3.12.46 \(\int \frac {1+x^4}{\sqrt [4]{-1+x^4} (1+x^4+x^8)} \, dx\)

Optimal. Leaf size=85 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8-3 \text {$\#$1}^4+3\& ,\frac {-\text {$\#$1}^4 \log \left (\sqrt [4]{x^4-1}-\text {$\#$1} x\right )+\text {$\#$1}^4 \log (x)+2 \log \left (\sqrt [4]{x^4-1}-\text {$\#$1} x\right )-2 \log (x)}{2 \text {$\#$1}^5-3 \text {$\#$1}}\& \right ] \]

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Rubi [C]  time = 0.43, antiderivative size = 369, normalized size of antiderivative = 4.34, number of steps used = 10, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6728, 377, 212, 208, 205} \begin {gather*} \frac {\sqrt [4]{\frac {-\sqrt {3}+i}{-\sqrt {3}+3 i}} \left (-\sqrt {3}+3 i\right ) \tan ^{-1}\left (\frac {x}{\sqrt [4]{\frac {-\sqrt {3}+i}{-\sqrt {3}+3 i}} \sqrt [4]{x^4-1}}\right )}{6 \left (-\sqrt {3}+i\right )}+\frac {\sqrt [4]{\frac {\sqrt {3}+i}{\sqrt {3}+3 i}} \left (\sqrt {3}+3 i\right ) \tan ^{-1}\left (\frac {x}{\sqrt [4]{\frac {\sqrt {3}+i}{\sqrt {3}+3 i}} \sqrt [4]{x^4-1}}\right )}{6 \left (\sqrt {3}+i\right )}+\frac {\sqrt [4]{\frac {-\sqrt {3}+i}{-\sqrt {3}+3 i}} \left (-\sqrt {3}+3 i\right ) \tanh ^{-1}\left (\frac {x}{\sqrt [4]{\frac {-\sqrt {3}+i}{-\sqrt {3}+3 i}} \sqrt [4]{x^4-1}}\right )}{6 \left (-\sqrt {3}+i\right )}+\frac {\sqrt [4]{\frac {\sqrt {3}+i}{\sqrt {3}+3 i}} \left (\sqrt {3}+3 i\right ) \tanh ^{-1}\left (\frac {x}{\sqrt [4]{\frac {\sqrt {3}+i}{\sqrt {3}+3 i}} \sqrt [4]{x^4-1}}\right )}{6 \left (\sqrt {3}+i\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((I - Sqrt[3])/(3*I - Sqrt[3]))^(1/4)*(3*I - Sqrt[3])*ArcTan[x/(((I - Sqrt[3])/(3*I - Sqrt[3]))^(1/4)*(-1 + x
^4)^(1/4))])/(6*(I - Sqrt[3])) + (((I + Sqrt[3])/(3*I + Sqrt[3]))^(1/4)*(3*I + Sqrt[3])*ArcTan[x/(((I + Sqrt[3
])/(3*I + Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4))])/(6*(I + Sqrt[3])) + (((I - Sqrt[3])/(3*I - Sqrt[3]))^(1/4)*(3*I
- Sqrt[3])*ArcTanh[x/(((I - Sqrt[3])/(3*I - Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4))])/(6*(I - Sqrt[3])) + (((I + Sqr
t[3])/(3*I + Sqrt[3]))^(1/4)*(3*I + Sqrt[3])*ArcTanh[x/(((I + Sqrt[3])/(3*I + Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4)
)])/(6*(I + Sqrt[3]))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+x^4}{\sqrt [4]{-1+x^4} \left (1+x^4+x^8\right )} \, dx &=\int \left (\frac {1-\frac {i}{\sqrt {3}}}{\sqrt [4]{-1+x^4} \left (1-i \sqrt {3}+2 x^4\right )}+\frac {1+\frac {i}{\sqrt {3}}}{\sqrt [4]{-1+x^4} \left (1+i \sqrt {3}+2 x^4\right )}\right ) \, dx\\ &=\frac {1}{3} \left (3-i \sqrt {3}\right ) \int \frac {1}{\sqrt [4]{-1+x^4} \left (1-i \sqrt {3}+2 x^4\right )} \, dx+\frac {1}{3} \left (3+i \sqrt {3}\right ) \int \frac {1}{\sqrt [4]{-1+x^4} \left (1+i \sqrt {3}+2 x^4\right )} \, dx\\ &=\frac {1}{3} \left (3-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{1-i \sqrt {3}-\left (3-i \sqrt {3}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {1}{3} \left (3+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{1+i \sqrt {3}-\left (3+i \sqrt {3}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=-\frac {\left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i-\sqrt {3}}-\sqrt {3 i-\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {3 \left (i-\sqrt {3}\right )}}-\frac {\left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i-\sqrt {3}}+\sqrt {3 i-\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {3 \left (i-\sqrt {3}\right )}}+\frac {\left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i+\sqrt {3}}-\sqrt {3 i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {3 \left (i+\sqrt {3}\right )}}+\frac {\left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i+\sqrt {3}}+\sqrt {3 i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {3 \left (i+\sqrt {3}\right )}}\\ &=\frac {\sqrt [4]{\frac {i-\sqrt {3}}{3 i-\sqrt {3}}} \left (3+i \sqrt {3}\right ) \tan ^{-1}\left (\frac {x}{\sqrt [4]{\frac {i-\sqrt {3}}{3 i-\sqrt {3}}} \sqrt [4]{-1+x^4}}\right )}{6 \left (1+i \sqrt {3}\right )}+\frac {\sqrt [4]{\frac {i+\sqrt {3}}{3 i+\sqrt {3}}} \left (3 i+\sqrt {3}\right ) \tan ^{-1}\left (\frac {x}{\sqrt [4]{\frac {i+\sqrt {3}}{3 i+\sqrt {3}}} \sqrt [4]{-1+x^4}}\right )}{6 \left (i+\sqrt {3}\right )}+\frac {\sqrt [4]{\frac {i-\sqrt {3}}{3 i-\sqrt {3}}} \left (3+i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt [4]{\frac {i-\sqrt {3}}{3 i-\sqrt {3}}} \sqrt [4]{-1+x^4}}\right )}{6 \left (1+i \sqrt {3}\right )}+\frac {\sqrt [4]{\frac {i+\sqrt {3}}{3 i+\sqrt {3}}} \left (3 i+\sqrt {3}\right ) \tanh ^{-1}\left (\frac {x}{\sqrt [4]{\frac {i+\sqrt {3}}{3 i+\sqrt {3}}} \sqrt [4]{-1+x^4}}\right )}{6 \left (i+\sqrt {3}\right )}\\ \end {align*}

