3.22.10 \(\int \frac {-100+50 x-4 x^2+e^{e^x} (-25+4 x^2+e^x (-25 x+20 x^2-4 x^3))}{16 x^2+e^{2 e^x} x^2-8 x^3+x^4+e^{e^x} (8 x^2-2 x^3)} \, dx\)

Optimal. Leaf size=26 \[ \frac {x \left (3-\frac {5+x}{x}\right )^2}{4+e^{e^x}-x} \]

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Rubi [F]  time = 3.11, 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 {-100+50 x-4 x^2+e^{e^x} \left (-25+4 x^2+e^x \left (-25 x+20 x^2-4 x^3\right )\right )}{16 x^2+e^{2 e^x} x^2-8 x^3+x^4+e^{e^x} \left (8 x^2-2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-100 + 50*x - 4*x^2 + E^E^x*(-25 + 4*x^2 + E^x*(-25*x + 20*x^2 - 4*x^3)))/(16*x^2 + E^(2*E^x)*x^2 - 8*x^3
+ x^4 + E^E^x*(8*x^2 - 2*x^3)),x]

[Out]

-20*Defer[Int][(4 + E^E^x - x)^(-2), x] + 20*Defer[Int][E^(E^x + x)/(4 + E^E^x - x)^2, x] + 4*Defer[Int][(4 +
E^E^x - x)^(-1), x] - 25*Defer[Int][1/((4 + E^E^x - x)*x^2), x] + 25*Defer[Int][1/((4 + E^E^x - x)^2*x), x] -
25*Defer[Int][E^(E^x + x)/((4 + E^E^x - x)^2*x), x] + 4*Defer[Int][x/(4 + E^E^x - x)^2, x] - 4*Defer[Int][(E^(
E^x + x)*x)/(4 + E^E^x - x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(5-2 x) \left (2 (-10+x)+e^{e^x+x} x (-5+2 x)-e^{e^x} (5+2 x)\right )}{\left (4+e^{e^x}-x\right )^2 x^2} \, dx\\ &=\int \left (-\frac {e^{e^x+x} (-5+2 x)^2}{\left (4+e^{e^x}-x\right )^2 x}+\frac {(-5+2 x) \left (20+5 e^{e^x}-2 x+2 e^{e^x} x\right )}{\left (4+e^{e^x}-x\right )^2 x^2}\right ) \, dx\\ &=-\int \frac {e^{e^x+x} (-5+2 x)^2}{\left (4+e^{e^x}-x\right )^2 x} \, dx+\int \frac {(-5+2 x) \left (20+5 e^{e^x}-2 x+2 e^{e^x} x\right )}{\left (4+e^{e^x}-x\right )^2 x^2} \, dx\\ &=-\int \left (-\frac {20 e^{e^x+x}}{\left (4+e^{e^x}-x\right )^2}+\frac {25 e^{e^x+x}}{\left (4+e^{e^x}-x\right )^2 x}+\frac {4 e^{e^x+x} x}{\left (4+e^{e^x}-x\right )^2}\right ) \, dx+\int \left (\frac {(-5+2 x)^2}{x \left (-4-e^{e^x}+x\right )^2}+\frac {-25+4 x^2}{\left (4+e^{e^x}-x\right ) x^2}\right ) \, dx\\ &=-\left (4 \int \frac {e^{e^x+x} x}{\left (4+e^{e^x}-x\right )^2} \, dx\right )+20 \int \frac {e^{e^x+x}}{\left (4+e^{e^x}-x\right )^2} \, dx-25 \int \frac {e^{e^x+x}}{\left (4+e^{e^x}-x\right )^2 x} \, dx+\int \frac {(-5+2 x)^2}{x \left (-4-e^{e^x}+x\right )^2} \, dx+\int \frac {-25+4 x^2}{\left (4+e^{e^x}-x\right ) x^2} \, dx\\ &=-\left (4 \int \frac {e^{e^x+x} x}{\left (4+e^{e^x}-x\right )^2} \, dx\right )+20 \int \frac {e^{e^x+x}}{\left (4+e^{e^x}-x\right )^2} \, dx-25 \int \frac {e^{e^x+x}}{\left (4+e^{e^x}-x\right )^2 x} \, dx+\int \left (\frac {4}{4+e^{e^x}-x}-\frac {25}{\left (4+e^{e^x}-x\right ) x^2}\right ) \, dx+\int \left (-\frac {20}{\left (4+e^{e^x}-x\right )^2}+\frac {25}{\left (4+e^{e^x}-x\right )^2 x}+\frac {4 x}{\left (4+e^{e^x}-x\right )^2}\right ) \, dx\\ &=4 \int \frac {1}{4+e^{e^x}-x} \, dx+4 \int \frac {x}{\left (4+e^{e^x}-x\right )^2} \, dx-4 \int \frac {e^{e^x+x} x}{\left (4+e^{e^x}-x\right )^2} \, dx-20 \int \frac {1}{\left (4+e^{e^x}-x\right )^2} \, dx+20 \int \frac {e^{e^x+x}}{\left (4+e^{e^x}-x\right )^2} \, dx-25 \int \frac {1}{\left (4+e^{e^x}-x\right ) x^2} \, dx+25 \int \frac {1}{\left (4+e^{e^x}-x\right )^2 x} \, dx-25 \int \frac {e^{e^x+x}}{\left (4+e^{e^x}-x\right )^2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.73, size = 23, normalized size = 0.88 \begin {gather*} \frac {(-5+2 x)^2}{\left (4+e^{e^x}-x\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100 + 50*x - 4*x^2 + E^E^x*(-25 + 4*x^2 + E^x*(-25*x + 20*x^2 - 4*x^3)))/(16*x^2 + E^(2*E^x)*x^2 -
8*x^3 + x^4 + E^E^x*(8*x^2 - 2*x^3)),x]

