3.9.50 \(\int \frac {-2+x^4}{\sqrt [4]{1+x^4} (-1+x^4+2 x^8)} \, dx\)

Optimal. Leaf size=64 \[ \frac {x}{\sqrt [4]{x^4+1}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{3}} \]

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Rubi [A]  time = 0.05, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1454, 527, 12, 377, 212, 206, 203} \begin {gather*} \frac {x}{\sqrt [4]{x^4+1}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(1 + x^4)^(1/4) + ArcTan[(3^(1/4)*x)/(1 + x^4)^(1/4)]/(2*3^(1/4)) + ArcTanh[(3^(1/4)*x)/(1 + x^4)^(1/4)]/(2*
3^(1/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 527

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

Rule 1454

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^
(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, b, c, d, e, f,
g, n, q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {-2+x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+2 x^8\right )} \, dx &=\int \frac {-2+x^4}{\left (1+x^4\right )^{5/4} \left (-1+2 x^4\right )} \, dx\\ &=\frac {x}{\sqrt [4]{1+x^4}}+\frac {1}{3} \int -\frac {3}{\sqrt [4]{1+x^4} \left (-1+2 x^4\right )} \, dx\\ &=\frac {x}{\sqrt [4]{1+x^4}}-\int \frac {1}{\sqrt [4]{1+x^4} \left (-1+2 x^4\right )} \, dx\\ &=\frac {x}{\sqrt [4]{1+x^4}}-\operatorname {Subst}\left (\int \frac {1}{-1+3 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {x}{\sqrt [4]{1+x^4}}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {x}{\sqrt [4]{1+x^4}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt [4]{3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt [4]{3}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 82, normalized size = 1.28 \begin {gather*} \frac {x}{\sqrt [4]{x^4+1}}+\frac {-\log \left (1-\frac {\sqrt [4]{3} x}{\sqrt [4]{x^4+1}}\right )+\log \left (\frac {\sqrt [4]{3} x}{\sqrt [4]{x^4+1}}+1\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(1 + x^4)^(1/4) + (2*ArcTan[(3^(1/4)*x)/(1 + x^4)^(1/4)] - Log[1 - (3^(1/4)*x)/(1 + x^4)^(1/4)] + Log[1 + (3
^(1/4)*x)/(1 + x^4)^(1/4)])/(4*3^(1/4))

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IntegrateAlgebraic [A]  time = 0.33, size = 64, normalized size = 1.00 \begin {gather*} \frac {x}{\sqrt [4]{1+x^4}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt [4]{3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{3} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-2 + x^4)/((1 + x^4)^(1/4)*(-1 + x^4 + 2*x^8)),x]

[Out]

x/(1 + x^4)^(1/4) + ArcTan[(3^(1/4)*x)/(1 + x^4)^(1/4)]/(2*3^(1/4)) + ArcTanh[(3^(1/4)*x)/(1 + x^4)^(1/4)]/(2*
3^(1/4))

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fricas [B]  time = 5.36, size = 248, normalized size = 3.88 \begin {gather*} -\frac {4 \cdot 3^{\frac {3}{4}} {\left (x^{4} + 1\right )} \arctan \left (\frac {6 \cdot 3^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 6 \cdot 3^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 3^{\frac {3}{4}} {\left (2 \cdot 3^{\frac {3}{4}} \sqrt {x^{4} + 1} x^{2} + 3^{\frac {1}{4}} {\left (4 \, x^{4} + 1\right )}\right )}}{3 \, {\left (2 \, x^{4} - 1\right )}}\right ) - 3^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {6 \, \sqrt {3} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 6 \cdot 3^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 3^{\frac {3}{4}} {\left (4 \, x^{4} + 1\right )} + 6 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{2 \, x^{4} - 1}\right ) + 3^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {6 \, \sqrt {3} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 6 \cdot 3^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 3^{\frac {3}{4}} {\left (4 \, x^{4} + 1\right )} + 6 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{2 \, x^{4} - 1}\right ) - 24 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{24 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(4*3^(3/4)*(x^4 + 1)*arctan(1/3*(6*3^(3/4)*(x^4 + 1)^(1/4)*x^3 + 6*3^(1/4)*(x^4 + 1)^(3/4)*x + 3^(3/4)*(
2*3^(3/4)*sqrt(x^4 + 1)*x^2 + 3^(1/4)*(4*x^4 + 1)))/(2*x^4 - 1)) - 3^(3/4)*(x^4 + 1)*log((6*sqrt(3)*(x^4 + 1)^
(1/4)*x^3 + 6*3^(1/4)*sqrt(x^4 + 1)*x^2 + 3^(3/4)*(4*x^4 + 1) + 6*(x^4 + 1)^(3/4)*x)/(2*x^4 - 1)) + 3^(3/4)*(x
^4 + 1)*log((6*sqrt(3)*(x^4 + 1)^(1/4)*x^3 - 6*3^(1/4)*sqrt(x^4 + 1)*x^2 - 3^(3/4)*(4*x^4 + 1) + 6*(x^4 + 1)^(
3/4)*x)/(2*x^4 - 1)) - 24*(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 {x^{4} - 2}{{\left (2 \, 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^4-2)/(x^4+1)^(1/4)/(2*x^8+x^4-1),x, algorithm="giac")

