\(\int \frac {\sqrt [4]{x^3+x^5} (1+x^4+x^8)}{x^4 (-1+x^4)} \, dx\) [2382]

Optimal result

Integrand size = 30, antiderivative size = 190 \[ \int \frac {\sqrt [4]{x^3+x^5} \left (1+x^4+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\frac {4 \left (1+2 x^2+x^4\right ) \sqrt [4]{x^3+x^5}}{9 x^3}+\frac {3 \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )}{2\ 2^{3/4}}-\frac {3 \arctan \left (\frac {2^{3/4} x \sqrt [4]{x^3+x^5}}{\sqrt {2} x^2-\sqrt {x^3+x^5}}\right )}{4 \sqrt [4]{2}}-\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )}{2\ 2^{3/4}}-\frac {3 \text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^3+x^5}}{2^{3/4}}}{x \sqrt [4]{x^3+x^5}}\right )}{4 \sqrt [4]{2}} \]

[Out]

4/9*(x^4+2*x^2+1)*(x^5+x^3)^(1/4)/x^3+3/4*2^(1/4)*arctan(2^(1/4)*x/(x^5+x^3)^(1/4))-3/8*arctan(2^(3/4)*x*(x^5+
x^3)^(1/4)/(2^(1/2)*x^2-(x^5+x^3)^(1/2)))*2^(3/4)-3/4*2^(1/4)*arctanh(2^(1/4)*x/(x^5+x^3)^(1/4))-3/8*arctanh((
1/2*x^2*2^(3/4)+1/2*(x^5+x^3)^(1/2)*2^(1/4))/x/(x^5+x^3)^(1/4))*2^(3/4)

Rubi [C] (verified)

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

Time = 0.37 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2081, 1600, 6857, 371, 477, 524} \[ \int \frac {\sqrt [4]{x^3+x^5} \left (1+x^4+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\frac {4 \sqrt [4]{x^5+x^3} \operatorname {AppellF1}\left (-\frac {9}{8},1,\frac {3}{4},-\frac {1}{8},x^2,-x^2\right )}{3 \sqrt [4]{x^2+1} x^3}+\frac {4 \sqrt [4]{x^5+x^3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {15}{8},\frac {23}{8},-x^2\right )}{15 \sqrt [4]{x^2+1}}+\frac {4 \sqrt [4]{x^5+x^3} x \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{8},\frac {15}{8},-x^2\right )}{7 \sqrt [4]{x^2+1}}-\frac {8 \sqrt [4]{x^5+x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {3}{4},\frac {7}{8},-x^2\right )}{\sqrt [4]{x^2+1} x}-\frac {8 \sqrt [4]{x^5+x^3} \operatorname {Hypergeometric2F1}\left (-\frac {9}{8},\frac {3}{4},-\frac {1}{8},-x^2\right )}{9 \sqrt [4]{x^2+1} x^3} \]

[In]

Int[((x^3 + x^5)^(1/4)*(1 + x^4 + x^8))/(x^4*(-1 + x^4)),x]

[Out]

(4*(x^3 + x^5)^(1/4)*AppellF1[-9/8, 1, 3/4, -1/8, x^2, -x^2])/(3*x^3*(1 + x^2)^(1/4)) - (8*(x^3 + x^5)^(1/4)*H
ypergeometric2F1[-9/8, 3/4, -1/8, -x^2])/(9*x^3*(1 + x^2)^(1/4)) - (8*(x^3 + x^5)^(1/4)*Hypergeometric2F1[-1/8
, 3/4, 7/8, -x^2])/(x*(1 + x^2)^(1/4)) + (4*x*(x^3 + x^5)^(1/4)*Hypergeometric2F1[3/4, 7/8, 15/8, -x^2])/(7*(1
+ x^2)^(1/4)) + (4*x^3*(x^3 + x^5)^(1/4)*Hypergeometric2F1[3/4, 15/8, 23/8, -x^2])/(15*(1 + x^2)^(1/4))

