\(\int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx\) [102]

Optimal result

Integrand size = 29, antiderivative size = 160 \[ \int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx=-\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{2 \sqrt {2+\sqrt {2}}}+\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2}}+2 x}{\sqrt {2-\sqrt {2}}}\right )}{2 \sqrt {2+\sqrt {2}}}-\frac {1}{4} \sqrt {1+\frac {1}{\sqrt {2}}} \log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )+\frac {1}{4} \sqrt {1+\frac {1}{\sqrt {2}}} \log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right ) \]

[Out]

-1/8*ln(1+x^2-x*(2+2^(1/2))^(1/2))*(4+2*2^(1/2))^(1/2)+1/8*ln(1+x^2+x*(2+2^(1/2))^(1/2))*(4+2*2^(1/2))^(1/2)-1
/2*arctan((-2*x+(2+2^(1/2))^(1/2))/(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)+1/2*arctan((2*x+(2+2^(1/2))^(1/2))/(2-
2^(1/2))^(1/2))/(2+2^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1183, 648, 632, 210, 642} \[ \int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx=-\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{2 \sqrt {2+\sqrt {2}}}+\frac {\arctan \left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{2 \sqrt {2+\sqrt {2}}}-\frac {1}{4} \sqrt {1+\frac {1}{\sqrt {2}}} \log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )+\frac {1}{4} \sqrt {1+\frac {1}{\sqrt {2}}} \log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right ) \]

[In]

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

[Out]

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

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 1183

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx=\frac {\sqrt {-1-i} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {-1-i}}\right )+\sqrt {-1+i} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {-1+i}}\right )}{2^{3/4}} \]

[In]

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

[Out]

(Sqrt[-1 - I]*ArcTan[(2^(1/4)*x)/Sqrt[-1 - I]] + Sqrt[-1 + I]*ArcTan[(2^(1/4)*x)/Sqrt[-1 + I]])/2^(3/4)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.34

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}-2, \operatorname {index} =1\right )+1\right )}{\sum }\frac {\left (\textit {\_R}^{2}-\sqrt {2}\right ) \ln \left (x -\textit {\_R} \right )}{-2 \textit {\_R}^{3}+\textit {\_R} \sqrt {2}}\right )}{2}\)	\(55\)
default	\(\frac {\sqrt {2}\, \left (\frac {\sqrt {2+\sqrt {2}}\, \ln \left (1+x^{2}+x \sqrt {2+\sqrt {2}}\right )}{2}+\frac {2 \left (1-\frac {\sqrt {2}}{2}\right ) \arctan \left (\frac {2 x +\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{\sqrt {2-\sqrt {2}}}\right )}{4}+\frac {\sqrt {2}\, \left (-\frac {\sqrt {2+\sqrt {2}}\, \ln \left (1+x^{2}-x \sqrt {2+\sqrt {2}}\right )}{2}+\frac {2 \left (1-\frac {\sqrt {2}}{2}\right ) \arctan \left (\frac {2 x -\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{\sqrt {2-\sqrt {2}}}\right )}{4}\)	\(145\)

[In]

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

[Out]

1/2*sum((_R^2-2^(1/2))/(-2*_R^3+_R*2^(1/2))*ln(x-_R),_R=RootOf(_Z^4-_Z^2*RootOf(_Z^2-2,index=1)+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx=\frac {1}{4} \, \sqrt {\left (i + 1\right ) \, \sqrt {2}} \log \left (2 \, x + \sqrt {2} \sqrt {\left (i + 1\right ) \, \sqrt {2}}\right ) - \frac {1}{4} \, \sqrt {\left (i + 1\right ) \, \sqrt {2}} \log \left (2 \, x - \sqrt {2} \sqrt {\left (i + 1\right ) \, \sqrt {2}}\right ) + \frac {1}{4} \, \sqrt {-\left (i - 1\right ) \, \sqrt {2}} \log \left (2 \, x + \sqrt {2} \sqrt {-\left (i - 1\right ) \, \sqrt {2}}\right ) - \frac {1}{4} \, \sqrt {-\left (i - 1\right ) \, \sqrt {2}} \log \left (2 \, x - \sqrt {2} \sqrt {-\left (i - 1\right ) \, \sqrt {2}}\right ) \]

