3.16.100 \(\int \frac {\sqrt [4]{-1+x^4} (2-x^4+2 x^8)}{x^{10} (-1+2 x^4)} \, dx\)

Optimal. Leaf size=109 \[ 2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{x^4-1}}{\sqrt {x^4-1}-x^2}\right )-2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{x^4-1}}{\sqrt {x^4-1}+x^2}\right )+\frac {\sqrt [4]{x^4-1} \left (65 x^8+5 x^4+2\right )}{9 x^9} \]

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Rubi [C]  time = 0.38, antiderivative size = 121, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 10, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6725, 271, 264, 277, 331, 298, 203, 206, 511, 510} \begin {gather*} -\frac {16 \sqrt [4]{x^4-1} x^3 F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,2 x^4\right )}{3 \sqrt [4]{1-x^4}}+\frac {8 \sqrt [4]{x^4-1}}{x}+4 \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-4 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {2 \left (x^4-1\right )^{5/4}}{9 x^9}-\frac {7 \left (x^4-1\right )^{5/4}}{9 x^5} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[((-1 + x^4)^(1/4)*(2 - x^4 + 2*x^8))/(x^10*(-1 + 2*x^4)),x]

[Out]

(8*(-1 + x^4)^(1/4))/x - (2*(-1 + x^4)^(5/4))/(9*x^9) - (7*(-1 + x^4)^(5/4))/(9*x^5) - (16*x^3*(-1 + x^4)^(1/4
)*AppellF1[3/4, -1/4, 1, 7/4, x^4, 2*x^4])/(3*(1 - x^4)^(1/4)) + 4*ArcTan[x/(-1 + x^4)^(1/4)] - 4*ArcTanh[x/(-
1 + x^4)^(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 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 271

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

Rule 277

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

Rule 298

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

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

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 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+x^4} \left (2-x^4+2 x^8\right )}{x^{10} \left (-1+2 x^4\right )} \, dx &=\int \left (-\frac {2 \sqrt [4]{-1+x^4}}{x^{10}}-\frac {3 \sqrt [4]{-1+x^4}}{x^6}-\frac {8 \sqrt [4]{-1+x^4}}{x^2}+\frac {16 x^2 \sqrt [4]{-1+x^4}}{-1+2 x^4}\right ) \, dx\\ &=-\left (2 \int \frac {\sqrt [4]{-1+x^4}}{x^{10}} \, dx\right )-3 \int \frac {\sqrt [4]{-1+x^4}}{x^6} \, dx-8 \int \frac {\sqrt [4]{-1+x^4}}{x^2} \, dx+16 \int \frac {x^2 \sqrt [4]{-1+x^4}}{-1+2 x^4} \, dx\\ &=\frac {8 \sqrt [4]{-1+x^4}}{x}-\frac {2 \left (-1+x^4\right )^{5/4}}{9 x^9}-\frac {3 \left (-1+x^4\right )^{5/4}}{5 x^5}-\frac {8}{9} \int \frac {\sqrt [4]{-1+x^4}}{x^6} \, dx-8 \int \frac {x^2}{\left (-1+x^4\right )^{3/4}} \, dx+\frac {\left (16 \sqrt [4]{-1+x^4}\right ) \int \frac {x^2 \sqrt [4]{1-x^4}}{-1+2 x^4} \, dx}{\sqrt [4]{1-x^4}}\\ &=\frac {8 \sqrt [4]{-1+x^4}}{x}-\frac {2 \left (-1+x^4\right )^{5/4}}{9 x^9}-\frac {7 \left (-1+x^4\right )^{5/4}}{9 x^5}-\frac {16 x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,2 x^4\right )}{3 \sqrt [4]{1-x^4}}-8 \operatorname {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\frac {8 \sqrt [4]{-1+x^4}}{x}-\frac {2 \left (-1+x^4\right )^{5/4}}{9 x^9}-\frac {7 \left (-1+x^4\right )^{5/4}}{9 x^5}-\frac {16 x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,2 x^4\right )}{3 \sqrt [4]{1-x^4}}-4 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+4 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=\frac {8 \sqrt [4]{-1+x^4}}{x}-\frac {2 \left (-1+x^4\right )^{5/4}}{9 x^9}-\frac {7 \left (-1+x^4\right )^{5/4}}{9 x^5}-\frac {16 x^3 \sqrt [4]{-1+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^4,2 x^4\right )}{3 \sqrt [4]{1-x^4}}+4 \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-4 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 97, normalized size = 0.89 \begin {gather*} \frac {-24 \sqrt [4]{1-2 x^4} \left (1-x^4\right )^{3/4} x^{12} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};\frac {x^4}{2 x^4-1}\right )+130 x^{16}-185 x^{12}+54 x^8-x^4+2}{9 x^9 \left (x^4-1\right )^{3/4} \left (2 x^4-1\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((-1 + x^4)^(1/4)*(2 - x^4 + 2*x^8))/(x^10*(-1 + 2*x^4)),x]

