3.1.93 \(\int \frac {\sqrt {2}+x^2}{1+b x^2+x^4} \, dx\)

Optimal. Leaf size=160 \[ \frac {\left (1-\sqrt {2}\right ) \log \left (-\sqrt {2-b} x+x^2+1\right )}{4 \sqrt {2-b}}-\frac {\left (1-\sqrt {2}\right ) \log \left (\sqrt {2-b} x+x^2+1\right )}{4 \sqrt {2-b}}-\frac {\left (1+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {2-b}-2 x}{\sqrt {b+2}}\right )}{2 \sqrt {b+2}}+\frac {\left (1+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {2-b}+2 x}{\sqrt {b+2}}\right )}{2 \sqrt {b+2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1169, 634, 618, 204, 628} \begin {gather*} \frac {\left (1-\sqrt {2}\right ) \log \left (-\sqrt {2-b} x+x^2+1\right )}{4 \sqrt {2-b}}-\frac {\left (1-\sqrt {2}\right ) \log \left (\sqrt {2-b} x+x^2+1\right )}{4 \sqrt {2-b}}-\frac {\left (1+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {2-b}-2 x}{\sqrt {b+2}}\right )}{2 \sqrt {b+2}}+\frac {\left (1+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {2-b}+2 x}{\sqrt {b+2}}\right )}{2 \sqrt {b+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[2] + x^2)/(1 + b*x^2 + x^4),x]

[Out]

-((1 + Sqrt[2])*ArcTan[(Sqrt[2 - b] - 2*x)/Sqrt[2 + b]])/(2*Sqrt[2 + b]) + ((1 + Sqrt[2])*ArcTan[(Sqrt[2 - b]
+ 2*x)/Sqrt[2 + b]])/(2*Sqrt[2 + b]) + ((1 - Sqrt[2])*Log[1 - Sqrt[2 - b]*x + x^2])/(4*Sqrt[2 - b]) - ((1 - Sq
rt[2])*Log[1 + Sqrt[2 - b]*x + x^2])/(4*Sqrt[2 - b])

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 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 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]

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 1169

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

Rubi steps

\begin {align*} \int \frac {\sqrt {2}+x^2}{1+b x^2+x^4} \, dx &=\frac {\int \frac {\sqrt {2} \sqrt {2-b}-\left (-1+\sqrt {2}\right ) x}{1-\sqrt {2-b} x+x^2} \, dx}{2 \sqrt {2-b}}+\frac {\int \frac {\sqrt {2} \sqrt {2-b}+\left (-1+\sqrt {2}\right ) x}{1+\sqrt {2-b} x+x^2} \, dx}{2 \sqrt {2-b}}\\ &=\frac {1}{4} \left (1+\sqrt {2}\right ) \int \frac {1}{1-\sqrt {2-b} x+x^2} \, dx+\frac {1}{4} \left (1+\sqrt {2}\right ) \int \frac {1}{1+\sqrt {2-b} x+x^2} \, dx+\frac {\left (1-\sqrt {2}\right ) \int \frac {-\sqrt {2-b}+2 x}{1-\sqrt {2-b} x+x^2} \, dx}{4 \sqrt {2-b}}-\frac {\left (1-\sqrt {2}\right ) \int \frac {\sqrt {2-b}+2 x}{1+\sqrt {2-b} x+x^2} \, dx}{4 \sqrt {2-b}}\\ &=\frac {\left (1-\sqrt {2}\right ) \log \left (1-\sqrt {2-b} x+x^2\right )}{4 \sqrt {2-b}}-\frac {\left (1-\sqrt {2}\right ) \log \left (1+\sqrt {2-b} x+x^2\right )}{4 \sqrt {2-b}}+\frac {1}{2} \left (-1-\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{-2-b-x^2} \, dx,x,-\sqrt {2-b}+2 x\right )+\frac {1}{2} \left (-1-\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{-2-b-x^2} \, dx,x,\sqrt {2-b}+2 x\right )\\ &=-\frac {\left (1+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {2-b}-2 x}{\sqrt {2+b}}\right )}{2 \sqrt {2+b}}+\frac {\left (1+\sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {2-b}+2 x}{\sqrt {2+b}}\right )}{2 \sqrt {2+b}}+\frac {\left (1-\sqrt {2}\right ) \log \left (1-\sqrt {2-b} x+x^2\right )}{4 \sqrt {2-b}}-\frac {\left (1-\sqrt {2}\right ) \log \left (1+\sqrt {2-b} x+x^2\right )}{4 \sqrt {2-b}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 136, normalized size = 0.85 \begin {gather*} \frac {\frac {\left (\sqrt {b^2-4}-b+2 \sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {b-\sqrt {b^2-4}}}\right )}{\sqrt {b-\sqrt {b^2-4}}}+\frac {\left (\sqrt {b^2-4}+b-2 \sqrt {2}\right ) \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {\sqrt {b^2-4}+b}}\right )}{\sqrt {\sqrt {b^2-4}+b}}}{\sqrt {2} \sqrt {b^2-4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[2] + x^2)/(1 + b*x^2 + x^4),x]

