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

Optimal. Leaf size=180 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^6+x^2}}\right )}{2^{3/4}}+\frac {\tan ^{-1}\left (\frac {2^{3/4} x \sqrt [4]{x^6+x^2}}{\sqrt {2} x^2-\sqrt {x^6+x^2}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^6+x^2}}\right )}{2^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^6+x^2}}{2^{3/4}}}{x \sqrt [4]{x^6+x^2}}\right )}{2 \sqrt [4]{2}}+\frac {2 \sqrt [4]{x^6+x^2} \left (x^4+1\right )}{5 x^3} \]

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Rubi [C]  time = 0.52, antiderivative size = 121, normalized size of antiderivative = 0.67, number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2056, 6725, 277, 329, 364, 279, 466, 510} \begin {gather*} \frac {4 \sqrt [4]{x^6+x^2} F_1\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 \, _2F_1\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} \end {gather*}

Warning: Unable to verify antiderivative.

[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 277

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 279

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 329

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 364

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

Rule 466

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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 2056

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 6725

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*} \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} \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 ) \operatorname {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} F_1\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 ) \operatorname {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} F_1\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} \, _2F_1\left (\frac {3}{8},\frac {3}{4};\frac {11}{8};-x^4\right )}{15 \sqrt [4]{1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 63, normalized size = 0.35 \begin {gather*} \frac {2 \sqrt [4]{x^6+x^2} \left (3 \left (x^4+1\right )^{5/4}-10 x^4 F_1\left (\frac {3}{8};-\frac {1}{4},1;\frac {11}{8};-x^4,x^4\right )\right )}{15 x^3 \sqrt [4]{x^4+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(x^2 + x^6)^(1/4)*(3*(1 + x^4)^(5/4) - 10*x^4*AppellF1[3/8, -1/4, 1, 11/8, -x^4, x^4]))/(15*x^3*(1 + x^4)^(
1/4))

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IntegrateAlgebraic [A]  time = 0.73, size = 180, normalized size = 1.00 \begin {gather*} \frac {2 \left (1+x^4\right ) \sqrt [4]{x^2+x^6}}{5 x^3}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{2^{3/4}}+\frac {\tan ^{-1}\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 {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{2^{3/4}}+\frac {\tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 13.09, size = 1099, normalized size = 6.11

