3.1221 \(\int \frac {1}{\sqrt [4]{a-b x^4}} \, dx\)

Optimal. Leaf size=209 \[ -\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\log \left (-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{4 \sqrt {2} \sqrt [4]{b}}+\frac {\log \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{4 \sqrt {2} \sqrt [4]{b}} \]

[Out]

1/4*arctan(-1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4))/b^(1/4)*2^(1/2)+1/4*arctan(1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1
/4))/b^(1/4)*2^(1/2)-1/8*ln(1-b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4)+x^2*b^(1/2)/(-b*x^4+a)^(1/2))/b^(1/4)*2^(1/2)
+1/8*ln(1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4)+x^2*b^(1/2)/(-b*x^4+a)^(1/2))/b^(1/4)*2^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {240, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\log \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{4 \sqrt {2} \sqrt [4]{b}}+\frac {\log \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{4 \sqrt {2} \sqrt [4]{b}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{2 \sqrt {2} \sqrt [4]{b}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^4)^(-1/4),x]

[Out]

-ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)]/(2*Sqrt[2]*b^(1/4)) + ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/(a - b
*x^4)^(1/4)]/(2*Sqrt[2]*b^(1/4)) - Log[1 + (Sqrt[b]*x^2)/Sqrt[a - b*x^4] - (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/
4)]/(4*Sqrt[2]*b^(1/4)) + Log[1 + (Sqrt[b]*x^2)/Sqrt[a - b*x^4] + (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)]/(4*Sq
rt[2]*b^(1/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

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

Rule 1165

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 173, normalized size = 0.83 \[ \frac {-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )-\log \left (-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )+\log \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{4 \sqrt {2} \sqrt [4]{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^4)^(-1/4),x]

[Out]

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

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fricas [A]  time = 0.63, size = 148, normalized size = 0.71 \[ \left (-\frac {1}{b}\right )^{\frac {1}{4}} \arctan \left (\frac {x \left (-\frac {1}{b}\right )^{\frac {1}{4}} \sqrt {-\frac {b x^{2} \sqrt {-\frac {1}{b}} - \sqrt {-b x^{4} + a}}{x^{2}}} - {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-\frac {1}{b}\right )^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \log \left (\frac {b x \left (-\frac {1}{b}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \log \left (-\frac {b x \left (-\frac {1}{b}\right )^{\frac {3}{4}} - {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/b)^(1/4)*arctan((x*(-1/b)^(1/4)*sqrt(-(b*x^2*sqrt(-1/b) - sqrt(-b*x^4 + a))/x^2) - (-b*x^4 + a)^(1/4)*(-1/
b)^(1/4))/x) - 1/4*(-1/b)^(1/4)*log((b*x*(-1/b)^(3/4) + (-b*x^4 + a)^(1/4))/x) + 1/4*(-1/b)^(1/4)*log(-(b*x*(-
1/b)^(3/4) - (-b*x^4 + a)^(1/4))/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-b*x^4 + a)^(-1/4), x)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.32, size = 176, normalized size = 0.84 \[ -\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{4 \, b^{\frac {1}{4}}} - \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{4 \, b^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} + \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{8 \, b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {b} - \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{8 \, b^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(1/4) - 1/4*sqrt(2)*arct
an(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(1/4) + 1/8*sqrt(2)*log(sqrt(b) + sqrt(2
)*(-b*x^4 + a)^(1/4)*b^(1/4)/x + sqrt(-b*x^4 + a)/x^2)/b^(1/4) - 1/8*sqrt(2)*log(sqrt(b) - sqrt(2)*(-b*x^4 + a
)^(1/4)*b^(1/4)/x + sqrt(-b*x^4 + a)/x^2)/b^(1/4)

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mupad [B]  time = 1.10, size = 38, normalized size = 0.18 \[ \frac {x\,{\left (1-\frac {b\,x^4}{a}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ \frac {b\,x^4}{a}\right )}{{\left (a-b\,x^4\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - b*x^4)^(1/4),x)

[Out]

(x*(1 - (b*x^4)/a)^(1/4)*hypergeom([1/4, 1/4], 5/4, (b*x^4)/a))/(a - b*x^4)^(1/4)

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sympy [C]  time = 1.58, size = 37, normalized size = 0.18 \[ \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*gamma(1/4)*hyper((1/4, 1/4), (5/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*a**(1/4)*gamma(5/4))

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