3.16.45 \(\int \frac {\sqrt [4]{-1+2 x^4} (-1+x^4+x^8)}{x^6 (-1+x^4)} \, dx\)

Optimal. Leaf size=106 \[ \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{2 x^4-1}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{2 x^4-1}}\right )}{2^{3/4}}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{2 x^4-1}}\right )+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{2 x^4-1}}\right )}{2^{3/4}}+\frac {\left (2 x^4-1\right )^{5/4}}{5 x^5} \]

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Rubi [C]  time = 0.65, antiderivative size = 67, normalized size of antiderivative = 0.63, number of steps used = 23, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6725, 264, 1240, 407, 409, 1213, 537, 511, 510} \begin {gather*} \frac {\left (2 x^4-1\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{2 x^4-1} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};2 x^4,x^4\right )}{3 \sqrt [4]{1-2 x^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((-1 + 2*x^4)^(1/4)*(-1 + x^4 + x^8))/(x^6*(-1 + x^4)),x]

[Out]

(-1 + 2*x^4)^(5/4)/(5*x^5) - (x^3*(-1 + 2*x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, 2*x^4, x^4])/(3*(1 - 2*x^4)^(
1/4))

Rule 264

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

Rule 407

Int[((a_) + (b_.)*(x_)^4)^(1/4)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[Sqrt[a + b*x^4]*Sqrt[a/(a + b*x^4)],
Subst[Int[1/(Sqrt[1 - b*x^4]*(c - (b*c - a*d)*x^4)), x], x, x/(a + b*x^4)^(1/4)], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[b*c - a*d, 0]

Rule 409

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

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

Rule 1240

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^4)^p, (d/
(d^2 - e^2*x^4) - (e*x^2)/(d^2 - e^2*x^4))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&&  !IntegerQ[p] && ILtQ[q, 0]

Rule 6725

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

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-1+2 x^4} \left (-1+x^4+x^8\right )}{x^6 \left (-1+x^4\right )} \, dx &=\int \left (\frac {\sqrt [4]{-1+2 x^4}}{x^6}+\frac {\sqrt [4]{-1+2 x^4}}{2 \left (-1+x^2\right )}+\frac {\sqrt [4]{-1+2 x^4}}{2 \left (1+x^2\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {\sqrt [4]{-1+2 x^4}}{-1+x^2} \, dx+\frac {1}{2} \int \frac {\sqrt [4]{-1+2 x^4}}{1+x^2} \, dx+\int \frac {\sqrt [4]{-1+2 x^4}}{x^6} \, dx\\ &=\frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} \int \left (\frac {\sqrt [4]{-1+2 x^4}}{1-x^4}+\frac {x^2 \sqrt [4]{-1+2 x^4}}{-1+x^4}\right ) \, dx+\frac {1}{2} \int \left (\frac {\sqrt [4]{-1+2 x^4}}{-1+x^4}+\frac {x^2 \sqrt [4]{-1+2 x^4}}{-1+x^4}\right ) \, dx\\ &=\frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} \int \frac {\sqrt [4]{-1+2 x^4}}{1-x^4} \, dx+\frac {1}{2} \int \frac {\sqrt [4]{-1+2 x^4}}{-1+x^4} \, dx+2 \left (\frac {1}{2} \int \frac {x^2 \sqrt [4]{-1+2 x^4}}{-1+x^4} \, dx\right )\\ &=\frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}+2 \frac {\sqrt [4]{-1+2 x^4} \int \frac {x^2 \sqrt [4]{1-2 x^4}}{-1+x^4} \, dx}{2 \sqrt [4]{1-2 x^4}}+\frac {1}{2} \left (\sqrt {-\frac {1}{-1+2 x^4}} \sqrt {-1+2 x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x^4} \left (1-x^4\right )} \, dx,x,\frac {x}{\sqrt [4]{-1+2 x^4}}\right )+\frac {1}{2} \left (\sqrt {-\frac {1}{-1+2 x^4}} \sqrt {-1+2 x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-2 x^4} \left (-1+x^4\right )} \, dx,x,\frac {x}{\sqrt [4]{-1+2 x^4}}\right )\\ &=\frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}-\frac {x^3 \sqrt [4]{-1+2 x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};2 x^4,x^4\right )}{3 \sqrt [4]{1-2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 67, normalized size = 0.63 \begin {gather*} \frac {\sqrt [4]{2 x^4-1} \left (-5 x^8 F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};2 x^4,x^4\right )-3 \left (1-2 x^4\right )^{5/4}\right )}{15 x^5 \sqrt [4]{1-2 x^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((-1 + 2*x^4)^(1/4)*(-1 + x^4 + x^8))/(x^6*(-1 + x^4)),x]

[Out]

