3.57.97 \(\int \frac {1400 x^2-200 x^3+25 x^4+e^{2 x} (56-8 x+x^2)+e^x (16-584 x+89 x^2-11 x^3)}{400 x^2-200 x^3+25 x^4+e^{2 x} (16-8 x+x^2)+e^x (-160 x+80 x^2-10 x^3)} \, dx\)

Optimal. Leaf size=22 \[ 2+x+\frac {x}{e^x-5 x}-\frac {10 x}{-4+x} \]

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Rubi [F]  time = 0.77, 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 {1400 x^2-200 x^3+25 x^4+e^{2 x} \left (56-8 x+x^2\right )+e^x \left (16-584 x+89 x^2-11 x^3\right )}{400 x^2-200 x^3+25 x^4+e^{2 x} \left (16-8 x+x^2\right )+e^x \left (-160 x+80 x^2-10 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1400*x^2 - 200*x^3 + 25*x^4 + E^(2*x)*(56 - 8*x + x^2) + E^x*(16 - 584*x + 89*x^2 - 11*x^3))/(400*x^2 - 2
00*x^3 + 25*x^4 + E^(2*x)*(16 - 8*x + x^2) + E^x*(-160*x + 80*x^2 - 10*x^3)),x]

[Out]

40/(4 - x) + x + Defer[Int][(E^x - 5*x)^(-1), x] + 5*Defer[Int][x/(E^x - 5*x)^2, x] - Defer[Int][x/(E^x - 5*x)
, x] - 5*Defer[Int][x^2/(E^x - 5*x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (56-8 x+x^2\right )+25 x^2 \left (56-8 x+x^2\right )+e^x \left (16-584 x+89 x^2-11 x^3\right )}{\left (e^x-5 x\right )^2 (4-x)^2} \, dx\\ &=\int \left (-\frac {-1+x}{e^x-5 x}-\frac {5 (-1+x) x}{\left (e^x-5 x\right )^2}+\frac {56-8 x+x^2}{(-4+x)^2}\right ) \, dx\\ &=-\left (5 \int \frac {(-1+x) x}{\left (e^x-5 x\right )^2} \, dx\right )-\int \frac {-1+x}{e^x-5 x} \, dx+\int \frac {56-8 x+x^2}{(-4+x)^2} \, dx\\ &=-\left (5 \int \left (-\frac {x}{\left (e^x-5 x\right )^2}+\frac {x^2}{\left (e^x-5 x\right )^2}\right ) \, dx\right )+\int \left (1+\frac {40}{(-4+x)^2}\right ) \, dx-\int \left (-\frac {1}{e^x-5 x}+\frac {x}{e^x-5 x}\right ) \, dx\\ &=\frac {40}{4-x}+x+5 \int \frac {x}{\left (e^x-5 x\right )^2} \, dx-5 \int \frac {x^2}{\left (e^x-5 x\right )^2} \, dx+\int \frac {1}{e^x-5 x} \, dx-\int \frac {x}{e^x-5 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 20, normalized size = 0.91 \begin {gather*} -\frac {40}{-4+x}+x+\frac {x}{e^x-5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1400*x^2 - 200*x^3 + 25*x^4 + E^(2*x)*(56 - 8*x + x^2) + E^x*(16 - 584*x + 89*x^2 - 11*x^3))/(400*x
^2 - 200*x^3 + 25*x^4 + E^(2*x)*(16 - 8*x + x^2) + E^x*(-160*x + 80*x^2 - 10*x^3)),x]

[Out]

-40/(-4 + x) + x + x/(E^x - 5*x)

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fricas [A]  time = 0.81, size = 45, normalized size = 2.05 \begin {gather*} \frac {5 \, x^{3} - 21 \, x^{2} - {\left (x^{2} - 4 \, x - 40\right )} e^{x} - 196 \, x}{5 \, x^{2} - {\left (x - 4\right )} e^{x} - 20 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-8*x+56)*exp(x)^2+(-11*x^3+89*x^2-584*x+16)*exp(x)+25*x^4-200*x^3+1400*x^2)/((x^2-8*x+16)*exp(x
)^2+(-10*x^3+80*x^2-160*x)*exp(x)+25*x^4-200*x^3+400*x^2),x, algorithm="fricas")

