3.13.18 \(\int \frac {1}{x^{11} \sqrt [4]{a-b x^4}} \, dx\) [1218]

Optimal. Leaf size=133 \[ -\frac {\left (a-b x^4\right )^{3/4}}{10 a x^{10}}-\frac {7 b \left (a-b x^4\right )^{3/4}}{60 a^2 x^6}-\frac {7 b^2 \left (a-b x^4\right )^{3/4}}{40 a^3 x^2}-\frac {7 b^{5/2} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 a^{5/2} \sqrt [4]{a-b x^4}} \]

[Out]

-1/10*(-b*x^4+a)^(3/4)/a/x^10-7/60*b*(-b*x^4+a)^(3/4)/a^2/x^6-7/40*b^2*(-b*x^4+a)^(3/4)/a^3/x^2-7/40*b^(5/2)*(
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^(5/2)/(-b*x^4+a)^(1/4)

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Rubi [A]
time = 0.07, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {281, 331, 235, 234} \begin {gather*} -\frac {7 b^{5/2} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \text {ArcSin}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 a^{5/2} \sqrt [4]{a-b x^4}}-\frac {7 b^2 \left (a-b x^4\right )^{3/4}}{40 a^3 x^2}-\frac {7 b \left (a-b x^4\right )^{3/4}}{60 a^2 x^6}-\frac {\left (a-b x^4\right )^{3/4}}{10 a x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^11*(a - b*x^4)^(1/4)),x]

[Out]

-1/10*(a - b*x^4)^(3/4)/(a*x^10) - (7*b*(a - b*x^4)^(3/4))/(60*a^2*x^6) - (7*b^2*(a - b*x^4)^(3/4))/(40*a^3*x^
2) - (7*b^(5/2)*(1 - (b*x^4)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(40*a^(5/2)*(a - b*x^4)^(
1/4))

Rule 234

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

Rule 235

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 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*} \int \frac {1}{x^{11} \sqrt [4]{a-b x^4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^6 \sqrt [4]{a-b x^2}} \, dx,x,x^2\right )\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{10 a x^{10}}+\frac {(7 b) \text {Subst}\left (\int \frac {1}{x^4 \sqrt [4]{a-b x^2}} \, dx,x,x^2\right )}{20 a}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{10 a x^{10}}-\frac {7 b \left (a-b x^4\right )^{3/4}}{60 a^2 x^6}+\frac {\left (7 b^2\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{a-b x^2}} \, dx,x,x^2\right )}{40 a^2}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{10 a x^{10}}-\frac {7 b \left (a-b x^4\right )^{3/4}}{60 a^2 x^6}-\frac {7 b^2 \left (a-b x^4\right )^{3/4}}{40 a^3 x^2}-\frac {\left (7 b^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{a-b x^2}} \, dx,x,x^2\right )}{80 a^3}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{10 a x^{10}}-\frac {7 b \left (a-b x^4\right )^{3/4}}{60 a^2 x^6}-\frac {7 b^2 \left (a-b x^4\right )^{3/4}}{40 a^3 x^2}-\frac {\left (7 b^3 \sqrt [4]{1-\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}} \, dx,x,x^2\right )}{80 a^3 \sqrt [4]{a-b x^4}}\\ &=-\frac {\left (a-b x^4\right )^{3/4}}{10 a x^{10}}-\frac {7 b \left (a-b x^4\right )^{3/4}}{60 a^2 x^6}-\frac {7 b^2 \left (a-b x^4\right )^{3/4}}{40 a^3 x^2}-\frac {7 b^{5/2} \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 )}{40 a^{5/2} \sqrt [4]{a-b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 52, normalized size = 0.39 \begin {gather*} -\frac {\sqrt [4]{1-\frac {b x^4}{a}} \, _2F_1\left (-\frac {5}{2},\frac {1}{4};-\frac {3}{2};\frac {b x^4}{a}\right )}{10 x^{10} \sqrt [4]{a-b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*((1 - (b*x^4)/a)^(1/4)*Hypergeometric2F1[-5/2, 1/4, -3/2, (b*x^4)/a])/(x^10*(a - b*x^4)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{11} \left (-b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^11/(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.08, size = 28, normalized size = 0.21 \begin {gather*} {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{b x^{15} - a x^{11}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^11/(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^4 + a)^(3/4)/(b*x^15 - a*x^11), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.79, size = 34, normalized size = 0.26 \begin {gather*} - \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {1}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{10 \sqrt [4]{a} x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**11/(-b*x**4+a)**(1/4),x)

[Out]

-hyper((-5/2, 1/4), (-3/2,), b*x**4*exp_polar(2*I*pi)/a)/(10*a**(1/4)*x**10)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^11/(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{11}\,{\left (a-b\,x^4\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^11*(a - b*x^4)^(1/4)),x)

[Out]

int(1/(x^11*(a - b*x^4)^(1/4)), x)

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