[Out]

(((2*Sqrt[2] - b + Sqrt[-4 + b^2])*ArcTan[(Sqrt[2]*x)/Sqrt[b - Sqrt[-4 + b^2]]])/Sqrt[b - Sqrt[-4 + b^2]] + ((
-2*Sqrt[2] + b + Sqrt[-4 + b^2])*ArcTan[(Sqrt[2]*x)/Sqrt[b + Sqrt[-4 + b^2]]])/Sqrt[b + Sqrt[-4 + b^2]])/(Sqrt
[2]*Sqrt[-4 + b^2])

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2}+x^2}{1+b x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(Sqrt[2] + x^2)/(1 + b*x^2 + x^4),x]

[Out]

IntegrateAlgebraic[(Sqrt[2] + x^2)/(1 + b*x^2 + x^4), x]

________________________________________________________________________________________

fricas [B]  time = 1.10, size = 455, normalized size = 2.84 \begin {gather*} \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {3 \, b - 4 \, \sqrt {2} + \sqrt {b^{2} - 4}}{b^{2} - 4}} \log \left (\frac {1}{2} \, {\left (2 \, b - 3 \, \sqrt {2}\right )} x + \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - \frac {b^{3} - \sqrt {2} b^{2} - 4 \, b + 4 \, \sqrt {2}}{\sqrt {b^{2} - 4}} - 4\right )} \sqrt {-\frac {3 \, b - 4 \, \sqrt {2} + \sqrt {b^{2} - 4}}{b^{2} - 4}}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {3 \, b - 4 \, \sqrt {2} + \sqrt {b^{2} - 4}}{b^{2} - 4}} \log \left (\frac {1}{2} \, {\left (2 \, b - 3 \, \sqrt {2}\right )} x - \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} - \frac {b^{3} - \sqrt {2} b^{2} - 4 \, b + 4 \, \sqrt {2}}{\sqrt {b^{2} - 4}} - 4\right )} \sqrt {-\frac {3 \, b - 4 \, \sqrt {2} + \sqrt {b^{2} - 4}}{b^{2} - 4}}\right ) + \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {3 \, b - 4 \, \sqrt {2} - \sqrt {b^{2} - 4}}{b^{2} - 4}} \log \left (\frac {1}{2} \, {\left (2 \, b - 3 \, \sqrt {2}\right )} x + \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} + \frac {b^{3} - \sqrt {2} b^{2} - 4 \, b + 4 \, \sqrt {2}}{\sqrt {b^{2} - 4}} - 4\right )} \sqrt {-\frac {3 \, b - 4 \, \sqrt {2} - \sqrt {b^{2} - 4}}{b^{2} - 4}}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {3 \, b - 4 \, \sqrt {2} - \sqrt {b^{2} - 4}}{b^{2} - 4}} \log \left (\frac {1}{2} \, {\left (2 \, b - 3 \, \sqrt {2}\right )} x - \frac {1}{2} \, \sqrt {\frac {1}{2}} {\left (b^{2} + \frac {b^{3} - \sqrt {2} b^{2} - 4 \, b + 4 \, \sqrt {2}}{\sqrt {b^{2} - 4}} - 4\right )} \sqrt {-\frac {3 \, b - 4 \, \sqrt {2} - \sqrt {b^{2} - 4}}{b^{2} - 4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(1/2)*sqrt(-(3*b - 4*sqrt(2) + sqrt(b^2 - 4))/(b^2 - 4))*log(1/2*(2*b - 3*sqrt(2))*x + 1/2*sqrt(1/2)*(
b^2 - (b^3 - sqrt(2)*b^2 - 4*b + 4*sqrt(2))/sqrt(b^2 - 4) - 4)*sqrt(-(3*b - 4*sqrt(2) + sqrt(b^2 - 4))/(b^2 -
4))) - 1/2*sqrt(1/2)*sqrt(-(3*b - 4*sqrt(2) + sqrt(b^2 - 4))/(b^2 - 4))*log(1/2*(2*b - 3*sqrt(2))*x - 1/2*sqrt
(1/2)*(b^2 - (b^3 - sqrt(2)*b^2 - 4*b + 4*sqrt(2))/sqrt(b^2 - 4) - 4)*sqrt(-(3*b - 4*sqrt(2) + sqrt(b^2 - 4))/
(b^2 - 4))) + 1/2*sqrt(1/2)*sqrt(-(3*b - 4*sqrt(2) - sqrt(b^2 - 4))/(b^2 - 4))*log(1/2*(2*b - 3*sqrt(2))*x + 1
/2*sqrt(1/2)*(b^2 + (b^3 - sqrt(2)*b^2 - 4*b + 4*sqrt(2))/sqrt(b^2 - 4) - 4)*sqrt(-(3*b - 4*sqrt(2) - sqrt(b^2
- 4))/(b^2 - 4))) - 1/2*sqrt(1/2)*sqrt(-(3*b - 4*sqrt(2) - sqrt(b^2 - 4))/(b^2 - 4))*log(1/2*(2*b - 3*sqrt(2)
)*x - 1/2*sqrt(1/2)*(b^2 + (b^3 - sqrt(2)*b^2 - 4*b + 4*sqrt(2))/sqrt(b^2 - 4) - 4)*sqrt(-(3*b - 4*sqrt(2) - s
qrt(b^2 - 4))/(b^2 - 4)))

