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

Optimal. Leaf size=67 \[ -\frac {x}{2 \sqrt [4]{x^4+1}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1404, 382, 377, 212, 206, 203} \begin {gather*} -\frac {x}{2 \sqrt [4]{x^4+1}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 + x^4)^(1/4)*(-1 + x^8)),x]

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 1404

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(a/d
+ (c*x^n)/e)^p, x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx &=\int \frac {1}{\left (-1+x^4\right ) \left (1+x^4\right )^{5/4}} \, dx\\ &=-\frac {x}{2 \sqrt [4]{1+x^4}}+\frac {1}{2} \int \frac {1}{\left (-1+x^4\right ) \sqrt [4]{1+x^4}} \, dx\\ &=-\frac {x}{2 \sqrt [4]{1+x^4}}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-1+2 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {x}{2 \sqrt [4]{1+x^4}}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=-\frac {x}{2 \sqrt [4]{1+x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 85, normalized size = 1.27 \begin {gather*} -\frac {x}{2 \sqrt [4]{x^4+1}}-\frac {-\log \left (1-\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )+\log \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}+1\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{8 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x/(1 + x^4)^(1/4) - (2*ArcTan[(2^(1/4)*x)/(1 + x^4)^(1/4)] - Log[1 - (2^(1/4)*x)/(1 + x^4)^(1/4)] + Log[1
+ (2^(1/4)*x)/(1 + x^4)^(1/4)])/(8*2^(1/4))

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IntegrateAlgebraic [A]  time = 0.32, size = 67, normalized size = 1.00 \begin {gather*} -\frac {x}{2 \sqrt [4]{1+x^4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 5.19, size = 242, normalized size = 3.61 \begin {gather*} \frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {1}{4}} {\left (3 \, x^{4} + 1\right )}\right )}}{2 \, {\left (x^{4} - 1\right )}}\right ) - 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) + 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{32 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*2^(3/4)*(x^4 + 1)*arctan(1/2*(4*2^(3/4)*(x^4 + 1)^(1/4)*x^3 + 4*2^(1/4)*(x^4 + 1)^(3/4)*x + 2^(3/4)*(2
*2^(3/4)*sqrt(x^4 + 1)*x^2 + 2^(1/4)*(3*x^4 + 1)))/(x^4 - 1)) - 2^(3/4)*(x^4 + 1)*log((4*sqrt(2)*(x^4 + 1)^(1/
4)*x^3 + 4*2^(1/4)*sqrt(x^4 + 1)*x^2 + 2^(3/4)*(3*x^4 + 1) + 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) + 2^(3/4)*(x^4 +
1)*log((4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 4*2^(1/4)*sqrt(x^4 + 1)*x^2 - 2^(3/4)*(3*x^4 + 1) + 4*(x^4 + 1)^(3/4)*
x)/(x^4 - 1)) - 16*(x^4 + 1)^(3/4)*x)/(x^4 + 1)

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

[Out]

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

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maple [C]  time = 2.45, size = 238, normalized size = 3.55

method	result	size
trager	\(-\frac {x}{2 \left (x^{4}+1\right )^{\frac {1}{4}}}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}-\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \RootOf \left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{4}-8\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}\)	\(238\)
risch	\(-\frac {x}{2 \left (x^{4}+1\right )^{\frac {1}{4}}}+\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {-\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 x^{4} \RootOf \left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}-8\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {-\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}\)	\(240\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(x^4+1)^(1/4)+1/16*RootOf(_Z^2+RootOf(_Z^4-8)^2)*ln(-((x^4+1)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-8)^2)*RootO
f(_Z^4-8)^2*x^2-2*(x^4+1)^(1/4)*RootOf(_Z^4-8)^2*x^3-3*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^4+4*(x^4+1)^(3/4)*x-Roo
tOf(_Z^2+RootOf(_Z^4-8)^2))/(-1+x)/(1+x)/(x^2+1))-1/16*RootOf(_Z^4-8)*ln(-((x^4+1)^(1/2)*RootOf(_Z^4-8)^3*x^2+
2*(x^4+1)^(1/4)*RootOf(_Z^4-8)^2*x^3+3*x^4*RootOf(_Z^4-8)+4*(x^4+1)^(3/4)*x+RootOf(_Z^4-8))/(-1+x)/(1+x)/(x^2+
1))

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

[Out]

integrate(1/((x^8 - 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 {1}{{\left (x^4+1\right )}^{1/4}\,\left (x^8-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^4 + 1)^(1/4)*(x^8 - 1)),x)

[Out]

int(1/((x^4 + 1)^(1/4)*(x^8 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)**(5/4)), x)

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