\(\int \frac {x^2}{(-1+x^4) \sqrt [4]{x^2+x^4}} \, dx\) [1049]

Optimal result

Integrand size = 22, antiderivative size = 79 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {\left (x^2+x^4\right )^{3/4}}{x \left (1+x^2\right )}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}} \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.57, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2081, 1268, 477, 524} \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=-\frac {2 x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {9}{4},\frac {2 x^2}{x^2+1}\right )}{5 \left (x^2+1\right ) \sqrt [4]{x^4+x^2}} \]

[In]

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

[Out]

(-2*x^3*Hypergeometric2F1[1, 5/4, 9/4, (2*x^2)/(1 + x^2)])/(5*(1 + x^2)*(x^2 + x^4)^(1/4))

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 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*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {x^{3/2}}{\sqrt [4]{1+x^2} \left (-1+x^4\right )} \, dx}{\sqrt [4]{x^2+x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {x^{3/2}}{\left (-1+x^2\right ) \left (1+x^2\right )^{5/4}} \, dx}{\sqrt [4]{x^2+x^4}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^2}\right ) \text {Subst}\left (\int \frac {x^4}{\left (-1+x^4\right ) \left (1+x^4\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}} \\ & = -\frac {2 x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {9}{4},\frac {2 x^2}{1+x^2}\right )}{5 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.29 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {\sqrt {x} \left (4 \sqrt {x}-2^{3/4} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )-2^{3/4} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )\right )}{4 \sqrt [4]{x^2+x^4}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 8.90 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.43

method	result	size
pseudoelliptic	\(\frac {2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+8 x}{8 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(113\)
risch	\(\frac {x}{\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\sqrt {x^{4}+x^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x^{3}-4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x}{\left (1+x \right ) x \left (x -1\right )}\right )}{8}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}+x^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}-4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x}{\left (1+x \right ) x \left (x -1\right )}\right )}{8}\)	\(242\)
trager	\(\frac {\left (x^{4}+x^{2}\right )^{\frac {3}{4}}}{x \left (x^{2}+1\right )}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\sqrt {x^{4}+x^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}+4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x}{\left (1+x \right ) x \left (x -1\right )}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\sqrt {x^{4}+x^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x^{3}-4 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x}{\left (1+x \right ) x \left (x -1\right )}\right )}{8}\)	\(248\)

[In]

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

[Out]

1/8*(2*arctan(1/2*2^(3/4)/x*(x^2*(x^2+1))^(1/4))*2^(3/4)*(x^2*(x^2+1))^(1/4)-ln((-2^(1/4)*x-(x^2*(x^2+1))^(1/4
))/(2^(1/4)*x-(x^2*(x^2+1))^(1/4)))*2^(3/4)*(x^2*(x^2+1))^(1/4)+8*x)/(x^2*(x^2+1))^(1/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.98 (sec) , antiderivative size = 341, normalized size of antiderivative = 4.32 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=-\frac {2^{\frac {3}{4}} {\left (x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 2^{\frac {3}{4}} {\left (x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 2^{\frac {3}{4}} {\left (-i \, x^{3} - i \, x\right )} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 i \, x^{3} + i \, x\right )} + 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 2^{\frac {3}{4}} {\left (i \, x^{3} + i \, x\right )} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (-3 i \, x^{3} - i \, x\right )} - 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 16 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{16 \, {\left (x^{3} + x\right )}} \]

[In]

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

[Out]

-1/16*(2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 + 2^(3/4)*(3*x^3 + x) + 4*2^(1/4)*sqrt(x^4 + x^2
)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) - 2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(3*x^
3 + x) - 4*2^(1/4)*sqrt(x^4 + x^2)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) - 2^(3/4)*(-I*x^3 - I*x)*log(-(4*sqrt(2
)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(3*I*x^3 + I*x) + 4*I*2^(1/4)*sqrt(x^4 + x^2)*x - 4*(x^4 + x^2)^(3/4))/(x^3
- x)) - 2^(3/4)*(I*x^3 + I*x)*log(-(4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(-3*I*x^3 - I*x) - 4*I*2^(1/4)*s
qrt(x^4 + x^2)*x - 4*(x^4 + x^2)^(3/4))/(x^3 - x)) - 16*(x^4 + x^2)^(3/4))/(x^3 + x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\int { \frac {x^{2}}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}} \,d x } \]

[In]

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

[Out]

-2/3*(x^3 + x)*x^(3/2)/((x^4 - 1)*(x^2 + 1)^(1/4)) - integrate(8/3*(x^2 + 1)^(3/4)*x^(3/2)/(x^8 - 2*x^4 + 1),
x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.80 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {1}{4} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{8} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{8} \cdot 2^{\frac {3}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} \right |}\right ) + \frac {1}{{\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

1/4*2^(3/4)*arctan(1/2*2^(3/4)*(1/x^2 + 1)^(1/4)) - 1/8*2^(3/4)*log(2^(1/4) + (1/x^2 + 1)^(1/4)) + 1/8*2^(3/4)
*log(abs(-2^(1/4) + (1/x^2 + 1)^(1/4))) + 1/(1/x^2 + 1)^(1/4)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\int \frac {x^2}{{\left (x^4+x^2\right )}^{1/4}\,\left (x^4-1\right )} \,d x \]

[In]

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

[Out]

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