3.9 \(\int \frac{1+x^4}{1+b x^4+x^8} \, dx\)

Optimal. Leaf size=411 \[ -\frac{\log \left (-\sqrt{2-\sqrt{2-b}} x+x^2+1\right )}{8 \sqrt{2-\sqrt{2-b}}}+\frac{\log \left (\sqrt{2-\sqrt{2-b}} x+x^2+1\right )}{8 \sqrt{2-\sqrt{2-b}}}-\frac{\log \left (-\sqrt{\sqrt{2-b}+2} x+x^2+1\right )}{8 \sqrt{\sqrt{2-b}+2}}+\frac{\log \left (\sqrt{\sqrt{2-b}+2} x+x^2+1\right )}{8 \sqrt{\sqrt{2-b}+2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2-b}}-2 x}{\sqrt{\sqrt{2-b}+2}}\right )}{4 \sqrt{\sqrt{2-b}+2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{2-b}+2}-2 x}{\sqrt{2-\sqrt{2-b}}}\right )}{4 \sqrt{2-\sqrt{2-b}}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2-b}}+2 x}{\sqrt{\sqrt{2-b}+2}}\right )}{4 \sqrt{\sqrt{2-b}+2}}+\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{2-b}+2}+2 x}{\sqrt{2-\sqrt{2-b}}}\right )}{4 \sqrt{2-\sqrt{2-b}}} \]

[Out]

-ArcTan[(Sqrt[2 - Sqrt[2 - b]] - 2*x)/Sqrt[2 + Sqrt[2 - b]]]/(4*Sqrt[2 + Sqrt[2 - b]]) - ArcTan[(Sqrt[2 + Sqrt
[2 - b]] - 2*x)/Sqrt[2 - Sqrt[2 - b]]]/(4*Sqrt[2 - Sqrt[2 - b]]) + ArcTan[(Sqrt[2 - Sqrt[2 - b]] + 2*x)/Sqrt[2
+ Sqrt[2 - b]]]/(4*Sqrt[2 + Sqrt[2 - b]]) + ArcTan[(Sqrt[2 + Sqrt[2 - b]] + 2*x)/Sqrt[2 - Sqrt[2 - b]]]/(4*Sq
rt[2 - Sqrt[2 - b]]) - Log[1 - Sqrt[2 - Sqrt[2 - b]]*x + x^2]/(8*Sqrt[2 - Sqrt[2 - b]]) + Log[1 + Sqrt[2 - Sqr
t[2 - b]]*x + x^2]/(8*Sqrt[2 - Sqrt[2 - b]]) - Log[1 - Sqrt[2 + Sqrt[2 - b]]*x + x^2]/(8*Sqrt[2 + Sqrt[2 - b]]
) + Log[1 + Sqrt[2 + Sqrt[2 - b]]*x + x^2]/(8*Sqrt[2 + Sqrt[2 - b]])

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Rubi [A]  time = 0.291718, antiderivative size = 411, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1419, 1094, 634, 618, 204, 628} \[ -\frac{\log \left (-\sqrt{2-\sqrt{2-b}} x+x^2+1\right )}{8 \sqrt{2-\sqrt{2-b}}}+\frac{\log \left (\sqrt{2-\sqrt{2-b}} x+x^2+1\right )}{8 \sqrt{2-\sqrt{2-b}}}-\frac{\log \left (-\sqrt{\sqrt{2-b}+2} x+x^2+1\right )}{8 \sqrt{\sqrt{2-b}+2}}+\frac{\log \left (\sqrt{\sqrt{2-b}+2} x+x^2+1\right )}{8 \sqrt{\sqrt{2-b}+2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2-b}}-2 x}{\sqrt{\sqrt{2-b}+2}}\right )}{4 \sqrt{\sqrt{2-b}+2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{2-b}+2}-2 x}{\sqrt{2-\sqrt{2-b}}}\right )}{4 \sqrt{2-\sqrt{2-b}}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2-b}}+2 x}{\sqrt{\sqrt{2-b}+2}}\right )}{4 \sqrt{\sqrt{2-b}+2}}+\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{2-b}+2}+2 x}{\sqrt{2-\sqrt{2-b}}}\right )}{4 \sqrt{2-\sqrt{2-b}}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)/(1 + b*x^4 + x^8),x]

[Out]

-ArcTan[(Sqrt[2 - Sqrt[2 - b]] - 2*x)/Sqrt[2 + Sqrt[2 - b]]]/(4*Sqrt[2 + Sqrt[2 - b]]) - ArcTan[(Sqrt[2 + Sqrt
[2 - b]] - 2*x)/Sqrt[2 - Sqrt[2 - b]]]/(4*Sqrt[2 - Sqrt[2 - b]]) + ArcTan[(Sqrt[2 - Sqrt[2 - b]] + 2*x)/Sqrt[2
+ Sqrt[2 - b]]]/(4*Sqrt[2 + Sqrt[2 - b]]) + ArcTan[(Sqrt[2 + Sqrt[2 - b]] + 2*x)/Sqrt[2 - Sqrt[2 - b]]]/(4*Sq
rt[2 - Sqrt[2 - b]]) - Log[1 - Sqrt[2 - Sqrt[2 - b]]*x + x^2]/(8*Sqrt[2 - Sqrt[2 - b]]) + Log[1 + Sqrt[2 - Sqr
t[2 - b]]*x + x^2]/(8*Sqrt[2 - Sqrt[2 - b]]) - Log[1 - Sqrt[2 + Sqrt[2 - b]]*x + x^2]/(8*Sqrt[2 + Sqrt[2 - b]]
) + Log[1 + Sqrt[2 + Sqrt[2 - b]]*x + x^2]/(8*Sqrt[2 + Sqrt[2 - b]])

