3.14.23 \(\int \frac {-2 x-7 x^2-9 x^3+27 x^4-9 x^5+e^3 (6 x^2-6 x^3)+(-2+6 e^3 x+9 x^3) \log (x)}{9 x^4-9 x^5+9 x^3 \log (x)} \, dx\)

Optimal. Leaf size=28 \[ 2+\left (e^3-\frac {1}{3 x}\right )^2+x-\log \left (x-x^2+\log (x)\right ) \]

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Rubi [A]  time = 0.57, antiderivative size = 31, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 5, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {6741, 12, 6742, 14, 6684} \begin {gather*} \frac {1}{9 x^2}-\log \left (-x^2+x+\log (x)\right )+x-\frac {2 e^3}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x - 7*x^2 - 9*x^3 + 27*x^4 - 9*x^5 + E^3*(6*x^2 - 6*x^3) + (-2 + 6*E^3*x + 9*x^3)*Log[x])/(9*x^4 - 9*x
^5 + 9*x^3*Log[x]),x]

[Out]

1/(9*x^2) - (2*E^3)/(3*x) + x - Log[x - x^2 + Log[x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x-7 x^2-9 x^3+27 x^4-9 x^5+e^3 \left (6 x^2-6 x^3\right )+\left (-2+6 e^3 x+9 x^3\right ) \log (x)}{9 x^3 \left (x-x^2+\log (x)\right )} \, dx\\ &=\frac {1}{9} \int \frac {-2 x-7 x^2-9 x^3+27 x^4-9 x^5+e^3 \left (6 x^2-6 x^3\right )+\left (-2+6 e^3 x+9 x^3\right ) \log (x)}{x^3 \left (x-x^2+\log (x)\right )} \, dx\\ &=\frac {1}{9} \int \left (\frac {-2+6 e^3 x+9 x^3}{x^3}-\frac {9 \left (-1-x+2 x^2\right )}{x \left (-x+x^2-\log (x)\right )}\right ) \, dx\\ &=\frac {1}{9} \int \frac {-2+6 e^3 x+9 x^3}{x^3} \, dx-\int \frac {-1-x+2 x^2}{x \left (-x+x^2-\log (x)\right )} \, dx\\ &=-\log \left (x-x^2+\log (x)\right )+\frac {1}{9} \int \left (9-\frac {2}{x^3}+\frac {6 e^3}{x^2}\right ) \, dx\\ &=\frac {1}{9 x^2}-\frac {2 e^3}{3 x}+x-\log \left (x-x^2+\log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 33, normalized size = 1.18 \begin {gather*} \frac {1}{9} \left (\frac {1}{x^2}-\frac {6 e^3}{x}+9 x-9 \log \left (-x+x^2-\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x - 7*x^2 - 9*x^3 + 27*x^4 - 9*x^5 + E^3*(6*x^2 - 6*x^3) + (-2 + 6*E^3*x + 9*x^3)*Log[x])/(9*x^4
- 9*x^5 + 9*x^3*Log[x]),x]

[Out]

(x^(-2) - (6*E^3)/x + 9*x - 9*Log[-x + x^2 - Log[x]])/9

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fricas [A]  time = 0.58, size = 32, normalized size = 1.14 \begin {gather*} \frac {9 \, x^{3} - 9 \, x^{2} \log \left (-x^{2} + x + \log \relax (x)\right ) - 6 \, x e^{3} + 1}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x*exp(3)+9*x^3-2)*log(x)+(-6*x^3+6*x^2)*exp(3)-9*x^5+27*x^4-9*x^3-7*x^2-2*x)/(9*x^3*log(x)-9*x^5
+9*x^4),x, algorithm="fricas")

[Out]

1/9*(9*x^3 - 9*x^2*log(-x^2 + x + log(x)) - 6*x*e^3 + 1)/x^2

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giac [A]  time = 0.34, size = 34, normalized size = 1.21 \begin {gather*} \frac {9 \, x^{3} - 9 \, x^{2} \log \left (x^{2} - x - \log \relax (x)\right ) - 6 \, x e^{3} + 1}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x*exp(3)+9*x^3-2)*log(x)+(-6*x^3+6*x^2)*exp(3)-9*x^5+27*x^4-9*x^3-7*x^2-2*x)/(9*x^3*log(x)-9*x^5
+9*x^4),x, algorithm="giac")

[Out]

1/9*(9*x^3 - 9*x^2*log(x^2 - x - log(x)) - 6*x*e^3 + 1)/x^2

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maple [A]  time = 0.05, size = 30, normalized size = 1.07

method	result	size
norman	\(\frac {\frac {1}{9}+x^{3}-\frac {2 x \,{\mathrm e}^{3}}{3}}{x^{2}}-\ln \left (x^{2}-x -\ln \relax (x )\right )\)	\(30\)
risch	\(-\frac {-9 x^{3}+6 x \,{\mathrm e}^{3}-1}{9 x^{2}}-\ln \left (\ln \relax (x )+x -x^{2}\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x*exp(3)+9*x^3-2)*ln(x)+(-6*x^3+6*x^2)*exp(3)-9*x^5+27*x^4-9*x^3-7*x^2-2*x)/(9*x^3*ln(x)-9*x^5+9*x^4),
x,method=_RETURNVERBOSE)

[Out]

(1/9+x^3-2/3*x*exp(3))/x^2-ln(x^2-x-ln(x))

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maxima [A]  time = 0.41, size = 30, normalized size = 1.07 \begin {gather*} \frac {9 \, x^{3} - 6 \, x e^{3} + 1}{9 \, x^{2}} - \log \left (-x^{2} + x + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x*exp(3)+9*x^3-2)*log(x)+(-6*x^3+6*x^2)*exp(3)-9*x^5+27*x^4-9*x^3-7*x^2-2*x)/(9*x^3*log(x)-9*x^5
+9*x^4),x, algorithm="maxima")

[Out]

1/9*(9*x^3 - 6*x*e^3 + 1)/x^2 - log(-x^2 + x + log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {2\,x-\ln \relax (x)\,\left (9\,x^3+6\,{\mathrm {e}}^3\,x-2\right )-{\mathrm {e}}^3\,\left (6\,x^2-6\,x^3\right )+7\,x^2+9\,x^3-27\,x^4+9\,x^5}{9\,x^3\,\ln \relax (x)+9\,x^4-9\,x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - log(x)*(6*x*exp(3) + 9*x^3 - 2) - exp(3)*(6*x^2 - 6*x^3) + 7*x^2 + 9*x^3 - 27*x^4 + 9*x^5)/(9*x^3*
log(x) + 9*x^4 - 9*x^5),x)

[Out]

int(-(2*x - log(x)*(6*x*exp(3) + 9*x^3 - 2) - exp(3)*(6*x^2 - 6*x^3) + 7*x^2 + 9*x^3 - 27*x^4 + 9*x^5)/(9*x^3*
log(x) + 9*x^4 - 9*x^5), x)

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sympy [A]  time = 0.20, size = 24, normalized size = 0.86 \begin {gather*} x - \log {\left (- x^{2} + x + \log {\relax (x )} \right )} + \frac {- 6 x e^{3} + 1}{9 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x*exp(3)+9*x**3-2)*ln(x)+(-6*x**3+6*x**2)*exp(3)-9*x**5+27*x**4-9*x**3-7*x**2-2*x)/(9*x**3*ln(x)
-9*x**5+9*x**4),x)

[Out]

x - log(-x**2 + x + log(x)) + (-6*x*exp(3) + 1)/(9*x**2)

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