3.297 \(\int \frac {3+x^2+x^3}{(2+x^2)^2} \, dx\)

Optimal. Leaf size=42 \[ \frac {x+4}{4 \left (x^2+2\right )}+\frac {1}{2} \log \left (x^2+2\right )+\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

[Out]

1/4*(4+x)/(x^2+2)+1/2*ln(x^2+2)+5/8*arctan(1/2*x*2^(1/2))*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1814, 635, 203, 260} \[ \frac {x+4}{4 \left (x^2+2\right )}+\frac {1}{2} \log \left (x^2+2\right )+\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(3 + x^2 + x^3)/(2 + x^2)^2,x]

[Out]

(4 + x)/(4*(2 + x^2)) + (5*ArcTan[x/Sqrt[2]])/(4*Sqrt[2]) + Log[2 + x^2]/2

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 260

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 635

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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {3+x^2+x^3}{\left (2+x^2\right )^2} \, dx &=\frac {4+x}{4 \left (2+x^2\right )}-\frac {1}{4} \int \frac {-5-4 x}{2+x^2} \, dx\\ &=\frac {4+x}{4 \left (2+x^2\right )}+\frac {5}{4} \int \frac {1}{2+x^2} \, dx+\int \frac {x}{2+x^2} \, dx\\ &=\frac {4+x}{4 \left (2+x^2\right )}+\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {1}{2} \log \left (2+x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 42, normalized size = 1.00 \[ \frac {x+4}{4 \left (x^2+2\right )}+\frac {1}{2} \log \left (x^2+2\right )+\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4 + x)/(4*(2 + x^2)) + (5*ArcTan[x/Sqrt[2]])/(4*Sqrt[2]) + Log[2 + x^2]/2

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 44, normalized size = 1.05 \[ \frac {5 \, \sqrt {2} {\left (x^{2} + 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + 4 \, {\left (x^{2} + 2\right )} \log \left (x^{2} + 2\right ) + 2 \, x + 8}{8 \, {\left (x^{2} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(5*sqrt(2)*(x^2 + 2)*arctan(1/2*sqrt(2)*x) + 4*(x^2 + 2)*log(x^2 + 2) + 2*x + 8)/(x^2 + 2)

________________________________________________________________________________________

giac [A]  time = 0.30, size = 33, normalized size = 0.79 \[ \frac {5}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {x + 4}{4 \, {\left (x^{2} + 2\right )}} + \frac {1}{2} \, \log \left (x^{2} + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/4*(x + 4)/(x^2 + 2) + 1/2*log(x^2 + 2)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 35, normalized size = 0.83 \[ \frac {5 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right )}{8}+\frac {\ln \left (x^{2}+2\right )}{2}+\frac {\frac {x}{4}+1}{x^{2}+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+x^2+3)/(x^2+2)^2,x)

[Out]

(1/4*x+1)/(x^2+2)+1/2*ln(x^2+2)+5/8*2^(1/2)*arctan(1/2*2^(1/2)*x)

________________________________________________________________________________________

maxima [A]  time = 2.36, size = 33, normalized size = 0.79 \[ \frac {5}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) + \frac {x + 4}{4 \, {\left (x^{2} + 2\right )}} + \frac {1}{2} \, \log \left (x^{2} + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*sqrt(2)*arctan(1/2*sqrt(2)*x) + 1/4*(x + 4)/(x^2 + 2) + 1/2*log(x^2 + 2)

________________________________________________________________________________________

mupad [B]  time = 2.19, size = 39, normalized size = 0.93 \[ \frac {\ln \left (x^2+2\right )}{2}+\frac {5\,\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{8}+\frac {x}{4\,\left (x^2+2\right )}+\frac {1}{x^2+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + x^3 + 3)/(x^2 + 2)^2,x)

[Out]

log(x^2 + 2)/2 + (5*2^(1/2)*atan((2^(1/2)*x)/2))/8 + x/(4*(x^2 + 2)) + 1/(x^2 + 2)

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 36, normalized size = 0.86 \[ \frac {x + 4}{4 x^{2} + 8} + \frac {\log {\left (x^{2} + 2 \right )}}{2} + \frac {5 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + 4)/(4*x**2 + 8) + log(x**2 + 2)/2 + 5*sqrt(2)*atan(sqrt(2)*x/2)/8

________________________________________________________________________________________