3.31.12 \(\int \frac {-8+10 e^3 x^2+e^{-2+x} (4+e^3 (-10 x+5 x^2))+x^{x^2} (-10+e^{-2+x} (5-5 x)+10 x^2+(-10 e^{-2+x} x+20 x^2) \log (x))}{e^{-4+2 x}-4 e^{-2+x} x+4 x^2} \, dx\)

Optimal. Leaf size=31 \[ \frac {4+5 \left (e^3 x^2+x^{x^2}\right )}{-e^{-2+x}+2 x} \]

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Rubi [F]  time = 2.55, 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 {-8+10 e^3 x^2+e^{-2+x} \left (4+e^3 \left (-10 x+5 x^2\right )\right )+x^{x^2} \left (-10+e^{-2+x} (5-5 x)+10 x^2+\left (-10 e^{-2+x} x+20 x^2\right ) \log (x)\right )}{e^{-4+2 x}-4 e^{-2+x} x+4 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 + 10*E^3*x^2 + E^(-2 + x)*(4 + E^3*(-10*x + 5*x^2)) + x^x^2*(-10 + E^(-2 + x)*(5 - 5*x) + 10*x^2 + (-1
0*E^(-2 + x)*x + 20*x^2)*Log[x]))/(E^(-4 + 2*x) - 4*E^(-2 + x)*x + 4*x^2),x]

[Out]