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(20*8^(3/4)*sqrt(2)*x^3*arctan(-1/8*(8*x^9 + 32*x^7 + 48*x^5 + 4*8^(3/4)*sqrt(2)*(x^6 + x^2)^(3/4)*(x^4
- 6*x^2 + 1) + 32*x^3 + 16*8^(1/4)*sqrt(2)*(3*x^6 - 2*x^4 + 3*x^2)*(x^6 + x^2)^(1/4) + 32*sqrt(2)*sqrt(x^6 + x
^2)*(x^5 + 2*x^3 + x) - sqrt(2)*(128*sqrt(2)*(x^6 + x^2)^(3/4)*x^2 + 8^(3/4)*sqrt(2)*(x^9 - 16*x^7 - 2*x^5 - 1
6*x^3 + x) + 8*8^(1/4)*sqrt(2)*sqrt(x^6 + x^2)*(x^5 - 6*x^3 + x) + 32*(x^6 + 2*x^4 + x^2)*(x^6 + x^2)^(1/4))*s
qrt((8^(3/4)*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + 2*8^(1/4)*sqrt(2)*(x^6 + x^2)^(3/4) + sqrt(2)*(x^5 + 2*x^3 + x) +
8*sqrt(x^6 + x^2)*x)/(x^5 + 2*x^3 + x)) + 8*x)/(x^9 - 28*x^7 + 6*x^5 - 28*x^3 + x)) - 20*8^(3/4)*sqrt(2)*x^3*
arctan(-1/8*(8*x^9 + 32*x^7 + 48*x^5 - 4*8^(3/4)*sqrt(2)*(x^6 + x^2)^(3/4)*(x^4 - 6*x^2 + 1) + 32*x^3 - 16*8^(
1/4)*sqrt(2)*(3*x^6 - 2*x^4 + 3*x^2)*(x^6 + x^2)^(1/4) + 32*sqrt(2)*sqrt(x^6 + x^2)*(x^5 + 2*x^3 + x) - sqrt(2
)*(128*sqrt(2)*(x^6 + x^2)^(3/4)*x^2 - 8^(3/4)*sqrt(2)*(x^9 - 16*x^7 - 2*x^5 - 16*x^3 + x) - 8*8^(1/4)*sqrt(2)
*sqrt(x^6 + x^2)*(x^5 - 6*x^3 + x) + 32*(x^6 + 2*x^4 + x^2)*(x^6 + x^2)^(1/4))*sqrt(-(8^(3/4)*sqrt(2)*(x^6 + x
^2)^(1/4)*x^2 + 2*8^(1/4)*sqrt(2)*(x^6 + x^2)^(3/4) - sqrt(2)*(x^5 + 2*x^3 + x) - 8*sqrt(x^6 + x^2)*x)/(x^5 +
2*x^3 + x)) + 8*x)/(x^9 - 28*x^7 + 6*x^5 - 28*x^3 + x)) + 5*8^(3/4)*sqrt(2)*x^3*log(8*(8^(3/4)*sqrt(2)*(x^6 +
x^2)^(1/4)*x^2 + 2*8^(1/4)*sqrt(2)*(x^6 + x^2)^(3/4) + sqrt(2)*(x^5 + 2*x^3 + x) + 8*sqrt(x^6 + x^2)*x)/(x^5 +
2*x^3 + x)) - 5*8^(3/4)*sqrt(2)*x^3*log(-8*(8^(3/4)*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + 2*8^(1/4)*sqrt(2)*(x^6 +
x^2)^(3/4) - sqrt(2)*(x^5 + 2*x^3 + x) - 8*sqrt(x^6 + x^2)*x)/(x^5 + 2*x^3 + x)) + 40*8^(3/4)*x^3*arctan(1/8*(
16*8^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 2^(3/4)*(8^(3/4)*(x^5 + 2*x^3 + x) + 8*8^(1/4)*sqrt(x^6 + x^2)*x) + 4*8^(3/
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)) + 10*8^(3/4)*x^3*log(-(4*s
qrt(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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 53.89, size = 658, normalized size = 3.66

