3.62.62 \(\int \frac {-4+16 x^2}{x+5 x^2+4 x^3} \, dx\)

Optimal. Leaf size=23 \[ \log \left (\frac {e^{4 e^9} \left ((-1-2 x)^2+x\right )^4}{x^4}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1594, 1628} \begin {gather*} -4 \log (x)+4 \log (x+1)+4 \log (4 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 16*x^2)/(x + 5*x^2 + 4*x^3),x]

[Out]

-4*Log[x] + 4*Log[1 + x] + 4*Log[1 + 4*x]

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 {-4+16 x^2}{x \left (1+5 x+4 x^2\right )} \, dx\\ &=\int \left (-\frac {4}{x}+\frac {4}{1+x}+\frac {16}{1+4 x}\right ) \, dx\\ &=-4 \log (x)+4 \log (1+x)+4 \log (1+4 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.78 \begin {gather*} -4 \log (x)+4 \log \left (1+5 x+4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 16*x^2)/(x + 5*x^2 + 4*x^3),x]

[Out]

-4*Log[x] + 4*Log[1 + 5*x + 4*x^2]

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fricas [A]  time = 0.50, size = 18, normalized size = 0.78 \begin {gather*} 4 \, \log \left (4 \, x^{2} + 5 \, x + 1\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^2-4)/(4*x^3+5*x^2+x),x, algorithm="fricas")

[Out]

4*log(4*x^2 + 5*x + 1) - 4*log(x)

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giac [A]  time = 0.14, size = 22, normalized size = 0.96 \begin {gather*} 4 \, \log \left ({\left | 4 \, x + 1 \right |}\right ) + 4 \, \log \left ({\left | x + 1 \right |}\right ) - 4 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^2-4)/(4*x^3+5*x^2+x),x, algorithm="giac")

[Out]

4*log(abs(4*x + 1)) + 4*log(abs(x + 1)) - 4*log(abs(x))

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maple [A]  time = 0.03, size = 19, normalized size = 0.83

method	result	size
risch	\(-4 \ln \relax (x )+4 \ln \left (4 x^{2}+5 x +1\right )\)	\(19\)
default	\(4 \ln \left (4 x +1\right )-4 \ln \relax (x )+4 \ln \left (x +1\right )\)	\(20\)
norman	\(4 \ln \left (4 x +1\right )-4 \ln \relax (x )+4 \ln \left (x +1\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^2-4)/(4*x^3+5*x^2+x),x,method=_RETURNVERBOSE)

[Out]

-4*ln(x)+4*ln(4*x^2+5*x+1)

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maxima [A]  time = 0.34, size = 19, normalized size = 0.83 \begin {gather*} 4 \, \log \left (4 \, x + 1\right ) + 4 \, \log \left (x + 1\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^2-4)/(4*x^3+5*x^2+x),x, algorithm="maxima")

[Out]

4*log(4*x + 1) + 4*log(x + 1) - 4*log(x)

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mupad [B]  time = 4.12, size = 16, normalized size = 0.70 \begin {gather*} 4\,\ln \left (x^2+\frac {5\,x}{4}+\frac {1}{4}\right )-4\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^2 - 4)/(x + 5*x^2 + 4*x^3),x)

[Out]

4*log((5*x)/4 + x^2 + 1/4) - 4*log(x)

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sympy [A]  time = 0.09, size = 17, normalized size = 0.74 \begin {gather*} - 4 \log {\relax (x )} + 4 \log {\left (4 x^{2} + 5 x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x**2-4)/(4*x**3+5*x**2+x),x)

[Out]

-4*log(x) + 4*log(4*x**2 + 5*x + 1)

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