3.5.7 \(\int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} (-b-a x^2+x^4)} \, dx\)

Optimal. Leaf size=33 \[ \tan ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{x}\right )-\tanh ^{-1}\left (\frac {x}{\sqrt [4]{a x^2+b}}\right ) \]

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Rubi [C]  time = 0.64, antiderivative size = 489, normalized size of antiderivative = 14.82, number of steps used = 10, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1692, 399, 490, 1218} \begin {gather*} \frac {\sqrt [4]{b} \left (a-\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {\sqrt {2} \sqrt {b}}{\sqrt {a^2-\sqrt {a^2+4 b} a+2 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{\sqrt {2} x \sqrt {-a \sqrt {a^2+4 b}+a^2+2 b}}-\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {\sqrt {2} \sqrt {b}}{\sqrt {a^2-\sqrt {a^2+4 b} a+2 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{\sqrt {2} x \sqrt {-a \sqrt {a^2+4 b}+a^2+2 b}}+\frac {\sqrt [4]{b} \left (\sqrt {a^2+4 b}+a\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {\sqrt {2} \sqrt {b}}{\sqrt {a^2+\sqrt {a^2+4 b} a+2 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{\sqrt {2} x \sqrt {a \sqrt {a^2+4 b}+a^2+2 b}}-\frac {\sqrt [4]{b} \left (\sqrt {a^2+4 b}+a\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {\sqrt {2} \sqrt {b}}{\sqrt {a^2+\sqrt {a^2+4 b} a+2 b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{\sqrt {2} x \sqrt {a \sqrt {a^2+4 b}+a^2+2 b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^(1/4)*(a - Sqrt[a^2 + 4*b])*Sqrt[-((a*x^2)/b)]*EllipticPi[-((Sqrt[2]*Sqrt[b])/Sqrt[a^2 + 2*b - a*Sqrt[a^2 +
4*b]]), ArcSin[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(Sqrt[2]*Sqrt[a^2 + 2*b - a*Sqrt[a^2 + 4*b]]*x) - (b^(1/4)*(a
- Sqrt[a^2 + 4*b])*Sqrt[-((a*x^2)/b)]*EllipticPi[(Sqrt[2]*Sqrt[b])/Sqrt[a^2 + 2*b - a*Sqrt[a^2 + 4*b]], ArcSi
n[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(Sqrt[2]*Sqrt[a^2 + 2*b - a*Sqrt[a^2 + 4*b]]*x) + (b^(1/4)*(a + Sqrt[a^2 +
4*b])*Sqrt[-((a*x^2)/b)]*EllipticPi[-((Sqrt[2]*Sqrt[b])/Sqrt[a^2 + 2*b + a*Sqrt[a^2 + 4*b]]), ArcSin[(b + a*x^
2)^(1/4)/b^(1/4)], -1])/(Sqrt[2]*Sqrt[a^2 + 2*b + a*Sqrt[a^2 + 4*b]]*x) - (b^(1/4)*(a + Sqrt[a^2 + 4*b])*Sqrt[
-((a*x^2)/b)]*EllipticPi[(Sqrt[2]*Sqrt[b])/Sqrt[a^2 + 2*b + a*Sqrt[a^2 + 4*b]], ArcSin[(b + a*x^2)^(1/4)/b^(1/
4)], -1])/(Sqrt[2]*Sqrt[a^2 + 2*b + a*Sqrt[a^2 + 4*b]]*x)

Rule 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1218

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

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-b-a x^2+x^4\right )} \, dx &=\int \left (\frac {a+\sqrt {a^2+4 b}}{\left (-a-\sqrt {a^2+4 b}+2 x^2\right ) \sqrt [4]{b+a x^2}}+\frac {a-\sqrt {a^2+4 b}}{\left (-a+\sqrt {a^2+4 b}+2 x^2\right ) \sqrt [4]{b+a x^2}}\right ) \, dx\\ &=\left (a-\sqrt {a^2+4 b}\right ) \int \frac {1}{\left (-a+\sqrt {a^2+4 b}+2 x^2\right ) \sqrt [4]{b+a x^2}} \, dx+\left (a+\sqrt {a^2+4 b}\right ) \int \frac {1}{\left (-a-\sqrt {a^2+4 b}+2 x^2\right ) \sqrt [4]{b+a x^2}} \, dx\\ &=\frac {\left (2 \left (a-\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-2 b+a \left (-a+\sqrt {a^2+4 b}\right )+2 x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{x}+\frac {\left (2 \left (a+\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-2 b+a \left (-a-\sqrt {a^2+4 b}\right )+2 x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{x}\\ &=-\frac {\left (\left (a-\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+2 b-a \sqrt {a^2+4 b}}-\sqrt {2} x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{\sqrt {2} x}+\frac {\left (\left (a-\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+2 b-a \sqrt {a^2+4 b}}+\sqrt {2} x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{\sqrt {2} x}-\frac {\left (\left (a+\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+2 b+a \sqrt {a^2+4 b}}-\sqrt {2} x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{\sqrt {2} x}+\frac {\left (\left (a+\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+2 b+a \sqrt {a^2+4 b}}+\sqrt {2} x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{\sqrt {2} x}\\ &=\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {\sqrt {2} \sqrt {b}}{\sqrt {a^2+2 b-a \sqrt {a^2+4 b}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{\sqrt {2} \sqrt {a^2+2 b-a \sqrt {a^2+4 b}} x}-\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {\sqrt {2} \sqrt {b}}{\sqrt {a^2+2 b-a \sqrt {a^2+4 b}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{\sqrt {2} \sqrt {a^2+2 b-a \sqrt {a^2+4 b}} x}+\frac {\sqrt [4]{b} \left (a+\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {\sqrt {2} \sqrt {b}}{\sqrt {a^2+2 b+a \sqrt {a^2+4 b}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{\sqrt {2} \sqrt {a^2+2 b+a \sqrt {a^2+4 b}} x}-\frac {\sqrt [4]{b} \left (a+\sqrt {a^2+4 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {\sqrt {2} \sqrt {b}}{\sqrt {a^2+2 b+a \sqrt {a^2+4 b}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{\sqrt {2} \sqrt {a^2+2 b+a \sqrt {a^2+4 b}} x}\\ \end {align*}

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Mathematica [F]  time = 0.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-b-a x^2+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 33, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{x}\right )-\tanh ^{-1}\left (\frac {x}{\sqrt [4]{b+a x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(b + a*x^2)^(1/4)/x] - ArcTanh[x/(b + a*x^2)^(1/4)]

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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((a*x^2+2*b)/(a*x^2+b)^(1/4)/(x^4-a*x^2-b),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 {a x^{2} + 2 \, b}{{\left (x^{4} - a x^{2} - b\right )} {\left (a x^{2} + b\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {a\,x^2+2\,b}{{\left (a\,x^2+b\right )}^{1/4}\,\left (-x^4+a\,x^2+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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