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

Optimal. Leaf size=141 \[ \frac {9}{8} \sqrt [4]{\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{2 x^4+1}}\right )-\sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{2 x^4+1}}\right )-\frac {9}{8} \sqrt [4]{\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{2 x^4+1}}\right )+\sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{2 x^4+1}}\right )+\frac {\sqrt [4]{2 x^4+1} \left (9 x^4+2\right )}{20 x^5} \]

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Rubi [C]  time = 0.37, antiderivative size = 123, normalized size of antiderivative = 0.87, number of steps used = 9, number of rules used = 8, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {6725, 264, 277, 331, 298, 203, 206, 510} \begin {gather*} \frac {3}{8} x^3 F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {x^4}{2},-2 x^4\right )+\frac {\sqrt [4]{2 x^4+1}}{4 x}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{2 x^4+1}}\right )}{4\ 2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{2 x^4+1}}\right )}{4\ 2^{3/4}}+\frac {\left (2 x^4+1\right )^{5/4}}{10 x^5} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(1 + 2*x^4)^(1/4)/(4*x) + (1 + 2*x^4)^(5/4)/(10*x^5) + (3*x^3*AppellF1[3/4, 1, -1/4, 7/4, -1/2*x^4, -2*x^4])/8
+ ArcTan[(2^(1/4)*x)/(1 + 2*x^4)^(1/4)]/(4*2^(3/4)) - ArcTanh[(2^(1/4)*x)/(1 + 2*x^4)^(1/4)]/(4*2^(3/4))

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

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Mathematica [C]  time = 0.14, size = 99, normalized size = 0.70 \begin {gather*} \frac {2}{7} x^7 F_1\left (\frac {7}{4};\frac {3}{4},1;\frac {11}{4};-2 x^4,-\frac {x^4}{2}\right )+\frac {5 x^3 \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-\frac {3 x^4}{x^4+2}\right )}{12 \sqrt [4]{2} \left (x^4+2\right )^{3/4}}+\frac {\sqrt [4]{2 x^4+1} \left (9 x^4+2\right )}{20 x^5} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 + 2*x^4)^(1/4)*(2 + 9*x^4))/(20*x^5) + (2*x^7*AppellF1[7/4, 3/4, 1, 11/4, -2*x^4, -1/2*x^4])/7 + (5*x^3*Hy
pergeometric2F1[3/4, 3/4, 7/4, (-3*x^4)/(2 + x^4)])/(12*2^(1/4)*(2 + x^4)^(3/4))

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IntegrateAlgebraic [A]  time = 0.44, size = 141, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{1+2 x^4} \left (2+9 x^4\right )}{20 x^5}+\frac {9}{8} \sqrt [4]{\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+2 x^4}}\right )-\sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+2 x^4}}\right )-\frac {9}{8} \sqrt [4]{\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+2 x^4}}\right )+\sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+2 x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 2*x^4)^(1/4)*(2 + 9*x^4))/(20*x^5) + (9*(3/2)^(1/4)*ArcTan[((3/2)^(1/4)*x)/(1 + 2*x^4)^(1/4)])/8 - 2^(1/
4)*ArcTan[(2^(1/4)*x)/(1 + 2*x^4)^(1/4)] - (9*(3/2)^(1/4)*ArcTanh[((3/2)^(1/4)*x)/(1 + 2*x^4)^(1/4)])/8 + 2^(1
/4)*ArcTanh[(2^(1/4)*x)/(1 + 2*x^4)^(1/4)]