________________________________________________________________________________________

giac [B]  time = 0.35, size = 1501, normalized size = 9.38

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*sqrt(b + 2)*b^4 + sqrt(2)*sqrt(b - 2)*b^4 - sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2)*b^3 - sqrt(2)*sqrt(
b^2 - 4)*sqrt(b - 2)*b^3 - sqrt(2)*sqrt(b + 2)*sqrt(b - 2)*b^3 - 3*sqrt(2)*b^4 + 3*sqrt(2)*sqrt(b^2 - 4)*sqrt(
b + 2)*sqrt(b - 2)*b^2 + sqrt(2)*sqrt(b^2 - 4)*b^3 - sqrt(2)*sqrt(b + 2)*b^3 - sqrt(2)*sqrt(b - 2)*b^3 + sqrt(
2)*sqrt(b^2 - 4)*sqrt(b + 2)*b^2 + sqrt(2)*sqrt(b^2 - 4)*sqrt(b - 2)*b^2 + sqrt(2)*sqrt(b + 2)*sqrt(b - 2)*b^2
+ 3*sqrt(2)*b^3 - 3*sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2)*sqrt(b - 2)*b - sqrt(2)*sqrt(b^2 - 4)*b^2 - 10*sqrt(2)*
sqrt(b + 2)*b^2 + 2*sqrt(b^2 - 4)*sqrt(b + 2)*b^2 - 6*sqrt(2)*sqrt(b - 2)*b^2 + 2*sqrt(b^2 - 4)*sqrt(b - 2)*b^
2 + 2*sqrt(b + 2)*sqrt(b - 2)*b^2 + 4*sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2)*b + 4*sqrt(2)*sqrt(b^2 - 4)*sqrt(b - 2
)*b + 4*sqrt(2)*sqrt(b + 2)*sqrt(b - 2)*b + 24*sqrt(2)*b^2 - 2*sqrt(b^2 - 4)*b^2 - 12*sqrt(2)*sqrt(b^2 - 4)*sq
rt(b + 2)*sqrt(b - 2) - 4*sqrt(2)*sqrt(b^2 - 4)*b + 6*sqrt(2)*sqrt(b + 2)*b - 4*sqrt(b^2 - 4)*sqrt(b + 2)*b +
2*sqrt(2)*sqrt(b - 2)*b - 4*sqrt(b^2 - 4)*sqrt(b - 2)*b - 4*sqrt(b + 2)*sqrt(b - 2)*b - 6*b^2 + 4*sqrt(2)*sqrt
(b^2 - 4)*sqrt(b + 2) + 4*sqrt(2)*sqrt(b^2 - 4)*sqrt(b - 2) + 4*sqrt(2)*sqrt(b + 2)*sqrt(b - 2) + 6*sqrt(b^2 -
4)*sqrt(b + 2)*sqrt(b - 2) - 12*sqrt(2)*b + 4*sqrt(b^2 - 4)*b + 2*sqrt(b + 2)*b + 2*sqrt(b - 2)*b - 4*sqrt(2)
*sqrt(b^2 - 4) + 20*sqrt(2)*sqrt(b + 2) - 8*sqrt(b^2 - 4)*sqrt(b + 2) + 4*sqrt(2)*sqrt(b - 2) - 8*sqrt(b^2 - 4
)*sqrt(b - 2) - 8*sqrt(b + 2)*sqrt(b - 2) - 48*sqrt(2) + 8*sqrt(b^2 - 4) - 4*sqrt(b + 2) + 4*sqrt(b - 2) + 24)
*arctan(x/sqrt(1/2*b + 1/2*sqrt(b^2 - 4)))/(b^4 - 2*b^3 - 7*b^2 + 8*b + 12) + 1/4*(sqrt(2)*sqrt(b + 2)*b^4 - s
qrt(2)*sqrt(b - 2)*b^4 + sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2)*b^3 - sqrt(2)*sqrt(b^2 - 4)*sqrt(b - 2)*b^3 - sqrt(
2)*sqrt(b + 2)*sqrt(b - 2)*b^3 + 3*sqrt(2)*b^4 - 3*sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2)*sqrt(b - 2)*b^2 + sqrt(2)
*sqrt(b^2 - 4)*b^3 - sqrt(2)*sqrt(b + 2)*b^3 + sqrt(2)*sqrt(b - 2)*b^3 - sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2)*b^2
+ sqrt(2)*sqrt(b^2 - 4)*sqrt(b - 2)*b^2 + sqrt(2)*sqrt(b + 2)*sqrt(b - 2)*b^2 - 3*sqrt(2)*b^3 + 3*sqrt(2)*sqr
t(b^2 - 4)*sqrt(b + 2)*sqrt(b - 2)*b - sqrt(2)*sqrt(b^2 - 4)*b^2 - 10*sqrt(2)*sqrt(b + 2)*b^2 - 2*sqrt(b^2 - 4
