3.9.81 \(\int \frac {3-3 x^2+2 x^4}{\sqrt [4]{-1+x^2} (2-3 x^2+x^4)} \, dx\)

Optimal. Leaf size=67 \[ \frac {4 x}{\sqrt [4]{x^2-1}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x^2-1}}{x}\right )}{2 \sqrt {2}}-\frac {5 \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}} \]

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Rubi [A]  time = 0.39, antiderivative size = 65, normalized size of antiderivative = 0.97, number of steps used = 20, number of rules used = 10, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1688, 6725, 199, 230, 305, 220, 1196, 288, 403, 398} \begin {gather*} \frac {4 x}{\sqrt [4]{x^2-1}}-\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}}-\frac {5 \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 3*x^2 + 2*x^4)/((-1 + x^2)^(1/4)*(2 - 3*x^2 + x^4)),x]

[Out]

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 230

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/(b*x), Subst[Int[x^2/Sqrt[1 - x^4/a
], x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 288

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

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 398

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-(b^2/a), 4]}, Simp[(b*Ar
cTan[(q*x)/(Sqrt[2]*(a + b*x^2)^(1/4))])/(2*Sqrt[2]*a*d*q), x] + Simp[(b*ArcTanh[(q*x)/(Sqrt[2]*(a + b*x^2)^(1
/4))])/(2*Sqrt[2]*a*d*q), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && NegQ[b^2/a]

Rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/(b*c - a*d), Int[(a + b*x^2)^p, x],
x] - Dist[d/(b*c - a*d), Int[(a + b*x^2)^(p + 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*
d, 0] && LtQ[p, -1] && EqQ[Denominator[p], 4] && (EqQ[p, -5/4] || EqQ[p, -7/4])

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1688

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[Px*(d + e*x
^2)^(p + q)*(a/d + (c*x^2)/e)^p, x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*
e + a*e^2, 0] && IntegerQ[p] && (PolyQ[Px, x^2] || MatchQ[Px, ((f_) + (g_.)*x^2)^(r_.) /; FreeQ[{f, g, r}, 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 {3-3 x^2+2 x^4}{\sqrt [4]{-1+x^2} \left (2-3 x^2+x^4\right )} \, dx &=\int \frac {3-3 x^2+2 x^4}{\left (-2+x^2\right ) \left (-1+x^2\right )^{5/4}} \, dx\\ &=\int \left (\frac {1}{\left (-1+x^2\right )^{5/4}}+\frac {2 x^2}{\left (-1+x^2\right )^{5/4}}+\frac {5}{\left (-2+x^2\right ) \left (-1+x^2\right )^{5/4}}\right ) \, dx\\ &=2 \int \frac {x^2}{\left (-1+x^2\right )^{5/4}} \, dx+5 \int \frac {1}{\left (-2+x^2\right ) \left (-1+x^2\right )^{5/4}} \, dx+\int \frac {1}{\left (-1+x^2\right )^{5/4}} \, dx\\ &=-\frac {6 x}{\sqrt [4]{-1+x^2}}+4 \int \frac {1}{\sqrt [4]{-1+x^2}} \, dx-5 \int \frac {1}{\left (-1+x^2\right )^{5/4}} \, dx+5 \int \frac {1}{\left (-2+x^2\right ) \sqrt [4]{-1+x^2}} \, dx+\int \frac {1}{\sqrt [4]{-1+x^2}} \, dx\\ &=\frac {4 x}{\sqrt [4]{-1+x^2}}-\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}}-\frac {5 \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}}-5 \int \frac {1}{\sqrt [4]{-1+x^2}} \, dx+\frac {\left (2 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}+\frac {\left (8 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}\\ &=\frac {4 x}{\sqrt [4]{-1+x^2}}-\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}}-\frac {5 \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}}+\frac {\left (2 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}-\frac {\left (2 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}+\frac {\left (8 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}-\frac {\left (8 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}-\frac {\left (10 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}\\ &=\frac {4 x}{\sqrt [4]{-1+x^2}}+\frac {10 x \sqrt [4]{-1+x^2}}{1+\sqrt {-1+x^2}}-\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}}-\frac {5 \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}}-\frac {10 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+x^2}\right )^2}} \left (1+\sqrt {-1+x^2}\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{-1+x^2}\right )|\frac {1}{2}\right )}{x}+\frac {5 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+x^2}\right )^2}} \left (1+\sqrt {-1+x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+x^2}\right )|\frac {1}{2}\right )}{x}-\frac {\left (10 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}+\frac {\left (10 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x^2}\right )}{x}\\ &=\frac {4 x}{\sqrt [4]{-1+x^2}}-\frac {5 \tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}}-\frac {5 \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 119, normalized size = 1.78 \begin {gather*} \frac {2 x \left (\frac {15 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x^2,\frac {x^2}{2}\right )}{\left (x^2-2\right ) \left (x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};x^2,\frac {x^2}{2}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};x^2,\frac {x^2}{2}\right )\right )+6 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x^2,\frac {x^2}{2}\right )\right )}+2\right )}{\sqrt [4]{x^2-1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(3 - 3*x^2 + 2*x^4)/((-1 + x^2)^(1/4)*(2 - 3*x^2 + x^4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.44, size = 67, normalized size = 1.00 \begin {gather*} \frac {4 x}{\sqrt [4]{-1+x^2}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{-1+x^2}}{x}\right )}{2 \sqrt {2}}-\frac {5 \tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-1+x^2}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(3 - 3*x^2 + 2*x^4)/((-1 + x^2)^(1/4)*(2 - 3*x^2 + x^4)),x]

