3.8.23 \(\int \frac {5+8 x-10 x^2-2 x^3-2 x \log (x)}{e^{27/2} (9 x+6 x^3+x^5)+e^{27/2} (-6 x-2 x^3) \log (x)+e^{27/2} x \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {5+2 x}{e^{27/2} \left (3+x^2-\log (x)\right )} \]

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Rubi [F]  time = 0.69, 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 {5+8 x-10 x^2-2 x^3-2 x \log (x)}{e^{27/2} \left (9 x+6 x^3+x^5\right )+e^{27/2} \left (-6 x-2 x^3\right ) \log (x)+e^{27/2} x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5 + 8*x - 10*x^2 - 2*x^3 - 2*x*Log[x])/(E^(27/2)*(9*x + 6*x^3 + x^5) + E^(27/2)*(-6*x - 2*x^3)*Log[x] + E
^(27/2)*x*Log[x]^2),x]

[Out]

(2*Defer[Int][(3 + x^2 - Log[x])^(-2), x])/E^(27/2) + (5*Defer[Int][1/(x*(3 + x^2 - Log[x])^2), x])/E^(27/2) -
(10*Defer[Int][x/(3 + x^2 - Log[x])^2, x])/E^(27/2) - (4*Defer[Int][x^2/(3 + x^2 - Log[x])^2, x])/E^(27/2) +
(2*Defer[Int][(3 + x^2 - Log[x])^(-1), x])/E^(27/2)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5+8 x-10 x^2-2 x^3-2 x \log (x)}{e^{27/2} x \left (3+x^2-\log (x)\right )^2} \, dx\\ &=\frac {\int \frac {5+8 x-10 x^2-2 x^3-2 x \log (x)}{x \left (3+x^2-\log (x)\right )^2} \, dx}{e^{27/2}}\\ &=\frac {\int \left (\frac {5+2 x-10 x^2-4 x^3}{x \left (3+x^2-\log (x)\right )^2}+\frac {2}{3+x^2-\log (x)}\right ) \, dx}{e^{27/2}}\\ &=\frac {\int \frac {5+2 x-10 x^2-4 x^3}{x \left (3+x^2-\log (x)\right )^2} \, dx}{e^{27/2}}+\frac {2 \int \frac {1}{3+x^2-\log (x)} \, dx}{e^{27/2}}\\ &=\frac {\int \left (\frac {2}{\left (3+x^2-\log (x)\right )^2}+\frac {5}{x \left (3+x^2-\log (x)\right )^2}-\frac {10 x}{\left (3+x^2-\log (x)\right )^2}-\frac {4 x^2}{\left (3+x^2-\log (x)\right )^2}\right ) \, dx}{e^{27/2}}+\frac {2 \int \frac {1}{3+x^2-\log (x)} \, dx}{e^{27/2}}\\ &=\frac {2 \int \frac {1}{\left (3+x^2-\log (x)\right )^2} \, dx}{e^{27/2}}+\frac {2 \int \frac {1}{3+x^2-\log (x)} \, dx}{e^{27/2}}-\frac {4 \int \frac {x^2}{\left (3+x^2-\log (x)\right )^2} \, dx}{e^{27/2}}+\frac {5 \int \frac {1}{x \left (3+x^2-\log (x)\right )^2} \, dx}{e^{27/2}}-\frac {10 \int \frac {x}{\left (3+x^2-\log (x)\right )^2} \, dx}{e^{27/2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 22, normalized size = 1.00 \begin {gather*} \frac {-5-2 x}{e^{27/2} \left (-3-x^2+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + 8*x - 10*x^2 - 2*x^3 - 2*x*Log[x])/(E^(27/2)*(9*x + 6*x^3 + x^5) + E^(27/2)*(-6*x - 2*x^3)*Log[
x] + E^(27/2)*x*Log[x]^2),x]

[Out]

(-5 - 2*x)/(E^(27/2)*(-3 - x^2 + Log[x]))

