3.19.93 \(\int \frac {8-17 x+10 x^2-4 x^3+e^x (x-3 x^2+3 x^3-x^4)}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x (x^2-2 x^3+x^4)} \, dx\)

Optimal. Leaf size=31 \[ \log \left (\frac {x^2}{\frac {4-x}{-1+x}+\left (4-e^x-x\right ) x}\right ) \]

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Rubi [F]  time = 2.77, 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 {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{4 x-9 x^2+10 x^3-6 x^4+x^5+e^x \left (x^2-2 x^3+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8 - 17*x + 10*x^2 - 4*x^3 + E^x*(x - 3*x^2 + 3*x^3 - x^4))/(4*x - 9*x^2 + 10*x^3 - 6*x^4 + x^5 + E^x*(x^2
- 2*x^3 + x^4)),x]

[Out]

-x + Log[x] - 3*Defer[Int][(-4 + 5*x - E^x*x - 5*x^2 + E^x*x^2 + x^3)^(-1), x] - 3*Defer[Int][1/((-1 + x)*(-4
+ 5*x - E^x*x - 5*x^2 + E^x*x^2 + x^3)), x] - 4*Defer[Int][1/(x*(-4 + 5*x - E^x*x - 5*x^2 + E^x*x^2 + x^3)), x
] + 6*Defer[Int][x/(-4 + 5*x - E^x*x - 5*x^2 + E^x*x^2 + x^3), x] - 6*Defer[Int][x^2/(-4 + 5*x - E^x*x - 5*x^2
+ E^x*x^2 + x^3), x] + Defer[Int][x^3/(-4 + 5*x - E^x*x - 5*x^2 + E^x*x^2 + x^3), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8-17 x+10 x^2-4 x^3+e^x \left (x-3 x^2+3 x^3-x^4\right )}{(1-x) x \left (4-5 x+e^x x+5 x^2-e^x x^2-x^3\right )} \, dx\\ &=\int \left (\frac {1-x}{x}+\frac {4-4 x-9 x^2+12 x^3-7 x^4+x^5}{(-1+x) x \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )}\right ) \, dx\\ &=\int \frac {1-x}{x} \, dx+\int \frac {4-4 x-9 x^2+12 x^3-7 x^4+x^5}{(-1+x) x \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )} \, dx\\ &=\int \left (-1+\frac {1}{x}\right ) \, dx+\int \left (-\frac {3}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3}-\frac {3}{(-1+x) \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )}-\frac {4}{x \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )}+\frac {6 x}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3}-\frac {6 x^2}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3}+\frac {x^3}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3}\right ) \, dx\\ &=-x+\log (x)-3 \int \frac {1}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3} \, dx-3 \int \frac {1}{(-1+x) \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )} \, dx-4 \int \frac {1}{x \left (-4+5 x-e^x x-5 x^2+e^x x^2+x^3\right )} \, dx+6 \int \frac {x}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3} \, dx-6 \int \frac {x^2}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3} \, dx+\int \frac {x^3}{-4+5 x-e^x x-5 x^2+e^x x^2+x^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 42, normalized size = 1.35 \begin {gather*} \log (x)+\log ((1-x) x)-\log \left (4-5 x+e^x x+5 x^2-e^x x^2-x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 - 17*x + 10*x^2 - 4*x^3 + E^x*(x - 3*x^2 + 3*x^3 - x^4))/(4*x - 9*x^2 + 10*x^3 - 6*x^4 + x^5 + E^
x*(x^2 - 2*x^3 + x^4)),x]

[Out]

Log[x] + Log[(1 - x)*x] - Log[4 - 5*x + E^x*x + 5*x^2 - E^x*x^2 - x^3]

