\(\int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx\) [264]

Optimal result

Integrand size = 26, antiderivative size = 27 \[ \int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx=\frac {x^3}{3}-3 \arctan (3+x)+\frac {1}{2} \log \left (10+6 x+x^2\right ) \]

[Out]

1/3*x^3-3*arctan(3+x)+1/2*ln(x^2+6*x+10)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1671, 648, 632, 210, 642} \[ \int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx=-3 \arctan (x+3)+\frac {x^3}{3}+\frac {1}{2} \log \left (x^2+6 x+10\right ) \]

[In]

Int[(x + 10*x^2 + 6*x^3 + x^4)/(10 + 6*x + x^2),x]

[Out]

x^3/3 - 3*ArcTan[3 + x] + Log[10 + 6*x + x^2]/2

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1671

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \left (x^2+\frac {x}{10+6 x+x^2}\right ) \, dx \\ & = \frac {x^3}{3}+\int \frac {x}{10+6 x+x^2} \, dx \\ & = \frac {x^3}{3}+\frac {1}{2} \int \frac {6+2 x}{10+6 x+x^2} \, dx-3 \int \frac {1}{10+6 x+x^2} \, dx \\ & = \frac {x^3}{3}+\frac {1}{2} \log \left (10+6 x+x^2\right )+6 \text {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,6+2 x\right ) \\ & = \frac {x^3}{3}-3 \tan ^{-1}(3+x)+\frac {1}{2} \log \left (10+6 x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx=\frac {x^3}{3}-3 \arctan (3+x)+\frac {1}{2} \log \left (10+6 x+x^2\right ) \]

[In]

Integrate[(x + 10*x^2 + 6*x^3 + x^4)/(10 + 6*x + x^2),x]

[Out]

x^3/3 - 3*ArcTan[3 + x] + Log[10 + 6*x + x^2]/2

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {x^{3}}{3}-3 \arctan \left (3+x \right )+\frac {\ln \left (x^{2}+6 x +10\right )}{2}\)	\(24\)
risch	\(\frac {x^{3}}{3}-3 \arctan \left (3+x \right )+\frac {\ln \left (x^{2}+6 x +10\right )}{2}\)	\(24\)
parallelrisch	\(\frac {x^{3}}{3}+\frac {\ln \left (x +3-i\right )}{2}+\frac {3 i \ln \left (x +3-i\right )}{2}+\frac {\ln \left (x +3+i\right )}{2}-\frac {3 i \ln \left (x +3+i\right )}{2}\)	\(41\)

[In]

int((x^4+6*x^3+10*x^2+x)/(x^2+6*x+10),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3-3*arctan(3+x)+1/2*ln(x^2+6*x+10)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx=\frac {1}{3} \, x^{3} - 3 \, \arctan \left (x + 3\right ) + \frac {1}{2} \, \log \left (x^{2} + 6 \, x + 10\right ) \]

[In]

integrate((x^4+6*x^3+10*x^2+x)/(x^2+6*x+10),x, algorithm="fricas")

[Out]

1/3*x^3 - 3*arctan(x + 3) + 1/2*log(x^2 + 6*x + 10)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx=\frac {x^{3}}{3} + \frac {\log {\left (x^{2} + 6 x + 10 \right )}}{2} - 3 \operatorname {atan}{\left (x + 3 \right )} \]

[In]

integrate((x**4+6*x**3+10*x**2+x)/(x**2+6*x+10),x)

[Out]

x**3/3 + log(x**2 + 6*x + 10)/2 - 3*atan(x + 3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx=\frac {1}{3} \, x^{3} - 3 \, \arctan \left (x + 3\right ) + \frac {1}{2} \, \log \left (x^{2} + 6 \, x + 10\right ) \]

[In]

integrate((x^4+6*x^3+10*x^2+x)/(x^2+6*x+10),x, algorithm="maxima")

[Out]

1/3*x^3 - 3*arctan(x + 3) + 1/2*log(x^2 + 6*x + 10)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx=\frac {1}{3} \, x^{3} - 3 \, \arctan \left (x + 3\right ) + \frac {1}{2} \, \log \left (x^{2} + 6 \, x + 10\right ) \]

[In]

integrate((x^4+6*x^3+10*x^2+x)/(x^2+6*x+10),x, algorithm="giac")

[Out]

1/3*x^3 - 3*arctan(x + 3) + 1/2*log(x^2 + 6*x + 10)

Mupad [B] (verification not implemented)

Time = 9.44 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx=\frac {\ln \left (x^2+6\,x+10\right )}{2}-3\,\mathrm {atan}\left (x+3\right )+\frac {x^3}{3} \]

[In]

int((x + 10*x^2 + 6*x^3 + x^4)/(6*x + x^2 + 10),x)

[Out]

log(6*x + x^2 + 10)/2 - 3*atan(x + 3) + x^3/3