((-1 + 2*x^4)^(1/4)*(-3*(1 - 2*x^4)^(5/4) - 5*x^8*AppellF1[3/4, -1/4, 1, 7/4, 2*x^4, x^4]))/(15*x^5*(1 - 2*x^4
)^(1/4))

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IntegrateAlgebraic [A]  time = 0.32, size = 106, normalized size = 1.00 \begin {gather*} \frac {\left (-1+2 x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+2 x^4}}\right )-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+2 x^4}}\right )}{2^{3/4}}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+2 x^4}}\right )+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+2 x^4}}\right )}{2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + 2*x^4)^(1/4)*(-1 + x^4 + x^8))/(x^6*(-1 + x^4)),x]

[Out]

(-1 + 2*x^4)^(5/4)/(5*x^5) + ArcTan[x/(-1 + 2*x^4)^(1/4)]/2 - ArcTan[(2^(1/4)*x)/(-1 + 2*x^4)^(1/4)]/2^(3/4) -
ArcTanh[x/(-1 + 2*x^4)^(1/4)]/2 + ArcTanh[(2^(1/4)*x)/(-1 + 2*x^4)^(1/4)]/2^(3/4)

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fricas [B]  time = 22.76, size = 330, normalized size = 3.11 \begin {gather*} \frac {20 \cdot 2^{\frac {1}{4}} x^{5} \arctan \left (2 \cdot 2^{\frac {3}{4}} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \cdot 2^{\frac {1}{4}} {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x + \frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {2 \, x^{4} - 1} x^{2} + 2^{\frac {1}{4}} {\left (4 \, x^{4} - 1\right )}\right )}\right ) + 5 \cdot 2^{\frac {1}{4}} x^{5} \log \left (4 \, \sqrt {2} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {2 \, x^{4} - 1} x^{2} + 2^{\frac {3}{4}} {\left (4 \, x^{4} - 1\right )} + 4 \, {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x\right ) - 5 \cdot 2^{\frac {1}{4}} x^{5} \log \left (4 \, \sqrt {2} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {2 \, x^{4} - 1} x^{2} - 2^{\frac {3}{4}} {\left (4 \, x^{4} - 1\right )} + 4 \, {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x\right ) + 10 \, x^{5} \arctan \left (\frac {2 \, {\left ({\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x\right )}}{x^{4} - 1}\right ) + 10 \, x^{5} \log \left (-\frac {3 \, x^{4} - 2 \, {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {2 \, x^{4} - 1} x^{2} - 2 \, {\left (2 \, x^{4} - 1\right )}^{\frac {3}{4}} x - 1}{x^{4} - 1}\right ) + 8 \, {\left (2 \, x^{4} - 1\right )}^{\frac {5}{4}}}{40 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4-1)^(1/4)*(x^8+x^4-1)/x^6/(x^4-1),x, algorithm="fricas")

[Out]

1/40*(20*2^(1/4)*x^5*arctan(2*2^(3/4)*(2*x^4 - 1)^(1/4)*x^3 + 2*2^(1/4)*(2*x^4 - 1)^(3/4)*x + 1/2*2^(3/4)*(2*2
^(3/4)*sqrt(2*x^4 - 1)*x^2 + 2^(1/4)*(4*x^4 - 1))) + 5*2^(1/4)*x^5*log(4*sqrt(2)*(2*x^4 - 1)^(1/4)*x^3 + 4*2^(
1/4)*sqrt(2*x^4 - 1)*x^2 + 2^(3/4)*(4*x^4 - 1) + 4*(2*x^4 - 1)^(3/4)*x) - 5*2^(1/4)*x^5*log(4*sqrt(2)*(2*x^4 -
1)^(1/4)*x^3 - 4*2^(1/4)*sqrt(2*x^4 - 1)*x^2 - 2^(3/4)*(4*x^4 - 1) + 4*(2*x^4 - 1)^(3/4)*x) + 10*x^5*arctan(2
*((2*x^4 - 1)^(1/4)*x^3 + (2*x^4 - 1)^(3/4)*x)/(x^4 - 1)) + 10*x^5*log(-(3*x^4 - 2*(2*x^4 - 1)^(1/4)*x^3 + 2*s
qrt(2*x^4 - 1)*x^2 - 2*(2*x^4 - 1)^(3/4)*x - 1)/(x^4 - 1)) + 8*(2*x^4 - 1)^(5/4))/x^5

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giac [A]  time = 0.22, size = 143, normalized size = 1.35 \begin {gather*} -\frac {1}{2} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{2 \, x}\right ) - \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + \frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} - \frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}} {\left (\frac {1}{x^{4}} - 2\right )}}{5 \, x} + \frac {1}{2} \, \arctan \left (\frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {1}{4} \, \log \left ({\left | \frac {{\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4-1)^(1/4)*(x^8+x^4-1)/x^6/(x^4-1),x, algorithm="giac")