[Out]

(5*x^3 - 21*x^2 - (x^2 - 4*x - 40)*e^x - 196*x)/(5*x^2 - (x - 4)*e^x - 20*x)

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giac [B]  time = 0.20, size = 51, normalized size = 2.32 \begin {gather*} \frac {5 \, x^{3} - x^{2} e^{x} - 21 \, x^{2} + 4 \, x e^{x} - 196 \, x + 40 \, e^{x}}{5 \, x^{2} - x e^{x} - 20 \, x + 4 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-8*x+56)*exp(x)^2+(-11*x^3+89*x^2-584*x+16)*exp(x)+25*x^4-200*x^3+1400*x^2)/((x^2-8*x+16)*exp(x
)^2+(-10*x^3+80*x^2-160*x)*exp(x)+25*x^4-200*x^3+400*x^2),x, algorithm="giac")

[Out]

(5*x^3 - x^2*e^x - 21*x^2 + 4*x*e^x - 196*x + 40*e^x)/(5*x^2 - x*e^x - 20*x + 4*e^x)

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maple [A]  time = 0.09, size = 23, normalized size = 1.05

method	result	size
risch	\(x -\frac {40}{x -4}-\frac {x}{5 x -{\mathrm e}^{x}}\)	\(23\)
norman	\(\frac {-276 x -x^{2}+56 \,{\mathrm e}^{x}+5 x^{3}-{\mathrm e}^{x} x^{2}}{5 x^{2}-{\mathrm e}^{x} x -20 x +4 \,{\mathrm e}^{x}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2-8*x+56)*exp(x)^2+(-11*x^3+89*x^2-584*x+16)*exp(x)+25*x^4-200*x^3+1400*x^2)/((x^2-8*x+16)*exp(x)^2+(-
10*x^3+80*x^2-160*x)*exp(x)+25*x^4-200*x^3+400*x^2),x,method=_RETURNVERBOSE)

[Out]

x-40/(x-4)-x/(5*x-exp(x))

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maxima [A]  time = 0.41, size = 45, normalized size = 2.05 \begin {gather*} \frac {5 \, x^{3} - 21 \, x^{2} - {\left (x^{2} - 4 \, x - 40\right )} e^{x} - 196 \, x}{5 \, x^{2} - {\left (x - 4\right )} e^{x} - 20 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2-8*x+56)*exp(x)^2+(-11*x^3+89*x^2-584*x+16)*exp(x)+25*x^4-200*x^3+1400*x^2)/((x^2-8*x+16)*exp(x
)^2+(-10*x^3+80*x^2-160*x)*exp(x)+25*x^4-200*x^3+400*x^2),x, algorithm="maxima")

[Out]

(5*x^3 - 21*x^2 - (x^2 - 4*x - 40)*e^x - 196*x)/(5*x^2 - (x - 4)*e^x - 20*x)

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mupad [B]  time = 3.65, size = 22, normalized size = 1.00 \begin {gather*} x-\frac {40}{x-4}-\frac {x}{5\,x-{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(x^2 - 8*x + 56) + 1400*x^2 - 200*x^3 + 25*x^4 - exp(x)*(584*x - 89*x^2 + 11*x^3 - 16))/(exp(2*x
)*(x^2 - 8*x + 16) + 400*x^2 - 200*x^3 + 25*x^4 - exp(x)*(160*x - 80*x^2 + 10*x^3)),x)

[Out]

x - 40/(x - 4) - x/(5*x - exp(x))

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sympy [A]  time = 0.15, size = 14, normalized size = 0.64 \begin {gather*} x + \frac {x}{- 5 x + e^{x}} - \frac {40}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2-8*x+56)*exp(x)**2+(-11*x**3+89*x**2-584*x+16)*exp(x)+25*x**4-200*x**3+1400*x**2)/((x**2-8*x+1
6)*exp(x)**2+(-10*x**3+80*x**2-160*x)*exp(x)+25*x**4-200*x**3+400*x**2),x)

[Out]

x + x/(-5*x + exp(x)) - 40/(x - 4)

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