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fricas [A]  time = 0.49, size = 23, normalized size = 1.05 \begin {gather*} \frac {2 \, x + 5}{{\left (x^{2} + 3\right )} e^{\frac {27}{2}} - e^{\frac {27}{2}} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)-2*x^3-10*x^2+8*x+5)/(x*exp(27/2)*log(x)^2+(-2*x^3-6*x)*exp(27/2)*log(x)+(x^5+6*x^3+9*x)
*exp(27/2)),x, algorithm="fricas")

[Out]

(2*x + 5)/((x^2 + 3)*e^(27/2) - e^(27/2)*log(x))

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giac [A]  time = 0.36, size = 22, normalized size = 1.00 \begin {gather*} \frac {2 \, x e^{\left (-\frac {27}{2}\right )} + 5 \, e^{\left (-\frac {27}{2}\right )}}{x^{2} - \log \relax (x) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)-2*x^3-10*x^2+8*x+5)/(x*exp(27/2)*log(x)^2+(-2*x^3-6*x)*exp(27/2)*log(x)+(x^5+6*x^3+9*x)
*exp(27/2)),x, algorithm="giac")

[Out]

(2*x*e^(-27/2) + 5*e^(-27/2))/(x^2 - log(x) + 3)

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maple [A]  time = 0.15, size = 20, normalized size = 0.91

method	result	size
risch	\(\frac {\left (5+2 x \right ) {\mathrm e}^{-\frac {27}{2}}}{3-\ln \relax (x )+x^{2}}\)	\(20\)
norman	\(\frac {2 \,{\mathrm e}^{-\frac {27}{2}} x +5 \,{\mathrm e}^{-\frac {27}{2}}}{3-\ln \relax (x )+x^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(x)-2*x^3-10*x^2+8*x+5)/(x*exp(27/2)*ln(x)^2+(-2*x^3-6*x)*exp(27/2)*ln(x)+(x^5+6*x^3+9*x)*exp(27/2
)),x,method=_RETURNVERBOSE)

[Out]

(5+2*x)*exp(-27/2)/(3-ln(x)+x^2)

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maxima [A]  time = 0.72, size = 25, normalized size = 1.14 \begin {gather*} \frac {2 \, x + 5}{x^{2} e^{\frac {27}{2}} - e^{\frac {27}{2}} \log \relax (x) + 3 \, e^{\frac {27}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)-2*x^3-10*x^2+8*x+5)/(x*exp(27/2)*log(x)^2+(-2*x^3-6*x)*exp(27/2)*log(x)+(x^5+6*x^3+9*x)
*exp(27/2)),x, algorithm="maxima")

[Out]

(2*x + 5)/(x^2*e^(27/2) - e^(27/2)*log(x) + 3*e^(27/2))

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mupad [B]  time = 0.79, size = 35, normalized size = 1.59 \begin {gather*} \frac {\frac {{\mathrm {e}}^{-\frac {27}{2}}\,\left (6\,x-5\,x^2\right )}{3}+\frac {5\,{\mathrm {e}}^{-\frac {27}{2}}\,\left (x^2+3\right )}{3}}{x^2-\ln \relax (x)+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*log(x) - 8*x + 10*x^2 + 2*x^3 - 5)/(exp(27/2)*(9*x + 6*x^3 + x^5) - exp(27/2)*log(x)*(6*x + 2*x^3) +
x*exp(27/2)*log(x)^2),x)

[Out]

((exp(-27/2)*(6*x - 5*x^2))/3 + (5*exp(-27/2)*(x^2 + 3))/3)/(x^2 - log(x) + 3)

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sympy [A]  time = 0.13, size = 29, normalized size = 1.32 \begin {gather*} \frac {- 2 x - 5}{- x^{2} e^{\frac {27}{2}} + e^{\frac {27}{2}} \log {\relax (x )} - 3 e^{\frac {27}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(x)-2*x**3-10*x**2+8*x+5)/(x*exp(27/2)*ln(x)**2+(-2*x**3-6*x)*exp(27/2)*ln(x)+(x**5+6*x**3+9
*x)*exp(27/2)),x)

[Out]

(-2*x - 5)/(-x**2*exp(27/2) + exp(27/2)*log(x) - 3*exp(27/2))

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