3.4.46 \(\int \frac {9+x+3 x^2+x^3}{(1+x^2) (3+x^2)} \, dx\) [346]

Optimal. Leaf size=15 \[ 3 \tan ^{-1}(x)+\frac {1}{2} \log \left (3+x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6857, 209, 266} \begin {gather*} 3 \text {ArcTan}(x)+\frac {1}{2} \log \left (x^2+3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9 + x + 3*x^2 + x^3)/((1 + x^2)*(3 + x^2)),x]

[Out]

3*ArcTan[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 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {9+x+3 x^2+x^3}{\left (1+x^2\right ) \left (3+x^2\right )} \, dx &=\int \left (\frac {3}{1+x^2}+\frac {x}{3+x^2}\right ) \, dx\\ &=3 \int \frac {1}{1+x^2} \, dx+\int \frac {x}{3+x^2} \, dx\\ &=3 \tan ^{-1}(x)+\frac {1}{2} \log \left (3+x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} 3 \tan ^{-1}(x)+\frac {1}{2} \log \left (3+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 + x + 3*x^2 + x^3)/((1 + x^2)*(3 + x^2)),x]

[Out]

3*ArcTan[x] + Log[3 + x^2]/2

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Maple [A]
time = 0.20, size = 14, normalized size = 0.93

method	result	size
default	\(3 \arctan \left (x \right )+\frac {\ln \left (x^{2}+3\right )}{2}\)	\(14\)
risch	\(3 \arctan \left (x \right )+\frac {\ln \left (x^{2}+3\right )}{2}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+3*x^2+x+9)/(x^2+1)/(x^2+3),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 13, normalized size = 0.87 \begin {gather*} 3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+3*x^2+x+9)/(x^2+1)/(x^2+3),x, algorithm="maxima")

[Out]

3*arctan(x) + 1/2*log(x^2 + 3)

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Fricas [A]
time = 0.37, size = 13, normalized size = 0.87 \begin {gather*} 3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+3*x^2+x+9)/(x^2+1)/(x^2+3),x, algorithm="fricas")

[Out]

3*arctan(x) + 1/2*log(x^2 + 3)

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Sympy [A]
time = 0.04, size = 12, normalized size = 0.80 \begin {gather*} \frac {\log {\left (x^{2} + 3 \right )}}{2} + 3 \operatorname {atan}{\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+3*x**2+x+9)/(x**2+1)/(x**2+3),x)

[Out]

log(x**2 + 3)/2 + 3*atan(x)

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Giac [A]
time = 4.01, size = 13, normalized size = 0.87 \begin {gather*} 3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+3*x^2+x+9)/(x^2+1)/(x^2+3),x, algorithm="giac")

[Out]

3*arctan(x) + 1/2*log(x^2 + 3)

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Mupad [B]
time = 2.12, size = 13, normalized size = 0.87 \begin {gather*} \frac {\ln \left (x^2+3\right )}{2}+3\,\mathrm {atan}\left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 3*x^2 + x^3 + 9)/((x^2 + 1)*(x^2 + 3)),x)

[Out]

log(x^2 + 3)/2 + 3*atan(x)

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