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

Optimal. Leaf size=33 \[ -5+\log \left (-x+\frac {1}{\left (-e^2+x\right ) \left (x-\log \left (1+e^{2-x}\right )\right )}\right ) \]

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Rubi [F]  time = 18.87, 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 x+e^4 x^2+x^4+e^2 \left (-1-2 x^3\right )+e^{2-x} \left (3 x+e^4 x^2+x^4+e^2 \left (-2-2 x^3\right )\right )+\left (-1-2 e^4 x+4 e^2 x^2-2 x^3+e^{2-x} \left (-1-2 e^4 x+4 e^2 x^2-2 x^3\right )\right ) \log \left (1+e^{2-x}\right )+\left (e^4-2 e^2 x+x^2+e^{2-x} \left (e^4-2 e^2 x+x^2\right )\right ) \log ^2\left (1+e^{2-x}\right )}{-x^2+e^4 x^3+x^5+e^2 \left (x-2 x^4\right )+e^{2-x} \left (-x^2+e^4 x^3+x^5+e^2 \left (x-2 x^4\right )\right )+\left (x-2 e^4 x^2-2 x^4+e^2 \left (-1+4 x^3\right )+e^{2-x} \left (x-2 e^4 x^2-2 x^4+e^2 \left (-1+4 x^3\right )\right )\right ) \log \left (1+e^{2-x}\right )+\left (e^4 x-2 e^2 x^2+x^3+e^{2-x} \left (e^4 x-2 e^2 x^2+x^3\right )\right ) \log ^2\left (1+e^{2-x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [B]  time = 0.22, size = 124, normalized size = 3.76 \begin {gather*} -\log \left (e^2-x\right )-\log \left (-x+\log \left (1+e^{2-x}\right )\right )+\log \left (1+2 e^2 x^2-2 x^3-e^2 x \left (x+\log \left (1+e^{2-x}\right )-\log \left (e^2+e^x\right )\right )+x^2 \left (x+\log \left (1+e^{2-x}\right )-\log \left (e^2+e^x\right )\right )-e^2 x \log \left (e^2+e^x\right )+x^2 \log \left (e^2+e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.61, size = 64, normalized size = 1.94 \begin {gather*} \log \relax (x) - \log \left (-x + \log \left (e^{\left (-x + 2\right )} + 1\right )\right ) + \log \left (\frac {x^{3} - x^{2} e^{2} - {\left (x^{2} - x e^{2}\right )} \log \left (e^{\left (-x + 2\right )} + 1\right ) - 1}{x^{2} - x e^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.70, size = 67, normalized size = 2.03 \begin {gather*} \log \left (-x^{3} + x^{2} e^{2} + x^{2} \log \left (e^{\left (-x + 2\right )} + 1\right ) - x e^{2} \log \left (e^{\left (-x + 2\right )} + 1\right ) + 1\right ) - \log \left (x - e^{2}\right ) - \log \left (x - \log \left (e^{\left (-x + 2\right )} + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 57, normalized size = 1.73

