\(\int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} (-2-x^4+x^8)} \, dx\) [1519]

Optimal result

Integrand size = 34, antiderivative size = 105 \[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=-\frac {5 x}{3 \sqrt [4]{1+x^4}}+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \arctan \left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \]

[Out]

-5/3*x/(x^4+1)^(1/4)+arctan(x/(x^4+1)^(1/4))-5/36*arctan(1/2*3^(1/4)*2^(3/4)*x/(x^4+1)^(1/4))*3^(3/4)*2^(1/4)+
arctanh(x/(x^4+1)^(1/4))-5/36*arctanh(1/2*3^(1/4)*2^(3/4)*x/(x^4+1)^(1/4))*3^(3/4)*2^(1/4)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {6860, 246, 218, 212, 209, 1417, 390, 385} \[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=\arctan \left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {5 \arctan \left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}-\frac {5 x}{3 \sqrt [4]{x^4+1}} \]

[In]

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

[Out]

(-5*x)/(3*(1 + x^4)^(1/4)) + ArcTan[x/(1 + x^4)^(1/4)] - (5*ArcTan[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4
)*3^(1/4)) + ArcTanh[x/(1 + x^4)^(1/4)] - (5*ArcTanh[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(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 218

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 1417

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*
x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{\sqrt [4]{1+x^4}}+\frac {5}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )}\right ) \, dx \\ & = 2 \int \frac {1}{\sqrt [4]{1+x^4}} \, dx+5 \int \frac {1}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx \\ & = 2 \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+5 \int \frac {1}{\left (-2+x^4\right ) \left (1+x^4\right )^{5/4}} \, dx \\ & = -\frac {5 x}{3 \sqrt [4]{1+x^4}}+\frac {5}{3} \int \frac {1}{\left (-2+x^4\right ) \sqrt [4]{1+x^4}} \, dx+\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = -\frac {5 x}{3 \sqrt [4]{1+x^4}}+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {5}{3} \text {Subst}\left (\int \frac {1}{-2+3 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = -\frac {5 x}{3 \sqrt [4]{1+x^4}}+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2}-\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}-\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2}+\sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}} \\ & = -\frac {5 x}{3 \sqrt [4]{1+x^4}}+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \arctan \left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=-\frac {5 x}{3 \sqrt [4]{1+x^4}}+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \arctan \left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \]

[In]

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

[Out]

(-5*x)/(3*(1 + x^4)^(1/4)) + ArcTan[x/(1 + x^4)^(1/4)] - (5*ArcTan[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4
)*3^(1/4)) + ArcTanh[x/(1 + x^4)^(1/4)] - (5*ArcTanh[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(172\) vs. \(2(81)=162\).

Time = 1.38 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.65

method	result	size
pseudoelliptic	\(\frac {10 \arctan \left (\frac {3^{\frac {3}{4}} 2^{\frac {1}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}}{3 x}\right ) 2^{\frac {1}{4}} 3^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}-5 \ln \left (\frac {-2^{\frac {3}{4}} 3^{\frac {1}{4}} x -2 \left (x^{4}+1\right )^{\frac {1}{4}}}{2^{\frac {3}{4}} 3^{\frac {1}{4}} x -2 \left (x^{4}+1\right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} 3^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}+36 \ln \left (\frac {x +\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right ) \left (x^{4}+1\right )^{\frac {1}{4}}-36 \ln \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}-x}{x}\right ) \left (x^{4}+1\right )^{\frac {1}{4}}-72 \arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right ) \left (x^{4}+1\right )^{\frac {1}{4}}-120 x}{72 \left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(173\)

[In]

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

[Out]

1/72*(10*arctan(1/3*3^(3/4)*2^(1/4)/x*(x^4+1)^(1/4))*2^(1/4)*3^(3/4)*(x^4+1)^(1/4)-5*ln((-2^(3/4)*3^(1/4)*x-2*
(x^4+1)^(1/4))/(2^(3/4)*3^(1/4)*x-2*(x^4+1)^(1/4)))*2^(1/4)*3^(3/4)*(x^4+1)^(1/4)+36*ln((x+(x^4+1)^(1/4))/x)*(
x^4+1)^(1/4)-36*ln(((x^4+1)^(1/4)-x)/x)*(x^4+1)^(1/4)-72*arctan((x^4+1)^(1/4)/x)*(x^4+1)^(1/4)-120*x)/(x^4+1)^
(1/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.01 \[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=-\frac {5 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {24^{\frac {1}{4}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 5 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (-\frac {24^{\frac {1}{4}} x - 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 5 \cdot 24^{\frac {3}{4}} {\left (-i \, x^{4} - i\right )} \log \left (\frac {i \cdot 24^{\frac {1}{4}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 5 \cdot 24^{\frac {3}{4}} {\left (i \, x^{4} + i\right )} \log \left (\frac {-i \cdot 24^{\frac {1}{4}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 288 \, {\left (x^{4} + 1\right )} \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 144 \, {\left (x^{4} + 1\right )} \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 144 \, {\left (x^{4} + 1\right )} \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 480 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{288 \, {\left (x^{4} + 1\right )}} \]

[In]

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

[Out]

-1/288*(5*24^(3/4)*(x^4 + 1)*log((24^(1/4)*x + 2*(x^4 + 1)^(1/4))/x) - 5*24^(3/4)*(x^4 + 1)*log(-(24^(1/4)*x -
2*(x^4 + 1)^(1/4))/x) + 5*24^(3/4)*(-I*x^4 - I)*log((I*24^(1/4)*x + 2*(x^4 + 1)^(1/4))/x) + 5*24^(3/4)*(I*x^4
+ I)*log((-I*24^(1/4)*x + 2*(x^4 + 1)^(1/4))/x) + 288*(x^4 + 1)*arctan((x^4 + 1)^(1/4)/x) - 144*(x^4 + 1)*log
((x + (x^4 + 1)^(1/4))/x) + 144*(x^4 + 1)*log(-(x - (x^4 + 1)^(1/4))/x) + 480*(x^4 + 1)^(3/4)*x)/(x^4 + 1)

Sympy [F]

\[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=\int \frac {2 x^{8} - 2 x^{4} + 1}{\left (x^{4} - 2\right ) \left (x^{4} + 1\right )^{\frac {5}{4}}}\, dx \]

[In]

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

[Out]

Integral((2*x**8 - 2*x**4 + 1)/((x**4 - 2)*(x**4 + 1)**(5/4)), x)

Maxima [F]

\[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{8} - 2 \, x^{4} + 1}{{\left (x^{8} - x^{4} - 2\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{8} - 2 \, x^{4} + 1}{{\left (x^{8} - x^{4} - 2\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=\int -\frac {2\,x^8-2\,x^4+1}{{\left (x^4+1\right )}^{1/4}\,\left (-x^8+x^4+2\right )} \,d x \]

[In]

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

[Out]

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