\(\int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} (-1+x^8)} \, dx\) [2364]

Optimal result

Integrand size = 29, antiderivative size = 187 \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=-\frac {3 \left (x^2+x^6\right )^{3/4}}{2 x \left (1+x^4\right )}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{8 \sqrt [4]{2}}+\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^2+x^6}}{\sqrt {2} x^2-\sqrt {x^2+x^6}}\right )}{8\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{8 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^2+x^6}}{2^{3/4}}}{x \sqrt [4]{x^2+x^6}}\right )}{8\ 2^{3/4}} \]

[Out]

-3/2*(x^6+x^2)^(3/4)/x/(x^4+1)-1/16*arctan(2^(1/4)*x/(x^6+x^2)^(1/4))*2^(3/4)+1/16*arctan(2^(3/4)*x*(x^6+x^2)^
(1/4)/(2^(1/2)*x^2-(x^6+x^2)^(1/2)))*2^(1/4)-1/16*arctanh(2^(1/4)*x/(x^6+x^2)^(1/4))*2^(3/4)-1/16*arctanh((1/2
*x^2*2^(3/4)+1/2*(x^6+x^2)^(1/2)*2^(1/4))/x/(x^6+x^2)^(1/4))*2^(1/4)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.38 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.53, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2081, 6847, 6857, 251, 1469, 541, 544, 440} \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=-\frac {\sqrt [4]{x^4+1} x \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},x^4,-x^4\right )}{\sqrt [4]{x^6+x^2}}+\frac {\sqrt [4]{x^4+1} x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{2 \sqrt [4]{x^6+x^2}}-\frac {3 x}{2 \sqrt [4]{x^6+x^2}} \]

[In]

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

[Out]

(-3*x)/(2*(x^2 + x^6)^(1/4)) - (x*(1 + x^4)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, x^4, -x^4])/(x^2 + x^6)^(1/4) + (
x*(1 + x^4)^(1/4)*Hypergeometric2F1[1/8, 1/4, 9/8, -x^4])/(2*(x^2 + x^6)^(1/4))

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 541

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

Rule 544

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(d*e - c*f)/d, Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, p, n}, x]

Rule 1469

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :
> Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, f, g, n, q, r}, x] && Eq
Q[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[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]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6857

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*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{1+x^4}\right ) \int \frac {1-x^4+x^8}{\sqrt {x} \sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1-x^8+x^{16}}{\sqrt [4]{1+x^8} \left (-1+x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt [4]{1+x^8}}+\frac {2-x^8}{\sqrt [4]{1+x^8} \left (-1+x^{16}\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}+\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {2-x^8}{\sqrt [4]{1+x^8} \left (-1+x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}}+\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {2-x^8}{\left (-1+x^8\right ) \left (1+x^8\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = -\frac {3 x}{2 \sqrt [4]{x^2+x^6}}+\frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}}+\frac {\left (\sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {5-3 x^8}{\left (-1+x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{2 \sqrt [4]{x^2+x^6}} \\ & = -\frac {3 x}{2 \sqrt [4]{x^2+x^6}}+\frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}}+\frac {\left (\sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (-1+x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (3 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{2 \sqrt [4]{x^2+x^6}} \\ & = -\frac {3 x}{2 \sqrt [4]{x^2+x^6}}-\frac {x \sqrt [4]{1+x^4} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},x^4,-x^4\right )}{\sqrt [4]{x^2+x^6}}+\frac {x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{2 \sqrt [4]{x^2+x^6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.16 \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=-\frac {\sqrt {x} \left (24 \sqrt {x}+2^{3/4} \sqrt [4]{1+x^4} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )-\sqrt [4]{2} \sqrt [4]{1+x^4} \arctan \left (\frac {2^{3/4} \sqrt {x} \sqrt [4]{1+x^4}}{\sqrt {2} x-\sqrt {1+x^4}}\right )+2^{3/4} \sqrt [4]{1+x^4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+\sqrt [4]{2} \sqrt [4]{1+x^4} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt {x} \sqrt [4]{1+x^4}}{2 x+\sqrt {2} \sqrt {1+x^4}}\right )\right )}{16 \sqrt [4]{x^2+x^6}} \]

[In]

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

[Out]

-1/16*(Sqrt[x]*(24*Sqrt[x] + 2^(3/4)*(1 + x^4)^(1/4)*ArcTan[(2^(1/4)*Sqrt[x])/(1 + x^4)^(1/4)] - 2^(1/4)*(1 +
x^4)^(1/4)*ArcTan[(2^(3/4)*Sqrt[x]*(1 + x^4)^(1/4))/(Sqrt[2]*x - Sqrt[1 + x^4])] + 2^(3/4)*(1 + x^4)^(1/4)*Arc
Tanh[(2^(1/4)*Sqrt[x])/(1 + x^4)^(1/4)] + 2^(1/4)*(1 + x^4)^(1/4)*ArcTanh[(2*2^(1/4)*Sqrt[x]*(1 + x^4)^(1/4))/
(2*x + Sqrt[2]*Sqrt[1 + x^4])]))/(x^2 + x^6)^(1/4)

