3.14.34 \(\int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x (24+2 x-2 x^2)}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x (24-2 x^2)} \, dx\)

Optimal. Leaf size=24 \[ 2+x-\frac {x}{12+e^{-5+2 x} x-x^2} \]

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Rubi [F]  time = 1.08, 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 {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(132 - 25*x^2 + E^(-10 + 4*x)*x^2 + x^4 + E^(-5 + 2*x)*x*(24 + 2*x - 2*x^2))/(144 - 24*x^2 + E^(-10 + 4*x)
*x^2 + x^4 + E^(-5 + 2*x)*x*(24 - 2*x^2)),x]

[Out]

x - 12*E^10*Defer[Int][(-12*E^5 - E^(2*x)*x + E^5*x^2)^(-2), x] - 24*E^10*Defer[Int][x/(-12*E^5 - E^(2*x)*x +
E^5*x^2)^2, x] - E^10*Defer[Int][x^2/(-12*E^5 - E^(2*x)*x + E^5*x^2)^2, x] + 2*E^10*Defer[Int][x^3/(-12*E^5 -
E^(2*x)*x + E^5*x^2)^2, x] - 2*E^5*Defer[Int][x/(-12*E^5 - E^(2*x)*x + E^5*x^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{10} \left (132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )\right )}{\left (12 e^5+e^{2 x} x-e^5 x^2\right )^2} \, dx\\ &=e^{10} \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{\left (12 e^5+e^{2 x} x-e^5 x^2\right )^2} \, dx\\ &=e^{10} \int \left (\frac {1}{e^{10}}-\frac {2 x}{e^5 \left (-12 e^5-e^{2 x} x+e^5 x^2\right )}+\frac {-12-24 x-x^2+2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}\right ) \, dx\\ &=x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx+e^{10} \int \frac {-12-24 x-x^2+2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx\\ &=x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx+e^{10} \int \left (-\frac {12}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}-\frac {24 x}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}-\frac {x^2}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}+\frac {2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}\right ) \, dx\\ &=x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx-e^{10} \int \frac {x^2}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx+\left (2 e^{10}\right ) \int \frac {x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx-\left (12 e^{10}\right ) \int \frac {1}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx-\left (24 e^{10}\right ) \int \frac {x}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 27, normalized size = 1.12 \begin {gather*} x+\frac {e^5 x}{-e^{2 x} x+e^5 \left (-12+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(132 - 25*x^2 + E^(-10 + 4*x)*x^2 + x^4 + E^(-5 + 2*x)*x*(24 + 2*x - 2*x^2))/(144 - 24*x^2 + E^(-10
+ 4*x)*x^2 + x^4 + E^(-5 + 2*x)*x*(24 - 2*x^2)),x]

[Out]

x + (E^5*x)/(-(E^(2*x)*x) + E^5*(-12 + x^2))

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fricas [A]  time = 0.98, size = 36, normalized size = 1.50 \begin {gather*} \frac {x^{3} - x e^{\left (2 \, x + \log \relax (x) - 5\right )} - 11 \, x}{x^{2} - e^{\left (2 \, x + \log \relax (x) - 5\right )} - 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2*x+log(x)-5)^2+(-2*x^2+2*x+24)*exp(2*x+log(x)-5)+x^4-25*x^2+132)/(exp(2*x+log(x)-5)^2+(-2*x^2+
24)*exp(2*x+log(x)-5)+x^4-24*x^2+144),x, algorithm="fricas")

[Out]

(x^3 - x*e^(2*x + log(x) - 5) - 11*x)/(x^2 - e^(2*x + log(x) - 5) - 12)

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giac [A]  time = 0.41, size = 42, normalized size = 1.75 \begin {gather*} \frac {x^{3} e^{5} - x^{2} e^{\left (2 \, x\right )} - 11 \, x e^{5}}{x^{2} e^{5} - x e^{\left (2 \, x\right )} - 12 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2*x+log(x)-5)^2+(-2*x^2+2*x+24)*exp(2*x+log(x)-5)+x^4-25*x^2+132)/(exp(2*x+log(x)-5)^2+(-2*x^2+
24)*exp(2*x+log(x)-5)+x^4-24*x^2+144),x, algorithm="giac")

[Out]

(x^3*e^5 - x^2*e^(2*x) - 11*x*e^5)/(x^2*e^5 - x*e^(2*x) - 12*e^5)

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maple [A]  time = 0.16, size = 21, normalized size = 0.88

method	result	size
risch	\(x +\frac {x}{x^{2}-x \,{\mathrm e}^{2 x -5}-12}\)	\(21\)
norman	\(\frac {x^{3}-11 x -x \,{\mathrm e}^{2 x +\ln \relax (x )-5}}{x^{2}-{\mathrm e}^{2 x +\ln \relax (x )-5}-12}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x+ln(x)-5)^2+(-2*x^2+2*x+24)*exp(2*x+ln(x)-5)+x^4-25*x^2+132)/(exp(2*x+ln(x)-5)^2+(-2*x^2+24)*exp(2
*x+ln(x)-5)+x^4-24*x^2+144),x,method=_RETURNVERBOSE)

[Out]

x+x/(x^2-x*exp(2*x-5)-12)

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maxima [A]  time = 0.47, size = 42, normalized size = 1.75 \begin {gather*} \frac {x^{3} e^{5} - x^{2} e^{\left (2 \, x\right )} - 11 \, x e^{5}}{x^{2} e^{5} - x e^{\left (2 \, x\right )} - 12 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2*x+log(x)-5)^2+(-2*x^2+2*x+24)*exp(2*x+log(x)-5)+x^4-25*x^2+132)/(exp(2*x+log(x)-5)^2+(-2*x^2+
24)*exp(2*x+log(x)-5)+x^4-24*x^2+144),x, algorithm="maxima")

[Out]

(x^3*e^5 - x^2*e^(2*x) - 11*x*e^5)/(x^2*e^5 - x*e^(2*x) - 12*e^5)

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mupad [B]  time = 1.04, size = 22, normalized size = 0.92 \begin {gather*} x-\frac {x}{x\,{\mathrm {e}}^{2\,x-5}-x^2+12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x + 2*log(x) - 10) + exp(2*x + log(x) - 5)*(2*x - 2*x^2 + 24) - 25*x^2 + x^4 + 132)/(exp(4*x + 2*lo
g(x) - 10) - exp(2*x + log(x) - 5)*(2*x^2 - 24) - 24*x^2 + x^4 + 144),x)

[Out]

x - x/(x*exp(2*x - 5) - x^2 + 12)

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sympy [A]  time = 0.16, size = 15, normalized size = 0.62 \begin {gather*} x - \frac {x}{- x^{2} + x e^{2 x - 5} + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2*x+ln(x)-5)**2+(-2*x**2+2*x+24)*exp(2*x+ln(x)-5)+x**4-25*x**2+132)/(exp(2*x+ln(x)-5)**2+(-2*x*
*2+24)*exp(2*x+ln(x)-5)+x**4-24*x**2+144),x)

[Out]

x - x/(-x**2 + x*exp(2*x - 5) + 12)

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