3.13.3 \(\int \frac {\sqrt [4]{a-b x^4}}{x^{16}} \, dx\) [1203]

Optimal. Leaf size=158 \[ -\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{231 a^2 x^7}+\frac {4 b^3 \sqrt [4]{a-b x^4}}{231 a^3 x^3}+\frac {8 b^{9/2} \left (1-\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{231 a^{7/2} \left (a-b x^4\right )^{3/4}} \]

[Out]

-1/15*(-b*x^4+a)^(1/4)/x^15+1/165*b*(-b*x^4+a)^(1/4)/a/x^11+2/231*b^2*(-b*x^4+a)^(1/4)/a^2/x^7+4/231*b^3*(-b*x
^4+a)^(1/4)/a^3/x^3+8/231*b^(9/2)*(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^(7/2)/(-b*x^4+a)^(3/4)

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Rubi [A]
time = 0.06, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {283, 331, 243, 342, 281, 238} \begin {gather*} \frac {8 b^{9/2} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} F\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{231 a^{7/2} \left (a-b x^4\right )^{3/4}}+\frac {4 b^3 \sqrt [4]{a-b x^4}}{231 a^3 x^3}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{231 a^2 x^7}-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x^4)^(1/4)/x^16,x]

[Out]

-1/15*(a - b*x^4)^(1/4)/x^15 + (b*(a - b*x^4)^(1/4))/(165*a*x^11) + (2*b^2*(a - b*x^4)^(1/4))/(231*a^2*x^7) +
(4*b^3*(a - b*x^4)^(1/4))/(231*a^3*x^3) + (8*b^(9/2)*(1 - a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCsc[(Sqrt[b]*x^2)/
Sqrt[a]]/2, 2])/(231*a^(7/2)*(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*} \int \frac {\sqrt [4]{a-b x^4}}{x^{16}} \, dx &=-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}-\frac {1}{15} b \int \frac {1}{x^{12} \left (a-b x^4\right )^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}}-\frac {\left (2 b^2\right ) \int \frac {1}{x^8 \left (a-b x^4\right )^{3/4}} \, dx}{33 a}\\ &=-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{231 a^2 x^7}-\frac {\left (4 b^3\right ) \int \frac {1}{x^4 \left (a-b x^4\right )^{3/4}} \, dx}{77 a^2}\\ &=-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{231 a^2 x^7}+\frac {4 b^3 \sqrt [4]{a-b x^4}}{231 a^3 x^3}-\frac {\left (8 b^4\right ) \int \frac {1}{\left (a-b x^4\right )^{3/4}} \, dx}{231 a^3}\\ &=-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{231 a^2 x^7}+\frac {4 b^3 \sqrt [4]{a-b x^4}}{231 a^3 x^3}-\frac {\left (8 b^4 \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}{231 a^3 \left (a-b x^4\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{231 a^2 x^7}+\frac {4 b^3 \sqrt [4]{a-b x^4}}{231 a^3 x^3}+\frac {\left (8 b^4 \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 )}{231 a^3 \left (a-b x^4\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{231 a^2 x^7}+\frac {4 b^3 \sqrt [4]{a-b x^4}}{231 a^3 x^3}+\frac {\left (4 b^4 \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 )}{231 a^3 \left (a-b x^4\right )^{3/4}}\\ &=-\frac {\sqrt [4]{a-b x^4}}{15 x^{15}}+\frac {b \sqrt [4]{a-b x^4}}{165 a x^{11}}+\frac {2 b^2 \sqrt [4]{a-b x^4}}{231 a^2 x^7}+\frac {4 b^3 \sqrt [4]{a-b x^4}}{231 a^3 x^3}+\frac {8 b^{9/2} \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 )}{231 a^{7/2} \left (a-b x^4\right )^{3/4}}\\ \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 = 0.33 \begin {gather*} -\frac {\sqrt [4]{a-b x^4} \, _2F_1\left (-\frac {15}{4},-\frac {1}{4};-\frac {11}{4};\frac {b x^4}{a}\right )}{15 x^{15} \sqrt [4]{1-\frac {b x^4}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^4)^(1/4)/x^16,x]

[Out]

-1/15*((a - b*x^4)^(1/4)*Hypergeometric2F1[-15/4, -1/4, -11/4, (b*x^4)/a])/(x^15*(1 - (b*x^4)/a)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{x^{16}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x^4+a)^(1/4)/x^16,x)

[Out]

int((-b*x^4+a)^(1/4)/x^16,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((-b*x^4+a)^(1/4)/x^16,x, algorithm="maxima")

[Out]

integrate((-b*x^4 + a)^(1/4)/x^16, x)

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Fricas [F]
time = 0.07, size = 16, normalized size = 0.10 \begin {gather*} {\rm integral}\left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x^{16}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x^4+a)^(1/4)/x^16,x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(1/4)/x^16, x)

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Sympy [C] Result contains complex when optimal does not.
time = 1.20, size = 34, normalized size = 0.22 \begin {gather*} \frac {i \sqrt [4]{b} e^{\frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{14 x^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x**4+a)**(1/4)/x**16,x)

[Out]

I*b**(1/4)*exp(3*I*pi/4)*hyper((-1/4, 7/2), (9/2,), a/(b*x**4))/(14*x**14)

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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((-b*x^4+a)^(1/4)/x^16,x, algorithm="giac")

[Out]

integrate((-b*x^4 + a)^(1/4)/x^16, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a-b\,x^4\right )}^{1/4}}{x^{16}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^4)^(1/4)/x^16,x)

[Out]

int((a - b*x^4)^(1/4)/x^16, x)

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