)*sqrt(b + 2)*b^2 + 6*sqrt(2)*sqrt(b - 2)*b^2 + 2*sqrt(b^2 - 4)*sqrt(b - 2)*b^2 + 2*sqrt(b + 2)*sqrt(b - 2)*b^
2 - 4*sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2)*b + 4*sqrt(2)*sqrt(b^2 - 4)*sqrt(b - 2)*b + 4*sqrt(2)*sqrt(b + 2)*sqrt
(b - 2)*b - 24*sqrt(2)*b^2 - 2*sqrt(b^2 - 4)*b^2 + 12*sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2)*sqrt(b - 2) - 4*sqrt(2
)*sqrt(b^2 - 4)*b + 6*sqrt(2)*sqrt(b + 2)*b + 4*sqrt(b^2 - 4)*sqrt(b + 2)*b - 2*sqrt(2)*sqrt(b - 2)*b - 4*sqrt
(b^2 - 4)*sqrt(b - 2)*b - 4*sqrt(b + 2)*sqrt(b - 2)*b + 6*b^2 - 4*sqrt(2)*sqrt(b^2 - 4)*sqrt(b + 2) + 4*sqrt(2
)*sqrt(b^2 - 4)*sqrt(b - 2) + 4*sqrt(2)*sqrt(b + 2)*sqrt(b - 2) - 6*sqrt(b^2 - 4)*sqrt(b + 2)*sqrt(b - 2) + 12
*sqrt(2)*b + 4*sqrt(b^2 - 4)*b + 2*sqrt(b + 2)*b - 2*sqrt(b - 2)*b - 4*sqrt(2)*sqrt(b^2 - 4) + 20*sqrt(2)*sqrt
(b + 2) + 8*sqrt(b^2 - 4)*sqrt(b + 2) - 4*sqrt(2)*sqrt(b - 2) - 8*sqrt(b^2 - 4)*sqrt(b - 2) - 8*sqrt(b + 2)*sq
rt(b - 2) + 48*sqrt(2) + 8*sqrt(b^2 - 4) - 4*sqrt(b + 2) - 4*sqrt(b - 2) - 24)*arctan(x/sqrt(1/2*b - 1/2*sqrt(
b^2 - 4)))/(b^4 - 2*b^3 - 7*b^2 + 8*b + 12)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 283, normalized size = 1.77 \begin {gather*} -\frac {b \arctan \left (\frac {2 x}{\sqrt {2 b -2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}\right )}{\sqrt {\left (b -2\right ) \left (b +2\right )}\, \sqrt {2 b -2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}+\frac {b \arctan \left (\frac {2 x}{\sqrt {2 b +2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}\right )}{\sqrt {\left (b -2\right ) \left (b +2\right )}\, \sqrt {2 b +2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}+\frac {\arctan \left (\frac {2 x}{\sqrt {2 b -2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}\right )}{\sqrt {2 b -2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}+\frac {2 \sqrt {2}\, \arctan \left (\frac {2 x}{\sqrt {2 b -2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}\right )}{\sqrt {\left (b -2\right ) \left (b +2\right )}\, \sqrt {2 b -2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}+\frac {\arctan \left (\frac {2 x}{\sqrt {2 b +2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}\right )}{\sqrt {2 b +2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}-\frac {2 \sqrt {2}\, \arctan \left (\frac {2 x}{\sqrt {2 b +2 \sqrt {\left (b -2\right ) \left (b +2\right )}}}\right )}{\sqrt {\left (b -2\right ) \left (b +2\right )}\, \sqrt {2 b +2 \sqrt {\left (b -2\right ) \left (b +2\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+2^(1/2))/(x^4+b*x^2+1),x)