[Out]

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

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fricas [B]  time = 0.49, size = 120, normalized size = 1.79 \begin {gather*} \frac {10 \, \sqrt {2} {\left (x^{2} - 1\right )} \arctan \left (\frac {\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{x}\right ) + 5 \, \sqrt {2} {\left (x^{2} - 1\right )} \log \left (-\frac {x^{4} - 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{2} - 1} x^{2} - 4 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {3}{4}} x + 4 \, x^{2} - 4}{x^{4} - 4 \, x^{2} + 4}\right ) + 32 \, {\left (x^{2} - 1\right )}^{\frac {3}{4}} x}{8 \, {\left (x^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4-3*x^2+3)/(x^2-1)^(1/4)/(x^4-3*x^2+2),x, algorithm="fricas")

[Out]

1/8*(10*sqrt(2)*(x^2 - 1)*arctan(sqrt(2)*(x^2 - 1)^(1/4)/x) + 5*sqrt(2)*(x^2 - 1)*log(-(x^4 - 2*sqrt(2)*(x^2 -
1)^(1/4)*x^3 + 4*sqrt(x^2 - 1)*x^2 - 4*sqrt(2)*(x^2 - 1)^(3/4)*x + 4*x^2 - 4)/(x^4 - 4*x^2 + 4)) + 32*(x^2 -
1)^(3/4)*x)/(x^2 - 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{4} - 3 \, x^{2} + 3}{{\left (x^{4} - 3 \, x^{2} + 2\right )} {\left (x^{2} - 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4-3*x^2+3)/(x^2-1)^(1/4)/(x^4-3*x^2+2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 2.57, size = 127, normalized size = 1.90

method	result	size
trager	\(\frac {4 x}{\left (x^{2}-1\right )^{\frac {1}{4}}}-\frac {5 \RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\left (x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+2\right )+x \sqrt {x^{2}-1}-\RootOf \left (\textit {\_Z}^{2}+2\right ) \left (x^{2}-1\right )^{\frac {1}{4}}-x}{x^{2}-2}\right )}{4}-\frac {5 \RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\left (x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}-2\right )+x \sqrt {x^{2}-1}+\RootOf \left (\textit {\_Z}^{2}-2\right ) \left (x^{2}-1\right )^{\frac {1}{4}}+x}{x^{2}-2}\right )}{4}\)	\(127\)
risch	\(\frac {4 x}{\left (x^{2}-1\right )^{\frac {1}{4}}}+\frac {5 \RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {\left (x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+2\right )-x \sqrt {x^{2}-1}-\RootOf \left (\textit {\_Z}^{2}+2\right ) \left (x^{2}-1\right )^{\frac {1}{4}}+x}{x^{2}-2}\right )}{4}+\frac {5 \RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\left (x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}-2\right )-x \sqrt {x^{2}-1}+\RootOf \left (\textit {\_Z}^{2}-2\right ) \left (x^{2}-1\right )^{\frac {1}{4}}-x}{x^{2}-2}\right )}{4}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^4-3*x^2+3)/(x^2-1)^(1/4)/(x^4-3*x^2+2),x,method=_RETURNVERBOSE)

[Out]

4*x/(x^2-1)^(1/4)-5/4*RootOf(_Z^2+2)*ln(((x^2-1)^(3/4)*RootOf(_Z^2+2)+x*(x^2-1)^(1/2)-RootOf(_Z^2+2)*(x^2-1)^(
1/4)-x)/(x^2-2))-5/4*RootOf(_Z^2-2)*ln(((x^2-1)^(3/4)*RootOf(_Z^2-2)+x*(x^2-1)^(1/2)+RootOf(_Z^2-2)*(x^2-1)^(1
/4)+x)/(x^2-2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{4} - 3 \, x^{2} + 3}{{\left (x^{4} - 3 \, x^{2} + 2\right )} {\left (x^{2} - 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4-3*x^2+3)/(x^2-1)^(1/4)/(x^4-3*x^2+2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {2\,x^4-3\,x^2+3}{{\left (x^2-1\right )}^{1/4}\,\left (x^4-3\,x^2+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 x^{4} - 3 x^{2} + 3}{\sqrt [4]{\left (x - 1\right ) \left (x + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} - 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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