3.1217 \(\int \frac {1}{x^7 \sqrt [4]{a-b x^4}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {b^{3/2} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 a^{3/2} \sqrt [4]{a-b x^4}}-\frac {b \left (a-b x^4\right )^{3/4}}{4 a^2 x^2}-\frac {\left (a-b x^4\right )^{3/4}}{6 a x^6} \]

[Out]

-1/6*(-b*x^4+a)^(3/4)/a/x^6-1/4*b*(-b*x^4+a)^(3/4)/a^2/x^2-1/4*b^(3/2)*(1-b*x^4/a)^(1/4)*(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)))*EllipticE(sin(1/2*arcsin(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 = 0.07, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {275, 325, 229, 228} \[ -\frac {b^{3/2} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 a^{3/2} \sqrt [4]{a-b x^4}}-\frac {b \left (a-b x^4\right )^{3/4}}{4 a^2 x^2}-\frac {\left (a-b x^4\right )^{3/4}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a - b*x^4)^(1/4)),x]

[Out]

-(a - b*x^4)^(3/4)/(6*a*x^6) - (b*(a - b*x^4)^(3/4))/(4*a^2*x^2) - (b^(3/2)*(1 - (b*x^4)/a)^(1/4)*EllipticE[Ar
cSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*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 229

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

Rubi steps

\begin {align*} \int \frac {1}{x^7 \sqrt [4]{a-b x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt [4]{a-b x^2}} \, dx,x,x^2\right )\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{6 a x^6}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{a-b x^2}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{6 a x^6}-\frac {b \left (a-b x^4\right )^{3/4}}{4 a^2 x^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{a-b x^2}} \, dx,x,x^2\right )}{8 a^2}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{6 a x^6}-\frac {b \left (a-b x^4\right )^{3/4}}{4 a^2 x^2}-\frac {\left (b^2 \sqrt [4]{1-\frac {b x^4}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}} \, dx,x,x^2\right )}{8 a^2 \sqrt [4]{a-b x^4}}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{6 a x^6}-\frac {b \left (a-b x^4\right )^{3/4}}{4 a^2 x^2}-\frac {b^{3/2} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 a^{3/2} \sqrt [4]{a-b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 52, normalized size = 0.48 \[ -\frac {\sqrt [4]{1-\frac {b x^4}{a}} \, _2F_1\left (-\frac {3}{2},\frac {1}{4};-\frac {1}{2};\frac {b x^4}{a}\right )}{6 x^6 \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{b x^{11} - a x^{7}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(-b*x^4 + a)^(3/4)/(b*x^11 - a*x^7), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}} x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^7\,{\left (a-b\,x^4\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a - b*x^4)^(1/4)),x)

[Out]

int(1/(x^7*(a - b*x^4)^(1/4)), x)

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sympy [C]  time = 1.91, size = 34, normalized size = 0.31 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{6 \sqrt [4]{a} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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