method	result	size
risch	\(\ln \relax (x )+\ln \left (\ln \left ({\mathrm e}^{2-x}+1\right )-\frac {x^{2} {\mathrm e}^{2}-x^{3}+1}{x \left ({\mathrm e}^{2}-x \right )}\right )-\ln \left (-x +\ln \left ({\mathrm e}^{2-x}+1\right )\right )\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 61, normalized size = 1.85 \begin {gather*} \log \relax (x) - \log \left (-2 \, x + \log \left (e^{2} + e^{x}\right )\right ) + \log \left (-\frac {2 \, x^{3} - 2 \, x^{2} e^{2} - {\left (x^{2} - x e^{2}\right )} \log \left (e^{2} + e^{x}\right ) - 1}{x^{2} - x e^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {2\,x+{\mathrm {e}}^{2-x}\,\left (3\,x-{\mathrm {e}}^2\,\left (2\,x^3+2\right )+x^2\,{\mathrm {e}}^4+x^4\right )-{\mathrm {e}}^2\,\left (2\,x^3+1\right )+x^2\,{\mathrm {e}}^4-\ln \left ({\mathrm {e}}^{2-x}+1\right )\,\left (2\,x\,{\mathrm {e}}^4+{\mathrm {e}}^{2-x}\,\left (2\,x^3-4\,{\mathrm {e}}^2\,x^2+2\,{\mathrm {e}}^4\,x+1\right )-4\,x^2\,{\mathrm {e}}^2+2\,x^3+1\right )+{\ln \left ({\mathrm {e}}^{2-x}+1\right )}^2\,\left ({\mathrm {e}}^4-2\,x\,{\mathrm {e}}^2+{\mathrm {e}}^{2-x}\,\left (x^2-2\,{\mathrm {e}}^2\,x+{\mathrm {e}}^4\right )+x^2\right )+x^4}{{\ln \left ({\mathrm {e}}^{2-x}+1\right )}^2\,\left ({\mathrm {e}}^{2-x}\,\left (x^3-2\,{\mathrm {e}}^2\,x^2+{\mathrm {e}}^4\,x\right )+x\,{\mathrm {e}}^4-2\,x^2\,{\mathrm {e}}^2+x^3\right )+x^3\,{\mathrm {e}}^4+\ln \left ({\mathrm {e}}^{2-x}+1\right )\,\left (x+{\mathrm {e}}^{2-x}\,\left (x+{\mathrm {e}}^2\,\left (4\,x^3-1\right )-2\,x^2\,{\mathrm {e}}^4-2\,x^4\right )+{\mathrm {e}}^2\,\left (4\,x^3-1\right )-2\,x^2\,{\mathrm {e}}^4-2\,x^4\right )+{\mathrm {e}}^2\,\left (x-2\,x^4\right )-x^2+x^5+{\mathrm {e}}^{2-x}\,\left (x^3\,{\mathrm {e}}^4+{\mathrm {e}}^2\,\left (x-2\,x^4\right )-x^2+x^5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + exp(2 - x)*(3*x - exp(2)*(2*x^3 + 2) + x^2*exp(4) + x^4) - exp(2)*(2*x^3 + 1) + x^2*exp(4) - log(ex
p(2 - x) + 1)*(2*x*exp(4) + exp(2 - x)*(2*x*exp(4) - 4*x^2*exp(2) + 2*x^3 + 1) - 4*x^2*exp(2) + 2*x^3 + 1) + l
og(exp(2 - x) + 1)^2*(exp(4) - 2*x*exp(2) + exp(2 - x)*(exp(4) - 2*x*exp(2) + x^2) + x^2) + x^4)/(log(exp(2 -
x) + 1)^2*(exp(2 - x)*(x*exp(4) - 2*x^2*exp(2) + x^3) + x*exp(4) - 2*x^2*exp(2) + x^3) + x^3*exp(4) + log(exp(
2 - x) + 1)*(x + exp(2 - x)*(x + exp(2)*(4*x^3 - 1) - 2*x^2*exp(4) - 2*x^4) + exp(2)*(4*x^3 - 1) - 2*x^2*exp(4
) - 2*x^4) + exp(2)*(x - 2*x^4) - x^2 + x^5 + exp(2 - x)*(x^3*exp(4) + exp(2)*(x - 2*x^4) - x^2 + x^5)),x)

[Out]

int((2*x + exp(2 - x)*(3*x - exp(2)*(2*x^3 + 2) + x^2*exp(4) + x^4) - exp(2)*(2*x^3 + 1) + x^2*exp(4) - log(ex
p(2 - x) + 1)*(2*x*exp(4) + exp(2 - x)*(2*x*exp(4) - 4*x^2*exp(2) + 2*x^3 + 1) - 4*x^2*exp(2) + 2*x^3 + 1) + l
og(exp(2 - x) + 1)^2*(exp(4) - 2*x*exp(2) + exp(2 - x)*(exp(4) - 2*x*exp(2) + x^2) + x^2) + x^4)/(log(exp(2 -
x) + 1)^2*(exp(2 - x)*(x*exp(4) - 2*x^2*exp(2) + x^3) + x*exp(4) - 2*x^2*exp(2) + x^3) + x^3*exp(4) + log(exp(
2 - x) + 1)*(x + exp(2 - x)*(x + exp(2)*(4*x^3 - 1) - 2*x^2*exp(4) - 2*x^4) + exp(2)*(4*x^3 - 1) - 2*x^2*exp(4
) - 2*x^4) + exp(2)*(x - 2*x^4) - x^2 + x^5 + exp(2 - x)*(x^3*exp(4) + exp(2)*(x - 2*x^4) - x^2 + x^5)), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(2)**2-2*exp(2)*x+x**2)*exp(2-x)+exp(2)**2-2*exp(2)*x+x**2)*ln(exp(2-x)+1)**2+((-2*x*exp(2)**2
+4*x**2*exp(2)-2*x**3-1)*exp(2-x)-2*x*exp(2)**2+4*x**2*exp(2)-2*x**3-1)*ln(exp(2-x)+1)+(x**2*exp(2)**2+(-2*x**
3-2)*exp(2)+x**4+3*x)*exp(2-x)+x**2*exp(2)**2+(-2*x**3-1)*exp(2)+x**4+2*x)/(((x*exp(2)**2-2*x**2*exp(2)+x**3)*
exp(2-x)+x*exp(2)**2-2*x**2*exp(2)+x**3)*ln(exp(2-x)+1)**2+((-2*x**2*exp(2)**2+(4*x**3-1)*exp(2)-2*x**4+x)*exp
(2-x)-2*x**2*exp(2)**2+(4*x**3-1)*exp(2)-2*x**4+x)*ln(exp(2-x)+1)+(x**3*exp(2)**2+(-2*x**4+x)*exp(2)+x**5-x**2
)*exp(2-x)+x**3*exp(2)**2+(-2*x**4+x)*exp(2)+x**5-x**2),x)

[Out]

Exception raised: PolynomialError

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