\(\int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx\) [1202]

Optimal result

Integrand size = 16, antiderivative size = 133 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{77 a^2 x^3}+\frac {4 b^{7/2} \left (1-\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{77 a^{5/2} \left (a-b x^4\right )^{3/4}} \]

[Out]

-1/11*(-b*x^4+a)^(1/4)/x^11+1/77*b*(-b*x^4+a)^(1/4)/a/x^7+2/77*b^2*(-b*x^4+a)^(1/4)/a^2/x^3+4/77*b^(7/2)*(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))/a^(5/2)/(-b*x^4+a)^(3/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {283, 331, 243, 342, 281, 238} \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\frac {4 b^{7/2} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{77 a^{5/2} \left (a-b x^4\right )^{3/4}}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{77 a^2 x^3}-\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7} \]

[In]

Int[(a - b*x^4)^(1/4)/x^12,x]

[Out]

-1/11*(a - b*x^4)^(1/4)/x^11 + (b*(a - b*x^4)^(1/4))/(77*a*x^7) + (2*b^2*(a - b*x^4)^(1/4))/(77*a^2*x^3) + (4*
b^(7/2)*(1 - a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCsc[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(77*a^(5/2)*(a - b*x^4)^(3/4)
)

Rule 238

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2]))*EllipticF[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 243

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 281

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 283

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 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]

Rule 342

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*} \text {integral}& = -\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}-\frac {1}{11} b \int \frac {1}{x^8 \left (a-b x^4\right )^{3/4}} \, dx \\ & = -\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7}-\frac {\left (6 b^2\right ) \int \frac {1}{x^4 \left (a-b x^4\right )^{3/4}} \, dx}{77 a} \\ & = -\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{77 a^2 x^3}-\frac {\left (4 b^3\right ) \int \frac {1}{\left (a-b x^4\right )^{3/4}} \, dx}{77 a^2} \\ & = -\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{77 a^2 x^3}-\frac {\left (4 b^3 \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}{77 a^2 \left (a-b x^4\right )^{3/4}} \\ & = -\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{77 a^2 x^3}+\frac {\left (4 b^3 \left (1-\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {x}{\left (1-\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{77 a^2 \left (a-b x^4\right )^{3/4}} \\ & = -\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{77 a^2 x^3}+\frac {\left (2 b^3 \left (1-\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{77 a^2 \left (a-b x^4\right )^{3/4}} \\ & = -\frac {\sqrt [4]{a-b x^4}}{11 x^{11}}+\frac {b \sqrt [4]{a-b x^4}}{77 a x^7}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{77 a^2 x^3}+\frac {4 b^{7/2} \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 )}{77 a^{5/2} \left (a-b x^4\right )^{3/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=-\frac {\sqrt [4]{a-b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {11}{4},-\frac {1}{4},-\frac {7}{4},\frac {b x^4}{a}\right )}{11 x^{11} \sqrt [4]{1-\frac {b x^4}{a}}} \]

[In]

Integrate[(a - b*x^4)^(1/4)/x^12,x]

[Out]

-1/11*((a - b*x^4)^(1/4)*Hypergeometric2F1[-11/4, -1/4, -7/4, (b*x^4)/a])/(x^11*(1 - (b*x^4)/a)^(1/4))

Maple [F]

\[\int \frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{12}}d x\]

[In]

int((-b*x^4+a)^(1/4)/x^12,x)

[Out]

int((-b*x^4+a)^(1/4)/x^12,x)

Fricas [F]

\[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}} \,d x } \]

[In]

integrate((-b*x^4+a)^(1/4)/x^12,x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(1/4)/x^12, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.83 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.26 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=- \frac {i \sqrt [4]{b} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{10 x^{10}} \]

[In]

integrate((-b*x**4+a)**(1/4)/x**12,x)

[Out]

-I*b**(1/4)*exp(-I*pi/4)*hyper((-1/4, 5/2), (7/2,), a/(b*x**4))/(10*x**10)

Maxima [F]

\[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}} \,d x } \]

[In]

integrate((-b*x^4+a)^(1/4)/x^12,x, algorithm="maxima")

[Out]

integrate((-b*x^4 + a)^(1/4)/x^12, x)

Giac [F]

\[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{12}} \,d x } \]

[In]

integrate((-b*x^4+a)^(1/4)/x^12,x, algorithm="giac")

[Out]

integrate((-b*x^4 + a)^(1/4)/x^12, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a-b x^4}}{x^{12}} \, dx=\int \frac {{\left (a-b\,x^4\right )}^{1/4}}{x^{12}} \,d x \]

[In]

int((a - b*x^4)^(1/4)/x^12,x)

[Out]

int((a - b*x^4)^(1/4)/x^12, x)