\(\int \frac {1+x^3}{(13+4 x+x^2)^2} \, dx\) [365]

Optimal result

Integrand size = 16, antiderivative size = 45 \[ \int \frac {1+x^3}{\left (13+4 x+x^2\right )^2} \, dx=\frac {67+47 x}{18 \left (13+4 x+x^2\right )}-\frac {61}{54} \arctan \left (\frac {2+x}{3}\right )+\frac {1}{2} \log \left (13+4 x+x^2\right ) \]

[Out]

1/18*(67+47*x)/(x^2+4*x+13)-61/54*arctan(2/3+1/3*x)+1/2*ln(x^2+4*x+13)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1674, 648, 632, 210, 642} \[ \int \frac {1+x^3}{\left (13+4 x+x^2\right )^2} \, dx=-\frac {61}{54} \arctan \left (\frac {x+2}{3}\right )+\frac {47 x+67}{18 \left (x^2+4 x+13\right )}+\frac {1}{2} \log \left (x^2+4 x+13\right ) \]

[In]

Int[(1 + x^3)/(13 + 4*x + x^2)^2,x]

[Out]

(67 + 47*x)/(18*(13 + 4*x + x^2)) - (61*ArcTan[(2 + x)/3])/54 + Log[13 + 4*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 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 {67+47 x}{18 \left (13+4 x+x^2\right )}+\frac {1}{36} \int \frac {-50+36 x}{13+4 x+x^2} \, dx \\ & = \frac {67+47 x}{18 \left (13+4 x+x^2\right )}+\frac {1}{2} \int \frac {4+2 x}{13+4 x+x^2} \, dx-\frac {61}{18} \int \frac {1}{13+4 x+x^2} \, dx \\ & = \frac {67+47 x}{18 \left (13+4 x+x^2\right )}+\frac {1}{2} \log \left (13+4 x+x^2\right )+\frac {61}{9} \text {Subst}\left (\int \frac {1}{-36-x^2} \, dx,x,4+2 x\right ) \\ & = \frac {67+47 x}{18 \left (13+4 x+x^2\right )}-\frac {61}{54} \tan ^{-1}\left (\frac {2+x}{3}\right )+\frac {1}{2} \log \left (13+4 x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^3}{\left (13+4 x+x^2\right )^2} \, dx=\frac {67+47 x}{18 \left (13+4 x+x^2\right )}-\frac {61}{54} \arctan \left (\frac {2+x}{3}\right )+\frac {1}{2} \log \left (13+4 x+x^2\right ) \]

[In]

Integrate[(1 + x^3)/(13 + 4*x + x^2)^2,x]

[Out]

(67 + 47*x)/(18*(13 + 4*x + x^2)) - (61*ArcTan[(2 + x)/3])/54 + Log[13 + 4*x + x^2]/2

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\frac {47 x}{18}+\frac {67}{18}}{x^{2}+4 x +13}+\frac {\ln \left (x^{2}+4 x +13\right )}{2}-\frac {61 \arctan \left (\frac {2}{3}+\frac {x}{3}\right )}{54}\)	\(37\)
risch	\(\frac {\frac {47 x}{18}+\frac {67}{18}}{x^{2}+4 x +13}+\frac {\ln \left (x^{2}+4 x +13\right )}{2}-\frac {61 \arctan \left (\frac {2}{3}+\frac {x}{3}\right )}{54}\)	\(37\)
parallelrisch	\(\frac {-793 i \ln \left (x +2+3 i\right ) x^{2}+10309 i \ln \left (x +2-3 i\right )+3172 i \ln \left (x +2-3 i\right ) x +702 \ln \left (x +2-3 i\right ) x^{2}-3172 i \ln \left (x +2+3 i\right ) x +702 \ln \left (x +2+3 i\right ) x^{2}+793 i \ln \left (x +2-3 i\right ) x^{2}+2808 \ln \left (x +2-3 i\right ) x -10309 i \ln \left (x +2+3 i\right )+2808 \ln \left (x +2+3 i\right ) x -402 x^{2}+9126 \ln \left (x +2-3 i\right )+9126 \ln \left (x +2+3 i\right )+2058 x}{1404 x^{2}+5616 x +18252}\)	\(140\)

