\(\int \frac {\sqrt [4]{x^2+x^6} (1+x^8)}{x^4 (-1+x^4)} \, dx\) [2324]

Optimal result

Integrand size = 27, antiderivative size = 180 \[ \int \frac {\sqrt [4]{x^2+x^6} \left (1+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\frac {2 \left (1+x^4\right ) \sqrt [4]{x^2+x^6}}{5 x^3}+\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{2^{3/4}}+\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^2+x^6}}{\sqrt {2} x^2-\sqrt {x^2+x^6}}\right )}{2 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{2^{3/4}}+\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^2+x^6}}{2^{3/4}}}{x \sqrt [4]{x^2+x^6}}\right )}{2 \sqrt [4]{2}} \]

[Out]

2/5*(x^4+1)*(x^6+x^2)^(1/4)/x^3+1/2*2^(1/4)*arctan(2^(1/4)*x/(x^6+x^2)^(1/4))+1/4*arctan(2^(3/4)*x*(x^6+x^2)^(
1/4)/(2^(1/2)*x^2-(x^6+x^2)^(1/2)))*2^(3/4)-1/2*2^(1/4)*arctanh(2^(1/4)*x/(x^6+x^2)^(1/4))+1/4*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^(3/4)

Rubi [C] (verified)

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

Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.67, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2081, 6857, 283, 335, 371, 285, 477, 524} \[ \int \frac {\sqrt [4]{x^2+x^6} \left (1+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\frac {4 \sqrt [4]{x^6+x^2} \operatorname {AppellF1}\left (-\frac {5}{8},1,-\frac {1}{4},\frac {3}{8},x^4,-x^4\right )}{5 \sqrt [4]{x^4+1} x^3}+\frac {8 \sqrt [4]{x^6+x^2} x \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {3}{4},\frac {11}{8},-x^4\right )}{15 \sqrt [4]{x^4+1}}+\frac {2}{5} \sqrt [4]{x^6+x^2} x-\frac {2 \sqrt [4]{x^6+x^2}}{5 x^3} \]

[In]

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

[Out]

(-2*(x^2 + x^6)^(1/4))/(5*x^3) + (2*x*(x^2 + x^6)^(1/4))/5 + (4*(x^2 + x^6)^(1/4)*AppellF1[-5/8, 1, -1/4, 3/8,
x^4, -x^4])/(5*x^3*(1 + x^4)^(1/4)) + (8*x*(x^2 + x^6)^(1/4)*Hypergeometric2F1[3/8, 3/4, 11/8, -x^4])/(15*(1
+ x^4)^(1/4))

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

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 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 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 {\sqrt [4]{x^2+x^6} \int \frac {\sqrt [4]{1+x^4} \left (1+x^8\right )}{x^{7/2} \left (-1+x^4\right )} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = \frac {\sqrt [4]{x^2+x^6} \int \left (\frac {\sqrt [4]{1+x^4}}{x^{7/2}}+\sqrt {x} \sqrt [4]{1+x^4}+\frac {2 \sqrt [4]{1+x^4}}{x^{7/2} \left (-1+x^4\right )}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = \frac {\sqrt [4]{x^2+x^6} \int \frac {\sqrt [4]{1+x^4}}{x^{7/2}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\sqrt [4]{x^2+x^6} \int \sqrt {x} \sqrt [4]{1+x^4} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt [4]{1+x^4}}{x^{7/2} \left (-1+x^4\right )} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = -\frac {2 \sqrt [4]{x^2+x^6}}{5 x^3}+\frac {2}{5} x \sqrt [4]{x^2+x^6}+2 \frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (1+x^4\right )^{3/4}} \, dx}{5 \sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1+x^8}}{x^6 \left (-1+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}} \\ & = -\frac {2 \sqrt [4]{x^2+x^6}}{5 x^3}+\frac {2}{5} x \sqrt [4]{x^2+x^6}+\frac {4 \sqrt [4]{x^2+x^6} \operatorname {AppellF1}\left (-\frac {5}{8},1,-\frac {1}{4},\frac {3}{8},x^4,-x^4\right )}{5 x^3 \sqrt [4]{1+x^4}}+2 \frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {x} \sqrt [4]{1+x^4}} \\ & = -\frac {2 \sqrt [4]{x^2+x^6}}{5 x^3}+\frac {2}{5} x \sqrt [4]{x^2+x^6}+\frac {4 \sqrt [4]{x^2+x^6} \operatorname {AppellF1}\left (-\frac {5}{8},1,-\frac {1}{4},\frac {3}{8},x^4,-x^4\right )}{5 x^3 \sqrt [4]{1+x^4}}+\frac {8 x \sqrt [4]{x^2+x^6} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {3}{4},\frac {11}{8},-x^4\right )}{15 \sqrt [4]{1+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt [4]{x^2+x^6} \left (1+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\frac {\sqrt [4]{x^2+x^6} \left (8 \sqrt [4]{1+x^4}+8 x^4 \sqrt [4]{1+x^4}+10 \sqrt [4]{2} x^{5/2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+5\ 2^{3/4} x^{5/2} \arctan \left (\frac {2^{3/4} \sqrt {x} \sqrt [4]{1+x^4}}{\sqrt {2} x-\sqrt {1+x^4}}\right )-10 \sqrt [4]{2} x^{5/2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+5\ 2^{3/4} x^{5/2} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt {x} \sqrt [4]{1+x^4}}{2 x+\sqrt {2} \sqrt {1+x^4}}\right )\right )}{20 x^3 \sqrt [4]{1+x^4}} \]