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 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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^3+x^5} \int \frac {\sqrt [4]{1+x^2} \left (1+x^4+x^8\right )}{x^{13/4} \left (-1+x^4\right )} \, dx}{x^{3/4} \sqrt [4]{1+x^2}} \\ & = \frac {\sqrt [4]{x^3+x^5} \int \frac {1+x^4+x^8}{x^{13/4} \left (-1+x^2\right ) \left (1+x^2\right )^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x^2}} \\ & = \frac {\sqrt [4]{x^3+x^5} \int \left (\frac {2}{x^{13/4} \left (1+x^2\right )^{3/4}}+\frac {2}{x^{5/4} \left (1+x^2\right )^{3/4}}+\frac {x^{3/4}}{\left (1+x^2\right )^{3/4}}+\frac {x^{11/4}}{\left (1+x^2\right )^{3/4}}+\frac {3}{x^{13/4} \left (-1+x^2\right ) \left (1+x^2\right )^{3/4}}\right ) \, dx}{x^{3/4} \sqrt [4]{1+x^2}} \\ & = \frac {\sqrt [4]{x^3+x^5} \int \frac {x^{3/4}}{\left (1+x^2\right )^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x^2}}+\frac {\sqrt [4]{x^3+x^5} \int \frac {x^{11/4}}{\left (1+x^2\right )^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x^2}}+\frac {\left (2 \sqrt [4]{x^3+x^5}\right ) \int \frac {1}{x^{13/4} \left (1+x^2\right )^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x^2}}+\frac {\left (2 \sqrt [4]{x^3+x^5}\right ) \int \frac {1}{x^{5/4} \left (1+x^2\right )^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x^2}}+\frac {\left (3 \sqrt [4]{x^3+x^5}\right ) \int \frac {1}{x^{13/4} \left (-1+x^2\right ) \left (1+x^2\right )^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x^2}} \\ & = -\frac {8 \sqrt [4]{x^3+x^5} \operatorname {Hypergeometric2F1}\left (-\frac {9}{8},\frac {3}{4},-\frac {1}{8},-x^2\right )}{9 x^3 \sqrt [4]{1+x^2}}-\frac {8 \sqrt [4]{x^3+x^5} \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {3}{4},\frac {7}{8},-x^2\right )}{x \sqrt [4]{1+x^2}}+\frac {4 x \sqrt [4]{x^3+x^5} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{8},\frac {15}{8},-x^2\right )}{7 \sqrt [4]{1+x^2}}+\frac {4 x^3 \sqrt [4]{x^3+x^5} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {15}{8},\frac {23}{8},-x^2\right )}{15 \sqrt [4]{1+x^2}}+\frac {\left (12 \sqrt [4]{x^3+x^5}\right ) \text {Subst}\left (\int \frac {1}{x^{10} \left (-1+x^8\right ) \left (1+x^8\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{x^{3/4} \sqrt [4]{1+x^2}} \\ & = \frac {4 \sqrt [4]{x^3+x^5} \operatorname {AppellF1}\left (-\frac {9}{8},1,\frac {3}{4},-\frac {1}{8},x^2,-x^2\right )}{3 x^3 \sqrt [4]{1+x^2}}-\frac {8 \sqrt [4]{x^3+x^5} \operatorname {Hypergeometric2F1}\left (-\frac {9}{8},\frac {3}{4},-\frac {1}{8},-x^2\right )}{9 x^3 \sqrt [4]{1+x^2}}-\frac {8 \sqrt [4]{x^3+x^5} \operatorname {Hypergeometric2F1}\left (-\frac {1}{8},\frac {3}{4},\frac {7}{8},-x^2\right )}{x \sqrt [4]{1+x^2}}+\frac {4 x \sqrt [4]{x^3+x^5} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{8},\frac {15}{8},-x^2\right )}{7 \sqrt [4]{1+x^2}}+\frac {4 x^3 \sqrt [4]{x^3+x^5} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {15}{8},\frac {23}{8},-x^2\right )}{15 \sqrt [4]{1+x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt [4]{x^3+x^5} \left (1+x^4+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\frac {\sqrt [4]{x^3+x^5} \left (32 \sqrt [4]{1+x^2}+64 x^2 \sqrt [4]{1+x^2}+32 x^4 \sqrt [4]{1+x^2}+54 \sqrt [4]{2} x^{9/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )-27\ 2^{3/4} x^{9/4} \arctan \left (\frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{1+x^2}}{\sqrt {2} \sqrt {x}-\sqrt {1+x^2}}\right )-54 \sqrt [4]{2} x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )-27\ 2^{3/4} x^{9/4} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt [4]{x} \sqrt [4]{1+x^2}}{2 \sqrt {x}+\sqrt {2} \sqrt {1+x^2}}\right )\right )}{72 x^3 \sqrt [4]{1+x^2}} \]

[In]