[In]

int((x^3+1)/(x^2+4*x+13)^2,x,method=_RETURNVERBOSE)

[Out]

(47/18*x+67/18)/(x^2+4*x+13)+1/2*ln(x^2+4*x+13)-61/54*arctan(2/3+1/3*x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16 \[ \int \frac {1+x^3}{\left (13+4 x+x^2\right )^2} \, dx=-\frac {61 \, {\left (x^{2} + 4 \, x + 13\right )} \arctan \left (\frac {1}{3} \, x + \frac {2}{3}\right ) - 27 \, {\left (x^{2} + 4 \, x + 13\right )} \log \left (x^{2} + 4 \, x + 13\right ) - 141 \, x - 201}{54 \, {\left (x^{2} + 4 \, x + 13\right )}} \]

[In]

integrate((x^3+1)/(x^2+4*x+13)^2,x, algorithm="fricas")

[Out]

-1/54*(61*(x^2 + 4*x + 13)*arctan(1/3*x + 2/3) - 27*(x^2 + 4*x + 13)*log(x^2 + 4*x + 13) - 141*x - 201)/(x^2 +
4*x + 13)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {1+x^3}{\left (13+4 x+x^2\right )^2} \, dx=\frac {47 x + 67}{18 x^{2} + 72 x + 234} + \frac {\log {\left (x^{2} + 4 x + 13 \right )}}{2} - \frac {61 \operatorname {atan}{\left (\frac {x}{3} + \frac {2}{3} \right )}}{54} \]

[In]

integrate((x**3+1)/(x**2+4*x+13)**2,x)

[Out]

(47*x + 67)/(18*x**2 + 72*x + 234) + log(x**2 + 4*x + 13)/2 - 61*atan(x/3 + 2/3)/54

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {1+x^3}{\left (13+4 x+x^2\right )^2} \, dx=\frac {47 \, x + 67}{18 \, {\left (x^{2} + 4 \, x + 13\right )}} - \frac {61}{54} \, \arctan \left (\frac {1}{3} \, x + \frac {2}{3}\right ) + \frac {1}{2} \, \log \left (x^{2} + 4 \, x + 13\right ) \]

[In]

integrate((x^3+1)/(x^2+4*x+13)^2,x, algorithm="maxima")

[Out]

1/18*(47*x + 67)/(x^2 + 4*x + 13) - 61/54*arctan(1/3*x + 2/3) + 1/2*log(x^2 + 4*x + 13)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {1+x^3}{\left (13+4 x+x^2\right )^2} \, dx=\frac {47 \, x + 67}{18 \, {\left (x^{2} + 4 \, x + 13\right )}} - \frac {61}{54} \, \arctan \left (\frac {1}{3} \, x + \frac {2}{3}\right ) + \frac {1}{2} \, \log \left (x^{2} + 4 \, x + 13\right ) \]

[In]

integrate((x^3+1)/(x^2+4*x+13)^2,x, algorithm="giac")

[Out]

1/18*(47*x + 67)/(x^2 + 4*x + 13) - 61/54*arctan(1/3*x + 2/3) + 1/2*log(x^2 + 4*x + 13)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int \frac {1+x^3}{\left (13+4 x+x^2\right )^2} \, dx=\frac {\ln \left (x^2+4\,x+13\right )}{2}-\frac {61\,\mathrm {atan}\left (\frac {x}{3}+\frac {2}{3}\right )}{54}+\frac {47\,x}{18\,\left (x^2+4\,x+13\right )}+\frac {67}{18\,\left (x^2+4\,x+13\right )} \]

[In]

int((x^3 + 1)/(4*x + x^2 + 13)^2,x)

[Out]

log(4*x + x^2 + 13)/2 - (61*atan(x/3 + 2/3))/54 + (47*x)/(18*(4*x + x^2 + 13)) + 67/(18*(4*x + x^2 + 13))