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

Optimal. Leaf size=121 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8-3 \text {$\#$1}^4+3\& ,\frac {\text {$\#$1} \log \left (\sqrt [4]{x^4-x^2}-\text {$\#$1} x\right )-\text {$\#$1} \log (x)}{2 \text {$\#$1}^4-3}\& \right ]+\frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8-\text {$\#$1}^4+1\& ,\frac {\text {$\#$1} \log \left (\sqrt [4]{x^4-x^2}-\text {$\#$1} x\right )-\text {$\#$1} \log (x)}{2 \text {$\#$1}^4-1}\& \right ] \]

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Rubi [C]  time = 1.15, antiderivative size = 333, normalized size of antiderivative = 2.75, number of steps used = 17, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2056, 6728, 1270, 1529, 511, 510} \begin {gather*} \frac {\left (\sqrt {3}+3 i\right ) x \sqrt [4]{x^4-x^2} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,-\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )}{9 \left (\sqrt {3}+i\right ) \sqrt [4]{1-x^2}}+\frac {\left (\sqrt {3}+3 i\right ) x \sqrt [4]{x^4-x^2} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )}{9 \left (\sqrt {3}+i\right ) \sqrt [4]{1-x^2}}+\frac {\left (-\sqrt {3}+3 i\right ) x \sqrt [4]{x^4-x^2} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,-\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{9 \left (-\sqrt {3}+i\right ) \sqrt [4]{1-x^2}}+\frac {\left (-\sqrt {3}+3 i\right ) x \sqrt [4]{x^4-x^2} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{9 \left (-\sqrt {3}+i\right ) \sqrt [4]{1-x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((3*I + Sqrt[3])*x*(-x^2 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^2, -1/2*((1 - I*Sqrt[3])*x^2)])/(9*(I + Sq
rt[3])*(1 - x^2)^(1/4)) + ((3*I + Sqrt[3])*x*(-x^2 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^2, ((1 - I*Sqrt[
3])*x^2)/2])/(9*(I + Sqrt[3])*(1 - x^2)^(1/4)) + ((3*I - Sqrt[3])*x*(-x^2 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1,
7/4, x^2, -1/2*((1 + I*Sqrt[3])*x^2)])/(9*(I - Sqrt[3])*(1 - x^2)^(1/4)) + ((3*I - Sqrt[3])*x*(-x^2 + x^4)^(1/
4)*AppellF1[3/4, -1/4, 1, 7/4, x^2, ((1 + I*Sqrt[3])*x^2)/2])/(9*(I - Sqrt[3])*(1 - x^2)^(1/4))

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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{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])

Rule 1270

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominat
or[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + (e*x^(2*k))/f)^q*(a + (c*x^(4*k))/f)^p, x], x, (f*x)^(1/k)]
, x]] /; FreeQ[{a, c, d, e, f, p, q}, x] && FractionQ[m] && IntegerQ[p]

Rule 1529

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Int[ExpandInte
grand[(d + e*x^n)^q, (f*x)^m/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, f, q, n}, x] && EqQ[n2, 2*n] && IGt
Q[n, 0] &&  !IntegerQ[q] && IntegerQ[m]

