\(\int \frac {1}{x^7 (a-b x^4)^{3/4}} \, dx\) [1247]

Optimal result

Integrand size = 16, antiderivative size = 108 \[ \int \frac {1}{x^7 \left (a-b x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a-b x^4}}{6 a x^6}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}+\frac {5 b^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 a^{3/2} \left (a-b x^4\right )^{3/4}} \]

[Out]

-1/6*(-b*x^4+a)^(1/4)/a/x^6-5/12*b*(-b*x^4+a)^(1/4)/a^2/x^2+5/12*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))/a^(3/2)/(-b*x^4+a)^(3/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {281, 331, 239, 238} \[ \int \frac {1}{x^7 \left (a-b x^4\right )^{3/4}} \, dx=\frac {5 b^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 a^{3/2} \left (a-b x^4\right )^{3/4}}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}-\frac {\sqrt [4]{a-b x^4}}{6 a x^6} \]

[In]

Int[1/(x^7*(a - b*x^4)^(3/4)),x]

[Out]

-1/6*(a - b*x^4)^(1/4)/(a*x^6) - (5*b*(a - b*x^4)^(1/4))/(12*a^2*x^2) + (5*b^(3/2)*(1 - (b*x^4)/a)^(3/4)*Ellip
ticF[ArcSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(12*a^(3/2)*(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 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 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 {1}{x^4 \left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt [4]{a-b x^4}}{6 a x^6}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{x^2 \left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{12 a} \\ & = -\frac {\sqrt [4]{a-b x^4}}{6 a x^6}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{\left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{24 a^2} \\ & = -\frac {\sqrt [4]{a-b x^4}}{6 a x^6}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}+\frac {\left (5 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^2 \left (a-b x^4\right )^{3/4}} \\ & = -\frac {\sqrt [4]{a-b x^4}}{6 a x^6}-\frac {5 b \sqrt [4]{a-b x^4}}{12 a^2 x^2}+\frac {5 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 a^{3/2} \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 = 0.48 \[ \int \frac {1}{x^7 \left (a-b x^4\right )^{3/4}} \, dx=-\frac {\left (1-\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},-\frac {1}{2},\frac {b x^4}{a}\right )}{6 x^6 \left (a-b x^4\right )^{3/4}} \]

[In]

Integrate[1/(x^7*(a - b*x^4)^(3/4)),x]

[Out]

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

Maple [F]

\[\int \frac {1}{x^{7} \left (-b \,x^{4}+a \right )^{\frac {3}{4}}}d x\]

[In]

int(1/x^7/(-b*x^4+a)^(3/4),x)

[Out]

int(1/x^7/(-b*x^4+a)^(3/4),x)

Fricas [F]

\[ \int \frac {1}{x^7 \left (a-b x^4\right )^{3/4}} \, dx=\int { \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}} x^{7}} \,d x } \]

[In]

integrate(1/x^7/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^4 + a)^(1/4)/(b*x^11 - a*x^7), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^7 \left (a-b x^4\right )^{3/4}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{6 a^{\frac {3}{4}} x^{6}} \]

[In]

integrate(1/x**7/(-b*x**4+a)**(3/4),x)

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^7 \left (a-b x^4\right )^{3/4}} \, dx=\int { \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}} x^{7}} \,d x } \]

[In]

integrate(1/x^7/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^4 + a)^(3/4)*x^7), x)

Giac [F]

\[ \int \frac {1}{x^7 \left (a-b x^4\right )^{3/4}} \, dx=\int { \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}} x^{7}} \,d x } \]

[In]

integrate(1/x^7/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((-b*x^4 + a)^(3/4)*x^7), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^7 \left (a-b x^4\right )^{3/4}} \, dx=\int \frac {1}{x^7\,{\left (a-b\,x^4\right )}^{3/4}} \,d x \]

[In]

int(1/(x^7*(a - b*x^4)^(3/4)),x)

[Out]

int(1/(x^7*(a - b*x^4)^(3/4)), x)