[In]

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

[Out]

1/4*sqrt((I + 1)*sqrt(2))*log(2*x + sqrt(2)*sqrt((I + 1)*sqrt(2))) - 1/4*sqrt((I + 1)*sqrt(2))*log(2*x - sqrt(
2)*sqrt((I + 1)*sqrt(2))) + 1/4*sqrt(-(I - 1)*sqrt(2))*log(2*x + sqrt(2)*sqrt(-(I - 1)*sqrt(2))) - 1/4*sqrt(-(
I - 1)*sqrt(2))*log(2*x - sqrt(2)*sqrt(-(I - 1)*sqrt(2)))

Sympy [F(-2)]

Exception generated. \[ \int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx=\text {Exception raised: PolynomialError} \]

[In]

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

[Out]

Exception raised: PolynomialError >> 1/(128*_t**4 - 16*sqrt(2)*_t**2 + 1) contains an element of the set of ge
nerators.

Maxima [F]

\[ \int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx=\int { -\frac {x^{2} - \sqrt {2}}{x^{4} - \sqrt {2} x^{2} + 1} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx=\frac {1}{4} \, \sqrt {-2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{4} \, \sqrt {-2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {2} + 4} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{8} \, \sqrt {2 \, \sqrt {2} + 4} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} + 1\right ) \]

[In]

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

[Out]

1/4*sqrt(-2*sqrt(2) + 4)*arctan((2*x + sqrt(sqrt(2) + 2))/sqrt(-sqrt(2) + 2)) + 1/4*sqrt(-2*sqrt(2) + 4)*arcta
n((2*x - sqrt(sqrt(2) + 2))/sqrt(-sqrt(2) + 2)) + 1/8*sqrt(2*sqrt(2) + 4)*log(x^2 + x*sqrt(sqrt(2) + 2) + 1) -
1/8*sqrt(2*sqrt(2) + 4)*log(x^2 - x*sqrt(sqrt(2) + 2) + 1)

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx=-\mathrm {atan}\left (x\,\sqrt {\frac {\sqrt {2}}{16}-\frac {\sqrt {8}\,1{}\mathrm {i}}{32}}\,2{}\mathrm {i}-\frac {\sqrt {2}\,\sqrt {8}\,x\,\sqrt {\frac {\sqrt {2}}{16}-\frac {\sqrt {8}\,1{}\mathrm {i}}{32}}}{2}\right )\,\sqrt {\frac {\sqrt {2}}{16}-\frac {\sqrt {8}\,1{}\mathrm {i}}{32}}\,2{}\mathrm {i}-\mathrm {atan}\left (x\,\sqrt {\frac {\sqrt {2}}{16}+\frac {\sqrt {8}\,1{}\mathrm {i}}{32}}\,2{}\mathrm {i}+\frac {\sqrt {2}\,\sqrt {8}\,x\,\sqrt {\frac {\sqrt {2}}{16}+\frac {\sqrt {8}\,1{}\mathrm {i}}{32}}}{2}\right )\,\sqrt {\frac {\sqrt {2}}{16}+\frac {\sqrt {8}\,1{}\mathrm {i}}{32}}\,2{}\mathrm {i} \]

[In]

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

[Out]

- atan(x*(2^(1/2)/16 - (8^(1/2)*1i)/32)^(1/2)*2i - (2^(1/2)*8^(1/2)*x*(2^(1/2)/16 - (8^(1/2)*1i)/32)^(1/2))/2)
*(2^(1/2)/16 - (8^(1/2)*1i)/32)^(1/2)*2i - atan(x*(2^(1/2)/16 + (8^(1/2)*1i)/32)^(1/2)*2i + (2^(1/2)*8^(1/2)*x
*(2^(1/2)/16 + (8^(1/2)*1i)/32)^(1/2))/2)*(2^(1/2)/16 + (8^(1/2)*1i)/32)^(1/2)*2i