\(\int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x (24+2 x-2 x^2)}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x (24-2 x^2)} \, dx\) [1334]

Optimal result

Integrand size = 80, antiderivative size = 24 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=2+x-\frac {x}{12+e^{-5+2 x} x-x^2} \]

[Out]

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

Rubi [F]

\[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=\int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 3.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=x+\frac {e^5 x}{-e^{2 x} x+e^5 \left (-12+x^2\right )} \]

[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))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

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

[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(-5+2*x)-12)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=\frac {x^{3} - x e^{\left (2 \, x + \log \left (x\right ) - 5\right )} - 11 \, x}{x^{2} - e^{\left (2 \, x + \log \left (x\right ) - 5\right )} - 12} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=x - \frac {x}{- x^{2} + x e^{2 x - 5} + 12} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=\frac {x^{3} e^{5} - x^{2} e^{\left (2 \, x\right )} - 11 \, x e^{5}}{x^{2} e^{5} - x e^{\left (2 \, x\right )} - 12 \, e^{5}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=\frac {x^{3} e^{5} - x^{2} e^{\left (2 \, x\right )} - 11 \, x e^{5}}{x^{2} e^{5} - x e^{\left (2 \, x\right )} - 12 \, e^{5}} \]

[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)

Mupad [B] (verification not implemented)

Time = 8.93 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=x-\frac {x}{x\,{\mathrm {e}}^{2\,x-5}-x^2+12} \]

[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)