[Out]

1/(2*b+2*((b-2)*(b+2))^(1/2))^(1/2)*arctan(2/(2*b+2*((b-2)*(b+2))^(1/2))^(1/2)*x)+1/((b-2)*(b+2))^(1/2)/(2*b+2
*((b-2)*(b+2))^(1/2))^(1/2)*b*arctan(2/(2*b+2*((b-2)*(b+2))^(1/2))^(1/2)*x)-2/((b-2)*(b+2))^(1/2)/(2*b+2*((b-2
)*(b+2))^(1/2))^(1/2)*2^(1/2)*arctan(2/(2*b+2*((b-2)*(b+2))^(1/2))^(1/2)*x)+1/(2*b-2*((b-2)*(b+2))^(1/2))^(1/2
)*arctan(2/(2*b-2*((b-2)*(b+2))^(1/2))^(1/2)*x)-1/((b-2)*(b+2))^(1/2)/(2*b-2*((b-2)*(b+2))^(1/2))^(1/2)*b*arct
an(2/(2*b-2*((b-2)*(b+2))^(1/2))^(1/2)*x)+2/((b-2)*(b+2))^(1/2)/(2*b-2*((b-2)*(b+2))^(1/2))^(1/2)*2^(1/2)*arct
an(2/(2*b-2*((b-2)*(b+2))^(1/2))^(1/2)*x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} + \sqrt {2}}{x^{4} + b x^{2} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^2 + sqrt(2))/(x^4 + b*x^2 + 1), x)