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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^4}}{1+x^4+x^8} \, dx &=\frac {\sqrt [4]{-x^2+x^4} \int \frac {\sqrt {x} \sqrt [4]{-1+x^2} \left (1+x^4\right )}{1+x^4+x^8} \, dx}{\sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\sqrt [4]{-x^2+x^4} \int \left (\frac {\left (1-\frac {i}{\sqrt {3}}\right ) \sqrt {x} \sqrt [4]{-1+x^2}}{1-i \sqrt {3}+2 x^4}+\frac {\left (1+\frac {i}{\sqrt {3}}\right ) \sqrt {x} \sqrt [4]{-1+x^2}}{1+i \sqrt {3}+2 x^4}\right ) \, dx}{\sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (\left (3-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \int \frac {\sqrt {x} \sqrt [4]{-1+x^2}}{1-i \sqrt {3}+2 x^4} \, dx}{3 \sqrt {x} \sqrt [4]{-1+x^2}}+\frac {\left (\left (3+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \int \frac {\sqrt {x} \sqrt [4]{-1+x^2}}{1+i \sqrt {3}+2 x^4} \, dx}{3 \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (2 \left (3-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{1-i \sqrt {3}+2 x^8} \, dx,x,\sqrt {x}\right )}{3 \sqrt {x} \sqrt [4]{-1+x^2}}+\frac {\left (2 \left (3+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{1+i \sqrt {3}+2 x^8} \, dx,x,\sqrt {x}\right )}{3 \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (2 \left (3-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1+i \sqrt {3}\right )} \left (-\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4\right )}-\frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1+i \sqrt {3}\right )} \left (\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{3 \sqrt {x} \sqrt [4]{-1+x^2}}+\frac {\left (2 \left (3+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1-i \sqrt {3}\right )} \left (-\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4\right )}-\frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1-i \sqrt {3}\right )} \left (\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{3 \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (\left (3-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{-\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{3 \sqrt {\frac {1}{2} \left (-1+i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{-1+x^2}}-\frac {\left (\left (3-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{3 \sqrt {\frac {1}{2} \left (-1+i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{-1+x^2}}+\frac {\left (\left (3+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{-\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{3 \sqrt {\frac {1}{2} \left (-1-i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{-1+x^2}}-\frac {\left (\left (3+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{3 \sqrt {\frac {1}{2} \left (-1-i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (\left (3-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{-\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{3 \sqrt {\frac {1}{2} \left (-1+i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{1-x^2}}-\frac {\left (\left (3-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{3 \sqrt {\frac {1}{2} \left (-1+i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{1-x^2}}+\frac {\left (\left (3+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{-\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{3 \sqrt {\frac {1}{2} \left (-1-i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{1-x^2}}-\frac {\left (\left (3+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{3 \sqrt {\frac {1}{2} \left (-1-i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{1-x^2}}\\ &=\frac {\left (3 i+\sqrt {3}\right ) x \sqrt [4]{-x^2+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,-\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )}{9 \left (i+\sqrt {3}\right ) \sqrt [4]{1-x^2}}+\frac {\left (3 i+\sqrt {3}\right ) x \sqrt [4]{-x^2+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )}{9 \left (i+\sqrt {3}\right ) \sqrt [4]{1-x^2}}+\frac {\left (3 i-\sqrt {3}\right ) x \sqrt [4]{-x^2+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,-\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{9 \left (i-\sqrt {3}\right ) \sqrt [4]{1-x^2}}+\frac {\left (3 i-\sqrt {3}\right ) x \sqrt [4]{-x^2+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{9 \left (i-\sqrt {3}\right ) \sqrt [4]{1-x^2}}\\ \end {align*}

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Mathematica [F]  time = 7.46, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^2+x^4}}{1+x^4+x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.32, size = 121, normalized size = 1.00 \begin {gather*} \frac {1}{4} \text {RootSum}\left [3-3 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{-x^2+x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-3+2 \text {$\#$1}^4}\&\right ]+\frac {1}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{-x^2+x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-1+2 \text {$\#$1}^4}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

RootSum[3 - 3*#1^4 + #1^8 & , (-(Log[x]*#1) + Log[(-x^2 + x^4)^(1/4) - x*#1]*#1)/(-3 + 2*#1^4) & ]/4 + RootSum
[1 - #1^4 + #1^8 & , (-(Log[x]*#1) + Log[(-x^2 + x^4)^(1/4) - x*#1]*#1)/(-1 + 2*#1^4) & ]/4

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}}{x^{8} + x^{4} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 157.28, size = 11446, normalized size = 94.60

method	result	size
trager	\(\text {Expression too large to display}\)	\(11446\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}}{x^{8} + x^{4} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((x^4 + 1)*(x^4 - x^2)^(1/4))/(x^4 + x^8 + 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 - 1\right ) \left (x + 1\right )} \left (x^{4} + 1\right )}{\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right ) \left (x^{4} - x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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