3.15.82 \(\int \frac {-20-91 x+x^2+8 x^3+(5+18 x-6 x^2) \log (x)}{25 x+50 x^2-x^3-2 x^4+(-5 x-9 x^2+2 x^3) \log (x)} \, dx\)

Optimal. Leaf size=27 \[ 5+e^5-\log \left ((5-x) x \left (\frac {1}{2}+x\right ) (-5-x+\log (x))\right ) \]

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Rubi [A]  time = 0.58, antiderivative size = 31, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 4, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6741, 6728, 1612, 6684} \begin {gather*} -\log (5-x)-\log (x)-\log (2 x+1)-\log (x-\log (x)+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 - 91*x + x^2 + 8*x^3 + (5 + 18*x - 6*x^2)*Log[x])/(25*x + 50*x^2 - x^3 - 2*x^4 + (-5*x - 9*x^2 + 2*x^
3)*Log[x]),x]

[Out]

-Log[5 - x] - Log[x] - Log[1 + 2*x] - Log[5 + x - Log[x]]

Rule 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

Rule 6684

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

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{x \left (5+9 x-2 x^2\right ) (5+x-\log (x))} \, dx\\ &=\int \left (\frac {5+18 x-6 x^2}{(-5+x) x (1+2 x)}+\frac {1-x}{x (5+x-\log (x))}\right ) \, dx\\ &=\int \frac {5+18 x-6 x^2}{(-5+x) x (1+2 x)} \, dx+\int \frac {1-x}{x (5+x-\log (x))} \, dx\\ &=-\log (5+x-\log (x))+\int \left (\frac {1}{5-x}-\frac {1}{x}-\frac {2}{1+2 x}\right ) \, dx\\ &=-\log (5-x)-\log (x)-\log (1+2 x)-\log (5+x-\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 28, normalized size = 1.04 \begin {gather*} -\log (x)-\log \left (5+9 x-2 x^2\right )-\log (5+x-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 91*x + x^2 + 8*x^3 + (5 + 18*x - 6*x^2)*Log[x])/(25*x + 50*x^2 - x^3 - 2*x^4 + (-5*x - 9*x^2
+ 2*x^3)*Log[x]),x]

[Out]

-Log[x] - Log[5 + 9*x - 2*x^2] - Log[5 + x - Log[x]]

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fricas [A]  time = 0.67, size = 28, normalized size = 1.04 \begin {gather*} -\log \left (2 \, x^{3} - 9 \, x^{2} - 5 \, x\right ) - \log \left (-x + \log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2+18*x+5)*log(x)+8*x^3+x^2-91*x-20)/((2*x^3-9*x^2-5*x)*log(x)-2*x^4-x^3+50*x^2+25*x),x, algor
ithm="fricas")

[Out]

-log(2*x^3 - 9*x^2 - 5*x) - log(-x + log(x) - 5)

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giac [A]  time = 0.36, size = 28, normalized size = 1.04 \begin {gather*} -\log \left (2 \, x^{2} - 9 \, x - 5\right ) - \log \relax (x) - \log \left (-x + \log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2+18*x+5)*log(x)+8*x^3+x^2-91*x-20)/((2*x^3-9*x^2-5*x)*log(x)-2*x^4-x^3+50*x^2+25*x),x, algor
ithm="giac")

[Out]

-log(2*x^2 - 9*x - 5) - log(x) - log(-x + log(x) - 5)

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maple [A]  time = 0.04, size = 29, normalized size = 1.07

method	result	size
risch	\(-\ln \left (2 x^{3}-9 x^{2}-5 x \right )-\ln \left (\ln \relax (x )-5-x \right )\)	\(29\)
norman	\(-\ln \relax (x )-\ln \left (x -5\right )-\ln \left (2 x +1\right )-\ln \left (x -\ln \relax (x )+5\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x^2+18*x+5)*ln(x)+8*x^3+x^2-91*x-20)/((2*x^3-9*x^2-5*x)*ln(x)-2*x^4-x^3+50*x^2+25*x),x,method=_RETURN
VERBOSE)

[Out]

-ln(2*x^3-9*x^2-5*x)-ln(ln(x)-5-x)

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maxima [A]  time = 0.49, size = 29, normalized size = 1.07 \begin {gather*} -\log \left (2 \, x + 1\right ) - \log \left (x - 5\right ) - \log \relax (x) - \log \left (-x + \log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2+18*x+5)*log(x)+8*x^3+x^2-91*x-20)/((2*x^3-9*x^2-5*x)*log(x)-2*x^4-x^3+50*x^2+25*x),x, algor
ithm="maxima")

[Out]

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

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mupad [B]  time = 1.14, size = 26, normalized size = 0.96 \begin {gather*} -\ln \left (x-\ln \relax (x)+5\right )-\ln \left (x\,\left (-2\,x^2+9\,x+5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(18*x - 6*x^2 + 5) - 91*x + x^2 + 8*x^3 - 20)/(x^3 - 50*x^2 - 25*x + 2*x^4 + log(x)*(5*x + 9*x^2
- 2*x^3)),x)

[Out]

- log(x - log(x) + 5) - log(x*(9*x - 2*x^2 + 5))

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sympy [A]  time = 0.18, size = 24, normalized size = 0.89 \begin {gather*} - \log {\left (- x + \log {\relax (x )} - 5 \right )} - \log {\left (2 x^{3} - 9 x^{2} - 5 x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x**2+18*x+5)*ln(x)+8*x**3+x**2-91*x-20)/((2*x**3-9*x**2-5*x)*ln(x)-2*x**4-x**3+50*x**2+25*x),x)

[Out]

-log(-x + log(x) - 5) - log(2*x**3 - 9*x**2 - 5*x)

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