3.9.44 \(\int \frac {1}{x^4 \sqrt {a-b x^4}} \, dx\) [844]

Optimal. Leaf size=79 \[ -\frac {\sqrt {a-b x^4}}{3 a x^3}+\frac {b^{3/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{3 a^{3/4} \sqrt {a-b x^4}} \]

[Out]

-1/3*(-b*x^4+a)^(1/2)/a/x^3+1/3*b^(3/4)*EllipticF(b^(1/4)*x/a^(1/4),I)*(1-b*x^4/a)^(1/2)/a^(3/4)/(-b*x^4+a)^(1
/2)

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Rubi [A]
time = 0.02, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {331, 230, 227} \begin {gather*} \frac {b^{3/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\text {ArcSin}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{3 a^{3/4} \sqrt {a-b x^4}}-\frac {\sqrt {a-b x^4}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[a - b*x^4]),x]

[Out]

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

Rule 227

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

Rule 230

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + b*(x^4/a)]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + b*(x^4/
a)], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 331

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 52, normalized size = 0.66 \begin {gather*} -\frac {\sqrt {1-\frac {b x^4}{a}} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};\frac {b x^4}{a}\right )}{3 x^3 \sqrt {a-b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[a - b*x^4]),x]

[Out]

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

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Maple [A]
time = 0.15, size = 88, normalized size = 1.11

method	result	size
default	\(-\frac {\sqrt {-b \,x^{4}+a}}{3 a \,x^{3}}+\frac {b \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(88\)
risch	\(-\frac {\sqrt {-b \,x^{4}+a}}{3 a \,x^{3}}+\frac {b \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(88\)
elliptic	\(-\frac {\sqrt {-b \,x^{4}+a}}{3 a \,x^{3}}+\frac {b \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(-b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(-b*x^4+a)^(1/2)/a/x^3+1/3*b/a/(1/a^(1/2)*b^(1/2))^(1/2)*(1-x^2*b^(1/2)/a^(1/2))^(1/2)*(1+x^2*b^(1/2)/a^(
1/2))^(1/2)/(-b*x^4+a)^(1/2)*EllipticF(x*(1/a^(1/2)*b^(1/2))^(1/2),I)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.08, size = 47, normalized size = 0.59 \begin {gather*} \frac {\sqrt {a} x^{3} \left (\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - \sqrt {-b x^{4} + a}}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(a)*x^3*(b/a)^(3/4)*elliptic_f(arcsin(x*(b/a)^(1/4)), -1) - sqrt(-b*x^4 + a))/(a*x^3)

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Sympy [A]
time = 0.45, size = 42, normalized size = 0.53 \begin {gather*} \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} x^{3} \Gamma \left (\frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^4\,\sqrt {a-b\,x^4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a - b*x^4)^(1/2)),x)

[Out]

int(1/(x^4*(a - b*x^4)^(1/2)), x)

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