3.80.42 \(\int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x (192 x+96 x^2)+e^{2 x} (1024 x^2+1024 x^3+256 x^4)} \, dx\)

Optimal. Leaf size=19 \[ \frac {4}{2+\frac {3 e^{-x}}{16 x}+x} \]

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Rubi [F]  time = 2.12, 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 {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1024*E^(2*x)*x^2 + E^x*(192 + 192*x))/(9 + E^x*(192*x + 96*x^2) + E^(2*x)*(1024*x^2 + 1024*x^3 + 256*x^4
)),x]

[Out]

384*Defer[Int][E^x/(3 + 16*E^x*x*(2 + x))^2, x] + 192*Defer[Int][(E^x*x)/(3 + 16*E^x*x*(2 + x))^2, x] - 384*De
fer[Int][E^x/((2 + x)*(3 + 16*E^x*x*(2 + x))^2), x] - 64*Defer[Int][E^x/(3 + 16*E^x*x*(2 + x)), x] + 128*Defer
[Int][E^x/((2 + x)*(3 + 16*E^x*x*(2 + x))), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {64 e^x \left (3+3 x-16 e^x x^2\right )}{\left (3+16 e^x x (2+x)\right )^2} \, dx\\ &=64 \int \frac {e^x \left (3+3 x-16 e^x x^2\right )}{\left (3+16 e^x x (2+x)\right )^2} \, dx\\ &=64 \int \left (\frac {3 e^x \left (2+4 x+x^2\right )}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )^2}-\frac {e^x x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )}\right ) \, dx\\ &=-\left (64 \int \frac {e^x x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )} \, dx\right )+192 \int \frac {e^x \left (2+4 x+x^2\right )}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )^2} \, dx\\ &=-\left (64 \int \frac {e^x x}{(2+x) \left (3+16 e^x x (2+x)\right )} \, dx\right )+192 \int \frac {e^x \left (2+4 x+x^2\right )}{(2+x) \left (3+16 e^x x (2+x)\right )^2} \, dx\\ &=-\left (64 \int \left (\frac {e^x}{3+32 e^x x+16 e^x x^2}-\frac {2 e^x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )}\right ) \, dx\right )+192 \int \left (\frac {2 e^x}{\left (3+32 e^x x+16 e^x x^2\right )^2}+\frac {e^x x}{\left (3+32 e^x x+16 e^x x^2\right )^2}-\frac {2 e^x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )^2}\right ) \, dx\\ &=-\left (64 \int \frac {e^x}{3+32 e^x x+16 e^x x^2} \, dx\right )+128 \int \frac {e^x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )} \, dx+192 \int \frac {e^x x}{\left (3+32 e^x x+16 e^x x^2\right )^2} \, dx+384 \int \frac {e^x}{\left (3+32 e^x x+16 e^x x^2\right )^2} \, dx-384 \int \frac {e^x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )^2} \, dx\\ &=-\left (64 \int \frac {e^x}{3+16 e^x x (2+x)} \, dx\right )+128 \int \frac {e^x}{(2+x) \left (3+16 e^x x (2+x)\right )} \, dx+192 \int \frac {e^x x}{\left (3+16 e^x x (2+x)\right )^2} \, dx+384 \int \frac {e^x}{\left (3+16 e^x x (2+x)\right )^2} \, dx-384 \int \frac {e^x}{(2+x) \left (3+16 e^x x (2+x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 19, normalized size = 1.00 \begin {gather*} \frac {64 e^x x}{3+16 e^x x (2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1024*E^(2*x)*x^2 + E^x*(192 + 192*x))/(9 + E^x*(192*x + 96*x^2) + E^(2*x)*(1024*x^2 + 1024*x^3 + 2
56*x^4)),x]

[Out]

(64*E^x*x)/(3 + 16*E^x*x*(2 + x))

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fricas [A]  time = 0.79, size = 20, normalized size = 1.05 \begin {gather*} \frac {64 \, x e^{x}}{16 \, {\left (x^{2} + 2 \, x\right )} e^{x} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1024*exp(x)^2*x^2+(192*x+192)*exp(x))/((256*x^4+1024*x^3+1024*x^2)*exp(x)^2+(96*x^2+192*x)*exp(x)+
9),x, algorithm="fricas")

[Out]

64*x*e^x/(16*(x^2 + 2*x)*e^x + 3)

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giac [A]  time = 0.21, size = 21, normalized size = 1.11 \begin {gather*} \frac {64 \, x e^{x}}{16 \, x^{2} e^{x} + 32 \, x e^{x} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1024*exp(x)^2*x^2+(192*x+192)*exp(x))/((256*x^4+1024*x^3+1024*x^2)*exp(x)^2+(96*x^2+192*x)*exp(x)+
9),x, algorithm="giac")

[Out]

64*x*e^x/(16*x^2*e^x + 32*x*e^x + 3)

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maple [A]  time = 0.04, size = 22, normalized size = 1.16

method	result	size
norman	\(\frac {64 \,{\mathrm e}^{x} x}{16 \,{\mathrm e}^{x} x^{2}+32 \,{\mathrm e}^{x} x +3}\)	\(22\)
risch	\(\frac {4}{2+x}-\frac {12}{\left (2+x \right ) \left (16 \,{\mathrm e}^{x} x^{2}+32 \,{\mathrm e}^{x} x +3\right )}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1024*exp(x)^2*x^2+(192*x+192)*exp(x))/((256*x^4+1024*x^3+1024*x^2)*exp(x)^2+(96*x^2+192*x)*exp(x)+9),x,m
ethod=_RETURNVERBOSE)

[Out]

64*exp(x)*x/(16*exp(x)*x^2+32*exp(x)*x+3)

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maxima [A]  time = 0.41, size = 20, normalized size = 1.05 \begin {gather*} \frac {64 \, x e^{x}}{16 \, {\left (x^{2} + 2 \, x\right )} e^{x} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1024*exp(x)^2*x^2+(192*x+192)*exp(x))/((256*x^4+1024*x^3+1024*x^2)*exp(x)^2+(96*x^2+192*x)*exp(x)+
9),x, algorithm="maxima")

[Out]

64*x*e^x/(16*(x^2 + 2*x)*e^x + 3)

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mupad [B]  time = 0.15, size = 31, normalized size = 1.63 \begin {gather*} \frac {4}{x+2}-\frac {12}{\left ({\mathrm {e}}^x\,\left (16\,x^2+32\,x\right )+3\right )\,\left (x+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(192*x + 192) - 1024*x^2*exp(2*x))/(exp(2*x)*(1024*x^2 + 1024*x^3 + 256*x^4) + exp(x)*(192*x + 96*
x^2) + 9),x)

[Out]

4/(x + 2) - 12/((exp(x)*(32*x + 16*x^2) + 3)*(x + 2))

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sympy [B]  time = 0.22, size = 27, normalized size = 1.42 \begin {gather*} - \frac {12}{3 x + \left (16 x^{3} + 64 x^{2} + 64 x\right ) e^{x} + 6} + \frac {4}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1024*exp(x)**2*x**2+(192*x+192)*exp(x))/((256*x**4+1024*x**3+1024*x**2)*exp(x)**2+(96*x**2+192*x)*
exp(x)+9),x)

[Out]

-12/(3*x + (16*x**3 + 64*x**2 + 64*x)*exp(x) + 6) + 4/(x + 2)

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