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

Optimal. Leaf size=172 \[ -\frac {1}{4} \sqrt {1-\frac {1}{\sqrt {2}}} \log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )+\frac {1}{4} \sqrt {1-\frac {1}{\sqrt {2}}} \log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )-\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{2 \sqrt {2-\sqrt {2}}}+\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{2 \sqrt {2-\sqrt {2}}} \]

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Rubi [A]  time = 0.14, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1169, 634, 618, 204, 628} \begin {gather*} -\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 )-\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{2 \sqrt {2-\sqrt {2}}}+\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{2 \sqrt {2-\sqrt {2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[(Sqrt[2 - Sqrt[2]] - 2*x)/Sqrt[2 + Sqrt[2]]]/(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 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+\sqrt {2} x^2+x^4} \, dx &=\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 {\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 {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 {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}} \operatorname {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}} \operatorname {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*}

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Mathematica [C]  time = 0.04, size = 53, normalized size = 0.31 \begin {gather*} \frac {\sqrt {1-i} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1-i}}\right )+\sqrt {1+i} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+i}}\right )}{2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[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)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [C]  time = 1.13, size = 97, normalized size = 0.56 \begin {gather*} \frac {1}{4} \, \sqrt {\left (i - 1\right ) \, \sqrt {2}} \log \left (x + \frac {1}{2} \, \sqrt {2} \sqrt {\left (i - 1\right ) \, \sqrt {2}}\right ) - \frac {1}{4} \, \sqrt {\left (i - 1\right ) \, \sqrt {2}} \log \left (x - \frac {1}{2} \, \sqrt {2} \sqrt {\left (i - 1\right ) \, \sqrt {2}}\right ) + \frac {1}{4} \, \sqrt {-\left (i + 1\right ) \, \sqrt {2}} \log \left (x + \frac {1}{2} \, \sqrt {2} \sqrt {-\left (i + 1\right ) \, \sqrt {2}}\right ) - \frac {1}{4} \, \sqrt {-\left (i + 1\right ) \, \sqrt {2}} \log \left (x - \frac {1}{2} \, \sqrt {2} \sqrt {-\left (i + 1\right ) \, \sqrt {2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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giac [A]  time = 0.33, size = 126, normalized size = 0.73 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)*arctan(
(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)

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maple [A]  time = 0.09, size = 199, normalized size = 1.16 \begin {gather*} \frac {\arctan \left (\frac {2 x -\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{2 \sqrt {2+\sqrt {2}}}+\frac {\sqrt {2}\, \arctan \left (\frac {2 x -\sqrt {2-\sqrt {2}}}{\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 {\sqrt {2}\, \arctan \left (\frac {2 x +\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{2 \sqrt {2+\sqrt {2}}}-\frac {\sqrt {2}\, \sqrt {2-\sqrt {2}}\, \ln \left (x^{2}-\sqrt {2-\sqrt {2}}\, x +1\right )}{8}+\frac {\sqrt {2}\, \sqrt {2-\sqrt {2}}\, \ln \left (x^{2}+\sqrt {2-\sqrt {2}}\, x +1\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.95, size = 121, normalized size = 0.70 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2^(1/2) + x^2)/(2^(1/2)*x^2 + x^4 + 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*((8^(1/2)*1i)/32 - 2^(1/2)/16)^(1/2)*2i - (2^(1/2)*8^(1/
2)*x*((8^(1/2)*1i)/32 - 2^(1/2)/16)^(1/2))/2)*((8^(1/2)*1i)/32 - 2^(1/2)/16)^(1/2)*2i

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: PolynomialError

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