{\mathrm e}^{16}\right ) {\mathrm e}^{-x}}{x}}}{2 \left (x +1\right )}\)	\(27\)
norman	\(\frac {\left ({\mathrm e}^{x} x^{3}-{\mathrm e}^{x} x -\frac {{\mathrm e}^{x} x \,{\mathrm e}^{\frac {2 \left (2 \ln \relax (2)+{\mathrm e}^{16}\right ) {\mathrm e}^{-x}}{x}}}{2}\right ) {\mathrm e}^{-x}}{x \left (x +1\right )}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*x^2+2*(2*x^2+4*x+2)*ln(2)+(2*x^2+4*x+2)*exp(16))*exp((2*ln(2)+exp(16))/exp(x)/x)^2+(2*x^4+4*x^3+2
*x^2)*exp(x))/(2*x^4+4*x^3+2*x^2)/exp(x),x,method=_RETURNVERBOSE)

[Out]

x-1/2/(x+1)*exp(2*(2*ln(2)+exp(16))*exp(-x)/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {x^{2} + x - 1}{x + 1} + \frac {1}{x + 1} + \frac {1}{2} \, \int \frac {{\left (2 \, x^{2} {\left (e^{16} + 2 \, \log \relax (2)\right )} + x^{2} e^{x} + 4 \, x {\left (e^{16} + 2 \, \log \relax (2)\right )} + 2 \, e^{16} + 4 \, \log \relax (2)\right )} e^{\left (-x + \frac {4 \, e^{\left (-x\right )} \log \relax (2)}{x} + \frac {2 \, e^{\left (-x + 16\right )}}{x}\right )}}{x^{4} + 2 \, x^{3} + x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^2+2*(2*x^2+4*x+2)*log(2)+(2*x^2+4*x+2)*exp(16))*exp((2*log(2)+exp(16))/exp(x)/x)^2+(2*x^4
+4*x^3+2*x^2)*exp(x))/(2*x^4+4*x^3+2*x^2)/exp(x),x, algorithm="maxima")

[Out]

(x^2 + x - 1)/(x + 1) + 1/(x + 1) + 1/2*integrate((2*x^2*(e^16 + 2*log(2)) + x^2*e^x + 4*x*(e^16 + 2*log(2)) +
2*e^16 + 4*log(2))*e^(-x + 4*e^(-x)*log(2)/x + 2*e^(-x + 16)/x)/(x^4 + 2*x^3 + x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^x\,\left (2\,x^4+4\,x^3+2\,x^2\right )+{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{16}+2\,\ln \relax (2)\right )}{x}}\,\left (x^2\,{\mathrm {e}}^x+{\mathrm {e}}^{16}\,\left (2\,x^2+4\,x+2\right )+2\,\ln \relax (2)\,\left (2\,x^2+4\,x+2\right )\right )\right )}{2\,x^4+4\,x^3+2\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(exp(x)*(2*x^2 + 4*x^3 + 2*x^4) + exp((2*exp(-x)*(exp(16) + 2*log(2)))/x)*(x^2*exp(x) + exp(16)*(
4*x + 2*x^2 + 2) + 2*log(2)*(4*x + 2*x^2 + 2))))/(2*x^2 + 4*x^3 + 2*x^4),x)

[Out]

int((exp(-x)*(exp(x)*(2*x^2 + 4*x^3 + 2*x^4) + exp((2*exp(-x)*(exp(16) + 2*log(2)))/x)*(x^2*exp(x) + exp(16)*(
4*x + 2*x^2 + 2) + 2*log(2)*(4*x + 2*x^2 + 2))))/(2*x^2 + 4*x^3 + 2*x^4), x)

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sympy [A]  time = 0.28, size = 22, normalized size = 0.76 \begin {gather*} x - \frac {e^{\frac {2 \left (2 \log {\relax (2 )} + e^{16}\right ) e^{- x}}{x}}}{2 x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x**2+2*(2*x**2+4*x+2)*ln(2)+(2*x**2+4*x+2)*exp(16))*exp((2*ln(2)+exp(16))/exp(x)/x)**2+(2*x
**4+4*x**3+2*x**2)*exp(x))/(2*x**4+4*x**3+2*x**2)/exp(x),x)

[Out]

x - exp(2*(2*log(2) + exp(16))*exp(-x)/x)/(2*x + 2)

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