3.73.91 \(\int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} (-52 x+2 x^2+e^4 (-4 x+4 e^2 x)+e^2 (16 x-4 x^2))}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx\)

Optimal. Leaf size=32 \[ x-\frac {4 \left (4+e^4-x\right )}{-e^{x-e^2 x}+x}+16 \log (x) \]

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Rubi [F]  time = 6.91, 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 {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16*x + 4*E^4*x + 16*x^2 + x^3 + E^(2*x - 2*E^2*x)*(16 + x) + E^(x - E^2*x)*(-52*x + 2*x^2 + E^4*(-4*x + 4
*E^2*x) + E^2*(16*x - 4*x^2)))/(E^(2*x - 2*E^2*x)*x - 2*E^(x - E^2*x)*x^2 + x^3),x]

[Out]

-4/(E^((1 - E^2)*x)*(1 - E^2)) + x + (4*(5 - 4*E^2 + E^4 - E^6 - (1 - E^2)*x))/(E^((1 - E^2)*x)*(1 - E^2)) + 1
6*Log[x] + 4*(4 + E^4)*Defer[Int][E^(2*E^2*x)/(E^x - E^(E^2*x)*x)^2, x] - 4*(5 - 4*E^2 + E^4 - E^6)*Defer[Int]
[(E^(2*E^2*x)*x)/(-E^x + E^(E^2*x)*x)^2, x] + 4*(1 - E)*(1 + E)*Defer[Int][(E^(2*E^2*x)*x^2)/(-E^x + E^(E^2*x)
*x)^2, x] + 4*(5 - 4*E^2 + E^4 - E^6)*Defer[Int][x/(E^((1 - 2*E^2)*x)*(-E^x + E^(E^2*x)*x)), x] - 4*(1 - E)*(1
+ E)*Defer[Int][x^2/(E^((1 - 2*E^2)*x)*(-E^x + E^(E^2*x)*x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (16+4 e^4\right ) x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx\\ &=\int \frac {e^{2 e^2 x} \left (\left (16+4 e^4\right ) x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )\right )}{x \left (e^x-e^{e^2 x} x\right )^2} \, dx\\ &=\int \left (\frac {16+x}{x}+4 e^{2 e^2 x-\left (1+e^2\right ) x} \left (-5+4 e^2-e^4+e^6+\left (1-e^2\right ) x\right )+\frac {4 e^{-x+2 e^2 x} x \left (-5+4 e^2-e^4+e^6+\left (1-e^2\right ) x\right )}{e^x-e^{e^2 x} x}+\frac {4 e^{2 e^2 x} \left (4+e^4-\left (5-4 e^2+e^4-e^6\right ) x+\left (1-e^2\right ) x^2\right )}{\left (e^x-e^{e^2 x} x\right )^2}\right ) \, dx\\ &=4 \int e^{2 e^2 x-\left (1+e^2\right ) x} \left (-5+4 e^2-e^4+e^6+\left (1-e^2\right ) x\right ) \, dx+4 \int \frac {e^{-x+2 e^2 x} x \left (-5+4 e^2-e^4+e^6+\left (1-e^2\right ) x\right )}{e^x-e^{e^2 x} x} \, dx+4 \int \frac {e^{2 e^2 x} \left (4+e^4-\left (5-4 e^2+e^4-e^6\right ) x+\left (1-e^2\right ) x^2\right )}{\left (e^x-e^{e^2 x} x\right )^2} \, dx+\int \frac {16+x}{x} \, dx\\ &=4 \int e^{-\left (\left (1-e^2\right ) x\right )} \left (-5+4 e^2-e^4+e^6+\left (1-e^2\right ) x\right ) \, dx+4 \int \frac {e^{-\left (\left (1-2 e^2\right ) x\right )} x \left (-5+4 e^2-e^4+e^6+\left (1-e^2\right ) x\right )}{e^x-e^{e^2 x} x} \, dx+4 \int \left (\frac {4 e^{2 e^2 x} \left (1+\frac {e^4}{4}\right )}{\left (e^x-e^{e^2 x} x\right )^2}+\frac {e^{2 e^2 x} \left (-5+4 e^2-e^4+e^6\right ) x}{\left (-e^x+e^{e^2 x} x\right )^2}-\frac {(-1+e) e^{2 e^2 x} (1+e) x^2}{\left (-e^x+e^{e^2 x} x\right )^2}\right ) \, dx+\int \left (1+\frac {16}{x}\right ) \, dx\\ &=x+\frac {4 e^{-\left (\left (1-e^2\right ) x\right )} \left (5-4 e^2+e^4-e^6-\left (1-e^2\right ) x\right )}{1-e^2}+16 \log (x)+4 \int e^{\left (-1+e^2\right ) x} \, dx+4 \int \left (-\frac {e^{-\left (\left (1-2 e^2\right ) x\right )} \left (-5+4 e^2-e^4+e^6\right ) x}{-e^x+e^{e^2 x} x}+\frac {(-1+e) e^{-\left (\left (1-2 e^2\right ) x\right )} (1+e) x^2}{-e^x+e^{e^2 x} x}\right ) \, dx-(4 (-1+e) (1+e)) \int \frac {e^{2 e^2 x} x^2}{\left (-e^x+e^{e^2 x} x\right )^2} \, dx+\left (4 \left (4+e^4\right )\right ) \int \frac {e^{2 e^2 x}}{\left (e^x-e^{e^2 x} x\right )^2} \, dx-\left (4 \left (5-4 e^2+e^4-e^6\right )\right ) \int \frac {e^{2 e^2 x} x}{\left (-e^x+e^{e^2 x} x\right )^2} \, dx\\ &=-\frac {4 e^{-\left (\left (1-e^2\right ) x\right )}}{1-e^2}+x+\frac {4 e^{-\left (\left (1-e^2\right ) x\right )} \left (5-4 e^2+e^4-e^6-\left (1-e^2\right ) x\right )}{1-e^2}+16 \log (x)-(4 (-1+e) (1+e)) \int \frac {e^{2 e^2 x} x^2}{\left (-e^x+e^{e^2 x} x\right )^2} \, dx+(4 (-1+e) (1+e)) \int \frac {e^{-\left (\left (1-2 e^2\right ) x\right )} x^2}{-e^x+e^{e^2 x} x} \, dx+\left (4 \left (4+e^4\right )\right ) \int \frac {e^{2 e^2 x}}{\left (e^x-e^{e^2 x} x\right )^2} \, dx-\left (4 \left (5-4 e^2+e^4-e^6\right )\right ) \int \frac {e^{2 e^2 x} x}{\left (-e^x+e^{e^2 x} x\right )^2} \, dx+\left (4 \left (5-4 e^2+e^4-e^6\right )\right ) \int \frac {e^{-\left (\left (1-2 e^2\right ) x\right )} x}{-e^x+e^{e^2 x} x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 49, normalized size = 1.53 \begin {gather*} -\frac {4 \left (4+e^4\right )}{x}+x+\frac {4 e^x \left (-4-e^4+x\right )}{x \left (-e^x+e^{e^2 x} x\right )}+16 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x + 4*E^4*x + 16*x^2 + x^3 + E^(2*x - 2*E^2*x)*(16 + x) + E^(x - E^2*x)*(-52*x + 2*x^2 + E^4*(-4
*x + 4*E^2*x) + E^2*(16*x - 4*x^2)))/(E^(2*x - 2*E^2*x)*x - 2*E^(x - E^2*x)*x^2 + x^3),x]

