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

Optimal. Leaf size=54 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a x^2+b}}{x}\right )}{2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{a x^2+b}}\right )}{2^{3/4}} \]

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Rubi [C]  time = 0.60, antiderivative size = 443, normalized size of antiderivative = 8.20, 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+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {\sqrt {b}}{\sqrt {a^2-\sqrt {a^2+2 b} a+b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {-a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {\sqrt {b}}{\sqrt {a^2-\sqrt {a^2+2 b} a+b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {-a \sqrt {a^2+2 b}+a^2+b}}+\frac {\sqrt [4]{b} \left (\sqrt {a^2+2 b}+a\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {\sqrt {b}}{\sqrt {a^2+\sqrt {a^2+2 b} a+b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{b} \left (\sqrt {a^2+2 b}+a\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {\sqrt {b}}{\sqrt {a^2+\sqrt {a^2+2 b} a+b}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 x \sqrt {a \sqrt {a^2+2 b}+a^2+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^(1/4)*(a - Sqrt[a^2 + 2*b])*Sqrt[-((a*x^2)/b)]*EllipticPi[-(Sqrt[b]/Sqrt[a^2 + b - a*Sqrt[a^2 + 2*b]]), Arc
Sin[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(2*Sqrt[a^2 + b - a*Sqrt[a^2 + 2*b]]*x) - (b^(1/4)*(a - Sqrt[a^2 + 2*b])*
Sqrt[-((a*x^2)/b)]*EllipticPi[Sqrt[b]/Sqrt[a^2 + b - a*Sqrt[a^2 + 2*b]], ArcSin[(b + a*x^2)^(1/4)/b^(1/4)], -1
])/(2*Sqrt[a^2 + b - a*Sqrt[a^2 + 2*b]]*x) + (b^(1/4)*(a + Sqrt[a^2 + 2*b])*Sqrt[-((a*x^2)/b)]*EllipticPi[-(Sq
rt[b]/Sqrt[a^2 + b + a*Sqrt[a^2 + 2*b]]), ArcSin[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(2*Sqrt[a^2 + b + a*Sqrt[a^2
+ 2*b]]*x) - (b^(1/4)*(a + Sqrt[a^2 + 2*b])*Sqrt[-((a*x^2)/b)]*EllipticPi[Sqrt[b]/Sqrt[a^2 + b + a*Sqrt[a^2 +
2*b]], ArcSin[(b + a*x^2)^(1/4)/b^(1/4)], -1])/(2*Sqrt[a^2 + b + a*Sqrt[a^2 + 2*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 (-2 b-2 a x^2+x^4\right )} \, dx &=\int \left (\frac {a+\sqrt {a^2+2 b}}{\left (-2 a-2 \sqrt {a^2+2 b}+2 x^2\right ) \sqrt [4]{b+a x^2}}+\frac {a-\sqrt {a^2+2 b}}{\left (-2 a+2 \sqrt {a^2+2 b}+2 x^2\right ) \sqrt [4]{b+a x^2}}\right ) \, dx\\ &=\left (a-\sqrt {a^2+2 b}\right ) \int \frac {1}{\left (-2 a+2 \sqrt {a^2+2 b}+2 x^2\right ) \sqrt [4]{b+a x^2}} \, dx+\left (a+\sqrt {a^2+2 b}\right ) \int \frac {1}{\left (-2 a-2 \sqrt {a^2+2 b}+2 x^2\right ) \sqrt [4]{b+a x^2}} \, dx\\ &=\frac {\left (2 \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-2 b+a \left (-2 a+2 \sqrt {a^2+2 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+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-2 b+a \left (-2 a-2 \sqrt {a^2+2 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+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b-a \sqrt {a^2+2 b}}-x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}+\frac {\left (\left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b-a \sqrt {a^2+2 b}}+x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}-\frac {\left (\left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b+a \sqrt {a^2+2 b}}-x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}+\frac {\left (\left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b+a \sqrt {a^2+2 b}}+x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}\\ &=\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {\sqrt {b}}{\sqrt {a^2+b-a \sqrt {a^2+2 b}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {a^2+b-a \sqrt {a^2+2 b}} x}-\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {\sqrt {b}}{\sqrt {a^2+b-a \sqrt {a^2+2 b}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {a^2+b-a \sqrt {a^2+2 b}} x}+\frac {\sqrt [4]{b} \left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (-\frac {\sqrt {b}}{\sqrt {a^2+b+a \sqrt {a^2+2 b}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {a^2+b+a \sqrt {a^2+2 b}} x}-\frac {\sqrt [4]{b} \left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \Pi \left (\frac {\sqrt {b}}{\sqrt {a^2+b+a \sqrt {a^2+2 b}}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )\right |-1\right )}{2 \sqrt {a^2+b+a \sqrt {a^2+2 b}} x}\\ \end {align*}

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Mathematica [F]  time = 0.23, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 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)*(-2*b - 2*a*x^2 + x^4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 54, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{b+a x^2}}{x}\right )}{2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{b+a x^2}}\right )}{2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(2^(1/4)*(b + a*x^2)^(1/4))/x]/2^(3/4) - ArcTanh[x/(2^(1/4)*(b + a*x^2)^(1/4))]/2^(3/4)

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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^4-2*a*x^2-2*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} - 2 \, a x^{2} - 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^4-2*a*x^2-2*b),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*x^2+2*b)/(a*x^2+b)^(1/4)/(x^4-2*a*x^2-2*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} - 2 \, a x^{2} - 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^4-2*a*x^2-2*b),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int -\frac {a\,x^2+2\,b}{{\left (a\,x^2+b\right )}^{1/4}\,\left (-x^4+2\,a\,x^2+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)*(2*b + 2*a*x^2 - x^4)),x)

[Out]

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

[Out]

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

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