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

Optimal. Leaf size=118 \[ \frac {2 \sqrt [4]{x^6+x^2}}{x}-\frac {\tan ^{-1}\left (\frac {2^{3/4} x \sqrt [4]{x^6+x^2}}{\sqrt {2} x^2-\sqrt {x^6+x^2}}\right )}{\sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^6+x^2}}{2^{3/4}}}{x \sqrt [4]{x^6+x^2}}\right )}{\sqrt [4]{2}} \]

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Rubi [C]  time = 0.59, antiderivative size = 147, normalized size of antiderivative = 1.25, number of steps used = 20, number of rules used = 11, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {2056, 6725, 277, 329, 364, 1312, 1336, 325, 1337, 466, 510} \begin {gather*} \frac {8 \sqrt [4]{x^6+x^2} x F_1\left (\frac {3}{8};1,\frac {3}{4};\frac {11}{8};x^4,-x^4\right )}{3 \sqrt [4]{x^4+1}}-\frac {8 \sqrt [4]{x^6+x^2} x^3 F_1\left (\frac {7}{8};1,\frac {3}{4};\frac {15}{8};x^4,-x^4\right )}{7 \sqrt [4]{x^4+1}}-\frac {4 \sqrt [4]{x^6+x^2} x \, _2F_1\left (\frac {3}{8},\frac {3}{4};\frac {11}{8};-x^4\right )}{3 \sqrt [4]{x^4+1}}+\frac {2 \sqrt [4]{x^6+x^2}}{x} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(x^2 + x^6)^(1/4))/x + (8*x*(x^2 + x^6)^(1/4)*AppellF1[3/8, 1, 3/4, 11/8, x^4, -x^4])/(3*(1 + x^4)^(1/4)) -
(8*x^3*(x^2 + x^6)^(1/4)*AppellF1[7/8, 1, 3/4, 15/8, x^4, -x^4])/(7*(1 + x^4)^(1/4)) - (4*x*(x^2 + x^6)^(1/4)
*Hypergeometric2F1[3/8, 3/4, 11/8, -x^4])/(3*(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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 1312

Int[(((f_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^4)^(p_.))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(d*e), Int[(f*
x)^m*(a*e + c*d*x^2)*(a + c*x^4)^(p - 1), x], x] - Dist[(c*d^2 + a*e^2)/(d*e*f^2), Int[((f*x)^(m + 2)*(a + c*x
^4)^(p - 1))/(d + e*x^2), x], x] /; FreeQ[{a, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, 0]

Rule 1336

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && (IGtQ[p, 0] || IGtQ[q,
0] || IntegersQ[m, q])

Rule 1337

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(f*x)^m/x^m, I
nt[ExpandIntegrand[x^m*(a + c*x^4)^p, (d/(d^2 - e^2*x^4) - (e*x^2)/(d^2 - e^2*x^4))^(-q), x], x], x] /; FreeQ[
{a, c, d, e, f, m, p}, x] &&  !IntegerQ[p] && ILtQ[q, 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 {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx &=\frac {\sqrt [4]{x^2+x^6} \int \frac {\left (-1+x^2\right ) \sqrt [4]{1+x^4}}{x^{3/2} \left (1+x^2\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^{3/2}}-\frac {2 \sqrt [4]{1+x^4}}{x^{3/2} \left (1+x^2\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^{3/2}} \, 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^{3/2} \left (1+x^2\right )} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=-\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {1+x^2}{x^{3/2} \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (1+x^2\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=-\frac {2 \sqrt [4]{x^2+x^6}}{x}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \left (\frac {1}{x^{3/2} \left (1+x^4\right )^{3/4}}+\frac {\sqrt {x}}{\left (1+x^4\right )^{3/4}}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \left (\frac {\sqrt {x}}{\left (1-x^4\right ) \left (1+x^4\right )^{3/4}}+\frac {x^{5/2}}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}}\right ) \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=-\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{4},\frac {7}{8};\frac {15}{8};-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {1}{x^{3/2} \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {\sqrt {x}}{\left (1-x^4\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{4},\frac {7}{8};\frac {15}{8};-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (2 \sqrt [4]{x^2+x^6}\right ) \int \frac {x^{5/2}}{\left (1+x^4\right )^{3/4}} \, dx}{\sqrt {x} \sqrt [4]{1+x^4}}-\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 )}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (8 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1-x^8\right ) \left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}+\frac {\left (8 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (-1+x^8\right ) \left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {8 x \sqrt [4]{x^2+x^6} F_1\left (\frac {3}{8};1,\frac {3}{4};\frac {11}{8};x^4,-x^4\right )}{3 \sqrt [4]{1+x^4}}-\frac {8 x^3 \sqrt [4]{x^2+x^6} F_1\left (\frac {7}{8};1,\frac {3}{4};\frac {15}{8};x^4,-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {4 x \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{8},\frac {3}{4};\frac {11}{8};-x^4\right )}{3 \sqrt [4]{1+x^4}}+\frac {4 x^3 \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{4},\frac {7}{8};\frac {15}{8};-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {\left (4 \sqrt [4]{x^2+x^6}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^8\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{1+x^4}}\\ &=\frac {2 \sqrt [4]{x^2+x^6}}{x}+\frac {8 x \sqrt [4]{x^2+x^6} F_1\left (\frac {3}{8};1,\frac {3}{4};\frac {11}{8};x^4,-x^4\right )}{3 \sqrt [4]{1+x^4}}-\frac {8 x^3 \sqrt [4]{x^2+x^6} F_1\left (\frac {7}{8};1,\frac {3}{4};\frac {15}{8};x^4,-x^4\right )}{7 \sqrt [4]{1+x^4}}-\frac {4 x \sqrt [4]{x^2+x^6} \, _2F_1\left (\frac {3}{8},\frac {3}{4};\frac {11}{8};-x^4\right )}{3 \sqrt [4]{1+x^4}}\\ \end {align*}