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Mathematica [C]  time = 0.56, size = 257, normalized size = 3.02 \begin {gather*} \frac {1}{6} \left (\frac {\tan ^{-1}\left (\frac {x}{\sqrt [4]{\frac {\sqrt {3}-i}{\sqrt {3}-3 i}} \sqrt [4]{x^4-1}}\right )}{\left (\frac {\sqrt {3}-i}{\sqrt {3}-3 i}\right )^{3/4}}+\frac {\tan ^{-1}\left (\frac {x}{\sqrt [4]{\frac {\sqrt {3}+i}{\sqrt {3}+3 i}} \sqrt [4]{x^4-1}}\right )}{\left (\frac {\sqrt {3}+i}{\sqrt {3}+3 i}\right )^{3/4}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{\frac {\sqrt {3}-i}{\sqrt {3}-3 i}} \sqrt [4]{x^4-1}}\right )}{\left (\frac {\sqrt {3}-i}{\sqrt {3}-3 i}\right )^{3/4}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{\frac {\sqrt {3}+i}{\sqrt {3}+3 i}} \sqrt [4]{x^4-1}}\right )}{\left (\frac {\sqrt {3}+i}{\sqrt {3}+3 i}\right )^{3/4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[x/(((-I + Sqrt[3])/(-3*I + Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4))]/((-I + Sqrt[3])/(-3*I + Sqrt[3]))^(3/4)
+ ArcTan[x/(((I + Sqrt[3])/(3*I + Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4))]/((I + Sqrt[3])/(3*I + Sqrt[3]))^(3/4) + A
rcTanh[x/(((-I + Sqrt[3])/(-3*I + Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4))]/((-I + Sqrt[3])/(-3*I + Sqrt[3]))^(3/4) +
ArcTanh[x/(((I + Sqrt[3])/(3*I + Sqrt[3]))^(1/4)*(-1 + x^4)^(1/4))]/((I + Sqrt[3])/(3*I + Sqrt[3]))^(3/4))/6

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IntegrateAlgebraic [A]  time = 0.27, size = 85, normalized size = 1.00 \begin {gather*} \frac {1}{4} \text {RootSum}\left [3-3 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 \log (x)+2 \log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-1+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 \text {$\#$1}+2 \text {$\#$1}^5}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^4)/((-1 + x^4)^(1/4)*(1 + x^4 + x^8)),x]

[Out]

RootSum[3 - 3*#1^4 + #1^8 & , (-2*Log[x] + 2*Log[(-1 + x^4)^(1/4) - x*#1] + Log[x]*#1^4 - Log[(-1 + x^4)^(1/4)
- x*#1]*#1^4)/(-3*#1 + 2*#1^5) & ]/4

