3.84.23 \(\int \frac {16}{1+64 x+1024 x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {x^2}{\frac {x}{16}+2 x^2}+\log (\log (4)) \]

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Rubi [A]  time = 0.00, antiderivative size = 11, normalized size of antiderivative = 0.52, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 27, 32} \begin {gather*} -\frac {1}{2 (32 x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[16/(1 + 64*x + 1024*x^2),x]

[Out]

-1/2*1/(1 + 32*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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=16 \int \frac {1}{1+64 x+1024 x^2} \, dx\\ &=16 \int \frac {1}{(1+32 x)^2} \, dx\\ &=-\frac {1}{2 (1+32 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.52 \begin {gather*} -\frac {1}{2 (1+32 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[16/(1 + 64*x + 1024*x^2),x]

[Out]

-1/2*1/(1 + 32*x)

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fricas [A]  time = 0.61, size = 9, normalized size = 0.43 \begin {gather*} -\frac {1}{2 \, {\left (32 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16/(1024*x^2+64*x+1),x, algorithm="fricas")

[Out]

-1/2/(32*x + 1)

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giac [A]  time = 0.21, size = 9, normalized size = 0.43 \begin {gather*} -\frac {1}{2 \, {\left (32 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16/(1024*x^2+64*x+1),x, algorithm="giac")

[Out]

-1/2/(32*x + 1)

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maple [A]  time = 0.36, size = 8, normalized size = 0.38

method	result	size
risch	\(-\frac {1}{64 \left (x +\frac {1}{32}\right )}\)	\(8\)
gosper	\(-\frac {1}{2 \left (32 x +1\right )}\)	\(10\)
default	\(-\frac {1}{2 \left (32 x +1\right )}\)	\(10\)
norman	\(\frac {16 x}{32 x +1}\)	\(11\)
meijerg	\(\frac {16 x}{32 x +1}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16/(1024*x^2+64*x+1),x,method=_RETURNVERBOSE)

[Out]

-1/64/(x+1/32)

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maxima [A]  time = 0.36, size = 9, normalized size = 0.43 \begin {gather*} -\frac {1}{2 \, {\left (32 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16/(1024*x^2+64*x+1),x, algorithm="maxima")

[Out]

-1/2/(32*x + 1)

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mupad [B]  time = 0.03, size = 9, normalized size = 0.43 \begin {gather*} -\frac {1}{64\,x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16/(64*x + 1024*x^2 + 1),x)

[Out]

-1/(64*x + 2)

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sympy [A]  time = 0.08, size = 7, normalized size = 0.33 \begin {gather*} - \frac {16}{1024 x + 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16/(1024*x**2+64*x+1),x)

[Out]

-16/(1024*x + 32)

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