3.1246 \(\int \frac{1}{x^3 (a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=85 \[ \frac{\sqrt{b} \left (1-\frac{b x^4}{a}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{2 \sqrt{a} \left (a-b x^4\right )^{3/4}}-\frac{\sqrt [4]{a-b x^4}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.0463563, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {275, 325, 233, 232} \[ \frac{\sqrt{b} \left (1-\frac{b x^4}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{2 \sqrt{a} \left (a-b x^4\right )^{3/4}}-\frac{\sqrt [4]{a-b x^4}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a - b*x^4)^(3/4)),x]

[Out]

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

Mathematica [C]  time = 0.0112614, size = 52, normalized size = 0.61 \[ -\frac{\left (1-\frac{b x^4}{a}\right )^{3/4} \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{1}{2};\frac{b x^4}{a}\right )}{2 x^2 \left (a-b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ( -b{x}^{4}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(-b*x^4+a)^(3/4),x)

[Out]

int(1/x^3/(-b*x^4+a)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-b x^{4} + a\right )}^{\frac{3}{4}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^4 + a)^(3/4)*x^3), 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}}}{b x^{7} - a x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^4 + a)^(1/4)/(b*x^7 - a*x^3), x)

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Sympy [C]  time = 1.2458, size = 32, normalized size = 0.38 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{4} e^{2 i \pi }}{a}} \right )}}{2 a^{\frac{3}{4}} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(-b*x**4+a)**(3/4),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-b x^{4} + a\right )}^{\frac{3}{4}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((-b*x^4 + a)^(3/4)*x^3), x)