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} + 1}{{\left (x^{8} + x^{4} + 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 16.57, size = 1514, normalized size = 17.81

method	result	size
trager	\(\text {Expression too large to display}\)	\(1514\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+1)/(x^4-1)^(1/4)/(x^8+x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/12*RootOf(_Z^2+RootOf(_Z^8+27)^2)*ln(-36*(-2*RootOf(_Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^8+27)^8*x^4+6*(x^4-1)
^(1/2)*RootOf(_Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^8+27)^6*x^2-24*(x^4-1)^(1/4)*RootOf(_Z^8+27)^6*x^3-39*RootOf(_
Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^8+27)^4*x^4+54*RootOf(_Z^8+27)^4*(x^4-1)^(3/4)*x+90*(x^4-1)^(1/2)*RootOf(_Z^2
+RootOf(_Z^8+27)^2)*RootOf(_Z^8+27)^2*x^2-108*(x^4-1)^(1/4)*RootOf(_Z^8+27)^2*x^3+12*RootOf(_Z^2+RootOf(_Z^8+2
7)^2)*RootOf(_Z^8+27)^4-135*x^4*RootOf(_Z^2+RootOf(_Z^8+27)^2)+54*(x^4-1)^(3/4)*x+54*RootOf(_Z^2+RootOf(_Z^8+2
7)^2))/(RootOf(_Z^8+27)^4*x^2+6-3*x^2)/(RootOf(_Z^8+27)^4*x+3*x-6)/(RootOf(_Z^8+27)^4*x+3*x+6))-1/12*RootOf(_Z
^8+27)*ln(-36*(-2*RootOf(_Z^8+27)^9*x^4-6*(x^4-1)^(1/2)*RootOf(_Z^8+27)^7*x^2+24*(x^4-1)^(1/4)*RootOf(_Z^8+27)
^6*x^3-39*RootOf(_Z^8+27)^5*x^4+54*RootOf(_Z^8+27)^4*(x^4-1)^(3/4)*x-90*(x^4-1)^(1/2)*RootOf(_Z^8+27)^3*x^2+10
8*(x^4-1)^(1/4)*RootOf(_Z^8+27)^2*x^3+12*RootOf(_Z^8+27)^5-135*RootOf(_Z^8+27)*x^4+54*(x^4-1)^(3/4)*x+54*RootO
f(_Z^8+27))/(RootOf(_Z^8+27)^4*x^2+6-3*x^2)/(RootOf(_Z^8+27)^4*x+3*x-6)/(RootOf(_Z^8+27)^4*x+3*x+6))+1/324*Roo
tOf(_Z^8+27)^7*RootOf(_Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*ln(3
6*(RootOf(_Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*RootOf(_Z^8+27)^
11*x^4-12*x^4*RootOf(_Z^8+27)^7*RootOf(_Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(
_Z^8+27)^2))-36*(x^4-1)^(1/4)*RootOf(_Z^8+27)^5*RootOf(_Z^2+RootOf(_Z^8+27)^2)*x^3+6*RootOf(_Z^8+27)^7*RootOf(
_Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))-45*RootOf(_Z^2+RootOf(_Z^8
+27)^2)*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*RootOf(_Z^8+27)^3*x^4+108*RootOf(_Z^8+27)^
4*(x^4-1)^(3/4)*x+216*(x^4-1)^(1/2)*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*RootOf(_Z^8+27
)^2*x^2+324*(x^4-1)^(1/4)*RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2)*x^3+18*RootOf(_Z^2+RootOf(_Z^8+27)^2)
*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*RootOf(_Z^8+27)^3-324*(x^4-1)^(3/4)*x)/(RootOf(_Z
^8+27)^4*x^2+3*x^2-6)/(RootOf(_Z^8+27)^4*x-3*x+6)/(RootOf(_Z^8+27)^4*x-3*x-6))-1/12*RootOf(_Z^2-RootOf(_Z^8+27
)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*ln(36*(RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*RootOf(_Z
^8+27)^8*x^4+4*(x^4-1)^(1/2)*RootOf(_Z^8+27)^5*RootOf(_Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^2-RootOf(_Z^8+27)*Root
Of(_Z^2+RootOf(_Z^8+27)^2))*x^2+12*(x^4-1)^(1/4)*RootOf(_Z^8+27)^5*RootOf(_Z^2+RootOf(_Z^8+27)^2)*x^3-30*RootO
f(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*RootOf(_Z^8+27)^4*x^4+48*RootOf(_Z^8+27)^4*(x^4-1)^(3/4
)*x+108*(x^4-1)^(1/2)*RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2)*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+R
ootOf(_Z^8+27)^2))*x^2-180*(x^4-1)^(1/4)*RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2)*x^3+6*RootOf(_Z^2-Root
Of(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2))*RootOf(_Z^8+27)^4+225*x^4*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+
RootOf(_Z^8+27)^2))-216*(x^4-1)^(3/4)*x-90*RootOf(_Z^2-RootOf(_Z^8+27)*RootOf(_Z^2+RootOf(_Z^8+27)^2)))/(RootO
f(_Z^8+27)^4*x^2+3*x^2-6)/(RootOf(_Z^8+27)^4*x-3*x+6)/(RootOf(_Z^8+27)^4*x-3*x-6))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} + 1}{{\left (x^{8} + x^{4} + 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4+1}{{\left (x^4-1\right )}^{1/4}\,\left (x^8+x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} + 1}{\sqrt [4]{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right ) \left (x^{4} - x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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