3.9.47 \(\int \frac {45-8 x+10 x^2-4 x^3-3 x^4+e^{3+x} (9-x+2 x^2+x^3-x^4)}{-15 x+2 x^2-14 x^3+2 x^4+x^5+e^{3+x} (-3 x+x^2-3 x^3+x^4)} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\frac {5 \left (\frac {1}{x}+x\right )}{(-3+x) x^2 \left (5+e^{3+x}+x\right )}\right ) \]

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Rubi [F]  time = 1.31, 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 {45-8 x+10 x^2-4 x^3-3 x^4+e^{3+x} \left (9-x+2 x^2+x^3-x^4\right )}{-15 x+2 x^2-14 x^3+2 x^4+x^5+e^{3+x} \left (-3 x+x^2-3 x^3+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(45 - 8*x + 10*x^2 - 4*x^3 - 3*x^4 + E^(3 + x)*(9 - x + 2*x^2 + x^3 - x^4))/(-15*x + 2*x^2 - 14*x^3 + 2*x^
4 + x^5 + E^(3 + x)*(-3*x + x^2 - 3*x^3 + x^4)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-45+8 x-10 x^2+4 x^3+3 x^4-e^{3+x} \left (9-x+2 x^2+x^3-x^4\right )}{x \left (5+e^{3+x}+x\right ) \left (3-x+3 x^2-x^3\right )} \, dx\\ &=\int \left (\frac {4+x}{5+e^{3+x}+x}+\frac {9-x+2 x^2+x^3-x^4}{(-3+x) x \left (1+x^2\right )}\right ) \, dx\\ &=\int \frac {4+x}{5+e^{3+x}+x} \, dx+\int \frac {9-x+2 x^2+x^3-x^4}{(-3+x) x \left (1+x^2\right )} \, dx\\ &=\int \left (\frac {4}{5+e^{3+x}+x}+\frac {x}{5+e^{3+x}+x}\right ) \, dx+\int \left (-1+\frac {1}{3-x}-\frac {3}{x}+\frac {2 x}{1+x^2}\right ) \, dx\\ &=-x-\log (3-x)-3 \log (x)+2 \int \frac {x}{1+x^2} \, dx+4 \int \frac {1}{5+e^{3+x}+x} \, dx+\int \frac {x}{5+e^{3+x}+x} \, dx\\ &=-x-\log (3-x)-3 \log (x)+\log \left (1+x^2\right )+4 \int \frac {1}{5+e^{3+x}+x} \, dx+\int \frac {x}{5+e^{3+x}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 30, normalized size = 1.15 \begin {gather*} -\log (3-x)-3 \log (x)-\log \left (5+e^{3+x}+x\right )+\log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(45 - 8*x + 10*x^2 - 4*x^3 - 3*x^4 + E^(3 + x)*(9 - x + 2*x^2 + x^3 - x^4))/(-15*x + 2*x^2 - 14*x^3
+ 2*x^4 + x^5 + E^(3 + x)*(-3*x + x^2 - 3*x^3 + x^4)),x]

[Out]

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

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fricas [A]  time = 0.51, size = 27, normalized size = 1.04 \begin {gather*} \log \left (x^{2} + 1\right ) - \log \left (x + e^{\left (x + 3\right )} + 5\right ) - \log \left (x - 3\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+x^3+2*x^2-x+9)*exp(3+x)-3*x^4-4*x^3+10*x^2-8*x+45)/((x^4-3*x^3+x^2-3*x)*exp(3+x)+x^5+2*x^4-14
*x^3+2*x^2-15*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.45, size = 27, normalized size = 1.04 \begin {gather*} \log \left (x^{2} + 1\right ) - \log \left (x + e^{\left (x + 3\right )} + 5\right ) - \log \left (x - 3\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+x^3+2*x^2-x+9)*exp(3+x)-3*x^4-4*x^3+10*x^2-8*x+45)/((x^4-3*x^3+x^2-3*x)*exp(3+x)+x^5+2*x^4-14
*x^3+2*x^2-15*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 28, normalized size = 1.08

method	result	size
norman	\(-3 \ln \relax (x )-\ln \left (x -3\right )-\ln \left ({\mathrm e}^{3+x}+x +5\right )+\ln \left (x^{2}+1\right )\)	\(28\)
risch	\(-\ln \left (x -3\right )-3 \ln \relax (x )+\ln \left (x^{2}+1\right )+3-\ln \left ({\mathrm e}^{3+x}+x +5\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^4+x^3+2*x^2-x+9)*exp(3+x)-3*x^4-4*x^3+10*x^2-8*x+45)/((x^4-3*x^3+x^2-3*x)*exp(3+x)+x^5+2*x^4-14*x^3+2
*x^2-15*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.74, size = 30, normalized size = 1.15 \begin {gather*} \log \left (x^{2} + 1\right ) - \log \left ({\left (x + e^{\left (x + 3\right )} + 5\right )} e^{\left (-3\right )}\right ) - \log \left (x - 3\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+x^3+2*x^2-x+9)*exp(3+x)-3*x^4-4*x^3+10*x^2-8*x+45)/((x^4-3*x^3+x^2-3*x)*exp(3+x)+x^5+2*x^4-14
*x^3+2*x^2-15*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.16, size = 27, normalized size = 1.04 \begin {gather*} \ln \left (x^2+1\right )-\ln \left (x+{\mathrm {e}}^{x+3}+5\right )-\ln \left (x-3\right )-3\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - exp(x + 3)*(2*x^2 - x + x^3 - x^4 + 9) - 10*x^2 + 4*x^3 + 3*x^4 - 45)/(15*x + exp(x + 3)*(3*x - x^2
+ 3*x^3 - x^4) - 2*x^2 + 14*x^3 - 2*x^4 - x^5),x)

[Out]

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

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sympy [A]  time = 0.21, size = 26, normalized size = 1.00 \begin {gather*} - 3 \log {\relax (x )} - \log {\left (x - 3 \right )} + \log {\left (x^{2} + 1 \right )} - \log {\left (x + e^{x + 3} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**4+x**3+2*x**2-x+9)*exp(3+x)-3*x**4-4*x**3+10*x**2-8*x+45)/((x**4-3*x**3+x**2-3*x)*exp(3+x)+x**
5+2*x**4-14*x**3+2*x**2-15*x),x)

[Out]

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

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