\(\int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx\) [270]

Optimal result

Integrand size = 22, antiderivative size = 13 \[ \int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx=\arctan (x)+\frac {1}{2} \log \left (2+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1687, 1163, 209, 1261, 640, 31} \[ \int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx=\arctan (x)+\frac {1}{2} \log \left (x^2+2\right ) \]

[In]

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

[Out]

ArcTan[x] + Log[2 + x^2]/2

Rule 31

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

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 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rule 1163

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(d + e*x^2)^(p +
q)*(a/d + (c/e)*x^2)^p, x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2
, 0] && IntegerQ[p]

Rule 1261

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 1687

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx=\arctan (x)+\frac {1}{2} \log \left (2+x^2\right ) \]

[In]

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

[Out]

ArcTan[x] + Log[2 + x^2]/2

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
default	\(\arctan \left (x \right )+\frac {\ln \left (x^{2}+2\right )}{2}\)	\(12\)
risch	\(\arctan \left (x \right )+\frac {\ln \left (x^{2}+2\right )}{2}\)	\(12\)
parallelrisch	\(\frac {i \ln \left (x +i\right )}{2}-\frac {i \ln \left (x -i\right )}{2}+\frac {\ln \left (x^{2}+2\right )}{2}\)	\(26\)

[In]

int((x^3+x^2+x+2)/(x^4+3*x^2+2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx=\arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 2\right ) \]

[In]

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

[Out]

arctan(x) + 1/2*log(x^2 + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx=\frac {\log {\left (x^{2} + 2 \right )}}{2} + \operatorname {atan}{\left (x \right )} \]

[In]

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

[Out]

log(x**2 + 2)/2 + atan(x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx=\arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 2\right ) \]

[In]

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

[Out]

arctan(x) + 1/2*log(x^2 + 2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx=\arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 2\right ) \]

[In]

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

[Out]

arctan(x) + 1/2*log(x^2 + 2)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {2+x+x^2+x^3}{2+3 x^2+x^4} \, dx=\frac {\ln \left (x^2+2\right )}{2}+\mathrm {atan}\left (x\right ) \]

[In]

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

[Out]

log(x^2 + 2)/2 + atan(x)