3.351 \(\int \frac {1+x+4 x^2}{x+4 x^3} \, dx\)

Optimal. Leaf size=11 \[ \log (x)+\frac {1}{2} \tan ^{-1}(2 x) \]

[Out]

1/2*arctan(2*x)+ln(x)

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Rubi [A]  time = 0.03, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1593, 1802, 203} \[ \log (x)+\frac {1}{2} \tan ^{-1}(2 x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x + 4*x^2)/(x + 4*x^3),x]

[Out]

ArcTan[2*x]/2 + Log[x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1593

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

Rule 1802

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

Rubi steps

\begin {align*} \int \frac {1+x+4 x^2}{x+4 x^3} \, dx &=\int \frac {1+x+4 x^2}{x \left (1+4 x^2\right )} \, dx\\ &=\int \left (\frac {1}{x}+\frac {1}{1+4 x^2}\right ) \, dx\\ &=\log (x)+\int \frac {1}{1+4 x^2} \, dx\\ &=\frac {1}{2} \tan ^{-1}(2 x)+\log (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 1.00 \[ \log (x)+\frac {1}{2} \tan ^{-1}(2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + 4*x^2)/(x + 4*x^3),x]

[Out]

ArcTan[2*x]/2 + Log[x]

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fricas [A]  time = 0.65, size = 9, normalized size = 0.82 \[ \frac {1}{2} \, \arctan \left (2 \, x\right ) + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+x+1)/(4*x^3+x),x, algorithm="fricas")

[Out]

1/2*arctan(2*x) + log(x)

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giac [A]  time = 0.29, size = 10, normalized size = 0.91 \[ \frac {1}{2} \, \arctan \left (2 \, x\right ) + \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+x+1)/(4*x^3+x),x, algorithm="giac")

[Out]

1/2*arctan(2*x) + log(abs(x))

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maple [A]  time = 0.01, size = 10, normalized size = 0.91 \[ \frac {\arctan \left (2 x \right )}{2}+\ln \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+x+1)/(4*x^3+x),x)

[Out]

1/2*arctan(2*x)+ln(x)

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maxima [A]  time = 1.61, size = 9, normalized size = 0.82 \[ \frac {1}{2} \, \arctan \left (2 \, x\right ) + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+x+1)/(4*x^3+x),x, algorithm="maxima")

[Out]

1/2*arctan(2*x) + log(x)

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mupad [B]  time = 2.17, size = 17, normalized size = 1.55 \[ \ln \relax (x)-\frac {\mathrm {atan}\left (\frac {17}{32\,\left (\frac {x}{16}-\frac {1}{8}\right )}+4\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 4*x^2 + 1)/(x + 4*x^3),x)

[Out]

log(x) - atan(17/(32*(x/16 - 1/8)) + 4)/2

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sympy [A]  time = 0.13, size = 8, normalized size = 0.73 \[ \log {\relax (x )} + \frac {\operatorname {atan}{\left (2 x \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+x+1)/(4*x**3+x),x)

[Out]

log(x) + atan(2*x)/2

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