Maple [A] (verified)

Time = 55.08 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.50

method	result	size
pseudoelliptic	\(\frac {2 \arctan \left (\frac {\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+\ln \left (\frac {-2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-48 x}{32 \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\)	\(281\)
risch	\(\text {Expression too large to display}\)	\(648\)
trager	\(\text {Expression too large to display}\)	\(656\)

[In]

int((x^8-x^4+1)/(x^6+x^2)^(1/4)/(x^8-1),x,method=_RETURNVERBOSE)

[Out]

1/32*(2*arctan(1/2*(x^2*(x^4+1))^(1/4)/x*2^(3/4))*2^(3/4)*(x^2*(x^4+1))^(1/4)-ln((-2^(1/4)*x-(x^2*(x^4+1))^(1/
4))/(2^(1/4)*x-(x^2*(x^4+1))^(1/4)))*2^(3/4)*(x^2*(x^4+1))^(1/4)+ln((-2^(3/4)*(x^2*(x^4+1))^(1/4)*x+2^(1/2)*x^
2+(x^2*(x^4+1))^(1/2))/(2^(3/4)*(x^2*(x^4+1))^(1/4)*x+2^(1/2)*x^2+(x^2*(x^4+1))^(1/2)))*2^(1/4)*(x^2*(x^4+1))^
(1/4)+2*arctan((2^(1/4)*(x^2*(x^4+1))^(1/4)+x)/x)*2^(1/4)*(x^2*(x^4+1))^(1/4)+2*arctan((2^(1/4)*(x^2*(x^4+1))^
(1/4)-x)/x)*2^(1/4)*(x^2*(x^4+1))^(1/4)-48*x)/(x^2*(x^4+1))^(1/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 32.53 (sec) , antiderivative size = 733, normalized size of antiderivative = 3.92 \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/64*(2^(3/4)*(x^5 + x)*log(-(4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + 2^(3/4)*(x^5 + 2*x^3 + x) + 4*2^(1/4)*sqrt(x^
6 + x^2)*x + 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) - 2^(3/4)*(x^5 + x)*log(-(4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2
- 2^(3/4)*(x^5 + 2*x^3 + x) - 4*2^(1/4)*sqrt(x^6 + x^2)*x + 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 2^(3/4)
*(-I*x^5 - I*x)*log((4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + 2^(3/4)*(I*x^5 + 2*I*x^3 + I*x) - 4*I*2^(1/4)*sqrt(x^6
+ x^2)*x - 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 2^(3/4)*(I*x^5 + I*x)*log((4*sqrt(2)*(x^6 + x^2)^(1/4)*x^
2 + 2^(3/4)*(-I*x^5 - 2*I*x^3 - I*x) + 4*I*2^(1/4)*sqrt(x^6 + x^2)*x - 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x))
- 2^(1/4)*((I + 1)*x^5 + (I + 1)*x)*log(-2*(4*I*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 - (2*I - 2)*2^(3/4)*sqrt(x^6 +
x^2)*x - 2^(1/4)*(-(I + 1)*x^5 + (2*I + 2)*x^3 - (I + 1)*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) - 2^(1/4
)*(-(I + 1)*x^5 - (I + 1)*x)*log(-2*(4*I*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + (2*I - 2)*2^(3/4)*sqrt(x^6 + x^2)*x -
2^(1/4)*((I + 1)*x^5 - (2*I + 2)*x^3 + (I + 1)*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) - 2^(1/4)*(-(I -
1)*x^5 - (I - 1)*x)*log(-2*(-4*I*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + (2*I + 2)*2^(3/4)*sqrt(x^6 + x^2)*x - 2^(1/4)
*((I - 1)*x^5 - (2*I - 2)*x^3 + (I - 1)*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) - 2^(1/4)*((I - 1)*x^5 +
(I - 1)*x)*log(-2*(-4*I*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 - (2*I + 2)*2^(3/4)*sqrt(x^6 + x^2)*x - 2^(1/4)*(-(I - 1
)*x^5 + (2*I - 2)*x^3 - (I - 1)*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) + 96*(x^6 + x^2)^(3/4))/(x^5 + x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int \frac {x^8-x^4+1}{{\left (x^6+x^2\right )}^{1/4}\,\left (x^8-1\right )} \,d x \]

[In]

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

[Out]

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