________________________________________________________________________________________

mupad [B]  time = 5.25, size = 1227, normalized size = 7.67 \begin {gather*} -\mathrm {atan}\left (\frac {x\,\sqrt {\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,32{}\mathrm {i}-b\,x\,{\left (\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}\right )}^{3/2}\,256{}\mathrm {i}+b^2\,x\,\sqrt {\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,8{}\mathrm {i}-b^4\,x\,\sqrt {\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,4{}\mathrm {i}+b^3\,x\,{\left (\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}\right )}^{3/2}\,128{}\mathrm {i}-b^5\,x\,{\left (\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}\right )}^{3/2}\,16{}\mathrm {i}-\sqrt {2}\,b\,x\,\sqrt {\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,32{}\mathrm {i}+\sqrt {2}\,b^3\,x\,\sqrt {\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,8{}\mathrm {i}}{4\,\sqrt {2}\,b-\sqrt {2}\,b^3+\sqrt {2}\,\sqrt {b^6-12\,b^4+48\,b^2-64}+2\,b^2-8}\right )\,\sqrt {\frac {12\,b-16\,\sqrt {2}+4\,\sqrt {2}\,b^2-3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {x\,\sqrt {-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,32{}\mathrm {i}-b\,x\,{\left (-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}\right )}^{3/2}\,256{}\mathrm {i}+b^2\,x\,\sqrt {-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,8{}\mathrm {i}-b^4\,x\,\sqrt {-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,4{}\mathrm {i}+b^3\,x\,{\left (-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}\right )}^{3/2}\,128{}\mathrm {i}-b^5\,x\,{\left (-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}\right )}^{3/2}\,16{}\mathrm {i}-\sqrt {2}\,b\,x\,\sqrt {-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,32{}\mathrm {i}+\sqrt {2}\,b^3\,x\,\sqrt {-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,8{}\mathrm {i}}{\sqrt {2}\,b^3-4\,\sqrt {2}\,b+\sqrt {2}\,\sqrt {b^6-12\,b^4+48\,b^2-64}-2\,b^2+8}\right )\,\sqrt {-\frac {16\,\sqrt {2}-12\,b-4\,\sqrt {2}\,b^2+3\,b^3+\sqrt {b^6-12\,b^4+48\,b^2-64}}{8\,b^4-64\,b^2+128}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2^(1/2) + x^2)/(b*x^2 + x^4 + 1),x)

[Out]

atan((x*(-(16*2^(1/2) - 12*b - 4*2^(1/2)*b^2 + 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 1
28))^(1/2)*32i - b*x*(-(16*2^(1/2) - 12*b - 4*2^(1/2)*b^2 + 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4
- 64*b^2 + 128))^(3/2)*256i + b^2*x*(-(16*2^(1/2) - 12*b - 4*2^(1/2)*b^2 + 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 6
4)^(1/2))/(8*b^4 - 64*b^2 + 128))^(1/2)*8i - b^4*x*(-(16*2^(1/2) - 12*b - 4*2^(1/2)*b^2 + 3*b^3 + (48*b^2 - 12
*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(1/2)*4i + b^3*x*(-(16*2^(1/2) - 12*b - 4*2^(1/2)*b^2 + 3*b^3
+ (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(3/2)*128i - b^5*x*(-(16*2^(1/2) - 12*b - 4*2^(1
/2)*b^2 + 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(3/2)*16i - 2^(1/2)*b*x*(-(16*2^
(1/2) - 12*b - 4*2^(1/2)*b^2 + 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(1/2)*32i +
2^(1/2)*b^3*x*(-(16*2^(1/2) - 12*b - 4*2^(1/2)*b^2 + 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*
b^2 + 128))^(1/2)*8i)/(2^(1/2)*b^3 - 4*2^(1/2)*b + 2^(1/2)*(48*b^2 - 12*b^4 + b^6 - 64)^(1/2) - 2*b^2 + 8))*(-
(16*2^(1/2) - 12*b - 4*2^(1/2)*b^2 + 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(1/2)
*2i - atan((x*((12*b - 16*2^(1/2) + 4*2^(1/2)*b^2 - 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^
2 + 128))^(1/2)*32i - b*x*((12*b - 16*2^(1/2) + 4*2^(1/2)*b^2 - 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8
*b^4 - 64*b^2 + 128))^(3/2)*256i + b^2*x*((12*b - 16*2^(1/2) + 4*2^(1/2)*b^2 - 3*b^3 + (48*b^2 - 12*b^4 + b^6
- 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(1/2)*8i - b^4*x*((12*b - 16*2^(1/2) + 4*2^(1/2)*b^2 - 3*b^3 + (48*b^2 -
12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(1/2)*4i + b^3*x*((12*b - 16*2^(1/2) + 4*2^(1/2)*b^2 - 3*b^3
+ (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(3/2)*128i - b^5*x*((12*b - 16*2^(1/2) + 4*2^(1
/2)*b^2 - 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(3/2)*16i - 2^(1/2)*b*x*((12*b -
16*2^(1/2) + 4*2^(1/2)*b^2 - 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(1/2)*32i +
2^(1/2)*b^3*x*((12*b - 16*2^(1/2) + 4*2^(1/2)*b^2 - 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^
2 + 128))^(1/2)*8i)/(4*2^(1/2)*b - 2^(1/2)*b^3 + 2^(1/2)*(48*b^2 - 12*b^4 + b^6 - 64)^(1/2) + 2*b^2 - 8))*((12
*b - 16*2^(1/2) + 4*2^(1/2)*b^2 - 3*b^3 + (48*b^2 - 12*b^4 + b^6 - 64)^(1/2))/(8*b^4 - 64*b^2 + 128))^(1/2)*2i

