3.1188 \(\int \frac{\sqrt [4]{a-b x^4}}{x^{11}} \, dx\)

Optimal. Leaf size=130 \[ -\frac{b^{5/2} \left (1-\frac{b x^4}{a}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{24 a^{3/2} \left (a-b x^4\right )^{3/4}}+\frac{b^2 \sqrt [4]{a-b x^4}}{24 a^2 x^2}+\frac{b \sqrt [4]{a-b x^4}}{60 a x^6}-\frac{\sqrt [4]{a-b x^4}}{10 x^{10}} \]

[Out]

-(a - b*x^4)^(1/4)/(10*x^10) + (b*(a - b*x^4)^(1/4))/(60*a*x^6) + (b^2*(a - b*x^4)^(1/4))/(24*a^2*x^2) - (b^(5
/2)*(1 - (b*x^4)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(24*a^(3/2)*(a - b*x^4)^(3/4))

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Rubi [A]  time = 0.0854398, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {275, 277, 325, 233, 232} \[ \frac{b^2 \sqrt [4]{a-b x^4}}{24 a^2 x^2}-\frac{b^{5/2} \left (1-\frac{b x^4}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{24 a^{3/2} \left (a-b x^4\right )^{3/4}}+\frac{b \sqrt [4]{a-b x^4}}{60 a x^6}-\frac{\sqrt [4]{a-b x^4}}{10 x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^4)^(1/4)/x^11,x]

[Out]

-(a - b*x^4)^(1/4)/(10*x^10) + (b*(a - b*x^4)^(1/4))/(60*a*x^6) + (b^2*(a - b*x^4)^(1/4))/(24*a^2*x^2) - (b^(5
/2)*(1 - (b*x^4)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(24*a^(3/2)*(a - b*x^4)^(3/4))

Rule 275

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

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 325

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

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

\begin{align*} \int \frac{\sqrt [4]{a-b x^4}}{x^{11}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt [4]{a-b x^2}}{x^6} \, dx,x,x^2\right )\\ &=-\frac{\sqrt [4]{a-b x^4}}{10 x^{10}}-\frac{1}{20} b \operatorname{Subst}\left (\int \frac{1}{x^4 \left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt [4]{a-b x^4}}{10 x^{10}}+\frac{b \sqrt [4]{a-b x^4}}{60 a x^6}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{24 a}\\ &=-\frac{\sqrt [4]{a-b x^4}}{10 x^{10}}+\frac{b \sqrt [4]{a-b x^4}}{60 a x^6}+\frac{b^2 \sqrt [4]{a-b x^4}}{24 a^2 x^2}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{\left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{48 a^2}\\ &=-\frac{\sqrt [4]{a-b x^4}}{10 x^{10}}+\frac{b \sqrt [4]{a-b x^4}}{60 a x^6}+\frac{b^2 \sqrt [4]{a-b x^4}}{24 a^2 x^2}-\frac{\left (b^3 \left (1-\frac{b x^4}{a}\right )^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{48 a^2 \left (a-b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a-b x^4}}{10 x^{10}}+\frac{b \sqrt [4]{a-b x^4}}{60 a x^6}+\frac{b^2 \sqrt [4]{a-b x^4}}{24 a^2 x^2}-\frac{b^{5/2} \left (1-\frac{b x^4}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{24 a^{3/2} \left (a-b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0121116, size = 52, normalized size = 0.4 \[ -\frac{\sqrt [4]{a-b x^4} \, _2F_1\left (-\frac{5}{2},-\frac{1}{4};-\frac{3}{2};\frac{b x^4}{a}\right )}{10 x^{10} \sqrt [4]{1-\frac{b x^4}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^4)^(1/4)/x^11,x]

[Out]

-((a - b*x^4)^(1/4)*Hypergeometric2F1[-5/2, -1/4, -3/2, (b*x^4)/a])/(10*x^10*(1 - (b*x^4)/a)^(1/4))

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{11}}\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^11,x)

[Out]

int((-b*x^4+a)^(1/4)/x^11,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b x^{4} + a\right )}^{\frac{1}{4}}}{x^{11}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^11,x, algorithm="maxima")

[Out]

integrate((-b*x^4 + a)^(1/4)/x^11, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-b x^{4} + a\right )}^{\frac{1}{4}}}{x^{11}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^11,x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(1/4)/x^11, x)

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Sympy [C]  time = 2.76234, size = 36, normalized size = 0.28 \begin{align*} - \frac{\sqrt [4]{a}{{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{2}, - \frac{1}{4} \\ - \frac{3}{2} \end{matrix}\middle |{\frac{b x^{4} e^{2 i \pi }}{a}} \right )}}{10 x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**4+a)**(1/4)/x**11,x)

[Out]

-a**(1/4)*hyper((-5/2, -1/4), (-3/2,), b*x**4*exp_polar(2*I*pi)/a)/(10*x**10)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b x^{4} + a\right )}^{\frac{1}{4}}}{x^{11}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^11,x, algorithm="giac")

[Out]

integrate((-b*x^4 + a)^(1/4)/x^11, x)