[Out]

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

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fricas [A]  time = 0.92, size = 54, normalized size = 1.69 \begin {gather*} \frac {x^{2} - x e^{\left (-x e^{2} + x\right )} + 16 \, {\left (x - e^{\left (-x e^{2} + x\right )}\right )} \log \relax (x) + 4 \, x - 4 \, e^{4} - 16}{x - e^{\left (-x e^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x)*exp(2)+2*x^2-52*x)*exp(-exp(2)*x+x
)+4*x*exp(4)+x^3+16*x^2+16*x)/(x*exp(-exp(2)*x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x, algorithm="fricas")

[Out]

(x^2 - x*e^(-x*e^2 + x) + 16*(x - e^(-x*e^2 + x))*log(x) + 4*x - 4*e^4 - 16)/(x - e^(-x*e^2 + x))

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giac [A]  time = 0.29, size = 55, normalized size = 1.72 \begin {gather*} \frac {x^{2} - x e^{\left (-x e^{2} + x\right )} + 16 \, x \log \relax (x) - 16 \, e^{\left (-x e^{2} + x\right )} \log \relax (x) + 4 \, x - 4 \, e^{4} - 16}{x - e^{\left (-x e^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x)*exp(2)+2*x^2-52*x)*exp(-exp(2)*x+x
)+4*x*exp(4)+x^3+16*x^2+16*x)/(x*exp(-exp(2)*x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x, algorithm="giac")

[Out]

(x^2 - x*e^(-x*e^2 + x) + 16*x*log(x) - 16*e^(-x*e^2 + x)*log(x) + 4*x - 4*e^4 - 16)/(x - e^(-x*e^2 + x))

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maple [A]  time = 0.25, size = 30, normalized size = 0.94

method	result	size
risch	\(x +16 \ln \relax (x )-\frac {4 \left (4-x +{\mathrm e}^{4}\right )}{x -{\mathrm e}^{-\left ({\mathrm e}^{2}-1\right ) x}}\)	\(30\)
norman	\(\frac {x^{2}+4 \,{\mathrm e}^{-{\mathrm e}^{2} x +x}-x \,{\mathrm e}^{-{\mathrm e}^{2} x +x}-16-4 \,{\mathrm e}^{4}}{x -{\mathrm e}^{-{\mathrm e}^{2} x +x}}+16 \ln \relax (x )\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x)*exp(2)+2*x^2-52*x)*exp(-exp(2)*x+x)+4*x*
exp(4)+x^3+16*x^2+16*x)/(x*exp(-exp(2)*x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x,method=_RETURNVERBOSE)

[Out]

x+16*ln(x)-4*(4-x+exp(4))/(x-exp(-(exp(2)-1)*x))

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maxima [A]  time = 0.41, size = 43, normalized size = 1.34 \begin {gather*} \frac {{\left (x^{2} - 4 \, e^{4} - 16\right )} e^{\left (x e^{2}\right )} - {\left (x - 4\right )} e^{x}}{x e^{\left (x e^{2}\right )} - e^{x}} + 16 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x)*exp(2)+2*x^2-52*x)*exp(-exp(2)*x+x
)+4*x*exp(4)+x^3+16*x^2+16*x)/(x*exp(-exp(2)*x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x, algorithm="maxima")

[Out]

((x^2 - 4*e^4 - 16)*e^(x*e^2) - (x - 4)*e^x)/(x*e^(x*e^2) - e^x) + 16*log(x)

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mupad [B]  time = 0.27, size = 45, normalized size = 1.41 \begin {gather*} 16\,\ln \relax (x)-\frac {4\,{\mathrm {e}}^4-4\,x+x\,{\mathrm {e}}^{x-x\,{\mathrm {e}}^2}-x^2+16}{x-{\mathrm {e}}^{x-x\,{\mathrm {e}}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + exp(2*x - 2*x*exp(2))*(x + 16) + 4*x*exp(4) - exp(x - x*exp(2))*(52*x - exp(2)*(16*x - 4*x^2) + ex
p(4)*(4*x - 4*x*exp(2)) - 2*x^2) + 16*x^2 + x^3)/(x*exp(2*x - 2*x*exp(2)) - 2*x^2*exp(x - x*exp(2)) + x^3),x)

[Out]

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

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sympy [A]  time = 0.20, size = 26, normalized size = 0.81 \begin {gather*} x + 16 \log {\relax (x )} + \frac {- 4 x + 16 + 4 e^{4}}{- x + e^{- x e^{2} + x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+16)*exp(-exp(2)*x+x)**2+((4*exp(2)*x-4*x)*exp(4)+(-4*x**2+16*x)*exp(2)+2*x**2-52*x)*exp(-exp(2)*
x+x)+4*x*exp(4)+x**3+16*x**2+16*x)/(x*exp(-exp(2)*x+x)**2-2*x**2*exp(-exp(2)*x+x)+x**3),x)

[Out]

x + 16*log(x) + (-4*x + 16 + 4*exp(4))/(-x + exp(-x*exp(2) + x))

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