Rule 1419

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[(2*d)/e -
b/c, 2]}, Dist[e/(2*c), Int[1/Simp[d/e + q*x^(n/2) + x^n, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x^(n/2
) + x^n, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2,
0] && IGtQ[n/2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !LtQ[(2*d)/e - b/c, 0] && EqQ[d, e*Rt[a/c, 2]]))

Rule 1094

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

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

Mathematica [C]  time = 0.027104, size = 55, normalized size = 0.13 \[ \frac{1}{4} \text{RootSum}\left [\text{$\#$1}^4 b+\text{$\#$1}^8+1\& ,\frac{\text{$\#$1}^4 \log (x-\text{$\#$1})+\log (x-\text{$\#$1})}{\text{$\#$1}^3 b+2 \text{$\#$1}^7}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^4)/(1 + b*x^4 + x^8),x]

[Out]

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

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Maple [C]  time = 0.052, size = 42, normalized size = 0.1 \begin{align*}{\frac{1}{4}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}+b{{\it \_Z}}^{4}+1 \right ) }{\frac{ \left ({{\it \_R}}^{4}+1 \right ) \ln \left ( x-{\it \_R} \right ) }{2\,{{\it \_R}}^{7}+{{\it \_R}}^{3}b}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+1)/(x^8+b*x^4+1),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} + 1}{x^{8} + b x^{4} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + 1)/(x^8 + b*x^4 + 1), x)

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Fricas [B]  time = 1.5601, size = 3723, normalized size = 9.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4)))*arctan(1/2*
sqrt(1/2)*(b^2 + (b^3 + 6*b^2 + 12*b + 8)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + 4*b + 4)*sqrt(x^2 + 1/2*sqr
t(1/2)*(b^2 + (b^3 + 6*b^2 + 12*b + 8)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + 2*b)*sqrt(((b^2 + 4*b + 4)*sqr
t((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4)))*sqrt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/(
b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4)))*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) -
b)/(b^2 + 4*b + 4)) - 1/2*sqrt(1/2)*((b^3 + 6*b^2 + 12*b + 8)*x*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + (b^2
+ 4*b + 4)*x)*sqrt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4
)))*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4))) - sqrt(sqrt(1/2)*sqrt(
-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4)))*arctan(-1/2*(sqrt(1/2)*(b^2 -
(b^3 + 6*b^2 + 12*b + 8)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + 4*b + 4)*sqrt(x^2 + 1/2*sqrt(1/2)*(b^2 - (b^
3 + 6*b^2 + 12*b + 8)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + 2*b)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 +
6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4)))*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/
(b^2 + 4*b + 4)) + sqrt(1/2)*((b^3 + 6*b^2 + 12*b + 8)*x*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - (b^2 + 4*b +
4)*x)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4)))*sqrt(sqrt(1/2)*sqr
t(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4)))) - 1/4*sqrt(sqrt(1/2)*sqrt(-
((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4)))*log(1/2*((b^2 + 4*b + 4)*sqrt((
b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b - 2)*sqrt(sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12
*b + 8)) + b)/(b^2 + 4*b + 4))) + x) + 1/4*sqrt(sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 1
2*b + 8)) + b)/(b^2 + 4*b + 4)))*log(-1/2*((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b - 2)*sqr
t(sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4))) + x) + 1/4*sq
rt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4)))*log(1/2*((b^2
+ 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b + 2)*sqrt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/
(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4))) + x) - 1/4*sqrt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/
(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4)))*log(-1/2*((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b +
8)) + b + 2)*sqrt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4))
) + x)

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Sympy [A]  time = 2.24891, size = 75, normalized size = 0.18 \begin{align*} \operatorname{RootSum}{\left (t^{8} \left (65536 b^{4} + 524288 b^{3} + 1572864 b^{2} + 2097152 b + 1048576\right ) + t^{4} \left (256 b^{3} + 1024 b^{2} + 1024 b\right ) + 1, \left ( t \mapsto t \log{\left (1024 t^{5} b^{2} + 4096 t^{5} b + 4096 t^{5} + 4 t b + 4 t + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+1)/(x**8+b*x**4+1),x)

[Out]

RootSum(_t**8*(65536*b**4 + 524288*b**3 + 1572864*b**2 + 2097152*b + 1048576) + _t**4*(256*b**3 + 1024*b**2 +
1024*b) + 1, Lambda(_t, _t*log(1024*_t**5*b**2 + 4096*_t**5*b + 4096*_t**5 + 4*_t*b + 4*_t + x)))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} + 1}{x^{8} + b x^{4} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + 1)/(x^8 + b*x^4 + 1), x)