Integrate[((x^3 + x^5)^(1/4)*(1 + x^4 + x^8))/(x^4*(-1 + x^4)),x]

[Out]

((x^3 + x^5)^(1/4)*(32*(1 + x^2)^(1/4) + 64*x^2*(1 + x^2)^(1/4) + 32*x^4*(1 + x^2)^(1/4) + 54*2^(1/4)*x^(9/4)*
ArcTan[(2^(1/4)*x^(1/4))/(1 + x^2)^(1/4)] - 27*2^(3/4)*x^(9/4)*ArcTan[(2^(3/4)*x^(1/4)*(1 + x^2)^(1/4))/(Sqrt[
2]*Sqrt[x] - Sqrt[1 + x^2])] - 54*2^(1/4)*x^(9/4)*ArcTanh[(2^(1/4)*x^(1/4))/(1 + x^2)^(1/4)] - 27*2^(3/4)*x^(9
/4)*ArcTanh[(2*2^(1/4)*x^(1/4)*(1 + x^2)^(1/4))/(2*Sqrt[x] + Sqrt[2]*Sqrt[1 + x^2])]))/(72*x^3*(1 + x^2)^(1/4)
)

Maple [A] (verified)

Time = 26.96 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.24

method	result	size
pseudoelliptic	\(\frac {-54 x^{3} \left (2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2 x}\right )+\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\right )\right ) 2^{\frac {1}{4}}-27 x^{3} \left (\ln \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{3} \left (x^{2}+1\right )}}{-2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{3} \left (x^{2}+1\right )}}\right )+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+x}{x}\right )+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x}{x}\right )\right ) 2^{\frac {3}{4}}+64 \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} \left (x^{2}+1\right )^{2}}{144 x^{3}}\)	\(235\)
trager	\(\text {Expression too large to display}\)	\(735\)
risch	\(\text {Expression too large to display}\)	\(1774\)

[In]

int((x^5+x^3)^(1/4)*(x^8+x^4+1)/x^4/(x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/144*(-54*x^3*(2*arctan(1/2*2^(3/4)/x*(x^3*(x^2+1))^(1/4))+ln((-2^(1/4)*x-(x^3*(x^2+1))^(1/4))/(2^(1/4)*x-(x^
3*(x^2+1))^(1/4))))*2^(1/4)-27*x^3*(ln((2^(3/4)*(x^3*(x^2+1))^(1/4)*x+2^(1/2)*x^2+(x^3*(x^2+1))^(1/2))/(-2^(3/
4)*(x^3*(x^2+1))^(1/4)*x+2^(1/2)*x^2+(x^3*(x^2+1))^(1/2)))+2*arctan((2^(1/4)*(x^3*(x^2+1))^(1/4)+x)/x)+2*arcta
n((2^(1/4)*(x^3*(x^2+1))^(1/4)-x)/x))*2^(3/4)+64*(x^3*(x^2+1))^(1/4)*(x^2+1)^2)/x^3

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.68 (sec) , antiderivative size = 763, normalized size of antiderivative = 4.02 \[ \int \frac {\sqrt [4]{x^3+x^5} \left (1+x^4+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\text {Too large to display} \]

[In]

integrate((x^5+x^3)^(1/4)*(x^8+x^4+1)/x^4/(x^4-1),x, algorithm="fricas")

[Out]

