3.31 \(\int \frac {-1+\sqrt {3}+2 x^4}{1-x^4+x^8} \, dx\)

Optimal. Leaf size=135 \[ -\frac {\log \left (x^2-\sqrt {2-\sqrt {3}} x+1\right )}{2 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2-\sqrt {3}} x+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {3}}-2 x}{\sqrt {2-\sqrt {3}}}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {3}}}{\sqrt {2-\sqrt {3}}}\right )}{\sqrt {2}} \]

[Out]

-1/2*arctan((-2*x+1/2*6^(1/2)+1/2*2^(1/2))/(1/2*6^(1/2)-1/2*2^(1/2)))*2^(1/2)+1/2*arctan((2*x+1/2*6^(1/2)+1/2*
2^(1/2))/(1/2*6^(1/2)-1/2*2^(1/2)))*2^(1/2)-1/4*ln(1+x^2-x*(1/2*6^(1/2)-1/2*2^(1/2)))*2^(1/2)+1/4*ln(1+x^2+x*(
1/2*6^(1/2)-1/2*2^(1/2)))*2^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1423, 1161, 618, 204, 1164, 628} \[ -\frac {\log \left (x^2-\sqrt {2-\sqrt {3}} x+1\right )}{2 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2-\sqrt {3}} x+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {3}}-2 x}{\sqrt {2-\sqrt {3}}}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {3}}}{\sqrt {2-\sqrt {3}}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + Sqrt[3] + 2*x^4)/(1 - x^4 + x^8),x]

[Out]

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

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 1161

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

Rule 1164

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

Rule 1423

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), 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^(n/2))/(q - r*x^(n/2) + x^n), x], x] + D
ist[1/(2*c*q*r), Int[(d*r + (d - e*q)*x^(n/2))/(q + r*x^(n/2) + x^n), x], x]]] /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[n/2, 0] && NegQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {-1+\sqrt {3}+2 x^4}{1-x^4+x^8} \, dx &=\frac {\int \frac {\sqrt {3} \left (-1+\sqrt {3}\right )+\left (3-\sqrt {3}\right ) x^2}{1-\sqrt {3} x^2+x^4} \, dx}{2 \sqrt {3}}+\frac {\int \frac {\sqrt {3} \left (-1+\sqrt {3}\right )+\left (-3+\sqrt {3}\right ) x^2}{1+\sqrt {3} x^2+x^4} \, dx}{2 \sqrt {3}}\\ &=-\frac {\int \frac {\sqrt {2-\sqrt {3}}+2 x}{-1-\sqrt {2-\sqrt {3}} x-x^2} \, dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2-\sqrt {3}}-2 x}{-1+\sqrt {2-\sqrt {3}} x-x^2} \, dx}{2 \sqrt {2}}+\frac {1}{4} \left (-1+\sqrt {3}\right ) \int \frac {1}{1-\sqrt {2+\sqrt {3}} x+x^2} \, dx+\frac {1}{4} \left (-1+\sqrt {3}\right ) \int \frac {1}{1+\sqrt {2+\sqrt {3}} x+x^2} \, dx\\ &=-\frac {\log \left (1-\sqrt {2-\sqrt {3}} x+x^2\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2-\sqrt {3}} x+x^2\right )}{2 \sqrt {2}}+\frac {1}{2} \left (1-\sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{-2+\sqrt {3}-x^2} \, dx,x,-\sqrt {2+\sqrt {3}}+2 x\right )+\frac {1}{2} \left (1-\sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{-2+\sqrt {3}-x^2} \, dx,x,\sqrt {2+\sqrt {3}}+2 x\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {3}}-2 x}{\sqrt {2-\sqrt {3}}}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {3}}+2 x}{\sqrt {2-\sqrt {3}}}\right )}{\sqrt {2}}-\frac {\log \left (1-\sqrt {2-\sqrt {3}} x+x^2\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2-\sqrt {3}} x+x^2\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 71, normalized size = 0.53 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8-\text {$\#$1}^4+1\& ,\frac {2 \text {$\#$1}^4 \log (x-\text {$\#$1})+\sqrt {3} \log (x-\text {$\#$1})-\log (x-\text {$\#$1})}{2 \text {$\#$1}^7-\text {$\#$1}^3}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Sqrt[3] + 2*x^4)/(1 - x^4 + x^8),x]

[Out]

RootSum[1 - #1^4 + #1^8 & , (-Log[x - #1] + Sqrt[3]*Log[x - #1] + 2*Log[x - #1]*#1^4)/(-#1^3 + 2*#1^7) & ]/4

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fricas [A]  time = 0.89, size = 104, normalized size = 0.77 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, {\left (\sqrt {3} \sqrt {2} + \sqrt {2}\right )} x^{3} - \sqrt {2} x\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, {\left (\sqrt {3} \sqrt {2} + \sqrt {2}\right )} x\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\frac {{\left (\sqrt {3} \sqrt {2} - \sqrt {2}\right )} x + 2 \, x^{2} + 2}{{\left (\sqrt {3} \sqrt {2} - \sqrt {2}\right )} x - 2 \, x^{2} - 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+2*x^4+3^(1/2))/(x^8-x^4+1),x, algorithm="fricas")

[Out]

