3.276 \(\int \frac{15-5 x+x^2+x^3}{(5+x^2) (3+2 x+x^2)} \, dx\)

Optimal. Leaf size=46 \[ \frac{1}{2} \log \left (x^2+2 x+3\right )-\sqrt{5} \tan ^{-1}\left (\frac{x}{\sqrt{5}}\right )+\frac{5 \tan ^{-1}\left (\frac{x+1}{\sqrt{2}}\right )}{\sqrt{2}} \]

[Out]

-(Sqrt[5]*ArcTan[x/Sqrt[5]]) + (5*ArcTan[(1 + x)/Sqrt[2]])/Sqrt[2] + Log[3 + 2*x + x^2]/2

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Rubi [A]  time = 0.13324, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {6725, 203, 634, 618, 204, 628} \[ \frac{1}{2} \log \left (x^2+2 x+3\right )-\sqrt{5} \tan ^{-1}\left (\frac{x}{\sqrt{5}}\right )+\frac{5 \tan ^{-1}\left (\frac{x+1}{\sqrt{2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(15 - 5*x + x^2 + x^3)/((5 + x^2)*(3 + 2*x + x^2)),x]

[Out]

-(Sqrt[5]*ArcTan[x/Sqrt[5]]) + (5*ArcTan[(1 + x)/Sqrt[2]])/Sqrt[2] + Log[3 + 2*x + x^2]/2

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 203

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

Rule 634

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 618

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 204

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

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{15-5 x+x^2+x^3}{\left (5+x^2\right ) \left (3+2 x+x^2\right )} \, dx &=\int \left (-\frac{5}{5+x^2}+\frac{6+x}{3+2 x+x^2}\right ) \, dx\\ &=-\left (5 \int \frac{1}{5+x^2} \, dx\right )+\int \frac{6+x}{3+2 x+x^2} \, dx\\ &=-\sqrt{5} \tan ^{-1}\left (\frac{x}{\sqrt{5}}\right )+\frac{1}{2} \int \frac{2+2 x}{3+2 x+x^2} \, dx+5 \int \frac{1}{3+2 x+x^2} \, dx\\ &=-\sqrt{5} \tan ^{-1}\left (\frac{x}{\sqrt{5}}\right )+\frac{1}{2} \log \left (3+2 x+x^2\right )-10 \operatorname{Subst}\left (\int \frac{1}{-8-x^2} \, dx,x,2+2 x\right )\\ &=-\sqrt{5} \tan ^{-1}\left (\frac{x}{\sqrt{5}}\right )+\frac{5 \tan ^{-1}\left (\frac{1+x}{\sqrt{2}}\right )}{\sqrt{2}}+\frac{1}{2} \log \left (3+2 x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0200394, size = 46, normalized size = 1. \[ \frac{1}{2} \log \left (x^2+2 x+3\right )-\sqrt{5} \tan ^{-1}\left (\frac{x}{\sqrt{5}}\right )+\frac{5 \tan ^{-1}\left (\frac{x+1}{\sqrt{2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(15 - 5*x + x^2 + x^3)/((5 + x^2)*(3 + 2*x + x^2)),x]

[Out]

-(Sqrt[5]*ArcTan[x/Sqrt[5]]) + (5*ArcTan[(1 + x)/Sqrt[2]])/Sqrt[2] + Log[3 + 2*x + x^2]/2

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Maple [A]  time = 0.006, size = 41, normalized size = 0.9 \begin{align*}{\frac{\ln \left ({x}^{2}+2\,x+3 \right ) }{2}}+{\frac{5\,\sqrt{2}}{2}\arctan \left ({\frac{ \left ( 2+2\,x \right ) \sqrt{2}}{4}} \right ) }-\arctan \left ({\frac{x\sqrt{5}}{5}} \right ) \sqrt{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+x^2-5*x+15)/(x^2+5)/(x^2+2*x+3),x)

[Out]

1/2*ln(x^2+2*x+3)+5/2*2^(1/2)*arctan(1/4*(2+2*x)*2^(1/2))-arctan(1/5*x*5^(1/2))*5^(1/2)

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Maxima [A]  time = 1.51628, size = 51, normalized size = 1.11 \begin{align*} \frac{5}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x + 1\right )}\right ) - \sqrt{5} \arctan \left (\frac{1}{5} \, \sqrt{5} x\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x^2-5*x+15)/(x^2+5)/(x^2+2*x+3),x, algorithm="maxima")

[Out]

5/2*sqrt(2)*arctan(1/2*sqrt(2)*(x + 1)) - sqrt(5)*arctan(1/5*sqrt(5)*x) + 1/2*log(x^2 + 2*x + 3)

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Fricas [A]  time = 1.30083, size = 132, normalized size = 2.87 \begin{align*} \frac{5}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x + 1\right )}\right ) - \sqrt{5} \arctan \left (\frac{1}{5} \, \sqrt{5} x\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x^2-5*x+15)/(x^2+5)/(x^2+2*x+3),x, algorithm="fricas")

[Out]

5/2*sqrt(2)*arctan(1/2*sqrt(2)*(x + 1)) - sqrt(5)*arctan(1/5*sqrt(5)*x) + 1/2*log(x^2 + 2*x + 3)

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Sympy [A]  time = 0.186845, size = 51, normalized size = 1.11 \begin{align*} \frac{\log{\left (x^{2} + 2 x + 3 \right )}}{2} - \sqrt{5} \operatorname{atan}{\left (\frac{\sqrt{5} x}{5} \right )} + \frac{5 \sqrt{2} \operatorname{atan}{\left (\frac{\sqrt{2} x}{2} + \frac{\sqrt{2}}{2} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+x**2-5*x+15)/(x**2+5)/(x**2+2*x+3),x)

[Out]

log(x**2 + 2*x + 3)/2 - sqrt(5)*atan(sqrt(5)*x/5) + 5*sqrt(2)*atan(sqrt(2)*x/2 + sqrt(2)/2)/2

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Giac [A]  time = 1.13888, size = 51, normalized size = 1.11 \begin{align*} \frac{5}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (x + 1\right )}\right ) - \sqrt{5} \arctan \left (\frac{1}{5} \, \sqrt{5} x\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x^2-5*x+15)/(x^2+5)/(x^2+2*x+3),x, algorithm="giac")

[Out]

5/2*sqrt(2)*arctan(1/2*sqrt(2)*(x + 1)) - sqrt(5)*arctan(1/5*sqrt(5)*x) + 1/2*log(x^2 + 2*x + 3)