3.10.87 \(\int \frac {\sqrt [4]{x^2+x^4}}{x^4 (-1+x^4)} \, dx\) [987]

Optimal. Leaf size=75 \[ \frac {2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}}{5 x^3}+\frac {\text {ArcTan}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2^{3/4}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 129, normalized size of antiderivative = 1.72, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2081, 1268, 477, 508, 472, 304, 209, 212} \begin {gather*} \frac {\sqrt [4]{x^4+x^2} \text {ArcTan}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{x^2+1}}-\frac {\sqrt [4]{x^4+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2^{3/4} \sqrt {x} \sqrt [4]{x^2+1}}+\frac {2 \sqrt [4]{x^4+x^2} \left (x^2+1\right )}{5 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 212

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

Rule 304

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 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 508

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[k*(a^(p + (m + 1)/n)/n), Subst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p +
q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 1268

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(f*x)^m*(d +
e*x^2)^(q + p)*(a/d + (c/e)*x^2)^p, x] /; FreeQ[{a, c, d, e, f, q, m, q}, x] && EqQ[c*d^2 + a*e^2, 0] && Integ
erQ[p]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

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

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Mathematica [A]
time = 0.23, size = 105, normalized size = 1.40 \begin {gather*} \frac {\sqrt [4]{x^2+x^4} \left (4 \left (1+x^2\right )^{5/4}+5 \sqrt [4]{2} x^{5/2} \text {ArcTan}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )-5 \sqrt [4]{2} x^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )\right )}{10 x^3 \sqrt [4]{1+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2 + x^4)^(1/4)*(4*(1 + x^2)^(5/4) + 5*2^(1/4)*x^(5/2)*ArcTan[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)] - 5*2^(1/4
)*x^(5/2)*ArcTanh[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)]))/(10*x^3*(1 + x^2)^(1/4))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 6.92, size = 258, normalized size = 3.44

method	result	size
trager	\(\frac {2 \left (x^{2}+1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}+\frac {\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{3}+4 \left (x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \sqrt {x^{4}+x^{2}}\, \RootOf \left (\textit {\_Z}^{4}-2\right ) x -\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x +4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}}{\left (1+x \right ) \left (-1+x \right ) x}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}-4 \left (x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \sqrt {x^{4}+x^{2}}\, x +\RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x +4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}}{\left (1+x \right ) \left (-1+x \right ) x}\right )}{4}\)	\(258\)
risch	\(\frac {2 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} \left (x^{4}+2 x^{2}+1\right )}{5 x^{3} \left (x^{2}+1\right )}+\frac {\left (\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {-2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{4}+3 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{6}-4 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}+7 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{4}+4 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )-4 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}\, x^{2}-2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}+5 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}}{\left (x^{2}+1\right )^{2} \left (1+x \right ) \left (-1+x \right )}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{4}-3 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{6}+4 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{2}-7 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{4}+4 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )-4 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}\, x^{2}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{3}-5 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}-\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}}{\left (x^{2}+1\right )^{2} \left (1+x \right ) \left (-1+x \right )}\right )}{4}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} \left (x^{2} \left (x^{2}+1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x^{2}+1\right )}\)	\(606\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x^2+1)*(x^4+x^2)^(1/4)/x^3+1/4*RootOf(_Z^4-2)*ln((-3*RootOf(_Z^4-2)^3*x^3+4*(x^4+x^2)^(1/4)*RootOf(_Z^4-2
)^2*x^2-4*(x^4+x^2)^(1/2)*RootOf(_Z^4-2)*x-RootOf(_Z^4-2)^3*x+4*(x^4+x^2)^(3/4))/(1+x)/(-1+x)/x)+1/4*RootOf(_Z
^2+RootOf(_Z^4-2)^2)*ln((3*RootOf(_Z^2+RootOf(_Z^4-2)^2)*RootOf(_Z^4-2)^2*x^3-4*(x^4+x^2)^(1/4)*RootOf(_Z^4-2)
^2*x^2-4*RootOf(_Z^2+RootOf(_Z^4-2)^2)*(x^4+x^2)^(1/2)*x+RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x+4*(x
^4+x^2)^(3/4))/(1+x)/(-1+x)/x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/585*(32*x^7 - 8*x^5 + 5*x^3 + 45*x)*(x^2 + 1)^(1/4)/(x^(15/2) - x^(7/2)) - integrate(8/585*(32*x^6 - 8*x^4
+ 5*x^2 + 45)*(x^2 + 1)^(1/4)/(x^(23/2) - 2*x^(15/2) + x^(7/2)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (60) = 120\).
time = 1.52, size = 253, normalized size = 3.37 \begin {gather*} -\frac {20 \cdot 8^{\frac {3}{4}} x^{3} \arctan \left (\frac {16 \cdot 8^{\frac {1}{4}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (8^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} + 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x\right )} + 4 \cdot 8^{\frac {3}{4}} {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{8 \, {\left (x^{3} - x\right )}}\right ) + 5 \cdot 8^{\frac {3}{4}} x^{3} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 8^{\frac {3}{4}} \sqrt {x^{4} + x^{2}} x + 8^{\frac {1}{4}} {\left (3 \, x^{3} + x\right )} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 5 \cdot 8^{\frac {3}{4}} x^{3} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 8^{\frac {3}{4}} \sqrt {x^{4} + x^{2}} x - 8^{\frac {1}{4}} {\left (3 \, x^{3} + x\right )} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 64 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}}{160 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/160*(20*8^(3/4)*x^3*arctan(1/8*(16*8^(1/4)*(x^4 + x^2)^(1/4)*x^2 + 2^(3/4)*(8^(3/4)*(3*x^3 + x) + 8*8^(1/4)
*sqrt(x^4 + x^2)*x) + 4*8^(3/4)*(x^4 + x^2)^(3/4))/(x^3 - x)) + 5*8^(3/4)*x^3*log((4*sqrt(2)*(x^4 + x^2)^(1/4)
*x^2 + 8^(3/4)*sqrt(x^4 + x^2)*x + 8^(1/4)*(3*x^3 + x) + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) - 5*8^(3/4)*x^3*log((
4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 8^(3/4)*sqrt(x^4 + x^2)*x - 8^(1/4)*(3*x^3 + x) + 4*(x^4 + x^2)^(3/4))/(x^3
- x)) - 64*(x^4 + x^2)^(1/4)*(x^2 + 1))/x^3

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

[Out]

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

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Giac [A]
time = 0.42, size = 65, normalized size = 0.87 \begin {gather*} \frac {2}{5} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} - \frac {1}{2} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \cdot 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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