[Out]

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

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maple [C]  time = 2.74, size = 225, normalized size = 3.52

method	result	size
trager	\(\frac {x}{\left (x^{4}+1\right )^{\frac {1}{4}}}+\frac {\RootOf \left (\textit {\_Z}^{4}-27\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-27\right )^{3} x^{2}+6 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-27\right )^{2} x^{3}+12 \RootOf \left (\textit {\_Z}^{4}-27\right ) x^{4}+18 \left (x^{4}+1\right )^{\frac {3}{4}} x +3 \RootOf \left (\textit {\_Z}^{4}-27\right )}{2 x^{4}-1}\right )}{12}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) \ln \left (-\frac {-2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-27\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) x^{2}-6 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-27\right )^{2} x^{3}+12 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) x^{4}+18 \left (x^{4}+1\right )^{\frac {3}{4}} x +3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right )}{2 x^{4}-1}\right )}{12}\)	\(225\)
risch	\(\frac {x}{\left (x^{4}+1\right )^{\frac {1}{4}}}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-27\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) x^{2}-6 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-27\right )^{2} x^{3}-12 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right ) x^{4}+18 \left (x^{4}+1\right )^{\frac {3}{4}} x -3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-27\right )^{2}\right )}{2 x^{4}-1}\right )}{12}+\frac {\RootOf \left (\textit {\_Z}^{4}-27\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}-27\right )^{3} x^{2}+6 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-27\right )^{2} x^{3}+12 \RootOf \left (\textit {\_Z}^{4}-27\right ) x^{4}+18 \left (x^{4}+1\right )^{\frac {3}{4}} x +3 \RootOf \left (\textit {\_Z}^{4}-27\right )}{2 x^{4}-1}\right )}{12}\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-2)/(x^4+1)^(1/4)/(2*x^8+x^4-1),x,method=_RETURNVERBOSE)

[Out]

x/(x^4+1)^(1/4)+1/12*RootOf(_Z^4-27)*ln(-(2*(x^4+1)^(1/2)*RootOf(_Z^4-27)^3*x^2+6*(x^4+1)^(1/4)*RootOf(_Z^4-27
)^2*x^3+12*RootOf(_Z^4-27)*x^4+18*(x^4+1)^(3/4)*x+3*RootOf(_Z^4-27))/(2*x^4-1))+1/12*RootOf(_Z^2+RootOf(_Z^4-2
7)^2)*ln(-(-2*(x^4+1)^(1/2)*RootOf(_Z^4-27)^2*RootOf(_Z^2+RootOf(_Z^4-27)^2)*x^2-6*(x^4+1)^(1/4)*RootOf(_Z^4-2
7)^2*x^3+12*RootOf(_Z^2+RootOf(_Z^4-27)^2)*x^4+18*(x^4+1)^(3/4)*x+3*RootOf(_Z^2+RootOf(_Z^4-27)^2))/(2*x^4-1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} - 2}{{\left (2 \, 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^4-2)/(x^4+1)^(1/4)/(2*x^8+x^4-1),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^4-2}{{\left (x^4+1\right )}^{1/4}\,\left (2\,x^8+x^4-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^4 - 2)/((x^4 + 1)^(1/4)*(x^4 + 2*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} - 2}{\left (x^{4} + 1\right )^{\frac {5}{4}} \left (2 x^{4} - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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