-8*E^4*Defer[Int][(E^x - 2*E^2*x)^(-2), x] - 10*E^4*Defer[Int][x^x^2/(E^x - 2*E^2*x)^2, x] + 4*E^2*Defer[Int][
(E^x - 2*E^2*x)^(-1), x] + 5*E^2*Defer[Int][x^x^2/(E^x - 2*E^2*x), x] + 8*E^4*Defer[Int][x/(-E^x + 2*E^2*x)^2,
x] - 10*E^7*Defer[Int][x^2/(-E^x + 2*E^2*x)^2, x] + 10*E^7*Defer[Int][x^3/(-E^x + 2*E^2*x)^2, x] + 10*E^4*Def
er[Int][x^(1 + x^2)/(-E^x + 2*E^2*x)^2, x] + 10*E^5*Defer[Int][x/(-E^x + 2*E^2*x), x] - 5*E^5*Defer[Int][x^2/(
-E^x + 2*E^2*x), x] + 5*E^2*Defer[Int][x^(1 + x^2)/(-E^x + 2*E^2*x), x] + 10*E^2*Log[x]*Defer[Int][x^(1 + x^2)
/(-E^x + 2*E^2*x), x] - 10*E^2*Defer[Int][Defer[Int][x^(1 + x^2)/(-E^x + 2*E^2*x), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^4 \left (-8+10 e^3 x^2+e^{-2+x} \left (4+e^3 \left (-10 x+5 x^2\right )\right )+x^{x^2} \left (-10+e^{-2+x} (5-5 x)+10 x^2+\left (-10 e^{-2+x} x+20 x^2\right ) \log (x)\right )\right )}{\left (e^x-2 e^2 x\right )^2} \, dx\\ &=e^4 \int \frac {-8+10 e^3 x^2+e^{-2+x} \left (4+e^3 \left (-10 x+5 x^2\right )\right )+x^{x^2} \left (-10+e^{-2+x} (5-5 x)+10 x^2+\left (-10 e^{-2+x} x+20 x^2\right ) \log (x)\right )}{\left (e^x-2 e^2 x\right )^2} \, dx\\ &=e^4 \int \left (\frac {-8 e^2+4 e^x-10 e^{3+x} x+10 e^5 x^2+5 e^{3+x} x^2}{e^2 \left (-e^x+2 e^2 x\right )^2}+\frac {5 x^{x^2} \left (-2 e^2+e^x-e^x x+2 e^2 x^2-2 e^x x \log (x)+4 e^2 x^2 \log (x)\right )}{e^2 \left (-e^x+2 e^2 x\right )^2}\right ) \, dx\\ &=e^2 \int \frac {-8 e^2+4 e^x-10 e^{3+x} x+10 e^5 x^2+5 e^{3+x} x^2}{\left (-e^x+2 e^2 x\right )^2} \, dx+\left (5 e^2\right ) \int \frac {x^{x^2} \left (-2 e^2+e^x-e^x x+2 e^2 x^2-2 e^x x \log (x)+4 e^2 x^2 \log (x)\right )}{\left (-e^x+2 e^2 x\right )^2} \, dx\\ &=e^2 \int \left (\frac {2 e^2 (-1+x) \left (4+5 e^3 x^2\right )}{\left (-e^x+2 e^2 x\right )^2}+\frac {4-10 e^3 x+5 e^3 x^2}{e^x-2 e^2 x}\right ) \, dx+\left (5 e^2\right ) \int \left (\frac {2 e^2 (-1+x) x^{x^2}}{\left (-e^x+2 e^2 x\right )^2}-\frac {x^{x^2} (-1+x+2 x \log (x))}{e^x-2 e^2 x}\right ) \, dx\\ &=e^2 \int \frac {4-10 e^3 x+5 e^3 x^2}{e^x-2 e^2 x} \, dx-\left (5 e^2\right ) \int \frac {x^{x^2} (-1+x+2 x \log (x))}{e^x-2 e^2 x} \, dx+\left (2 e^4\right ) \int \frac {(-1+x) \left (4+5 e^3 x^2\right )}{\left (-e^x+2 e^2 x\right )^2} \, dx+\left (10 e^4\right ) \int \frac {(-1+x) x^{x^2}}{\left (-e^x+2 e^2 x\right )^2} \, dx\\ &=e^2 \int \left (\frac {4}{e^x-2 e^2 x}+\frac {10 e^3 x}{-e^x+2 e^2 x}-\frac {5 e^3 x^2}{-e^x+2 e^2 x}\right ) \, dx-\left (5 e^2\right ) \int \left (-\frac {x^{x^2}}{e^x-2 e^2 x}-\frac {x^{1+x^2}}{-e^x+2 e^2 x}-\frac {2 x^{1+x^2} \log (x)}{-e^x+2 e^2 x}\right ) \, dx+\left (2 e^4\right ) \int \left (-\frac {4}{\left (e^x-2 e^2 x\right )^2}+\frac {4 x}{\left (-e^x+2 e^2 x\right )^2}-\frac {5 e^3 x^2}{\left (-e^x+2 e^2 x\right )^2}+\frac {5 e^3 x^3}{\left (-e^x+2 e^2 x\right )^2}\right ) \, dx+\left (10 e^4\right ) \int \left (-\frac {x^{x^2}}{\left (e^x-2 e^2 x\right )^2}+\frac {x^{1+x^2}}{\left (-e^x+2 e^2 x\right )^2}\right ) \, dx\\ &=\left (4 e^2\right ) \int \frac {1}{e^x-2 e^2 x} \, dx+\left (5 e^2\right ) \int \frac {x^{x^2}}{e^x-2 e^2 x} \, dx+\left (5 e^2\right ) \int \frac {x^{1+x^2}}{-e^x+2 e^2 x} \, dx+\left (10 e^2\right ) \int \frac {x^{1+x^2} \log (x)}{-e^x+2 e^2 x} \, dx-\left (8 e^4\right ) \int \frac {1}{\left (e^x-2 e^2 x\right )^2} \, dx+\left (8 e^4\right ) \int \frac {x}{\left (-e^x+2 e^2 x\right )^2} \, dx-\left (10 e^4\right ) \int \frac {x^{x^2}}{\left (e^x-2 e^2 x\right )^2} \, dx+\left (10 e^4\right ) \int \frac {x^{1+x^2}}{\left (-e^x+2 e^2 x\right )^2} \, dx-\left (5 e^5\right ) \int \frac {x^2}{-e^x+2 e^2 x} \, dx+\left (10 e^5\right ) \int \frac {x}{-e^x+2 e^2 x} \, dx-\left (10 e^7\right ) \int \frac {x^2}{\left (-e^x+2 e^2 x\right )^2} \, dx+\left (10 e^7\right ) \int \frac {x^3}{\left (-e^x+2 e^2 x\right )^2} \, dx\\ &=\left (4 e^2\right ) \int \frac {1}{e^x-2 e^2 x} \, dx+\left (5 e^2\right ) \int \frac {x^{x^2}}{e^x-2 e^2 x} \, dx+\left (5 e^2\right ) \int \frac {x^{1+x^2}}{-e^x+2 e^2 x} \, dx-\left (10 e^2\right ) \int \frac {\int \frac {x^{1+x^2}}{-e^x+2 e^2 x} \, dx}{x} \, dx-\left (8 e^4\right ) \int \frac {1}{\left (e^x-2 e^2 x\right )^2} \, dx+\left (8 e^4\right ) \int \frac {x}{\left (-e^x+2 e^2 x\right )^2} \, dx-\left (10 e^4\right ) \int \frac {x^{x^2}}{\left (e^x-2 e^2 x\right )^2} \, dx+\left (10 e^4\right ) \int \frac {x^{1+x^2}}{\left (-e^x+2 e^2 x\right )^2} \, dx-\left (5 e^5\right ) \int \frac {x^2}{-e^x+2 e^2 x} \, dx+\left (10 e^5\right ) \int \frac {x}{-e^x+2 e^2 x} \, dx-\left (10 e^7\right ) \int \frac {x^2}{\left (-e^x+2 e^2 x\right )^2} \, dx+\left (10 e^7\right ) \int \frac {x^3}{\left (-e^x+2 e^2 x\right )^2} \, dx+\left (10 e^2 \log (x)\right ) \int \frac {x^{1+x^2}}{-e^x+2 e^2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 34, normalized size = 1.10 \begin {gather*} -\frac {e^2 \left (4+5 e^3 x^2+5 x^{x^2}\right )}{e^x-2 e^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 10*E^3*x^2 + E^(-2 + x)*(4 + E^3*(-10*x + 5*x^2)) + x^x^2*(-10 + E^(-2 + x)*(5 - 5*x) + 10*x^2
+ (-10*E^(-2 + x)*x + 20*x^2)*Log[x]))/(E^(-4 + 2*x) - 4*E^(-2 + x)*x + 4*x^2),x]

