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

Optimal. Leaf size=83 \[ -\frac {2 \sqrt [4]{a x^4-b x^2}}{x}-\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^2}}\right )+\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^2}}\right ) \]

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Rubi [A]  time = 0.17, antiderivative size = 153, normalized size of antiderivative = 1.84, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2020, 2032, 329, 331, 298, 203, 206} \begin {gather*} -\frac {2 \sqrt [4]{a x^4-b x^2}}{x}-\frac {\sqrt [4]{a} x^{3/2} \left (a x^2-b\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{\left (a x^4-b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} x^{3/2} \left (a x^2-b\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{\left (a x^4-b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(b*x^2) + a*x^4)^(1/4)/x^2,x]

[Out]

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

Rule 203

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

Rule 206

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(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), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && 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 2020

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

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^2} \, dx &=-\frac {2 \sqrt [4]{-b x^2+a x^4}}{x}+a \int \frac {x^2}{\left (-b x^2+a x^4\right )^{3/4}} \, dx\\ &=-\frac {2 \sqrt [4]{-b x^2+a x^4}}{x}+\frac {\left (a x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (-b+a x^2\right )^{3/4}} \, dx}{\left (-b x^2+a x^4\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{-b x^2+a x^4}}{x}+\frac {\left (2 a x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\left (-b x^2+a x^4\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{-b x^2+a x^4}}{x}+\frac {\left (2 a x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\left (-b x^2+a x^4\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{-b x^2+a x^4}}{x}+\frac {\left (\sqrt {a} x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\left (-b x^2+a x^4\right )^{3/4}}-\frac {\left (\sqrt {a} x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\left (-b x^2+a x^4\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{-b x^2+a x^4}}{x}-\frac {\sqrt [4]{a} x^{3/2} \left (-b+a x^2\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\left (-b x^2+a x^4\right )^{3/4}}+\frac {\sqrt [4]{a} x^{3/2} \left (-b+a x^2\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\left (-b x^2+a x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.65 \begin {gather*} -\frac {2 \sqrt [4]{a x^4-b x^2} \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};\frac {a x^2}{b}\right )}{x \sqrt [4]{1-\frac {a x^2}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-(b*x^2) + a*x^4)^(1/4)*Hypergeometric2F1[-1/4, -1/4, 3/4, (a*x^2)/b])/(x*(1 - (a*x^2)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.20, size = 83, normalized size = 1.00 \begin {gather*} -\frac {2 \sqrt [4]{-b x^2+a x^4}}{x}-\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )+\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-(b*x^2) + a*x^4)^(1/4))/x - a^(1/4)*ArcTan[(a^(1/4)*x)/(-(b*x^2) + a*x^4)^(1/4)] + a^(1/4)*ArcTanh[(a^(1
/4)*x)/(-(b*x^2) + a*x^4)^(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^4-b*x^2)^(1/4)/x^2,x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.19, size = 192, normalized size = 2.31 \begin {gather*} \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.03, size = 45, normalized size = 0.54 \begin {gather*} -\frac {2\,{\left (a\,x^4-b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ \frac {a\,x^2}{b}\right )}{x\,{\left (1-\frac {a\,x^2}{b}\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4 - b*x^2)^(1/4)/x^2,x)

[Out]

-(2*(a*x^4 - b*x^2)^(1/4)*hypergeom([-1/4, -1/4], 3/4, (a*x^2)/b))/(x*(1 - (a*x^2)/b)^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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