\(\int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x (-5-15 x+19 x^2)}{36 x^2+72 x^3+36 x^4+e^x (-36 x^2-72 x^3-36 x^4)+e^{2 x} (9 x^2+18 x^3+9 x^4)} \, dx\) [6478]

Optimal result

Integrand size = 95, antiderivative size = 30 \[ \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx=\frac {5-\frac {5}{\left (2-e^x\right ) x}-x}{3 (3+3 x)} \]

[Out]

1/3*(5-x-5/x/(-exp(x)+2))/(3*x+3)

Rubi [F]

\[ \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx=\int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx \]

[In]

Int[(10 + 20*x - 24*x^2 - 6*E^(2*x)*x^2 + E^x*(-5 - 15*x + 19*x^2))/(36*x^2 + 72*x^3 + 36*x^4 + E^x*(-36*x^2 -
72*x^3 - 36*x^4) + E^(2*x)*(9*x^2 + 18*x^3 + 9*x^4)),x]

[Out]

2/(3*(1 + x)) - (5*Defer[Int][1/((-2 + E^x)*x^2), x])/9 - (10*Defer[Int][1/((-2 + E^x)^2*x), x])/9 - (5*Defer[
Int][1/((-2 + E^x)*x), x])/9 + (5*Defer[Int][1/((-2 + E^x)*(1 + x)^2), x])/9 + (10*Defer[Int][1/((-2 + E^x)^2*
(1 + x)), x])/9 + (5*Defer[Int][1/((-2 + E^x)*(1 + x)), x])/9

Rubi steps \begin{align*} \text {integral}& = \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{9 \left (2-e^x\right )^2 x^2 (1+x)^2} \, dx \\ & = \frac {1}{9} \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{\left (2-e^x\right )^2 x^2 (1+x)^2} \, dx \\ & = \frac {1}{9} \int \left (-\frac {6}{(1+x)^2}-\frac {10}{\left (-2+e^x\right )^2 x (1+x)}-\frac {5 \left (1+3 x+x^2\right )}{\left (-2+e^x\right ) x^2 (1+x)^2}\right ) \, dx \\ & = \frac {2}{3 (1+x)}-\frac {5}{9} \int \frac {1+3 x+x^2}{\left (-2+e^x\right ) x^2 (1+x)^2} \, dx-\frac {10}{9} \int \frac {1}{\left (-2+e^x\right )^2 x (1+x)} \, dx \\ & = \frac {2}{3 (1+x)}-\frac {5}{9} \int \left (\frac {1}{\left (-2+e^x\right ) x^2}+\frac {1}{\left (-2+e^x\right ) x}-\frac {1}{\left (-2+e^x\right ) (1+x)^2}-\frac {1}{\left (-2+e^x\right ) (1+x)}\right ) \, dx-\frac {10}{9} \int \left (\frac {1}{\left (-2+e^x\right )^2 x}-\frac {1}{\left (-2+e^x\right )^2 (1+x)}\right ) \, dx \\ & = \frac {2}{3 (1+x)}-\frac {5}{9} \int \frac {1}{\left (-2+e^x\right ) x^2} \, dx-\frac {5}{9} \int \frac {1}{\left (-2+e^x\right ) x} \, dx+\frac {5}{9} \int \frac {1}{\left (-2+e^x\right ) (1+x)^2} \, dx+\frac {5}{9} \int \frac {1}{\left (-2+e^x\right ) (1+x)} \, dx-\frac {10}{9} \int \frac {1}{\left (-2+e^x\right )^2 x} \, dx+\frac {10}{9} \int \frac {1}{\left (-2+e^x\right )^2 (1+x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx=\frac {6+\frac {5}{\left (-2+e^x\right ) x}}{9 (1+x)} \]

[In]

Integrate[(10 + 20*x - 24*x^2 - 6*E^(2*x)*x^2 + E^x*(-5 - 15*x + 19*x^2))/(36*x^2 + 72*x^3 + 36*x^4 + E^x*(-36
*x^2 - 72*x^3 - 36*x^4) + E^(2*x)*(9*x^2 + 18*x^3 + 9*x^4)),x]

[Out]

(6 + 5/((-2 + E^x)*x))/(9*(1 + x))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {2}{3 \left (1+x \right )}+\frac {5}{9 x \left (1+x \right ) \left ({\mathrm e}^{x}-2\right )}\)	\(25\)
norman	\(\frac {\frac {5}{9}-\frac {4 x}{3}+\frac {2 \,{\mathrm e}^{x} x}{3}}{x \left ({\mathrm e}^{x}-2\right ) \left (1+x \right )}\)	\(26\)
parallelrisch	\(\frac {6 \,{\mathrm e}^{x} x -12 x +5}{9 x \left (1+x \right ) \left ({\mathrm e}^{x}-2\right )}\)	\(27\)

