3.1216 \(\int \frac {1}{x^3 \sqrt [4]{a-b x^4}} \, dx\)

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

[Out]

-1/2*(-b*x^4+a)^(3/4)/a/x^2-1/2*(1-b*x^4/a)^(1/4)*(cos(1/2*arcsin(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcsi
n(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arcsin(x^2*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)/(-b*x^4+a)^(1/4)/a^(1/
2)

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Rubi [A]  time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {275, 325, 229, 228} \[ -\frac {\left (a-b x^4\right )^{3/4}}{2 a x^2}-\frac {\sqrt {b} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a - b*x^4)^(1/4)),x]

[Out]

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

Rule 228

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

Rule 229

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

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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.61 \[ -\frac {\sqrt [4]{1-\frac {b x^4}{a}} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};\frac {b x^4}{a}\right )}{2 x^2 \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{b x^{7} - a x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}} x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^3\,{\left (a-b\,x^4\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a - b*x^4)^(1/4)),x)

[Out]

int(1/(x^3*(a - b*x^4)^(1/4)), x)

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sympy [C]  time = 1.41, size = 32, normalized size = 0.38 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{2 \sqrt [4]{a} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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