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fricas [A]  time = 0.65, size = 39, normalized size = 1.26 \begin {gather*} \log \relax (x) - \log \left (\frac {x^{3} - 5 \, x^{2} + {\left (x^{2} - x\right )} e^{x} + 5 \, x - 4}{x^{2} - x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+3*x^3-3*x^2+x)*exp(x)-4*x^3+10*x^2-17*x+8)/((x^4-2*x^3+x^2)*exp(x)+x^5-6*x^4+10*x^3-9*x^2+4*x
),x, algorithm="fricas")

[Out]

log(x) - log((x^3 - 5*x^2 + (x^2 - x)*e^x + 5*x - 4)/(x^2 - x))

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giac [A]  time = 0.23, size = 36, normalized size = 1.16 \begin {gather*} -\log \left (x^{3} + x^{2} e^{x} - 5 \, x^{2} - x e^{x} + 5 \, x - 4\right ) + \log \left (x - 1\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+3*x^3-3*x^2+x)*exp(x)-4*x^3+10*x^2-17*x+8)/((x^4-2*x^3+x^2)*exp(x)+x^5-6*x^4+10*x^3-9*x^2+4*x
),x, algorithm="giac")

[Out]

-log(x^3 + x^2*e^x - 5*x^2 - x*e^x + 5*x - 4) + log(x - 1) + 2*log(x)

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maple [A]  time = 0.06, size = 32, normalized size = 1.03

method	result	size
risch	\(\ln \relax (x )-\ln \left ({\mathrm e}^{x}+\frac {x^{3}-5 x^{2}+5 x -4}{x \left (x -1\right )}\right )\)	\(32\)
norman	\(2 \ln \relax (x )-\ln \left (x^{3}+{\mathrm e}^{x} x^{2}-5 x^{2}-{\mathrm e}^{x} x +5 x -4\right )+\ln \left (x -1\right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^4+3*x^3-3*x^2+x)*exp(x)-4*x^3+10*x^2-17*x+8)/((x^4-2*x^3+x^2)*exp(x)+x^5-6*x^4+10*x^3-9*x^2+4*x),x,me
thod=_RETURNVERBOSE)

[Out]

ln(x)-ln(exp(x)+(x^3-5*x^2+5*x-4)/x/(x-1))

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maxima [A]  time = 0.68, size = 39, normalized size = 1.26 \begin {gather*} \log \relax (x) - \log \left (\frac {x^{3} - 5 \, x^{2} + {\left (x^{2} - x\right )} e^{x} + 5 \, x - 4}{x^{2} - x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+3*x^3-3*x^2+x)*exp(x)-4*x^3+10*x^2-17*x+8)/((x^4-2*x^3+x^2)*exp(x)+x^5-6*x^4+10*x^3-9*x^2+4*x
),x, algorithm="maxima")

[Out]

log(x) - log((x^3 - 5*x^2 + (x^2 - x)*e^x + 5*x - 4)/(x^2 - x))

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mupad [B]  time = 1.22, size = 36, normalized size = 1.16 \begin {gather*} \ln \left (x-1\right )-\ln \left (5\,x+x^2\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x-5\,x^2+x^3-4\right )+2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x - 3*x^2 + 3*x^3 - x^4) - 17*x + 10*x^2 - 4*x^3 + 8)/(4*x + exp(x)*(x^2 - 2*x^3 + x^4) - 9*x^2 +
10*x^3 - 6*x^4 + x^5),x)

[Out]

log(x - 1) - log(5*x + x^2*exp(x) - x*exp(x) - 5*x^2 + x^3 - 4) + 2*log(x)

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sympy [A]  time = 0.35, size = 26, normalized size = 0.84 \begin {gather*} \log {\relax (x )} - \log {\left (e^{x} + \frac {x^{3} - 5 x^{2} + 5 x - 4}{x^{2} - x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**4+3*x**3-3*x**2+x)*exp(x)-4*x**3+10*x**2-17*x+8)/((x**4-2*x**3+x**2)*exp(x)+x**5-6*x**4+10*x**
3-9*x**2+4*x),x)

[Out]

log(x) - log(exp(x) + (x**3 - 5*x**2 + 5*x - 4)/(x**2 - x))

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