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

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

[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])

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Rubi [A]  time = 0.110774, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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 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 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 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 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 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 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]

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]

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*}

Mathematica [C]  time = 0.0089847, 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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Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}\sqrt [4]{-b{x}^{4}+a}}\, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Sympy [C]  time = 1.54517, size = 42, normalized size = 0.19 \begin{align*} \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 )} \end{align*}

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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Giac [A]  time = 1.17781, size = 258, normalized size = 1.14 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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