method	result	size
trager	\(\frac {2 \left (x^{4}+1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}-\frac {\RootOf \left (\textit {\_Z}^{4}+2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{4}+2\right )^{3} x^{5}-2 \RootOf \left (\textit {\_Z}^{4}+2\right )^{3} x^{3}+4 \left (x^{6}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x^{2}-4 \sqrt {x^{6}+x^{2}}\, \RootOf \left (\textit {\_Z}^{4}+2\right ) x +\RootOf \left (\textit {\_Z}^{4}+2\right )^{3} x +4 \left (x^{6}+x^{2}\right )^{\frac {3}{4}}}{x \left (x^{2}+1\right )^{2}}\right )}{4}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x^{5}+2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x^{3}-4 \left (x^{6}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} x^{2}-4 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \sqrt {x^{6}+x^{2}}\, x -\RootOf \left (\textit {\_Z}^{4}+2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) x +4 \left (x^{6}+x^{2}\right )^{\frac {3}{4}}}{x \left (x^{2}+1\right )^{2}}\right )}{4}-\frac {\ln \left (\frac {\RootOf \left (\textit {\_Z}^{4}+2\right )^{2} \sqrt {x^{6}+x^{2}}-2 \left (x^{6}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right ) x +2 x^{2}}{x \left (-1+x \right ) \left (1+x \right )}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{3}}{4}-\frac {\ln \left (\frac {\RootOf \left (\textit {\_Z}^{4}+2\right )^{2} \sqrt {x^{6}+x^{2}}-2 \left (x^{6}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right ) x +2 x^{2}}{x \left (-1+x \right ) \left (1+x \right )}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{4}+2\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{4}+2\right )^{3} \sqrt {x^{6}+x^{2}}\, x -2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+2\right )^{2} \sqrt {x^{6}+x^{2}}\, x -\RootOf \left (\textit {\_Z}^{4}+2\right ) x^{5}-\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) x^{5}+4 \left (x^{6}+x^{2}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+2\right ) \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) x^{2}-2 \RootOf \left (\textit {\_Z}^{4}+2\right ) x^{3}-2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) x^{3}+4 \left (x^{6}+x^{2}\right )^{\frac {3}{4}}-\RootOf \left (\textit {\_Z}^{4}+2\right ) x -\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+2\right )^{2}\right ) x}{\left (-1+x \right )^{2} \left (1+x \right )^{2} x}\right )}{4}\)	\(658\)
risch	\(\text {Expression too large to display}\)	\(1589\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x^4+1)*(x^6+x^2)^(1/4)/x^3-1/4*RootOf(_Z^4+2)*ln(-(RootOf(_Z^4+2)^3*x^5-2*RootOf(_Z^4+2)^3*x^3+4*(x^6+x^2
)^(1/4)*RootOf(_Z^4+2)^2*x^2-4*(x^6+x^2)^(1/2)*RootOf(_Z^4+2)*x+RootOf(_Z^4+2)^3*x+4*(x^6+x^2)^(3/4))/x/(x^2+1
)^2)-1/4*RootOf(_Z^2+RootOf(_Z^4+2)^2)*ln(-(-RootOf(_Z^2+RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2*x^5+2*RootOf(_Z^2+
RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2*x^3-4*(x^6+x^2)^(1/4)*RootOf(_Z^4+2)^2*x^2-4*RootOf(_Z^2+RootOf(_Z^4+2)^2)*
(x^6+x^2)^(1/2)*x-RootOf(_Z^4+2)^2*RootOf(_Z^2+RootOf(_Z^4+2)^2)*x+4*(x^6+x^2)^(3/4))/x/(x^2+1)^2)-1/4*ln((Roo
tOf(_Z^4+2)^2*(x^6+x^2)^(1/2)-2*(x^6+x^2)^(1/4)*RootOf(_Z^4+2)*x+2*x^2)/x/(-1+x)/(1+x))*RootOf(_Z^4+2)^3-1/4*l
n((RootOf(_Z^4+2)^2*(x^6+x^2)^(1/2)-2*(x^6+x^2)^(1/4)*RootOf(_Z^4+2)*x+2*x^2)/x/(-1+x)/(1+x))*RootOf(_Z^4+2)^2
*RootOf(_Z^2+RootOf(_Z^4+2)^2)+1/4*RootOf(_Z^4+2)^2*RootOf(_Z^2+RootOf(_Z^4+2)^2)*ln((2*RootOf(_Z^4+2)^3*(x^6+
x^2)^(1/2)*x-2*RootOf(_Z^2+RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2*(x^6+x^2)^(1/2)*x-RootOf(_Z^4+2)*x^5-RootOf(_Z^2
+RootOf(_Z^4+2)^2)*x^5+4*(x^6+x^2)^(1/4)*RootOf(_Z^4+2)*RootOf(_Z^2+RootOf(_Z^4+2)^2)*x^2-2*RootOf(_Z^4+2)*x^3
-2*RootOf(_Z^2+RootOf(_Z^4+2)^2)*x^3+4*(x^6+x^2)^(3/4)-RootOf(_Z^4+2)*x-RootOf(_Z^2+RootOf(_Z^4+2)^2)*x)/(-1+x
)^2/(1+x)^2/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^6+x^2\right )}^{1/4}\,\left (x^8+1\right )}{x^4\,\left (x^4-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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