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

Optimal. Leaf size=101 \[ \frac {\tan ^{-1}\left (\frac {2^{3/4} x \sqrt [4]{x^6+x^2}}{\sqrt {2} x^2-\sqrt {x^6+x^2}}\right )}{2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^6+x^2}}{2^{3/4}}}{x \sqrt [4]{x^6+x^2}}\right )}{2^{3/4}} \]

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Rubi [C]  time = 0.64, antiderivative size = 127, normalized size of antiderivative = 1.26, number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2056, 6715, 6725, 245, 1438, 429, 510} \begin {gather*} -\frac {4 \sqrt [4]{x^4+1} x F_1\left (\frac {1}{8};1,\frac {1}{4};\frac {9}{8};x^4,-x^4\right )}{\sqrt [4]{x^6+x^2}}+\frac {4 \sqrt [4]{x^4+1} x^3 F_1\left (\frac {5}{8};1,\frac {1}{4};\frac {13}{8};x^4,-x^4\right )}{5 \sqrt [4]{x^6+x^2}}+\frac {2 \sqrt [4]{x^4+1} x \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-x^4\right )}{\sqrt [4]{x^6+x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*x*(1 + x^4)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, x^4, -x^4])/(x^2 + x^6)^(1/4) + (4*x^3*(1 + x^4)^(1/4)*Appell
F1[5/8, 1, 1/4, 13/8, x^4, -x^4])/(5*(x^2 + x^6)^(1/4)) + (2*x*(1 + x^4)^(1/4)*Hypergeometric2F1[1/8, 1/4, 9/8
, -x^4])/(x^2 + x^6)^(1/4)

Rule 245

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

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

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 1438

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2
*n))^p, (d/(d^2 - e^2*x^(2*n)) - (e*x^n)/(d^2 - e^2*x^(2*n)))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] &&
EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !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 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

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 {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{1+x^4}\right ) \int \frac {-1+x^2}{\sqrt {x} \left (1+x^2\right ) \sqrt [4]{1+x^4}} \, dx}{\sqrt [4]{x^2+x^6}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {-1+x^4}{\left (1+x^4\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{\sqrt [4]{1+x^8}}-\frac {2}{\left (1+x^4\right ) \sqrt [4]{1+x^8}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^4\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}\\ &=\frac {2 x \sqrt [4]{1+x^4} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-x^4\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{\left (1-x^8\right ) \sqrt [4]{1+x^8}}+\frac {x^4}{\left (-1+x^8\right ) \sqrt [4]{1+x^8}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}\\ &=\frac {2 x \sqrt [4]{1+x^4} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-x^4\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}-\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^8\right ) \sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}\\ &=-\frac {4 x \sqrt [4]{1+x^4} F_1\left (\frac {1}{8};1,\frac {1}{4};\frac {9}{8};x^4,-x^4\right )}{\sqrt [4]{x^2+x^6}}+\frac {4 x^3 \sqrt [4]{1+x^4} F_1\left (\frac {5}{8};1,\frac {1}{4};\frac {13}{8};x^4,-x^4\right )}{5 \sqrt [4]{x^2+x^6}}+\frac {2 x \sqrt [4]{1+x^4} \, _2F_1\left (\frac {1}{8},\frac {1}{4};\frac {9}{8};-x^4\right )}{\sqrt [4]{x^2+x^6}}\\ \end {align*}

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Mathematica [F]  time = 0.42, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.48, size = 101, normalized size = 1.00 \begin {gather*} \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^{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^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 24.31, size = 526, normalized size = 5.21 \begin {gather*} \frac {1}{2} \cdot 2^{\frac {1}{4}} \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 ) + \frac {1}{2} \cdot 2^{\frac {1}{4}} \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 ) - \frac {1}{8} \cdot 2^{\frac {1}{4}} \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 ) + \frac {1}{8} \cdot 2^{\frac {1}{4}} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*2^(1/4)*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)) + 1/2*2^(1/4)*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)) - 1/8*2^(1/4)*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)) + 1/8*2^(1/4)*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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{2}-1}{\left (x^{2}+1\right ) \left (x^{6}+x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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