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

Optimal. Leaf size=20 \[ -\frac {x}{2}+\frac {1}{\frac {1}{2}+\frac {e^x x}{5}} \]

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Rubi [F]  time = 0.66, 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 {-25+e^x (-40-60 x)-4 e^{2 x} x^2}{50+40 e^x x+8 e^{2 x} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 20, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (-x+\frac {20}{5+2 e^x x}\right ) \end {gather*}

Antiderivative was successfully verified.

[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]

(-x + 20/(5 + 2*E^x*x))/2

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fricas [A]  time = 0.59, size = 23, normalized size = 1.15 \begin {gather*} -\frac {2 \, x^{2} e^{x} + 5 \, x - 20}{2 \, {\left (2 \, x e^{x} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.24, size = 23, normalized size = 1.15 \begin {gather*} -\frac {2 \, x^{2} e^{x} + 5 \, x - 20}{2 \, {\left (2 \, x e^{x} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.04, size = 16, normalized size = 0.80

method	result	size
risch	\(-\frac {x}{2}+\frac {10}{2 \,{\mathrm e}^{x} x +5}\)	\(16\)
norman	\(\frac {-\frac {5 x}{2}-{\mathrm e}^{x} x^{2}+10}{2 \,{\mathrm e}^{x} x +5}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.39, size = 23, normalized size = 1.15 \begin {gather*} -\frac {2 \, x^{2} e^{x} + 5 \, x - 20}{2 \, {\left (2 \, x e^{x} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.12, size = 15, normalized size = 0.75 \begin {gather*} \frac {10}{2\,x\,{\mathrm {e}}^x+5}-\frac {x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.13, size = 12, normalized size = 0.60 \begin {gather*} - \frac {x}{2} + \frac {10}{2 x e^{x} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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