[Out]

-1/2*2^(1/4)*arctan(1/2*2^(3/4)*(2*x^4 - 1)^(1/4)/x) - 1/4*2^(1/4)*log(2^(1/4) + (2*x^4 - 1)^(1/4)/x) + 1/4*2^
(1/4)*log(2^(1/4) - (2*x^4 - 1)^(1/4)/x) + 1/5*(2*x^4 - 1)^(1/4)*(1/x^4 - 2)/x + 1/2*arctan((2*x^4 - 1)^(1/4)/
x) + 1/4*log((2*x^4 - 1)^(1/4)/x + 1) - 1/4*log(abs((2*x^4 - 1)^(1/4)/x - 1))

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maple [C]  time = 17.94, size = 448, normalized size = 4.23

method	result	size
trager	\(\frac {\left (2 x^{4}-1\right )^{\frac {5}{4}}}{5 x^{5}}-\frac {\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (-4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{4}+4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \left (2 x^{4}-1\right )^{\frac {1}{4}} x^{3}-4 \RootOf \left (\textit {\_Z}^{4}-2\right ) \sqrt {2 x^{4}-1}\, x^{2}+4 \left (2 x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{4}-2\right )^{3}\right )}{4}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{4}-4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \left (2 x^{4}-1\right )^{\frac {1}{4}} x^{3}-4 \sqrt {2 x^{4}-1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{2}+4 \left (2 x^{4}-1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )\right )}{4}-\frac {\ln \left (\frac {2 \left (2 x^{4}-1\right )^{\frac {3}{4}} x +2 \sqrt {2 x^{4}-1}\, x^{2}+2 \left (2 x^{4}-1\right )^{\frac {1}{4}} x^{3}+3 x^{4}-1}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{4}-\frac {\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \sqrt {2 x^{4}-1}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{4}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )+4 \left (2 x^{4}-1\right )^{\frac {3}{4}} x -4 \left (2 x^{4}-1\right )^{\frac {1}{4}} x^{3}}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{8}\)	\(448\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^4-1)^(1/4)*(x^8+x^4-1)/x^6/(x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/5*(2*x^4-1)^(5/4)/x^5-1/4*RootOf(_Z^4-2)*ln(-4*RootOf(_Z^4-2)^3*x^4+4*RootOf(_Z^4-2)^2*(2*x^4-1)^(1/4)*x^3-4
*RootOf(_Z^4-2)*(2*x^4-1)^(1/2)*x^2+4*(2*x^4-1)^(3/4)*x+RootOf(_Z^4-2)^3)-1/4*RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln
(4*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^4-4*RootOf(_Z^4-2)^2*(2*x^4-1)^(1/4)*x^3-4*(2*x^4-1)^(1/2)
*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^2+4*(2*x^4-1)^(3/4)*x-RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2))-1/4*ln(
(2*(2*x^4-1)^(3/4)*x+2*(2*x^4-1)^(1/2)*x^2+2*(2*x^4-1)^(1/4)*x^3+3*x^4-1)/(-1+x)/(1+x)/(x^2+1))-1/8*RootOf(_Z^
4-2)^3*RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln((2*RootOf(_Z^4-2)^3*(2*x^4-1)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^2-
3*RootOf(_Z^2+RootOf(_Z^4-2)^2)*RootOf(_Z^4-2)^3*x^4+RootOf(_Z^4-2)^3*RootOf(_Z^2+RootOf(_Z^4-2)^2)+4*(2*x^4-1
)^(3/4)*x-4*(2*x^4-1)^(1/4)*x^3)/(-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 {{\left (x^{8} + x^{4} - 1\right )} {\left (2 \, x^{4} - 1\right )}^{\frac {1}{4}}}{{\left (x^{4} - 1\right )} x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4-1)^(1/4)*(x^8+x^4-1)/x^6/(x^4-1),x, algorithm="maxima")

[Out]

integrate((x^8 + x^4 - 1)*(2*x^4 - 1)^(1/4)/((x^4 - 1)*x^6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (2\,x^4-1\right )}^{1/4}\,\left (x^8+x^4-1\right )}{x^6\,\left (x^4-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^4 - 1)^(1/4)*(x^4 + x^8 - 1))/(x^6*(x^4 - 1)),x)

[Out]

int(((2*x^4 - 1)^(1/4)*(x^4 + x^8 - 1))/(x^6*(x^4 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{2 x^{4} - 1} \left (x^{8} + x^{4} - 1\right )}{x^{6} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**4-1)**(1/4)*(x**8+x**4-1)/x**6/(x**4-1),x)

[Out]

Integral((2*x**4 - 1)**(1/4)*(x**8 + x**4 - 1)/(x**6*(x - 1)*(x + 1)*(x**2 + 1)), x)

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