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fricas [B]  time = 34.57, size = 519, normalized size = 3.68 \begin {gather*} -\frac {180 \cdot 3^{\frac {1}{4}} 2^{\frac {3}{4}} x^{5} \arctan \left (-\frac {12 \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 12 \cdot 3^{\frac {1}{4}} 2^{\frac {3}{4}} {\left (2 \, x^{4} + 1\right )}^{\frac {3}{4}} x - \sqrt {3} {\left (4 \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} \sqrt {2 \, x^{4} + 1} x^{2} + 3^{\frac {1}{4}} 2^{\frac {3}{4}} {\left (7 \, x^{4} + 2\right )}\right )} \sqrt {\sqrt {3} \sqrt {2}}}{6 \, {\left (x^{4} + 2\right )}}\right ) + 45 \cdot 3^{\frac {1}{4}} 2^{\frac {3}{4}} x^{5} \log \left (\frac {6 \, \sqrt {3} \sqrt {2} {\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 6 \cdot 3^{\frac {1}{4}} 2^{\frac {3}{4}} \sqrt {2 \, x^{4} + 1} x^{2} + 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (7 \, x^{4} + 2\right )} + 12 \, {\left (2 \, x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - 45 \cdot 3^{\frac {1}{4}} 2^{\frac {3}{4}} x^{5} \log \left (\frac {6 \, \sqrt {3} \sqrt {2} {\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 6 \cdot 3^{\frac {1}{4}} 2^{\frac {3}{4}} \sqrt {2 \, x^{4} + 1} x^{2} - 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (7 \, x^{4} + 2\right )} + 12 \, {\left (2 \, x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - 320 \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 ) - 80 \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 ) + 80 \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 ) - 16 \, {\left (9 \, x^{4} + 2\right )} {\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}}}{320 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/320*(180*3^(1/4)*2^(3/4)*x^5*arctan(-1/6*(12*3^(3/4)*2^(1/4)*(2*x^4 + 1)^(1/4)*x^3 + 12*3^(1/4)*2^(3/4)*(2*
x^4 + 1)^(3/4)*x - sqrt(3)*(4*3^(3/4)*2^(1/4)*sqrt(2*x^4 + 1)*x^2 + 3^(1/4)*2^(3/4)*(7*x^4 + 2))*sqrt(sqrt(3)*
sqrt(2)))/(x^4 + 2)) + 45*3^(1/4)*2^(3/4)*x^5*log((6*sqrt(3)*sqrt(2)*(2*x^4 + 1)^(1/4)*x^3 + 6*3^(1/4)*2^(3/4)
*sqrt(2*x^4 + 1)*x^2 + 3^(3/4)*2^(1/4)*(7*x^4 + 2) + 12*(2*x^4 + 1)^(3/4)*x)/(x^4 + 2)) - 45*3^(1/4)*2^(3/4)*x
^5*log((6*sqrt(3)*sqrt(2)*(2*x^4 + 1)^(1/4)*x^3 - 6*3^(1/4)*2^(3/4)*sqrt(2*x^4 + 1)*x^2 - 3^(3/4)*2^(1/4)*(7*x
^4 + 2) + 12*(2*x^4 + 1)^(3/4)*x)/(x^4 + 2)) - 320*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))) - 80*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) + 80*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) - 16*(9*x^4 + 2)*(2*x^4 + 1)^(1/4))/x^5

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giac [A]  time = 0.20, size = 177, normalized size = 1.26 \begin {gather*} -\frac {1}{16} \cdot 54^{\frac {3}{4}} \arctan \left (\frac {24^{\frac {3}{4}} {\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}}}{12 \, x}\right ) - \frac {1}{32} \cdot 54^{\frac {3}{4}} \log \left (\frac {1}{2} \cdot 24^{\frac {1}{4}} + \frac {{\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{32} \cdot 54^{\frac {3}{4}} \log \left (-\frac {1}{2} \cdot 24^{\frac {1}{4}} + \frac {{\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 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}{2} \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}{2} \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 )}}{10 \, x} + \frac {{\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*54^(3/4)*arctan(1/12*24^(3/4)*(2*x^4 + 1)^(1/4)/x) - 1/32*54^(3/4)*log(1/2*24^(1/4) + (2*x^4 + 1)^(1/4)/
x) + 1/32*54^(3/4)*log(-1/2*24^(1/4) + (2*x^4 + 1)^(1/4)/x) + 2^(1/4)*arctan(1/2*2^(3/4)*(2*x^4 + 1)^(1/4)/x)
+ 1/2*2^(1/4)*log(2^(1/4) + (2*x^4 + 1)^(1/4)/x) - 1/2*2^(1/4)*log(-2^(1/4) + (2*x^4 + 1)^(1/4)/x) + 1/10*(2*x
^4 + 1)^(1/4)*(1/x^4 + 2)/x + 1/4*(2*x^4 + 1)^(1/4)/x

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (2 x^{4}+1\right )^{\frac {1}{4}} \left (2 x^{8}-x^{4}-1\right )}{x^{6} \left (x^{4}+2\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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} - 1\right )} {\left (2 \, x^{4} + 1\right )}^{\frac {1}{4}}}{{\left (x^{4} + 2\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)*(2*x^8-x^4-1)/x^6/(x^4+2),x, algorithm="maxima")

[Out]

integrate((2*x^8 - x^4 - 1)*(2*x^4 + 1)^(1/4)/((x^4 + 2)*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 (-2\,x^8+x^4+1\right )}{x^6\,\left (x^4+2\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 - 2*x^8 + 1))/(x^6*(x^4 + 2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (2 x^{4} + 1\right )^{\frac {5}{4}}}{x^{6} \left (x^{4} + 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)*(2*x**4 + 1)**(5/4)/(x**6*(x**4 + 2)), x)

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