[Out]

(-5 + 2*x)^2/((4 + E^E^x - x)*x)

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fricas [A]  time = 0.63, size = 27, normalized size = 1.04 \begin {gather*} -\frac {4 \, x^{2} - 20 \, x + 25}{x^{2} - x e^{\left (e^{x}\right )} - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+20*x^2-25*x)*exp(x)+4*x^2-25)*exp(exp(x))-4*x^2+50*x-100)/(x^2*exp(exp(x))^2+(-2*x^3+8*x^2
)*exp(exp(x))+x^4-8*x^3+16*x^2),x, algorithm="fricas")

[Out]

-(4*x^2 - 20*x + 25)/(x^2 - x*e^(e^x) - 4*x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+20*x^2-25*x)*exp(x)+4*x^2-25)*exp(exp(x))-4*x^2+50*x-100)/(x^2*exp(exp(x))^2+(-2*x^3+8*x^2
)*exp(exp(x))+x^4-8*x^3+16*x^2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.09, size = 25, normalized size = 0.96

method	result	size
norman	\(\frac {-4 x^{2}+20 x -25}{x \left (x -{\mathrm e}^{{\mathrm e}^{x}}-4\right )}\)	\(25\)
risch	\(-\frac {4 x^{2}-20 x +25}{x \left (x -{\mathrm e}^{{\mathrm e}^{x}}-4\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^3+20*x^2-25*x)*exp(x)+4*x^2-25)*exp(exp(x))-4*x^2+50*x-100)/(x^2*exp(exp(x))^2+(-2*x^3+8*x^2)*exp(
exp(x))+x^4-8*x^3+16*x^2),x,method=_RETURNVERBOSE)

[Out]

(-4*x^2+20*x-25)/x/(x-exp(exp(x))-4)

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maxima [A]  time = 0.45, size = 27, normalized size = 1.04 \begin {gather*} -\frac {4 \, x^{2} - 20 \, x + 25}{x^{2} - x e^{\left (e^{x}\right )} - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^3+20*x^2-25*x)*exp(x)+4*x^2-25)*exp(exp(x))-4*x^2+50*x-100)/(x^2*exp(exp(x))^2+(-2*x^3+8*x^2
)*exp(exp(x))+x^4-8*x^3+16*x^2),x, algorithm="maxima")

[Out]

-(4*x^2 - 20*x + 25)/(x^2 - x*e^(e^x) - 4*x)

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mupad [B]  time = 1.22, size = 67, normalized size = 2.58 \begin {gather*} \frac {25\,x-105\,x^2\,{\mathrm {e}}^x+36\,x^3\,{\mathrm {e}}^x-4\,x^4\,{\mathrm {e}}^x+100\,x\,{\mathrm {e}}^x-20\,x^2+4\,x^3}{x^2\,\left (4\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x+1\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}-x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x))*(exp(x)*(25*x - 20*x^2 + 4*x^3) - 4*x^2 + 25) - 50*x + 4*x^2 + 100)/(exp(exp(x))*(8*x^2 - 2*
x^3) + 16*x^2 - 8*x^3 + x^4 + x^2*exp(2*exp(x))),x)

[Out]

(25*x - 105*x^2*exp(x) + 36*x^3*exp(x) - 4*x^4*exp(x) + 100*x*exp(x) - 20*x^2 + 4*x^3)/(x^2*(4*exp(x) - x*exp(
x) + 1)*(exp(exp(x)) - x + 4))

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sympy [A]  time = 0.16, size = 22, normalized size = 0.85 \begin {gather*} \frac {4 x^{2} - 20 x + 25}{- x^{2} + x e^{e^{x}} + 4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**3+20*x**2-25*x)*exp(x)+4*x**2-25)*exp(exp(x))-4*x**2+50*x-100)/(x**2*exp(exp(x))**2+(-2*x**
3+8*x**2)*exp(exp(x))+x**4-8*x**3+16*x**2),x)

[Out]

(4*x**2 - 20*x + 25)/(-x**2 + x*exp(exp(x)) + 4*x)

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