\(\int \frac {4-x+2 x^2}{4 x+x^3} \, dx\) [307]

Optimal result

Integrand size = 20, antiderivative size = 23 \[ \int \frac {4-x+2 x^2}{4 x+x^3} \, dx=-\frac {1}{2} \arctan \left (\frac {x}{2}\right )+\log (x)+\frac {1}{2} \log \left (4+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1607, 1816, 649, 209, 266} \[ \int \frac {4-x+2 x^2}{4 x+x^3} \, dx=-\frac {1}{2} \arctan \left (\frac {x}{2}\right )+\frac {1}{2} \log \left (x^2+4\right )+\log (x) \]

[In]

Int[(4 - x + 2*x^2)/(4*x + x^3),x]

[Out]

-1/2*ArcTan[x/2] + Log[x] + Log[4 + 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 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]

Rule 1816

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {4-x+2 x^2}{4 x+x^3} \, dx=-\frac {1}{2} \arctan \left (\frac {x}{2}\right )+\log (x)+\frac {1}{2} \log \left (4+x^2\right ) \]

[In]

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

[Out]

-1/2*ArcTan[x/2] + Log[x] + Log[4 + x^2]/2

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {\arctan \left (\frac {x}{2}\right )}{2}+\ln \left (x \right )+\frac {\ln \left (x^{2}+4\right )}{2}\)	\(18\)
risch	\(-\frac {\arctan \left (\frac {x}{2}\right )}{2}+\ln \left (x \right )+\frac {\ln \left (x^{2}+4\right )}{2}\)	\(18\)
meijerg	\(\frac {\ln \left (1+\frac {x^{2}}{4}\right )}{2}+\ln \left (x \right )-\ln \left (2\right )-\frac {\arctan \left (\frac {x}{2}\right )}{2}\)	\(24\)
parallelrisch	\(\ln \left (x \right )+\frac {\ln \left (x -2 i\right )}{2}+\frac {i \ln \left (x -2 i\right )}{4}+\frac {\ln \left (x +2 i\right )}{2}-\frac {i \ln \left (x +2 i\right )}{4}\)	\(34\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {4-x+2 x^2}{4 x+x^3} \, dx=-\frac {1}{2} \, \arctan \left (\frac {1}{2} \, x\right ) + \frac {1}{2} \, \log \left (x^{2} + 4\right ) + \log \left (x\right ) \]

[In]

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

[Out]

-1/2*arctan(1/2*x) + 1/2*log(x^2 + 4) + log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {4-x+2 x^2}{4 x+x^3} \, dx=\log {\left (x \right )} + \frac {\log {\left (x^{2} + 4 \right )}}{2} - \frac {\operatorname {atan}{\left (\frac {x}{2} \right )}}{2} \]

[In]

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

[Out]

log(x) + log(x**2 + 4)/2 - atan(x/2)/2

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {4-x+2 x^2}{4 x+x^3} \, dx=-\frac {1}{2} \, \arctan \left (\frac {1}{2} \, x\right ) + \frac {1}{2} \, \log \left (x^{2} + 4\right ) + \log \left (x\right ) \]

[In]

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

[Out]

-1/2*arctan(1/2*x) + 1/2*log(x^2 + 4) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {4-x+2 x^2}{4 x+x^3} \, dx=-\frac {1}{2} \, \arctan \left (\frac {1}{2} \, x\right ) + \frac {1}{2} \, \log \left (x^{2} + 4\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

-1/2*arctan(1/2*x) + 1/2*log(x^2 + 4) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 9.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4-x+2 x^2}{4 x+x^3} \, dx=\ln \left (x\right )+\ln \left (x-2{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+2{}\mathrm {i}\right )\,\left (\frac {1}{2}-\frac {1}{4}{}\mathrm {i}\right ) \]

[In]

int((2*x^2 - x + 4)/(4*x + x^3),x)

[Out]

log(x - 2i)*(1/2 + 1i/4) + log(x + 2i)*(1/2 - 1i/4) + log(x)