3.1231 \(\int \frac {1}{x^6 \sqrt [4]{a-b x^4}} \, dx\)

Optimal. Leaf size=86 \[ -\frac {2 b^{3/2} x \sqrt [4]{1-\frac {a}{b x^4}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt [4]{a-b x^4}}-\frac {\left (a-b x^4\right )^{3/4}}{5 a x^5} \]

[Out]

-1/5*(-b*x^4+a)^(3/4)/a/x^5-2/5*b^(3/2)*(1-a/b/x^4)^(1/4)*x*(cos(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos
(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccsc(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(-b*x^4+a)^(
1/4)

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Rubi [A]  time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {325, 313, 335, 275, 228} \[ -\frac {2 b^{3/2} x \sqrt [4]{1-\frac {a}{b x^4}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt [4]{a-b x^4}}-\frac {\left (a-b x^4\right )^{3/4}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a - b*x^4)^(1/4)),x]

[Out]

-(a - b*x^4)^(3/4)/(5*a*x^5) - (2*b^(3/2)*(1 - a/(b*x^4))^(1/4)*x*EllipticE[ArcCsc[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2
])/(5*a^(3/2)*(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 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 313

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

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 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \sqrt [4]{a-b x^4}} \, dx &=-\frac {\left (a-b x^4\right )^{3/4}}{5 a x^5}+\frac {(2 b) \int \frac {1}{x^2 \sqrt [4]{a-b x^4}} \, dx}{5 a}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{5 a x^5}+\frac {\left (2 b \sqrt [4]{1-\frac {a}{b x^4}} x\right ) \int \frac {1}{\sqrt [4]{1-\frac {a}{b x^4}} x^3} \, dx}{5 a \sqrt [4]{a-b x^4}}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{5 a x^5}-\frac {\left (2 b \sqrt [4]{1-\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt [4]{1-\frac {a x^4}{b}}} \, dx,x,\frac {1}{x}\right )}{5 a \sqrt [4]{a-b x^4}}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{5 a x^5}-\frac {\left (b \sqrt [4]{1-\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {a x^2}{b}}} \, dx,x,\frac {1}{x^2}\right )}{5 a \sqrt [4]{a-b x^4}}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{5 a x^5}-\frac {2 b^{3/2} \sqrt [4]{1-\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt [4]{a-b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.60 \[ -\frac {\sqrt [4]{1-\frac {b x^4}{a}} \, _2F_1\left (-\frac {5}{4},\frac {1}{4};-\frac {1}{4};\frac {b x^4}{a}\right )}{5 x^5 \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a - b*x^4)^(1/4)),x]

[Out]

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

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{b x^{10} - a x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^4 + a)^(3/4)/(b*x^10 - a*x^6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((-b*x^4 + a)^(1/4)*x^6), x)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}} x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(-b*x^4+a)^(1/4),x)

[Out]

int(1/x^6/(-b*x^4+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^4 + a)^(1/4)*x^6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^6\,{\left (a-b\,x^4\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a - b*x^4)^(1/4)),x)

[Out]

int(1/(x^6*(a - b*x^4)^(1/4)), x)

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sympy [C]  time = 2.55, size = 34, normalized size = 0.40 \[ - \frac {i e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{6 \sqrt [4]{b} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(-b*x**4+a)**(1/4),x)

[Out]

-I*exp(-3*I*pi/4)*hyper((1/4, 3/2), (5/2,), a/(b*x**4))/(6*b**(1/4)*x**6)

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