3.3.2 \(\int \frac {3+2 x}{3 x+x^3} \, dx\) [202]

Optimal. Leaf size=28 \[ \frac {2 \tan ^{-1}\left (\frac {x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-\frac {1}{2} \log \left (3+x^2\right ) \]

[Out]

ln(x)-1/2*ln(x^2+3)+2/3*arctan(1/3*x*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1607, 815, 649, 209, 266} \begin {gather*} -\frac {1}{2} \log \left (x^2+3\right )+\log (x)+\frac {2 \tan ^{-1}\left (\frac {x}{\sqrt {3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 2*x)/(3*x + x^3),x]

[Out]

(2*ArcTan[x/Sqrt[3]])/Sqrt[3] + Log[x] - Log[3 + x^2]/2

Rule 209

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

Rule 266

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

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1607

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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 28, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-\frac {1}{2} \log \left (3+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 2*x)/(3*x + x^3),x]

[Out]

(2*ArcTan[x/Sqrt[3]])/Sqrt[3] + Log[x] - Log[3 + x^2]/2

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Mathics [A]
time = 1.84, size = 23, normalized size = 0.82 \begin {gather*} \text {Log}\left [x\right ]-\frac {\text {Log}\left [3+x^2\right ]}{2}+\frac {2 \sqrt {3} \text {ArcTan}\left [\frac {\sqrt {3} x}{3}\right ]}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(3 + 2*x)/(3*x + x^3),x]')

[Out]

Log[x] - Log[3 + x ^ 2] / 2 + 2 Sqrt[3] ArcTan[Sqrt[3] x / 3] / 3

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Maple [A]
time = 0.05, size = 24, normalized size = 0.86

method	result	size
default	\(\ln \left (x \right )-\frac {\ln \left (x^{2}+3\right )}{2}+\frac {2 \arctan \left (\frac {x \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(24\)
risch	\(\ln \left (x \right )-\frac {\ln \left (x^{2}+3\right )}{2}+\frac {2 \arctan \left (\frac {x \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(24\)
meijerg	\(\ln \left (x \right )-\frac {\ln \left (3\right )}{2}-\frac {\ln \left (\frac {x^{2}}{3}+1\right )}{2}+\frac {2 \arctan \left (\frac {x \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3+2*x)/(x^3+3*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)-1/2*ln(x^2+3)+2/3*arctan(1/3*x*3^(1/2))*3^(1/2)

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Maxima [A]
time = 0.38, size = 23, normalized size = 0.82 \begin {gather*} \frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} x\right ) - \frac {1}{2} \, \log \left (x^{2} + 3\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*x) - 1/2*log(x^2 + 3) + log(x)

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Fricas [A]
time = 0.33, size = 23, normalized size = 0.82 \begin {gather*} \frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} x\right ) - \frac {1}{2} \, \log \left (x^{2} + 3\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*x) - 1/2*log(x^2 + 3) + log(x)

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Sympy [A]
time = 0.07, size = 29, normalized size = 1.04 \begin {gather*} \log {\left (x \right )} - \frac {\log {\left (x^{2} + 3 \right )}}{2} + \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - log(x**2 + 3)/2 + 2*sqrt(3)*atan(sqrt(3)*x/3)/3

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Giac [A]
time = 0.00, size = 33, normalized size = 1.18 \begin {gather*} \ln \left |x\right |-\frac {\ln \left (x^{2}+3\right )}{2}+\frac {4 \arctan \left (\frac {x}{\sqrt {3}}\right )}{2 \sqrt {3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+2*x)/(x^3+3*x),x)

[Out]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*x) - 1/2*log(x^2 + 3) + log(abs(x))

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Mupad [B]
time = 0.30, size = 55, normalized size = 1.96 \begin {gather*} \ln \left (x\right )-\frac {\ln \left (x+\sqrt {3}\,1{}\mathrm {i}\right )}{2}-\frac {\ln \left (x-\sqrt {3}\,1{}\mathrm {i}\right )}{2}-\frac {\sqrt {3}\,\ln \left (x-\sqrt {3}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}+\frac {\sqrt {3}\,\ln \left (x+\sqrt {3}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 3)/(3*x + x^3),x)

[Out]

log(x) - log(x + 3^(1/2)*1i)/2 - log(x - 3^(1/2)*1i)/2 - (3^(1/2)*log(x - 3^(1/2)*1i)*1i)/3 + (3^(1/2)*log(x +
3^(1/2)*1i)*1i)/3

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