\(\int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx\) [1187]

Optimal result

Integrand size = 16, antiderivative size = 105 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx=-\frac {\sqrt [4]{a-b x^4}}{6 x^6}+\frac {b \sqrt [4]{a-b x^4}}{12 a x^2}-\frac {b^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 \sqrt {a} \left (a-b x^4\right )^{3/4}} \]

[Out]

-1/6*(-b*x^4+a)^(1/4)/x^6+1/12*b*(-b*x^4+a)^(1/4)/a/x^2-1/12*b^(3/2)*(1-b*x^4/a)^(3/4)*(cos(1/2*arcsin(x^2*b^(
1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcsin(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arcsin(x^2*b^(1/2)/a^(1/2))),2^
(1/2))/(-b*x^4+a)^(3/4)/a^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {281, 283, 331, 239, 238} \[ \int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx=-\frac {b^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 \sqrt {a} \left (a-b x^4\right )^{3/4}}-\frac {\sqrt [4]{a-b x^4}}{6 x^6}+\frac {b \sqrt [4]{a-b x^4}}{12 a x^2} \]

[In]

Int[(a - b*x^4)^(1/4)/x^7,x]

[Out]

-1/6*(a - b*x^4)^(1/4)/x^6 + (b*(a - b*x^4)^(1/4))/(12*a*x^2) - (b^(3/2)*(1 - (b*x^4)/a)^(3/4)*EllipticF[ArcSi
n[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(12*Sqrt[a]*(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 239

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\sqrt [4]{a-b x^2}}{x^4} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt [4]{a-b x^4}}{6 x^6}-\frac {1}{12} b \text {Subst}\left (\int \frac {1}{x^2 \left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt [4]{a-b x^4}}{6 x^6}+\frac {b \sqrt [4]{a-b x^4}}{12 a x^2}-\frac {b^2 \text {Subst}\left (\int \frac {1}{\left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{24 a} \\ & = -\frac {\sqrt [4]{a-b x^4}}{6 x^6}+\frac {b \sqrt [4]{a-b x^4}}{12 a x^2}-\frac {\left (b^2 \left (1-\frac {b x^4}{a}\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{24 a \left (a-b x^4\right )^{3/4}} \\ & = -\frac {\sqrt [4]{a-b x^4}}{6 x^6}+\frac {b \sqrt [4]{a-b x^4}}{12 a x^2}-\frac {b^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 \sqrt {a} \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 = 0.50 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx=-\frac {\sqrt [4]{a-b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},-\frac {1}{2},\frac {b x^4}{a}\right )}{6 x^6 \sqrt [4]{1-\frac {b x^4}{a}}} \]

[In]

Integrate[(a - b*x^4)^(1/4)/x^7,x]

[Out]

-1/6*((a - b*x^4)^(1/4)*Hypergeometric2F1[-3/2, -1/4, -1/2, (b*x^4)/a])/(x^6*(1 - (b*x^4)/a)^(1/4))

Maple [F]

\[\int \frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{7}}d x\]

[In]

int((-b*x^4+a)^(1/4)/x^7,x)

[Out]

int((-b*x^4+a)^(1/4)/x^7,x)

Fricas [F]

\[ \int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{7}} \,d x } \]

[In]

integrate((-b*x^4+a)^(1/4)/x^7,x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(1/4)/x^7, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.61 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.34 \[ \int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx=- \frac {\sqrt [4]{a} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{6 x^{6}} \]

[In]

integrate((-b*x**4+a)**(1/4)/x**7,x)

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{7}} \,d x } \]

[In]

integrate((-b*x^4+a)^(1/4)/x^7,x, algorithm="maxima")

[Out]

integrate((-b*x^4 + a)^(1/4)/x^7, x)

Giac [F]

\[ \int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx=\int { \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{7}} \,d x } \]

[In]

integrate((-b*x^4+a)^(1/4)/x^7,x, algorithm="giac")

[Out]

integrate((-b*x^4 + a)^(1/4)/x^7, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a-b x^4}}{x^7} \, dx=\int \frac {{\left (a-b\,x^4\right )}^{1/4}}{x^7} \,d x \]

[In]

int((a - b*x^4)^(1/4)/x^7,x)

[Out]

int((a - b*x^4)^(1/4)/x^7, x)