\(\int \frac {4+8 x+5 x^2+2 x^3}{(2+2 x+x^2)^2} \, dx\) [327]

Optimal result

Integrand size = 26, antiderivative size = 28 \[ \int \frac {4+8 x+5 x^2+2 x^3}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {1}{2+2 x+x^2}-\arctan (1+x)+\log \left (2+2 x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1674, 648, 631, 210, 642} \[ \int \frac {4+8 x+5 x^2+2 x^3}{\left (2+2 x+x^2\right )^2} \, dx=-\arctan (x+1)-\frac {1}{x^2+2 x+2}+\log \left (x^2+2 x+2\right ) \]

[In]

Int[(4 + 8*x + 5*x^2 + 2*x^3)/(2 + 2*x + x^2)^2,x]

[Out]

-(2 + 2*x + x^2)^(-1) - ArcTan[1 + x] + Log[2 + 2*x + x^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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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 1674

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2+2 x+x^2}+\frac {1}{4} \int \frac {4+8 x}{2+2 x+x^2} \, dx \\ & = -\frac {1}{2+2 x+x^2}-\int \frac {1}{2+2 x+x^2} \, dx+\int \frac {2+2 x}{2+2 x+x^2} \, dx \\ & = -\frac {1}{2+2 x+x^2}+\log \left (2+2 x+x^2\right )+\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+x\right ) \\ & = -\frac {1}{2+2 x+x^2}-\tan ^{-1}(1+x)+\log \left (2+2 x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {4+8 x+5 x^2+2 x^3}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {1}{2+2 x+x^2}-\arctan (1+x)+\log \left (2+2 x+x^2\right ) \]

[In]

Integrate[(4 + 8*x + 5*x^2 + 2*x^3)/(2 + 2*x + x^2)^2,x]

[Out]

-(2 + 2*x + x^2)^(-1) - ArcTan[1 + x] + Log[2 + 2*x + x^2]

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {1}{x^{2}+2 x +2}-\arctan \left (x +1\right )+\ln \left (x^{2}+2 x +2\right )\)	\(29\)
risch	\(-\frac {1}{x^{2}+2 x +2}-\arctan \left (x +1\right )+\ln \left (x^{2}+2 x +2\right )\)	\(29\)
parallelrisch	\(\frac {2 i \ln \left (x +1-i\right )-i \ln \left (x +1+i\right ) x^{2}-2 i \ln \left (x +1+i\right )+2 \ln \left (x +1-i\right ) x^{2}+2 i \ln \left (x +1-i\right ) x +2 \ln \left (x +1+i\right ) x^{2}-2+i \ln \left (x +1-i\right ) x^{2}+4 \ln \left (x +1-i\right ) x -2 i \ln \left (x +1+i\right ) x +4 \ln \left (x +1+i\right ) x +4 \ln \left (x +1-i\right )+4 \ln \left (x +1+i\right )}{2 x^{2}+4 x +4}\)	\(133\)

[In]

int((2*x^3+5*x^2+8*x+4)/(x^2+2*x+2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {4+8 x+5 x^2+2 x^3}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {{\left (x^{2} + 2 \, x + 2\right )} \arctan \left (x + 1\right ) - {\left (x^{2} + 2 \, x + 2\right )} \log \left (x^{2} + 2 \, x + 2\right ) + 1}{x^{2} + 2 \, x + 2} \]

[In]

integrate((2*x^3+5*x^2+8*x+4)/(x^2+2*x+2)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {4+8 x+5 x^2+2 x^3}{\left (2+2 x+x^2\right )^2} \, dx=\log {\left (x^{2} + 2 x + 2 \right )} - \operatorname {atan}{\left (x + 1 \right )} - \frac {1}{x^{2} + 2 x + 2} \]

[In]

integrate((2*x**3+5*x**2+8*x+4)/(x**2+2*x+2)**2,x)

[Out]

log(x**2 + 2*x + 2) - atan(x + 1) - 1/(x**2 + 2*x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {4+8 x+5 x^2+2 x^3}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {1}{x^{2} + 2 \, x + 2} - \arctan \left (x + 1\right ) + \log \left (x^{2} + 2 \, x + 2\right ) \]

[In]

integrate((2*x^3+5*x^2+8*x+4)/(x^2+2*x+2)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {4+8 x+5 x^2+2 x^3}{\left (2+2 x+x^2\right )^2} \, dx=-\frac {1}{x^{2} + 2 \, x + 2} - \arctan \left (x + 1\right ) + \log \left (x^{2} + 2 \, x + 2\right ) \]

[In]

integrate((2*x^3+5*x^2+8*x+4)/(x^2+2*x+2)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.43 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {4+8 x+5 x^2+2 x^3}{\left (2+2 x+x^2\right )^2} \, dx=\ln \left (x^2+2\,x+2\right )-\mathrm {atan}\left (x+1\right )-\frac {1}{x^2+2\,x+2} \]

[In]

int((8*x + 5*x^2 + 2*x^3 + 4)/(2*x + x^2 + 2)^2,x)

[Out]

log(2*x + x^2 + 2) - atan(x + 1) - 1/(2*x + x^2 + 2)