3.1191 \(\int \frac {\sqrt [4]{a-b x^4}}{x^2} \, dx\)

Optimal. Leaf size=226 \[ -\frac {\sqrt [4]{a-b x^4}}{x}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \log \left (-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{4 \sqrt {2}}+\frac {\sqrt [4]{b} \log \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{4 \sqrt {2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {277, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\sqrt [4]{a-b x^4}}{x}-\frac {\sqrt [4]{b} \log \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{4 \sqrt {2}}+\frac {\sqrt [4]{b} \log \left (\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{4 \sqrt {2}}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^4)^(1/4)/x^2,x]

[Out]

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

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 277

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

Rule 297

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

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 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 {\sqrt [4]{a-b x^4}}{x^2} \, dx &=-\frac {\sqrt [4]{a-b x^4}}{x}-b \int \frac {x^2}{\left (a-b x^4\right )^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{a-b x^4}}{x}-b \operatorname {Subst}\left (\int \frac {x^2}{1+b x^4} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )\\ &=-\frac {\sqrt [4]{a-b x^4}}{x}+\frac {1}{2} \sqrt {b} \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} \sqrt {b} \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 {\sqrt [4]{a-b x^4}}{x}-\frac {1}{4} \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 )-\frac {1}{4} \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 )-\frac {\sqrt [4]{b} \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}}-\frac {\sqrt [4]{b} \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}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{x}-\frac {\sqrt [4]{b} \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}}+\frac {\sqrt [4]{b} \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}}-\frac {\sqrt [4]{b} \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}}+\frac {\sqrt [4]{b} \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}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{x}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \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}}+\frac {\sqrt [4]{b} \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}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 50, normalized size = 0.22 \[ -\frac {\sqrt [4]{a-b x^4} \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};\frac {b x^4}{a}\right )}{x \sqrt [4]{1-\frac {b x^4}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^4)^(1/4)/x^2,x]

[Out]

-(((a - b*x^4)^(1/4)*Hypergeometric2F1[-1/4, -1/4, 3/4, (b*x^4)/a])/(x*(1 - (b*x^4)/a)^(1/4)))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-b*x^4 + a)^(1/4)/x^2, x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.05, size = 191, normalized size = 0.85 \[ \frac {1}{4} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right ) + \frac {1}{8} \, \sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {b} + \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right ) - \frac {1}{8} \, \sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {b} - \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right ) - \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.28, size = 41, normalized size = 0.18 \[ -\frac {{\left (a-b\,x^4\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ \frac {b\,x^4}{a}\right )}{x\,{\left (1-\frac {b\,x^4}{a}\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^4)^(1/4)/x^2,x)

[Out]

-((a - b*x^4)^(1/4)*hypergeom([-1/4, -1/4], 3/4, (b*x^4)/a))/(x*(1 - (b*x^4)/a)^(1/4))

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sympy [C]  time = 1.86, size = 42, normalized size = 0.19 \[ \frac {\sqrt [4]{a} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(1/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*x*gamma(3/4))

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