[In]

int((-6*exp(x)^2*x^2+(19*x^2-15*x-5)*exp(x)-24*x^2+20*x+10)/((9*x^4+18*x^3+9*x^2)*exp(x)^2+(-36*x^4-72*x^3-36*
x^2)*exp(x)+36*x^4+72*x^3+36*x^2),x,method=_RETURNVERBOSE)

[Out]

2/3/(1+x)+5/9/x/(1+x)/(exp(x)-2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx=-\frac {6 \, x e^{x} - 12 \, x + 5}{9 \, {\left (2 \, x^{2} - {\left (x^{2} + x\right )} e^{x} + 2 \, x\right )}} \]

[In]

integrate((-6*exp(x)^2*x^2+(19*x^2-15*x-5)*exp(x)-24*x^2+20*x+10)/((9*x^4+18*x^3+9*x^2)*exp(x)^2+(-36*x^4-72*x
^3-36*x^2)*exp(x)+36*x^4+72*x^3+36*x^2),x, algorithm="fricas")

[Out]

-1/9*(6*x*e^x - 12*x + 5)/(2*x^2 - (x^2 + x)*e^x + 2*x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx=\frac {5}{- 18 x^{2} - 18 x + \left (9 x^{2} + 9 x\right ) e^{x}} + \frac {2}{3 x + 3} \]

[In]

integrate((-6*exp(x)**2*x**2+(19*x**2-15*x-5)*exp(x)-24*x**2+20*x+10)/((9*x**4+18*x**3+9*x**2)*exp(x)**2+(-36*
x**4-72*x**3-36*x**2)*exp(x)+36*x**4+72*x**3+36*x**2),x)

[Out]

5/(-18*x**2 - 18*x + (9*x**2 + 9*x)*exp(x)) + 2/(3*x + 3)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx=-\frac {6 \, x e^{x} - 12 \, x + 5}{9 \, {\left (2 \, x^{2} - {\left (x^{2} + x\right )} e^{x} + 2 \, x\right )}} \]

[In]

integrate((-6*exp(x)^2*x^2+(19*x^2-15*x-5)*exp(x)-24*x^2+20*x+10)/((9*x^4+18*x^3+9*x^2)*exp(x)^2+(-36*x^4-72*x
^3-36*x^2)*exp(x)+36*x^4+72*x^3+36*x^2),x, algorithm="maxima")

[Out]

-1/9*(6*x*e^x - 12*x + 5)/(2*x^2 - (x^2 + x)*e^x + 2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx=\frac {6 \, x e^{x} - 12 \, x + 5}{9 \, {\left (x^{2} e^{x} - 2 \, x^{2} + x e^{x} - 2 \, x\right )}} \]

[In]

integrate((-6*exp(x)^2*x^2+(19*x^2-15*x-5)*exp(x)-24*x^2+20*x+10)/((9*x^4+18*x^3+9*x^2)*exp(x)^2+(-36*x^4-72*x
^3-36*x^2)*exp(x)+36*x^4+72*x^3+36*x^2),x, algorithm="giac")

[Out]

1/9*(6*x*e^x - 12*x + 5)/(x^2*e^x - 2*x^2 + x*e^x - 2*x)

Mupad [B] (verification not implemented)

Time = 13.94 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {10+20 x-24 x^2-6 e^{2 x} x^2+e^x \left (-5-15 x+19 x^2\right )}{36 x^2+72 x^3+36 x^4+e^x \left (-36 x^2-72 x^3-36 x^4\right )+e^{2 x} \left (9 x^2+18 x^3+9 x^4\right )} \, dx=\frac {5}{9\,x\,\left ({\mathrm {e}}^x-2\right )\,\left (x+1\right )}-\frac {2\,x}{3\,\left (x+1\right )} \]

[In]

int(-(6*x^2*exp(2*x) - 20*x + exp(x)*(15*x - 19*x^2 + 5) + 24*x^2 - 10)/(exp(2*x)*(9*x^2 + 18*x^3 + 9*x^4) - e
xp(x)*(36*x^2 + 72*x^3 + 36*x^4) + 36*x^2 + 72*x^3 + 36*x^4),x)

[Out]

5/(9*x*(exp(x) - 2)*(x + 1)) - (2*x)/(3*(x + 1))