1/1152*((27*I + 27)*8^(3/4)*sqrt(2)*x^3*log(((I + 1)*8^(3/4)*sqrt(2)*sqrt(x^5 + x^3)*x - 8*I*sqrt(2)*(x^5 + x^
3)^(1/4)*x^2 + 8^(1/4)*sqrt(2)*(-(I - 1)*x^4 + (2*I - 2)*x^3 - (I - 1)*x^2) - 8*(x^5 + x^3)^(3/4))/(x^4 + 2*x^
3 + x^2)) - (27*I - 27)*8^(3/4)*sqrt(2)*x^3*log((-(I - 1)*8^(3/4)*sqrt(2)*sqrt(x^5 + x^3)*x + 8*I*sqrt(2)*(x^5
+ x^3)^(1/4)*x^2 + 8^(1/4)*sqrt(2)*((I + 1)*x^4 - (2*I + 2)*x^3 + (I + 1)*x^2) - 8*(x^5 + x^3)^(3/4))/(x^4 +
2*x^3 + x^2)) + (27*I - 27)*8^(3/4)*sqrt(2)*x^3*log(((I - 1)*8^(3/4)*sqrt(2)*sqrt(x^5 + x^3)*x + 8*I*sqrt(2)*(
x^5 + x^3)^(1/4)*x^2 + 8^(1/4)*sqrt(2)*(-(I + 1)*x^4 + (2*I + 2)*x^3 - (I + 1)*x^2) - 8*(x^5 + x^3)^(3/4))/(x^
4 + 2*x^3 + x^2)) - (27*I + 27)*8^(3/4)*sqrt(2)*x^3*log((-(I + 1)*8^(3/4)*sqrt(2)*sqrt(x^5 + x^3)*x - 8*I*sqrt
(2)*(x^5 + x^3)^(1/4)*x^2 + 8^(1/4)*sqrt(2)*((I - 1)*x^4 - (2*I - 2)*x^3 + (I - 1)*x^2) - 8*(x^5 + x^3)^(3/4))
/(x^4 + 2*x^3 + x^2)) - 54*8^(3/4)*x^3*log((4*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 + 8^(3/4)*sqrt(x^5 + x^3)*x + 8^(1
/4)*(x^4 + 2*x^3 + x^2) + 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) + 54*I*8^(3/4)*x^3*log(-(4*sqrt(2)*(x^5 +
x^3)^(1/4)*x^2 + I*8^(3/4)*sqrt(x^5 + x^3)*x - 8^(1/4)*(I*x^4 + 2*I*x^3 + I*x^2) - 4*(x^5 + x^3)^(3/4))/(x^4 -
2*x^3 + x^2)) - 54*I*8^(3/4)*x^3*log(-(4*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 - I*8^(3/4)*sqrt(x^5 + x^3)*x - 8^(1/4
)*(-I*x^4 - 2*I*x^3 - I*x^2) - 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) + 54*8^(3/4)*x^3*log((4*sqrt(2)*(x^5
+ x^3)^(1/4)*x^2 - 8^(3/4)*sqrt(x^5 + x^3)*x - 8^(1/4)*(x^4 + 2*x^3 + x^2) + 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3
+ x^2)) + 512*(x^5 + x^3)^(1/4)*(x^4 + 2*x^2 + 1))/x^3

Sympy [F]

\[ \int \frac {\sqrt [4]{x^3+x^5} \left (1+x^4+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x^{2} + 1\right )} \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right ) \left (x^{4} - x^{2} + 1\right )}{x^{4} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \]

[In]

integrate((x**5+x**3)**(1/4)*(x**8+x**4+1)/x**4/(x**4-1),x)

[Out]

Integral((x**3*(x**2 + 1))**(1/4)*(x**2 - x + 1)*(x**2 + x + 1)*(x**4 - x**2 + 1)/(x**4*(x - 1)*(x + 1)*(x**2
+ 1)), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{x^3+x^5} \left (1+x^4+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\int { \frac {{\left (x^{8} + x^{4} + 1\right )} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}}}{{\left (x^{4} - 1\right )} x^{4}} \,d x } \]

[In]

integrate((x^5+x^3)^(1/4)*(x^8+x^4+1)/x^4/(x^4-1),x, algorithm="maxima")

[Out]

integrate((x^8 + x^4 + 1)*(x^5 + x^3)^(1/4)/((x^4 - 1)*x^4), x)

Giac [F]

\[ \int \frac {\sqrt [4]{x^3+x^5} \left (1+x^4+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\int { \frac {{\left (x^{8} + x^{4} + 1\right )} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}}}{{\left (x^{4} - 1\right )} x^{4}} \,d x } \]

[In]

integrate((x^5+x^3)^(1/4)*(x^8+x^4+1)/x^4/(x^4-1),x, algorithm="giac")

[Out]

integrate((x^8 + x^4 + 1)*(x^5 + x^3)^(1/4)/((x^4 - 1)*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{x^3+x^5} \left (1+x^4+x^8\right )}{x^4 \left (-1+x^4\right )} \, dx=\int \frac {{\left (x^5+x^3\right )}^{1/4}\,\left (x^8+x^4+1\right )}{x^4\,\left (x^4-1\right )} \,d x \]

[In]

int(((x^3 + x^5)^(1/4)*(x^4 + x^8 + 1))/(x^4*(x^4 - 1)),x)

[Out]

int(((x^3 + x^5)^(1/4)*(x^4 + x^8 + 1))/(x^4*(x^4 - 1)), x)