[In]

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

[Out]

((x^2 + x^6)^(1/4)*(8*(1 + x^4)^(1/4) + 8*x^4*(1 + x^4)^(1/4) + 10*2^(1/4)*x^(5/2)*ArcTan[(2^(1/4)*Sqrt[x])/(1
+ x^4)^(1/4)] + 5*2^(3/4)*x^(5/2)*ArcTan[(2^(3/4)*Sqrt[x]*(1 + x^4)^(1/4))/(Sqrt[2]*x - Sqrt[1 + x^4])] - 10*
2^(1/4)*x^(5/2)*ArcTanh[(2^(1/4)*Sqrt[x])/(1 + x^4)^(1/4)] + 5*2^(3/4)*x^(5/2)*ArcTanh[(2*2^(1/4)*Sqrt[x]*(1 +
x^4)^(1/4))/(2*x + Sqrt[2]*Sqrt[1 + x^4])]))/(20*x^3*(1 + x^4)^(1/4))

Maple [A] (verified)

Time = 47.08 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.44

method	result	size
pseudoelliptic	\(\frac {5 \ln \left (\frac {2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{-2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right ) 2^{\frac {3}{4}} x^{3}+10 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {3}{4}} x^{3}+10 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {3}{4}} x^{3}-10 \ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} x^{3}-20 \arctan \left (\frac {\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right ) 2^{\frac {1}{4}} x^{3}+16 x^{4} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+16 \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{40 x^{3}}\)	\(260\)
trager	\(\text {Expression too large to display}\)	\(656\)
risch	\(\text {Expression too large to display}\)	\(1587\)

[In]

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

[Out]

1/40/x^3*(5*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^(3/4)*x^3+10*arctan((2^(1/4)*(x^2*(x^4+1))^(1/4)+x)/x)*2^(3/4)*x^3+10*arc
tan((2^(1/4)*(x^2*(x^4+1))^(1/4)-x)/x)*2^(3/4)*x^3-10*ln((-2^(1/4)*x-(x^2*(x^4+1))^(1/4))/(2^(1/4)*x-(x^2*(x^4
+1))^(1/4)))*2^(1/4)*x^3-20*arctan(1/2*(x^2*(x^4+1))^(1/4)/x*2^(3/4))*2^(1/4)*x^3+16*x^4*(x^2*(x^4+1))^(1/4)+1
6*(x^2*(x^4+1))^(1/4))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 8.37 (sec) , antiderivative size = 724, normalized size of antiderivative = 4.02 \[ \int \frac {\sqrt [4]{x^2+x^6} \left (1+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/320*((5*I + 5)*8^(3/4)*sqrt(2)*x^3*log(((I + 1)*8^(3/4)*sqrt(2)*sqrt(x^6 + x^2)*x + 8*I*sqrt(2)*(x^6 + x^2)^
(1/4)*x^2 + 8^(1/4)*sqrt(2)*(-(I - 1)*x^5 + (2*I - 2)*x^3 - (I - 1)*x) + 8*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x
)) - (5*I - 5)*8^(3/4)*sqrt(2)*x^3*log((-(I - 1)*8^(3/4)*sqrt(2)*sqrt(x^6 + x^2)*x - 8*I*sqrt(2)*(x^6 + x^2)^(
1/4)*x^2 + 8^(1/4)*sqrt(2)*((I + 1)*x^5 - (2*I + 2)*x^3 + (I + 1)*x) + 8*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x))
+ (5*I - 5)*8^(3/4)*sqrt(2)*x^3*log(((I - 1)*8^(3/4)*sqrt(2)*sqrt(x^6 + x^2)*x - 8*I*sqrt(2)*(x^6 + x^2)^(1/4
)*x^2 + 8^(1/4)*sqrt(2)*(-(I + 1)*x^5 + (2*I + 2)*x^3 - (I + 1)*x) + 8*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) -
(5*I + 5)*8^(3/4)*sqrt(2)*x^3*log((-(I + 1)*8^(3/4)*sqrt(2)*sqrt(x^6 + x^2)*x + 8*I*sqrt(2)*(x^6 + x^2)^(1/4)
*x^2 + 8^(1/4)*sqrt(2)*((I - 1)*x^5 - (2*I - 2)*x^3 + (I - 1)*x) + 8*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) - 1
0*8^(3/4)*x^3*log(-(4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + 8^(3/4)*sqrt(x^6 + x^2)*x + 8^(1/4)*(x^5 + 2*x^3 + x) +
4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 10*I*8^(3/4)*x^3*log((4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + I*8^(3/4)*sq
rt(x^6 + x^2)*x + 8^(1/4)*(-I*x^5 - 2*I*x^3 - I*x) - 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) - 10*I*8^(3/4)*x^
3*log((4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 - I*8^(3/4)*sqrt(x^6 + x^2)*x + 8^(1/4)*(I*x^5 + 2*I*x^3 + I*x) - 4*(x^
6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 10*8^(3/4)*x^3*log(-(4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 - 8^(3/4)*sqrt(x^6 +
x^2)*x - 8^(1/4)*(x^5 + 2*x^3 + x) + 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 128*(x^6 + x^2)^(1/4)*(x^4 + 1
))/x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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