________________________________________________________________________________________

sympy [B]  time = 2.73, size = 1467, normalized size = 9.17

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+2**(1/2))/(x**4+b*x**2+1),x)

[Out]

RootSum(_t**4*(16*b**4 - 128*b**2 + 256) + _t**2*(12*b**3 - 16*sqrt(2)*b**2 - 48*b + 64*sqrt(2)) + 2*b**2 - 6*
sqrt(2)*b + 9, Lambda(_t, _t*log(_t**3*(64*b**12/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6
470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729) - 672*sqrt(2)*b**11/(8*b**10 -
88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**
3 - 3402*b**2 + 729) + 5760*b**10/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310
*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729) - 12064*sqrt(2)*b**9/(8*b**10 - 88*sqrt(2)*b*
*9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2
+ 729) + 17744*b**8/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5
+ 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729) + 27480*sqrt(2)*b**7/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8
- 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729) - 15460
8*b**6/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 +
2322*sqrt(2)*b**3 - 3402*b**2 + 729) + 141376*sqrt(2)*b**5/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(
2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729) - 69072*b**4/(8*b**
10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)
*b**3 - 3402*b**2 + 729) - 61704*sqrt(2)*b**3/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470
*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729) + 78192*b**2/(8*b**10 - 88*sqrt(2
)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b
**2 + 729) + 2592*sqrt(2)*b/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 2144*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(
2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729) - 15552/(8*b**10 - 88*sqrt(2)*b**9 + 828*b**8 - 214
4*sqrt(2)*b**7 + 6470*b**6 - 5310*sqrt(2)*b**5 + 2781*b**4 + 2322*sqrt(2)*b**3 - 3402*b**2 + 729)) + _t*(16*b*
*7/(4*b**6 - 28*sqrt(2)*b**5 + 152*b**4 - 192*sqrt(2)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81) - 116*sqrt(2)*b**6/
(4*b**6 - 28*sqrt(2)*b**5 + 152*b**4 - 192*sqrt(2)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81) + 668*b**5/(4*b**6 - 2
8*sqrt(2)*b**5 + 152*b**4 - 192*sqrt(2)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81) - 942*sqrt(2)*b**4/(4*b**6 - 28*s
qrt(2)*b**5 + 152*b**4 - 192*sqrt(2)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81) + 1226*b**3/(4*b**6 - 28*sqrt(2)*b**
5 + 152*b**4 - 192*sqrt(2)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81) - 144*sqrt(2)*b**2/(4*b**6 - 28*sqrt(2)*b**5 +
152*b**4 - 192*sqrt(2)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81) - 378*b/(4*b**6 - 28*sqrt(2)*b**5 + 152*b**4 - 19
2*sqrt(2)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81) + 108*sqrt(2)/(4*b**6 - 28*sqrt(2)*b**5 + 152*b**4 - 192*sqrt(2
)*b**3 + 189*b**2 + 27*sqrt(2)*b - 81)) + x)))

________________________________________________________________________________________