3.84.12 \(\int \frac {-500 x-100 x^2-5 x^3+e^x (-200 x+70 x^2+10 x^3)}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {5 x \left (x^2-2 x (5+x)\right )}{16 \left (10+4 e^x+x\right )} \]

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Rubi [F]  time = 0.93, 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 {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-500*x - 100*x^2 - 5*x^3 + E^x*(-200*x + 70*x^2 + 10*x^3))/(800 + 128*E^(2*x) + 160*x + 8*x^2 + E^x*(640
+ 64*x)),x]

[Out]

(-225*Defer[Int][x^2/(10 + 4*E^x + x)^2, x])/8 - (95*Defer[Int][x^3/(10 + 4*E^x + x)^2, x])/16 - (5*Defer[Int]
[x^4/(10 + 4*E^x + x)^2, x])/16 - (25*Defer[Int][x/(10 + 4*E^x + x), x])/4 + (35*Defer[Int][x^2/(10 + 4*E^x +
x), x])/16 + (5*Defer[Int][x^3/(10 + 4*E^x + x), x])/16

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x \left (-(10+x)^2+2 e^x \left (-20+7 x+x^2\right )\right )}{8 \left (10+4 e^x+x\right )^2} \, dx\\ &=\frac {5}{8} \int \frac {x \left (-(10+x)^2+2 e^x \left (-20+7 x+x^2\right )\right )}{\left (10+4 e^x+x\right )^2} \, dx\\ &=\frac {5}{8} \int \left (\frac {x \left (-20+7 x+x^2\right )}{2 \left (10+4 e^x+x\right )}-\frac {x^2 \left (90+19 x+x^2\right )}{2 \left (10+4 e^x+x\right )^2}\right ) \, dx\\ &=\frac {5}{16} \int \frac {x \left (-20+7 x+x^2\right )}{10+4 e^x+x} \, dx-\frac {5}{16} \int \frac {x^2 \left (90+19 x+x^2\right )}{\left (10+4 e^x+x\right )^2} \, dx\\ &=-\left (\frac {5}{16} \int \left (\frac {90 x^2}{\left (10+4 e^x+x\right )^2}+\frac {19 x^3}{\left (10+4 e^x+x\right )^2}+\frac {x^4}{\left (10+4 e^x+x\right )^2}\right ) \, dx\right )+\frac {5}{16} \int \left (-\frac {20 x}{10+4 e^x+x}+\frac {7 x^2}{10+4 e^x+x}+\frac {x^3}{10+4 e^x+x}\right ) \, dx\\ &=-\left (\frac {5}{16} \int \frac {x^4}{\left (10+4 e^x+x\right )^2} \, dx\right )+\frac {5}{16} \int \frac {x^3}{10+4 e^x+x} \, dx+\frac {35}{16} \int \frac {x^2}{10+4 e^x+x} \, dx-\frac {95}{16} \int \frac {x^3}{\left (10+4 e^x+x\right )^2} \, dx-\frac {25}{4} \int \frac {x}{10+4 e^x+x} \, dx-\frac {225}{8} \int \frac {x^2}{\left (10+4 e^x+x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 20, normalized size = 0.80 \begin {gather*} -\frac {5 x^2 (10+x)}{16 \left (10+4 e^x+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-500*x - 100*x^2 - 5*x^3 + E^x*(-200*x + 70*x^2 + 10*x^3))/(800 + 128*E^(2*x) + 160*x + 8*x^2 + E^x
*(640 + 64*x)),x]

[Out]

(-5*x^2*(10 + x))/(16*(10 + 4*E^x + x))

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fricas [A]  time = 0.71, size = 20, normalized size = 0.80 \begin {gather*} -\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+70*x^2-200*x)*exp(x)-5*x^3-100*x^2-500*x)/(128*exp(x)^2+(64*x+640)*exp(x)+8*x^2+160*x+800),
x, algorithm="fricas")

[Out]

-5/16*(x^3 + 10*x^2)/(x + 4*e^x + 10)

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giac [A]  time = 0.15, size = 20, normalized size = 0.80 \begin {gather*} -\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+70*x^2-200*x)*exp(x)-5*x^3-100*x^2-500*x)/(128*exp(x)^2+(64*x+640)*exp(x)+8*x^2+160*x+800),
x, algorithm="giac")

[Out]

-5/16*(x^3 + 10*x^2)/(x + 4*e^x + 10)

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maple [A]  time = 0.04, size = 18, normalized size = 0.72

method	result	size
risch	\(-\frac {5 \left (x +10\right ) x^{2}}{16 \left (4 \,{\mathrm e}^{x}+10+x \right )}\)	\(18\)
norman	\(\frac {-\frac {25}{8} x^{2}-\frac {5}{16} x^{3}}{4 \,{\mathrm e}^{x}+10+x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^3+70*x^2-200*x)*exp(x)-5*x^3-100*x^2-500*x)/(128*exp(x)^2+(64*x+640)*exp(x)+8*x^2+160*x+800),x,meth
od=_RETURNVERBOSE)

[Out]

-5/16*(x+10)*x^2/(4*exp(x)+10+x)

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maxima [A]  time = 0.40, size = 20, normalized size = 0.80 \begin {gather*} -\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+70*x^2-200*x)*exp(x)-5*x^3-100*x^2-500*x)/(128*exp(x)^2+(64*x+640)*exp(x)+8*x^2+160*x+800),
x, algorithm="maxima")

[Out]

-5/16*(x^3 + 10*x^2)/(x + 4*e^x + 10)

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mupad [B]  time = 5.12, size = 24, normalized size = 0.96 \begin {gather*} -\frac {\frac {5\,x^3}{2}+25\,x^2}{8\,x+32\,{\mathrm {e}}^x+80} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(500*x + 100*x^2 + 5*x^3 - exp(x)*(70*x^2 - 200*x + 10*x^3))/(160*x + 128*exp(2*x) + exp(x)*(64*x + 640)
+ 8*x^2 + 800),x)

[Out]

-(25*x^2 + (5*x^3)/2)/(8*x + 32*exp(x) + 80)

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sympy [A]  time = 0.13, size = 20, normalized size = 0.80 \begin {gather*} \frac {- 5 x^{3} - 50 x^{2}}{16 x + 64 e^{x} + 160} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**3+70*x**2-200*x)*exp(x)-5*x**3-100*x**2-500*x)/(128*exp(x)**2+(64*x+640)*exp(x)+8*x**2+160*x
+800),x)

[Out]

(-5*x**3 - 50*x**2)/(16*x + 64*exp(x) + 160)

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