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

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

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Rubi [A]  time = 0.24, 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 = {2056, 1254, 466, 494, 461, 298, 203, 206} \begin {gather*} \frac {\sqrt [4]{x^4+x^2} \tan ^{-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 {\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 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 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 461

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 466

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 494

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 1254

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*x^2)/e)^p, x] /; FreeQ[{a, c, d, e, f, q, m, q}, x] && EqQ[c*d^2 + a*e^2, 0] && Integ
erQ[p]

Rule 2056

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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \operatorname {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} \operatorname {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 [F]  time = 0.56, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x^2+x^4}}{x^4 \left (-1+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.33, size = 75, normalized size = 1.00 \begin {gather*} \frac {2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}}{5 x^3}+\frac {\tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [B]  time = 2.02, 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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giac [A]  time = 0.34, 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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maple [C]  time = 8.53, size = 259, normalized size = 3.45

method	result	size
trager	\(\frac {2 \left (x^{2}+1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{3}-4 \left (x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}+4 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \sqrt {x^{4}+x^{2}}\, x -\RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) x +4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}}{x \left (-1+x \right ) \left (1+x \right )}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x^{3}+4 \left (x^{4}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-4 \sqrt {x^{4}+x^{2}}\, \RootOf \left (\textit {\_Z}^{4}-2\right ) x -\RootOf \left (\textit {\_Z}^{4}-2\right )^{3} x +4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}}{x \left (-1+x \right ) \left (1+x \right )}\right )}{4}\)	\(259\)
risch	\(\frac {2 \left (x^{4}+2 x^{2}+1\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{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 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{4}+3 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{6}+4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}+7 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{4}+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 \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}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}-4 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{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 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{4}-3 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{6}+4 \RootOf \left (\textit {\_Z}^{4}-2\right )^{3} \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}-7 \RootOf \left (\textit {\_Z}^{4}-2\right )^{2} x^{4}+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 \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}-\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}-4 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{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^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))/x/(-1+x)/(1+x))+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))/x/(-1+x)/(1+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {2 \, {\left (32 \, x^{7} - 8 \, x^{5} + 5 \, x^{3} + 45 \, x\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{585 \, {\left (x^{\frac {15}{2}} - x^{\frac {7}{2}}\right )}} - \int \frac {8 \, {\left (32 \, x^{6} - 8 \, x^{4} + 5 \, x^{2} + 45\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{585 \, {\left (x^{\frac {23}{2}} - 2 \, x^{\frac {15}{2}} + x^{\frac {7}{2}}\right )}}\,{d x} \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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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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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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