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Mathematica [F]  time = 1.07, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^2+x^6}}{x^2 \left (1+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.48, size = 118, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt [4]{x^2+x^6}}{x}-\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 )}{\sqrt [4]{2}}-\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 )}{\sqrt [4]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 7.42, size = 545, normalized size = 4.62 \begin {gather*} -\frac {4 \cdot 2^{\frac {3}{4}} x \arctan \left (-\frac {4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - \sqrt {2} {\left (2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - \sqrt {2} {\left (x^{5} + 2 \, x^{3} + x\right )} - 4 \, \sqrt {x^{6} + x^{2}} x + 2 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}\right )} \sqrt {\frac {x^{5} + 2 \, x^{3} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 4 \, \sqrt {2} \sqrt {x^{6} + x^{2}} x + 2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + x}{x^{5} + 2 \, x^{3} + x}} + 2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{2 \, {\left (x^{5} - 2 \, x^{3} + x\right )}}\right ) + 4 \cdot 2^{\frac {3}{4}} x \arctan \left (-\frac {4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} - \sqrt {2} {\left (2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + \sqrt {2} {\left (x^{5} + 2 \, x^{3} + x\right )} + 4 \, \sqrt {x^{6} + x^{2}} x + 2 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}\right )} \sqrt {\frac {x^{5} + 2 \, x^{3} - 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 4 \, \sqrt {2} \sqrt {x^{6} + x^{2}} x - 2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + x}{x^{5} + 2 \, x^{3} + x}} + 2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{2 \, {\left (x^{5} - 2 \, x^{3} + x\right )}}\right ) + 2^{\frac {3}{4}} x \log \left (\frac {2 \, {\left (x^{5} + 2 \, x^{3} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 4 \, \sqrt {2} \sqrt {x^{6} + x^{2}} x + 2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + x\right )}}{x^{5} + 2 \, x^{3} + x}\right ) - 2^{\frac {3}{4}} x \log \left (\frac {2 \, {\left (x^{5} + 2 \, x^{3} - 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 4 \, \sqrt {2} \sqrt {x^{6} + x^{2}} x - 2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + x\right )}}{x^{5} + 2 \, x^{3} + x}\right ) - 16 \, {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*2^(3/4)*x*arctan(-1/2*(4*2^(1/4)*(x^6 + x^2)^(1/4)*x^2 - sqrt(2)*(2*2^(3/4)*(x^6 + x^2)^(1/4)*x^2 - sq
rt(2)*(x^5 + 2*x^3 + x) - 4*sqrt(x^6 + x^2)*x + 2*2^(1/4)*(x^6 + x^2)^(3/4))*sqrt((x^5 + 2*x^3 + 4*2^(1/4)*(x^
6 + x^2)^(1/4)*x^2 + 4*sqrt(2)*sqrt(x^6 + x^2)*x + 2*2^(3/4)*(x^6 + x^2)^(3/4) + x)/(x^5 + 2*x^3 + x)) + 2*2^(
3/4)*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 4*2^(3/4)*x*arctan(-1/2*(4*2^(1/4)*(x^6 + x^2)^(1/4)*x^2 - sqrt(2
)*(2*2^(3/4)*(x^6 + x^2)^(1/4)*x^2 + sqrt(2)*(x^5 + 2*x^3 + x) + 4*sqrt(x^6 + x^2)*x + 2*2^(1/4)*(x^6 + x^2)^(
3/4))*sqrt((x^5 + 2*x^3 - 4*2^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 4*sqrt(2)*sqrt(x^6 + x^2)*x - 2*2^(3/4)*(x^6 + x^2
)^(3/4) + x)/(x^5 + 2*x^3 + x)) + 2*2^(3/4)*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 2^(3/4)*x*log(2*(x^5 + 2*x
^3 + 4*2^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 4*sqrt(2)*sqrt(x^6 + x^2)*x + 2*2^(3/4)*(x^6 + x^2)^(3/4) + x)/(x^5 + 2
*x^3 + x)) - 2^(3/4)*x*log(2*(x^5 + 2*x^3 - 4*2^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 4*sqrt(2)*sqrt(x^6 + x^2)*x - 2*
2^(3/4)*(x^6 + x^2)^(3/4) + x)/(x^5 + 2*x^3 + x)) - 16*(x^6 + x^2)^(1/4))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{{\left (x^{2} + 1\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{2}-1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{4}}}{x^{2} \left (x^{2}+1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )}}{{\left (x^{2} + 1\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^6 + x^2)^(1/4)*(x^2 - 1)/((x^2 + 1)*x^2), 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^2-1\right )}{x^2\,\left (x^2+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^2 - 1))/(x^2*(x^2 + 1)),x)

[Out]

int(((x^2 + x^6)^(1/4)*(x^2 - 1))/(x^2*(x^2 + 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 - 1\right ) \left (x + 1\right )}{x^{2} \left (x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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