1/2*sqrt(2)*arctan(1/2*(sqrt(3)*sqrt(2) + sqrt(2))*x^3 - sqrt(2)*x) + 1/2*sqrt(2)*arctan(1/2*(sqrt(3)*sqrt(2)
+ sqrt(2))*x) + 1/4*sqrt(2)*log(-((sqrt(3)*sqrt(2) - sqrt(2))*x + 2*x^2 + 2)/((sqrt(3)*sqrt(2) - sqrt(2))*x -
2*x^2 - 2))

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giac [A]  time = 0.49, size = 107, normalized size = 0.79 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {4 \, x + \sqrt {6} + \sqrt {2}}{\sqrt {6} - \sqrt {2}}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {4 \, x - \sqrt {6} - \sqrt {2}}{\sqrt {6} - \sqrt {2}}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {6} - \sqrt {2}\right )} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {6} - \sqrt {2}\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+2*x^4+3^(1/2))/(x^8-x^4+1),x, algorithm="giac")

[Out]

1/2*sqrt(2)*arctan((4*x + sqrt(6) + sqrt(2))/(sqrt(6) - sqrt(2))) + 1/2*sqrt(2)*arctan((4*x - sqrt(6) - sqrt(2
))/(sqrt(6) - sqrt(2))) + 1/4*sqrt(2)*log(x^2 + 1/2*x*(sqrt(6) - sqrt(2)) + 1) - 1/4*sqrt(2)*log(x^2 - 1/2*x*(
sqrt(6) - sqrt(2)) + 1)

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maple [C]  time = 0.06, size = 47, normalized size = 0.35 \[ \frac {\left (2 \RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )^{4}-1+\sqrt {3}\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )+x \right )}{8 \RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )^{7}-4 \RootOf \left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+2*x^4+3^(1/2))/(x^8-x^4+1),x)

[Out]

1/4*sum(1/(2*_R^7-_R^3)*(-1+2*_R^4+3^(1/2))*ln(-_R+x),_R=RootOf(_Z^8-_Z^4+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {2 \, x^{4} + \sqrt {3} - 1}{x^{8} - x^{4} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+2*x^4+3^(1/2))/(x^8-x^4+1),x, algorithm="maxima")

[Out]

integrate((2*x^4 + sqrt(3) - 1)/(x^8 - x^4 + 1), x)

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mupad [B]  time = 2.24, size = 133, normalized size = 0.99 \[ \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {72\,\sqrt {2}\,x}{144\,\sqrt {3}-144\,\sqrt {3}\,x^2-288\,x^2+288}+\frac {72\,\sqrt {2}\,\sqrt {3}\,x}{144\,\sqrt {3}-144\,\sqrt {3}\,x^2-288\,x^2+288}\right )}{2}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {72\,\sqrt {2}\,x}{144\,\sqrt {3}+144\,\sqrt {3}\,x^2+288\,x^2+288}+\frac {72\,\sqrt {2}\,\sqrt {3}\,x}{144\,\sqrt {3}+144\,\sqrt {3}\,x^2+288\,x^2+288}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3^(1/2) + 2*x^4 - 1)/(x^8 - x^4 + 1),x)

[Out]

(2^(1/2)*atan((72*2^(1/2)*x)/(144*3^(1/2) - 144*3^(1/2)*x^2 - 288*x^2 + 288) + (72*2^(1/2)*3^(1/2)*x)/(144*3^(
1/2) - 144*3^(1/2)*x^2 - 288*x^2 + 288)))/2 + (2^(1/2)*atanh((72*2^(1/2)*x)/(144*3^(1/2) + 144*3^(1/2)*x^2 + 2
88*x^2 + 288) + (72*2^(1/2)*3^(1/2)*x)/(144*3^(1/2) + 144*3^(1/2)*x^2 + 288*x^2 + 288)))/2

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sympy [A]  time = 0.90, size = 163, normalized size = 1.21 \[ \frac {\sqrt {2} \left (2 \operatorname {atan}{\left (x \left (\frac {\sqrt {6}}{1 + \sqrt {3}} + \frac {2 \sqrt {2}}{1 + \sqrt {3}}\right ) \right )} + 2 \operatorname {atan}{\left (x^{3} \left (\frac {\sqrt {6}}{1 + \sqrt {3}} + \frac {2 \sqrt {2}}{1 + \sqrt {3}}\right ) - \sqrt {2} x \right )}\right )}{4} - \frac {\sqrt {2} \log {\left (x^{2} - \frac {\sqrt {2} x \left (\frac {2}{\sqrt {3} + 2} + \frac {2 \sqrt {3}}{\sqrt {3} + 2}\right )}{4} + 1 \right )}}{4} + \frac {\sqrt {2} \log {\left (x^{2} + \frac {\sqrt {2} x \left (\frac {2}{\sqrt {3} + 2} + \frac {2 \sqrt {3}}{\sqrt {3} + 2}\right )}{4} + 1 \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+2*x**4+3**(1/2))/(x**8-x**4+1),x)

[Out]

sqrt(2)*(2*atan(x*(sqrt(6)/(1 + sqrt(3)) + 2*sqrt(2)/(1 + sqrt(3)))) + 2*atan(x**3*(sqrt(6)/(1 + sqrt(3)) + 2*
sqrt(2)/(1 + sqrt(3))) - sqrt(2)*x))/4 - sqrt(2)*log(x**2 - sqrt(2)*x*(2/(sqrt(3) + 2) + 2*sqrt(3)/(sqrt(3) +
2))/4 + 1)/4 + sqrt(2)*log(x**2 + sqrt(2)*x*(2/(sqrt(3) + 2) + 2*sqrt(3)/(sqrt(3) + 2))/4 + 1)/4

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