3.1257 \(\int \frac {x^8}{(a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=109 \[ -\frac {5 a^{3/2} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} F\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 b^{3/2} \left (a-b x^4\right )^{3/4}}-\frac {5 a x \sqrt [4]{a-b x^4}}{12 b^2}-\frac {x^5 \sqrt [4]{a-b x^4}}{6 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {321, 237, 335, 275, 232} \[ -\frac {5 a^{3/2} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} F\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{12 b^{3/2} \left (a-b x^4\right )^{3/4}}-\frac {5 a x \sqrt [4]{a-b x^4}}{12 b^2}-\frac {x^5 \sqrt [4]{a-b x^4}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(a - b*x^4)^(3/4),x]

[Out]

(-5*a*x*(a - b*x^4)^(1/4))/(12*b^2) - (x^5*(a - b*x^4)^(1/4))/(6*b) - (5*a^(3/2)*(1 - a/(b*x^4))^(3/4)*x^3*Ell
ipticF[ArcCsc[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(12*b^(3/2)*(a - b*x^4)^(3/4))

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 237

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 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

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 {x^8}{\left (a-b x^4\right )^{3/4}} \, dx &=-\frac {x^5 \sqrt [4]{a-b x^4}}{6 b}+\frac {(5 a) \int \frac {x^4}{\left (a-b x^4\right )^{3/4}} \, dx}{6 b}\\ &=-\frac {5 a x \sqrt [4]{a-b x^4}}{12 b^2}-\frac {x^5 \sqrt [4]{a-b x^4}}{6 b}+\frac {\left (5 a^2\right ) \int \frac {1}{\left (a-b x^4\right )^{3/4}} \, dx}{12 b^2}\\ &=-\frac {5 a x \sqrt [4]{a-b x^4}}{12 b^2}-\frac {x^5 \sqrt [4]{a-b x^4}}{6 b}+\frac {\left (5 a^2 \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}{12 b^2 \left (a-b x^4\right )^{3/4}}\\ &=-\frac {5 a x \sqrt [4]{a-b x^4}}{12 b^2}-\frac {x^5 \sqrt [4]{a-b x^4}}{6 b}-\frac {\left (5 a^2 \left (1-\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1-\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{12 b^2 \left (a-b x^4\right )^{3/4}}\\ &=-\frac {5 a x \sqrt [4]{a-b x^4}}{12 b^2}-\frac {x^5 \sqrt [4]{a-b x^4}}{6 b}-\frac {\left (5 a^2 \left (1-\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{24 b^2 \left (a-b x^4\right )^{3/4}}\\ &=-\frac {5 a x \sqrt [4]{a-b x^4}}{12 b^2}-\frac {x^5 \sqrt [4]{a-b x^4}}{6 b}-\frac {5 a^{3/2} \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 )}{12 b^{3/2} \left (a-b x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 80, normalized size = 0.73 \[ \frac {5 a^2 x \left (1-\frac {b x^4}{a}\right )^{3/4} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};\frac {b x^4}{a}\right )-5 a^2 x+3 a b x^5+2 b^2 x^9}{12 b^2 \left (a-b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a - b*x^4)^(3/4),x]

[Out]

(-5*a^2*x + 3*a*b*x^5 + 2*b^2*x^9 + 5*a^2*x*(1 - (b*x^4)/a)^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, (b*x^4)/a])
/(12*b^2*(a - b*x^4)^(3/4))

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{8}}{b x^{4} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^4 + a)^(1/4)*x^8/(b*x^4 - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^8/(-b*x^4 + a)^(3/4), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\left (-b \,x^{4}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(-b*x^4+a)^(3/4),x)

[Out]

int(x^8/(-b*x^4+a)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(x^8/(-b*x^4 + a)^(3/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^8}{{\left (a-b\,x^4\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(a - b*x^4)^(3/4),x)

[Out]

int(x^8/(a - b*x^4)^(3/4), x)

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sympy [C]  time = 1.93, size = 39, normalized size = 0.36 \[ \frac {x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(-b*x**4+a)**(3/4),x)

[Out]

x**9*gamma(9/4)*hyper((3/4, 9/4), (13/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*a**(3/4)*gamma(13/4))

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