\(\int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx\) [8374]

Optimal result

Integrand size = 42, antiderivative size = 20 \[ \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx=-\frac {x}{2}+\frac {1}{\frac {1}{2}+\frac {e^x x}{5}} \]

[Out]

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

Rubi [F]

\[ \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx=\int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx \]

[In]

Int[(-25 + E^x*(-40 - 60*x) - 4*E^(2*x)*x^2)/(50 + 40*E^x*x + 8*E^(2*x)*x^2),x]

[Out]

-1/2*x + 50*Defer[Int][(5 + 2*E^x*x)^(-2), x] + 50*Defer[Int][1/(x*(5 + 2*E^x*x)^2), x] - 10*Defer[Int][(5 + 2
*E^x*x)^(-1), x] - 10*Defer[Int][1/(x*(5 + 2*E^x*x)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{2 \left (5+2 e^x x\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{\left (5+2 e^x x\right )^2} \, dx \\ & = \frac {1}{2} \int \left (-1+\frac {100 (1+x)}{x \left (5+2 e^x x\right )^2}-\frac {20 (1+x)}{x \left (5+2 e^x x\right )}\right ) \, dx \\ & = -\frac {x}{2}-10 \int \frac {1+x}{x \left (5+2 e^x x\right )} \, dx+50 \int \frac {1+x}{x \left (5+2 e^x x\right )^2} \, dx \\ & = -\frac {x}{2}-10 \int \left (\frac {1}{5+2 e^x x}+\frac {1}{x \left (5+2 e^x x\right )}\right ) \, dx+50 \int \left (\frac {1}{\left (5+2 e^x x\right )^2}+\frac {1}{x \left (5+2 e^x x\right )^2}\right ) \, dx \\ & = -\frac {x}{2}-10 \int \frac {1}{5+2 e^x x} \, dx-10 \int \frac {1}{x \left (5+2 e^x x\right )} \, dx+50 \int \frac {1}{\left (5+2 e^x x\right )^2} \, dx+50 \int \frac {1}{x \left (5+2 e^x x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx=-\frac {-20+5 x+2 e^x x^2}{2 \left (5+2 e^x x\right )} \]

[In]

Integrate[(-25 + E^x*(-40 - 60*x) - 4*E^(2*x)*x^2)/(50 + 40*E^x*x + 8*E^(2*x)*x^2),x]

[Out]

-1/2*(-20 + 5*x + 2*E^x*x^2)/(5 + 2*E^x*x)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {x}{2}+\frac {10}{2 \,{\mathrm e}^{x} x +5}\)	\(16\)
norman	\(\frac {-\frac {5 x}{2}-{\mathrm e}^{x} x^{2}+10}{2 \,{\mathrm e}^{x} x +5}\)	\(23\)
parallelrisch	\(-\frac {4 \,{\mathrm e}^{x} x^{2}-40+10 x}{4 \left (2 \,{\mathrm e}^{x} x +5\right )}\)	\(24\)

[In]

int((-4*exp(x)^2*x^2+(-60*x-40)*exp(x)-25)/(8*exp(x)^2*x^2+40*exp(x)*x+50),x,method=_RETURNVERBOSE)

[Out]

-1/2*x+10/(2*exp(x)*x+5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx=-\frac {2 \, x^{2} e^{x} + 5 \, x - 20}{2 \, {\left (2 \, x e^{x} + 5\right )}} \]

[In]

integrate((-4*exp(x)^2*x^2+(-60*x-40)*exp(x)-25)/(8*exp(x)^2*x^2+40*exp(x)*x+50),x, algorithm="fricas")

[Out]

-1/2*(2*x^2*e^x + 5*x - 20)/(2*x*e^x + 5)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx=- \frac {x}{2} + \frac {10}{2 x e^{x} + 5} \]

[In]

integrate((-4*exp(x)**2*x**2+(-60*x-40)*exp(x)-25)/(8*exp(x)**2*x**2+40*exp(x)*x+50),x)

[Out]

-x/2 + 10/(2*x*exp(x) + 5)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx=-\frac {2 \, x^{2} e^{x} + 5 \, x - 20}{2 \, {\left (2 \, x e^{x} + 5\right )}} \]

[In]

integrate((-4*exp(x)^2*x^2+(-60*x-40)*exp(x)-25)/(8*exp(x)^2*x^2+40*exp(x)*x+50),x, algorithm="maxima")

[Out]

-1/2*(2*x^2*e^x + 5*x - 20)/(2*x*e^x + 5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx=-\frac {2 \, x^{2} e^{x} + 5 \, x - 20}{2 \, {\left (2 \, x e^{x} + 5\right )}} \]

[In]

integrate((-4*exp(x)^2*x^2+(-60*x-40)*exp(x)-25)/(8*exp(x)^2*x^2+40*exp(x)*x+50),x, algorithm="giac")

[Out]

-1/2*(2*x^2*e^x + 5*x - 20)/(2*x*e^x + 5)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx=\frac {10}{2\,x\,{\mathrm {e}}^x+5}-\frac {x}{2} \]

[In]

int(-(exp(x)*(60*x + 40) + 4*x^2*exp(2*x) + 25)/(8*x^2*exp(2*x) + 40*x*exp(x) + 50),x)

[Out]

10/(2*x*exp(x) + 5) - x/2