[Out]

-((E^2*(4 + 5*E^3*x^2 + 5*x^x^2))/(E^x - 2*E^2*x))

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fricas [A]  time = 1.75, size = 29, normalized size = 0.94 \begin {gather*} \frac {5 \, x^{2} e^{3} + 5 \, x^{\left (x^{2}\right )} + 4}{2 \, x - e^{\left (x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x*exp(x-2)+20*x^2)*log(x)+(-5*x+5)*exp(x-2)+10*x^2-10)*exp(x^2*log(x))+((5*x^2-10*x)*exp(3)+4
)*exp(x-2)+10*x^2*exp(3)-8)/(exp(x-2)^2-4*x*exp(x-2)+4*x^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.90, size = 133, normalized size = 4.29 \begin {gather*} \frac {10 \, x^{3} e^{7}}{4 \, x^{2} e^{4} - 4 \, x e^{\left (x + 2\right )} + e^{\left (2 \, x\right )}} - \frac {5 \, x^{2} e^{\left (x + 5\right )}}{4 \, x^{2} e^{4} - 4 \, x e^{\left (x + 2\right )} + e^{\left (2 \, x\right )}} + \frac {8 \, x e^{4}}{4 \, x^{2} e^{4} - 4 \, x e^{\left (x + 2\right )} + e^{\left (2 \, x\right )}} + \frac {5 \, x^{\left (x^{2}\right )} e^{2}}{2 \, x e^{2} - e^{x}} - \frac {4 \, e^{\left (x + 2\right )}}{4 \, x^{2} e^{4} - 4 \, x e^{\left (x + 2\right )} + e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x*exp(x-2)+20*x^2)*log(x)+(-5*x+5)*exp(x-2)+10*x^2-10)*exp(x^2*log(x))+((5*x^2-10*x)*exp(3)+4
)*exp(x-2)+10*x^2*exp(3)-8)/(exp(x-2)^2-4*x*exp(x-2)+4*x^2),x, algorithm="giac")