[Out]

(2 - x^4 + 54*x^8 - 185*x^12 + 130*x^16 - 24*x^12*(1 - 2*x^4)^(1/4)*(1 - x^4)^(3/4)*Hypergeometric2F1[3/4, 3/4
, 7/4, x^4/(-1 + 2*x^4)])/(9*x^9*(-1 + x^4)^(3/4)*(-1 + 2*x^4))

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IntegrateAlgebraic [A]  time = 0.30, size = 109, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{-1+x^4} \left (2+5 x^4+65 x^8\right )}{9 x^9}+2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{-1+x^4}}{-x^2+\sqrt {-1+x^4}}\right )-2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{-1+x^4}}{x^2+\sqrt {-1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^4)^(1/4)*(2 - x^4 + 2*x^8))/(x^10*(-1 + 2*x^4)),x]

[Out]

((-1 + x^4)^(1/4)*(2 + 5*x^4 + 65*x^8))/(9*x^9) + 2*Sqrt[2]*ArcTan[(Sqrt[2]*x*(-1 + x^4)^(1/4))/(-x^2 + Sqrt[-
1 + x^4])] - 2*Sqrt[2]*ArcTanh[(Sqrt[2]*x*(-1 + x^4)^(1/4))/(x^2 + Sqrt[-1 + x^4])]

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fricas [B]  time = 2.67, size = 444, normalized size = 4.07 \begin {gather*} -\frac {36 \, \sqrt {2} x^{9} \arctan \left (-\frac {\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {5}{4}} + {\left (2 \, x^{5} - \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {5}{4}} - 2 \, x\right )} \sqrt {\frac {2 \, x^{4} + 2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{4} - 1} x^{2} + 2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 1}{2 \, x^{4} - 1}}}{2 \, {\left (x^{5} - x\right )}}\right ) + 36 \, \sqrt {2} x^{9} \arctan \left (-\frac {\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {5}{4}} - {\left (2 \, x^{5} + \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x^{2} + \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {5}{4}} - 2 \, x\right )} \sqrt {\frac {2 \, x^{4} - 2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{4} - 1} x^{2} - 2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 1}{2 \, x^{4} - 1}}}{2 \, {\left (x^{5} - x\right )}}\right ) + 9 \, \sqrt {2} x^{9} \log \left (\frac {2 \, x^{4} + 2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{4} - 1} x^{2} + 2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 1}{2 \, x^{4} - 1}\right ) - 9 \, \sqrt {2} x^{9} \log \left (\frac {2 \, x^{4} - 2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{4} - 1} x^{2} - 2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x - 1}{2 \, x^{4} - 1}\right ) - 2 \, {\left (65 \, x^{8} + 5 \, x^{4} + 2\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{18 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(1/4)*(2*x^8-x^4+2)/x^10/(2*x^4-1),x, algorithm="fricas")

[Out]

