3.1200 \(\int \frac {\sqrt [4]{a-b x^4}}{x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 85, 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 = {277, 237, 335, 275, 232} \[ \frac {b^{3/2} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} F\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} \left (a-b x^4\right )^{3/4}}-\frac {\sqrt [4]{a-b x^4}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^4)^(1/4)/x^4,x]

[Out]

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

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]

Rule 237

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

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

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.61 \[ -\frac {\sqrt [4]{a-b x^4} \, _2F_1\left (-\frac {3}{4},-\frac {1}{4};\frac {1}{4};\frac {b x^4}{a}\right )}{3 x^3 \sqrt [4]{1-\frac {b x^4}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^4)^(1/4)/x^4,x]

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-b*x^4 + a)^(1/4)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-b*x^4 + a)^(1/4)/x^4, x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-b*x^4 + a)^(1/4)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^4)^(1/4)/x^4,x)

[Out]

int((a - b*x^4)^(1/4)/x^4, x)

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sympy [C]  time = 1.99, size = 34, normalized size = 0.40 \[ - \frac {i \sqrt [4]{b} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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