3.23.42 \(\int \frac {-20-8 x^2}{5 x-x^2-2 x^3+x^2 \log (4)} \, dx\)

Optimal. Leaf size=20 \[ \log \left (\left (2+2 x-\frac {5+x+x \log (4)}{x}\right )^4\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 24, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6, 1594, 1628, 628} \begin {gather*} 4 \log \left (-2 x^2-x (1-\log (4))+5\right )-4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 - 8*x^2)/(5*x - x^2 - 2*x^3 + x^2*Log[4]),x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20-8 x^2}{5 x-2 x^3+x^2 (-1+\log (4))} \, dx\\ &=\int \frac {-20-8 x^2}{x \left (5-2 x^2+x (-1+\log (4))\right )} \, dx\\ &=\int \left (-\frac {4}{x}+\frac {4 (-1-4 x+\log (4))}{5-2 x^2-x (1-\log (4))}\right ) \, dx\\ &=-4 \log (x)+4 \int \frac {-1-4 x+\log (4)}{5-2 x^2+x (-1+\log (4))} \, dx\\ &=-4 \log (x)+4 \log \left (5-2 x^2-x (1-\log (4))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.10 \begin {gather*} 4 \left (-\log (x)+\log \left (5-x-2 x^2+x \log (4)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 8*x^2)/(5*x - x^2 - 2*x^3 + x^2*Log[4]),x]

[Out]

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

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fricas [A]  time = 1.07, size = 21, normalized size = 1.05 \begin {gather*} 4 \, \log \left (2 \, x^{2} - 2 \, x \log \relax (2) + x - 5\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^2-20)/(2*x^2*log(2)-2*x^3-x^2+5*x),x, algorithm="fricas")

[Out]

4*log(2*x^2 - 2*x*log(2) + x - 5) - 4*log(x)

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giac [A]  time = 0.25, size = 23, normalized size = 1.15 \begin {gather*} 4 \, \log \left ({\left | 2 \, x^{2} - 2 \, x \log \relax (2) + x - 5 \right |}\right ) - 4 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^2-20)/(2*x^2*log(2)-2*x^3-x^2+5*x),x, algorithm="giac")

[Out]

4*log(abs(2*x^2 - 2*x*log(2) + x - 5)) - 4*log(abs(x))

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maple [A]  time = 0.05, size = 22, normalized size = 1.10

method	result	size
default	\(4 \ln \left (-2 x \ln \relax (2)+2 x^{2}+x -5\right )-4 \ln \relax (x )\)	\(22\)
norman	\(-4 \ln \relax (x )+4 \ln \left (2 x \ln \relax (2)-2 x^{2}-x +5\right )\)	\(24\)
risch	\(-4 \ln \left (-x \right )+4 \ln \left (-5+2 x^{2}+\left (1-2 \ln \relax (2)\right ) x \right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*x^2-20)/(2*x^2*ln(2)-2*x^3-x^2+5*x),x,method=_RETURNVERBOSE)

[Out]

4*ln(-2*x*ln(2)+2*x^2+x-5)-4*ln(x)

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maxima [A]  time = 0.46, size = 24, normalized size = 1.20 \begin {gather*} 4 \, \log \left (2 \, x^{2} - x {\left (2 \, \log \relax (2) - 1\right )} - 5\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x^2-20)/(2*x^2*log(2)-2*x^3-x^2+5*x),x, algorithm="maxima")

[Out]

4*log(2*x^2 - x*(2*log(2) - 1) - 5) - 4*log(x)

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mupad [B]  time = 0.16, size = 21, normalized size = 1.05 \begin {gather*} 4\,\ln \left (\frac {x}{2}-x\,\ln \relax (2)+x^2-\frac {5}{2}\right )-4\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x^2 + 20)/(5*x + 2*x^2*log(2) - x^2 - 2*x^3),x)

[Out]

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

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sympy [A]  time = 0.37, size = 22, normalized size = 1.10 \begin {gather*} - 4 \log {\relax (x )} + 4 \log {\left (x^{2} + x \left (\frac {1}{2} - \log {\relax (2 )}\right ) - \frac {5}{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*x**2-20)/(2*x**2*ln(2)-2*x**3-x**2+5*x),x)

[Out]

-4*log(x) + 4*log(x**2 + x*(1/2 - log(2)) - 5/2)

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