-1/18*(36*sqrt(2)*x^9*arctan(-1/2*(sqrt(2)*(x^4 - 1)^(3/4)*x^2 - sqrt(2)*(x^4 - 1)^(5/4) + (2*x^5 - sqrt(2)*(x
^4 - 1)^(3/4)*x^2 - sqrt(2)*(x^4 - 1)^(5/4) - 2*x)*sqrt((2*x^4 + 2*sqrt(2)*(x^4 - 1)^(1/4)*x^3 + 4*sqrt(x^4 -
1)*x^2 + 2*sqrt(2)*(x^4 - 1)^(3/4)*x - 1)/(2*x^4 - 1)))/(x^5 - x)) + 36*sqrt(2)*x^9*arctan(-1/2*(sqrt(2)*(x^4
- 1)^(3/4)*x^2 - sqrt(2)*(x^4 - 1)^(5/4) - (2*x^5 + sqrt(2)*(x^4 - 1)^(3/4)*x^2 + sqrt(2)*(x^4 - 1)^(5/4) - 2*
x)*sqrt((2*x^4 - 2*sqrt(2)*(x^4 - 1)^(1/4)*x^3 + 4*sqrt(x^4 - 1)*x^2 - 2*sqrt(2)*(x^4 - 1)^(3/4)*x - 1)/(2*x^4
- 1)))/(x^5 - x)) + 9*sqrt(2)*x^9*log((2*x^4 + 2*sqrt(2)*(x^4 - 1)^(1/4)*x^3 + 4*sqrt(x^4 - 1)*x^2 + 2*sqrt(2
)*(x^4 - 1)^(3/4)*x - 1)/(2*x^4 - 1)) - 9*sqrt(2)*x^9*log((2*x^4 - 2*sqrt(2)*(x^4 - 1)^(1/4)*x^3 + 4*sqrt(x^4
- 1)*x^2 - 2*sqrt(2)*(x^4 - 1)^(3/4)*x - 1)/(2*x^4 - 1)) - 2*(65*x^8 + 5*x^4 + 2)*(x^4 - 1)^(1/4))/x^9

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giac [A]  time = 0.34, size = 172, normalized size = 1.58 \begin {gather*} 2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right )}\right ) + \sqrt {2} \log \left (\frac {\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {\sqrt {x^{4} - 1}}{x^{2}} + 1\right ) - \sqrt {2} \log \left (-\frac {\sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {\sqrt {x^{4} - 1}}{x^{2}} + 1\right ) - \frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}} {\left (\frac {1}{x^{4}} - 1\right )}}{x} - \frac {8 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - \frac {2 \, {\left (x^{8} - 2 \, x^{4} + 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{9 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(1/4)*(2*x^8-x^4+2)/x^10/(2*x^4-1),x, algorithm="giac")

[Out]

2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(x^4 - 1)^(1/4)/x)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(x^
4 - 1)^(1/4)/x)) + sqrt(2)*log(sqrt(2)*(x^4 - 1)^(1/4)/x + sqrt(x^4 - 1)/x^2 + 1) - sqrt(2)*log(-sqrt(2)*(x^4
- 1)^(1/4)/x + sqrt(x^4 - 1)/x^2 + 1) - (x^4 - 1)^(1/4)*(1/x^4 - 1)/x - 8*(x^4 - 1)^(1/4)/x - 2/9*(x^8 - 2*x^4
+ 1)*(x^4 - 1)^(1/4)/x^9

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maple [C]  time = 8.95, size = 182, normalized size = 1.67

