\(\int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x (-2400 x+90 x^2+50 x^3)}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx\) [9042]

Optimal result

Integrand size = 69, antiderivative size = 34 \[ \int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx=e^4+\frac {x^2}{2-\frac {x}{-4 e^x+\frac {1}{2} \left (12-\frac {x}{5}\right )}} \]

[Out]

exp(4)+x^2/(2-x/(6-1/10*x-4*exp(x)))

Rubi [F]

\[ \int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx=\int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx \]

[In]

Int[(1800*x + 800*E^(2*x)*x - 135*x^2 + 3*x^3 + E^x*(-2400*x + 90*x^2 + 50*x^3))/(1800 + 800*E^(2*x) - 360*x +
18*x^2 + E^x*(-2400 + 240*x)),x]

[Out]

x^2/2 + (165*Defer[Int][x^3/(-30 + 20*E^x + 3*x)^2, x])/4 - (15*Defer[Int][x^4/(-30 + 20*E^x + 3*x)^2, x])/4 -
(15*Defer[Int][x^2/(-30 + 20*E^x + 3*x), x])/4 + (5*Defer[Int][x^3/(-30 + 20*E^x + 3*x), x])/4

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (800 e^{2 x}+3 \left (600-45 x+x^2\right )+10 e^x \left (-240+9 x+5 x^2\right )\right )}{2 \left (30-20 e^x-3 x\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {x \left (800 e^{2 x}+3 \left (600-45 x+x^2\right )+10 e^x \left (-240+9 x+5 x^2\right )\right )}{\left (30-20 e^x-3 x\right )^2} \, dx \\ & = \frac {1}{2} \int \left (2 x-\frac {15 (-11+x) x^3}{2 \left (-30+20 e^x+3 x\right )^2}+\frac {5 (-3+x) x^2}{2 \left (-30+20 e^x+3 x\right )}\right ) \, dx \\ & = \frac {x^2}{2}+\frac {5}{4} \int \frac {(-3+x) x^2}{-30+20 e^x+3 x} \, dx-\frac {15}{4} \int \frac {(-11+x) x^3}{\left (-30+20 e^x+3 x\right )^2} \, dx \\ & = \frac {x^2}{2}+\frac {5}{4} \int \left (-\frac {3 x^2}{-30+20 e^x+3 x}+\frac {x^3}{-30+20 e^x+3 x}\right ) \, dx-\frac {15}{4} \int \left (-\frac {11 x^3}{\left (-30+20 e^x+3 x\right )^2}+\frac {x^4}{\left (-30+20 e^x+3 x\right )^2}\right ) \, dx \\ & = \frac {x^2}{2}+\frac {5}{4} \int \frac {x^3}{-30+20 e^x+3 x} \, dx-\frac {15}{4} \int \frac {x^4}{\left (-30+20 e^x+3 x\right )^2} \, dx-\frac {15}{4} \int \frac {x^2}{-30+20 e^x+3 x} \, dx+\frac {165}{4} \int \frac {x^3}{\left (-30+20 e^x+3 x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx=\frac {1}{2} \left (x^2-\frac {5 x^3}{2 \left (-30+20 e^x+3 x\right )}\right ) \]

[In]

Integrate[(1800*x + 800*E^(2*x)*x - 135*x^2 + 3*x^3 + E^x*(-2400*x + 90*x^2 + 50*x^3))/(1800 + 800*E^(2*x) - 3
60*x + 18*x^2 + E^x*(-2400 + 240*x)),x]

[Out]

(x^2 - (5*x^3)/(2*(-30 + 20*E^x + 3*x)))/2

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68

method	result	size
risch	\(\frac {x^{2}}{2}-\frac {5 x^{3}}{4 \left (-30+20 \,{\mathrm e}^{x}+3 x \right )}\)	\(23\)
norman	\(\frac {-15 x^{2}+\frac {x^{3}}{4}+10 \,{\mathrm e}^{x} x^{2}}{-30+20 \,{\mathrm e}^{x}+3 x}\)	\(31\)
parallelrisch	\(\frac {10 x^{3}+400 \,{\mathrm e}^{x} x^{2}-600 x^{2}}{-1200+800 \,{\mathrm e}^{x}+120 x}\)	\(32\)

[In]

int((800*x*exp(x)^2+(50*x^3+90*x^2-2400*x)*exp(x)+3*x^3-135*x^2+1800*x)/(800*exp(x)^2+(240*x-2400)*exp(x)+18*x
^2-360*x+1800),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2-5/4*x^3/(-30+20*exp(x)+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx=\frac {x^{3} + 40 \, x^{2} e^{x} - 60 \, x^{2}}{4 \, {\left (3 \, x + 20 \, e^{x} - 30\right )}} \]

[In]

integrate((800*x*exp(x)^2+(50*x^3+90*x^2-2400*x)*exp(x)+3*x^3-135*x^2+1800*x)/(800*exp(x)^2+(240*x-2400)*exp(x
)+18*x^2-360*x+1800),x, algorithm="fricas")

[Out]

1/4*(x^3 + 40*x^2*e^x - 60*x^2)/(3*x + 20*e^x - 30)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx=- \frac {5 x^{3}}{12 x + 80 e^{x} - 120} + \frac {x^{2}}{2} \]

[In]

integrate((800*x*exp(x)**2+(50*x**3+90*x**2-2400*x)*exp(x)+3*x**3-135*x**2+1800*x)/(800*exp(x)**2+(240*x-2400)
*exp(x)+18*x**2-360*x+1800),x)

[Out]

-5*x**3/(12*x + 80*exp(x) - 120) + x**2/2

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx=\frac {x^{3} + 40 \, x^{2} e^{x} - 60 \, x^{2}}{4 \, {\left (3 \, x + 20 \, e^{x} - 30\right )}} \]

[In]

integrate((800*x*exp(x)^2+(50*x^3+90*x^2-2400*x)*exp(x)+3*x^3-135*x^2+1800*x)/(800*exp(x)^2+(240*x-2400)*exp(x
)+18*x^2-360*x+1800),x, algorithm="maxima")

[Out]

1/4*(x^3 + 40*x^2*e^x - 60*x^2)/(3*x + 20*e^x - 30)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx=\frac {x^{3} + 40 \, x^{2} e^{x} - 60 \, x^{2}}{4 \, {\left (3 \, x + 20 \, e^{x} - 30\right )}} \]

[In]

integrate((800*x*exp(x)^2+(50*x^3+90*x^2-2400*x)*exp(x)+3*x^3-135*x^2+1800*x)/(800*exp(x)^2+(240*x-2400)*exp(x
)+18*x^2-360*x+1800),x, algorithm="giac")

[Out]

1/4*(x^3 + 40*x^2*e^x - 60*x^2)/(3*x + 20*e^x - 30)

Mupad [B] (verification not implemented)

Time = 13.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {1800 x+800 e^{2 x} x-135 x^2+3 x^3+e^x \left (-2400 x+90 x^2+50 x^3\right )}{1800+800 e^{2 x}-360 x+18 x^2+e^x (-2400+240 x)} \, dx=\frac {x^2}{2}-\frac {5\,x^3}{2\,\left (6\,x+40\,{\mathrm {e}}^x-60\right )} \]

[In]

int((1800*x + 800*x*exp(2*x) - 135*x^2 + 3*x^3 + exp(x)*(90*x^2 - 2400*x + 50*x^3))/(800*exp(2*x) - 360*x + ex
p(x)*(240*x - 2400) + 18*x^2 + 1800),x)

[Out]

x^2/2 - (5*x^3)/(2*(6*x + 40*exp(x) - 60))