3.1244 \(\int \frac {x^5}{(a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=85 \[ \frac {2 a^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{3 b^{3/2} \left (a-b x^4\right )^{3/4}}-\frac {x^2 \sqrt [4]{a-b x^4}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {275, 321, 233, 232} \[ \frac {2 a^{3/2} \left (1-\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{3 b^{3/2} \left (a-b x^4\right )^{3/4}}-\frac {x^2 \sqrt [4]{a-b x^4}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a - b*x^4)^(3/4),x]

[Out]

-(x^2*(a - b*x^4)^(1/4))/(3*b) + (2*a^(3/2)*(1 - (b*x^4)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2
])/(3*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 233

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 66, normalized size = 0.78 \[ \frac {x^2 \left (a \left (1-\frac {b x^4}{a}\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {b x^4}{a}\right )-a+b x^4\right )}{3 b \left (a-b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a - b*x^4)^(3/4),x]

[Out]

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

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{5}}{b x^{4} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^4 + a)^(1/4)*x^5/(b*x^4 - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^5/(-b*x^4 + a)^(3/4), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\left (-b \,x^{4}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(-b*x^4+a)^(3/4),x)

[Out]

int(x^5/(-b*x^4+a)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{{\left (-b x^{4} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(x^5/(-b*x^4 + a)^(3/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5}{{\left (a-b\,x^4\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a - b*x^4)^(3/4),x)

[Out]

int(x^5/(a - b*x^4)^(3/4), x)

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sympy [C]  time = 1.88, size = 29, normalized size = 0.34 \[ \frac {x^{6} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{6 a^{\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(-b*x**4+a)**(3/4),x)

[Out]

x**6*hyper((3/4, 3/2), (5/2,), b*x**4*exp_polar(2*I*pi)/a)/(6*a**(3/4))

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