method	result	size
trager	\(\frac {\left (x^{4}-1\right )^{\frac {1}{4}} \left (65 x^{8}+5 x^{4}+2\right )}{9 x^{9}}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {-2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{3}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}+1\right )}{2 x^{4}-1}\right )-2 \RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{3}+2 \sqrt {x^{4}-1}\, \RootOf \left (\textit {\_Z}^{4}+1\right ) x^{2}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{4}+1\right )^{3}}{2 x^{4}-1}\right )\)	\(182\)
risch	\(\frac {65 x^{12}-60 x^{8}-3 x^{4}-2}{9 x^{9} \left (x^{4}-1\right )^{\frac {3}{4}}}+\frac {\left (2 \RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \left (x^{12}-3 x^{8}+3 x^{4}-1\right )^{\frac {1}{4}} x^{9}+\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{8}-4 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \left (x^{12}-3 x^{8}+3 x^{4}-1\right )^{\frac {1}{4}} x^{5}+2 \sqrt {x^{12}-3 x^{8}+3 x^{4}-1}\, x^{6}-2 \RootOf \left (\textit {\_Z}^{4}+1\right ) \left (x^{12}-3 x^{8}+3 x^{4}-1\right )^{\frac {3}{4}} x^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{4}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \left (x^{12}-3 x^{8}+3 x^{4}-1\right )^{\frac {1}{4}} x -2 \sqrt {x^{12}-3 x^{8}+3 x^{4}-1}\, x^{2}+\RootOf \left (\textit {\_Z}^{4}+1\right )^{2}}{\left (-1+x \right )^{2} \left (1+x \right )^{2} \left (x^{2}+1\right )^{2} \left (2 x^{4}-1\right )}\right )-2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{4}+1\right ) \left (x^{12}-3 x^{8}+3 x^{4}-1\right )^{\frac {1}{4}} x^{9}-\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{8}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \left (x^{12}-3 x^{8}+3 x^{4}-1\right )^{\frac {3}{4}} x^{3}+2 \sqrt {x^{12}-3 x^{8}+3 x^{4}-1}\, x^{6}+4 \RootOf \left (\textit {\_Z}^{4}+1\right ) \left (x^{12}-3 x^{8}+3 x^{4}-1\right )^{\frac {1}{4}} x^{5}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{4}-2 \sqrt {x^{12}-3 x^{8}+3 x^{4}-1}\, x^{2}-2 \RootOf \left (\textit {\_Z}^{4}+1\right ) \left (x^{12}-3 x^{8}+3 x^{4}-1\right )^{\frac {1}{4}} x -\RootOf \left (\textit {\_Z}^{4}+1\right )^{2}}{\left (-1+x \right )^{2} \left (1+x \right )^{2} \left (x^{2}+1\right )^{2} \left (2 x^{4}-1\right )}\right )\right ) \left (\left (x^{4}-1\right )^{3}\right )^{\frac {1}{4}}}{\left (x^{4}-1\right )^{\frac {3}{4}}}\)	\(512\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)^(1/4)*(2*x^8-x^4+2)/x^10/(2*x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/9*(x^4-1)^(1/4)*(65*x^8+5*x^4+2)/x^9+2*RootOf(_Z^4+1)^3*ln((-2*(x^4-1)^(1/2)*RootOf(_Z^4+1)^3*x^2-2*(x^4-1)^
(1/4)*RootOf(_Z^4+1)^2*x^3+2*(x^4-1)^(3/4)*x-RootOf(_Z^4+1))/(2*x^4-1))-2*RootOf(_Z^4+1)*ln((2*(x^4-1)^(1/4)*R
ootOf(_Z^4+1)^2*x^3+2*(x^4-1)^(1/2)*RootOf(_Z^4+1)*x^2+2*(x^4-1)^(3/4)*x+RootOf(_Z^4+1)^3)/(2*x^4-1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{8} - x^{4} + 2\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{{\left (2 \, x^{4} - 1\right )} x^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(1/4)*(2*x^8-x^4+2)/x^10/(2*x^4-1),x, algorithm="maxima")

[Out]

integrate((2*x^8 - x^4 + 2)*(x^4 - 1)^(1/4)/((2*x^4 - 1)*x^10), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^4-1\right )}^{1/4}\,\left (2\,x^8-x^4+2\right )}{x^{10}\,\left (2\,x^4-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 - 1)^(1/4)*(2*x^8 - x^4 + 2))/(x^10*(2*x^4 - 1)),x)

[Out]

int(((x^4 - 1)^(1/4)*(2*x^8 - x^4 + 2))/(x^10*(2*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]{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (2 x^{8} - x^{4} + 2\right )}{x^{10} \left (2 x^{4} - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-1)**(1/4)*(2*x**8-x**4+2)/x**10/(2*x**4-1),x)

[Out]

Integral(((x - 1)*(x + 1)*(x**2 + 1))**(1/4)*(2*x**8 - x**4 + 2)/(x**10*(2*x**4 - 1)), x)

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