[Out]

10*x^3*e^7/(4*x^2*e^4 - 4*x*e^(x + 2) + e^(2*x)) - 5*x^2*e^(x + 5)/(4*x^2*e^4 - 4*x*e^(x + 2) + e^(2*x)) + 8*x
*e^4/(4*x^2*e^4 - 4*x*e^(x + 2) + e^(2*x)) + 5*x^(x^2)*e^2/(2*x*e^2 - e^x) - 4*e^(x + 2)/(4*x^2*e^4 - 4*x*e^(x
+ 2) + e^(2*x))

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maple [A]  time = 0.17, size = 43, normalized size = 1.39

method	result	size
risch	\(\frac {4+5 x^{2} {\mathrm e}^{3}}{2 x -{\mathrm e}^{x -2}}+\frac {5 x^{x^{2}}}{2 x -{\mathrm e}^{x -2}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-10*x*exp(x-2)+20*x^2)*ln(x)+(-5*x+5)*exp(x-2)+10*x^2-10)*exp(x^2*ln(x))+((5*x^2-10*x)*exp(3)+4)*exp(x-
2)+10*x^2*exp(3)-8)/(exp(x-2)^2-4*x*exp(x-2)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

(4+5*x^2*exp(3))/(2*x-exp(x-2))+5/(2*x-exp(x-2))*x^(x^2)

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maxima [A]  time = 0.52, size = 36, normalized size = 1.16 \begin {gather*} \frac {5 \, x^{2} e^{5} + 4 \, e^{2} + 5 \, e^{\left (x^{2} \log \relax (x) + 2\right )}}{2 \, x e^{2} - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x*exp(x-2)+20*x^2)*log(x)+(-5*x+5)*exp(x-2)+10*x^2-10)*exp(x^2*log(x))+((5*x^2-10*x)*exp(3)+4
)*exp(x-2)+10*x^2*exp(3)-8)/(exp(x-2)^2-4*x*exp(x-2)+4*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.00, size = 29, normalized size = 0.94 \begin {gather*} \frac {5\,x^2\,{\mathrm {e}}^3+5\,x^{x^2}+4}{2\,x-{\mathrm {e}}^{x-2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 2)*(exp(3)*(10*x - 5*x^2) - 4) + exp(x^2*log(x))*(exp(x - 2)*(5*x - 5) - 10*x^2 + log(x)*(10*x*e
xp(x - 2) - 20*x^2) + 10) - 10*x^2*exp(3) + 8)/(exp(2*x - 4) - 4*x*exp(x - 2) + 4*x^2),x)

[Out]

(5*x^2*exp(3) + 5*x^(x^2) + 4)/(2*x - exp(x - 2))

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sympy [A]  time = 0.40, size = 29, normalized size = 0.94 \begin {gather*} \frac {- 5 x^{2} e^{3} - 5 e^{x^{2} \log {\relax (x )}} - 4}{- 2 x + e^{x - 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x*exp(x-2)+20*x**2)*ln(x)+(-5*x+5)*exp(x-2)+10*x**2-10)*exp(x**2*ln(x))+((5*x**2-10*x)*exp(3)
+4)*exp(x-2)+10*x**2*exp(3)-8)/(exp(x-2)**2-4*x*exp(x-2)+4*x**2),x)

[Out]

(-5*x**2*exp(3) - 5*exp(x**2*log(x)) - 4)/(-2*x + exp(x - 2))

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