3.837 \(\int \frac {x^4}{(a-b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=104 \[ \frac {8 a^{5/2} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{7 b^{5/2} \left (a-b x^2\right )^{3/4}}-\frac {4 a x \sqrt [4]{a-b x^2}}{7 b^2}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b} \]

[Out]

-4/7*a*x*(-b*x^2+a)^(1/4)/b^2-2/7*x^3*(-b*x^2+a)^(1/4)/b+8/7*a^(5/2)*(1-b*x^2/a)^(3/4)*(cos(1/2*arcsin(x*b^(1/
2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcsin(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arcsin(x*b^(1/2)/a^(1/2))),2^(1/2))
/b^(5/2)/(-b*x^2+a)^(3/4)

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Rubi [A]  time = 0.03, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {321, 233, 232} \[ \frac {8 a^{5/2} \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{7 b^{5/2} \left (a-b x^2\right )^{3/4}}-\frac {4 a x \sqrt [4]{a-b x^2}}{7 b^2}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a - b*x^2)^(3/4),x]

[Out]

(-4*a*x*(a - b*x^2)^(1/4))/(7*b^2) - (2*x^3*(a - b*x^2)^(1/4))/(7*b) + (8*a^(5/2)*(1 - (b*x^2)/a)^(3/4)*Ellipt
icF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(7*b^(5/2)*(a - b*x^2)^(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 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^4}{\left (a-b x^2\right )^{3/4}} \, dx &=-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b}+\frac {(6 a) \int \frac {x^2}{\left (a-b x^2\right )^{3/4}} \, dx}{7 b}\\ &=-\frac {4 a x \sqrt [4]{a-b x^2}}{7 b^2}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b}+\frac {\left (4 a^2\right ) \int \frac {1}{\left (a-b x^2\right )^{3/4}} \, dx}{7 b^2}\\ &=-\frac {4 a x \sqrt [4]{a-b x^2}}{7 b^2}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b}+\frac {\left (4 a^2 \left (1-\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/4}} \, dx}{7 b^2 \left (a-b x^2\right )^{3/4}}\\ &=-\frac {4 a x \sqrt [4]{a-b x^2}}{7 b^2}-\frac {2 x^3 \sqrt [4]{a-b x^2}}{7 b}+\frac {8 a^{5/2} \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{7 b^{5/2} \left (a-b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 77, normalized size = 0.74 \[ \frac {2 x \left (2 a^2 \left (1-\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {b x^2}{a}\right )-2 a^2+a b x^2+b^2 x^4\right )}{7 b^2 \left (a-b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a - b*x^2)^(3/4),x]

[Out]

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

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}} x^{4}}{b x^{2} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^4/(-b*x^2 + a)^(3/4), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (-b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (-b x^{2} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(x^4/(-b*x^2 + a)^(3/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (a-b\,x^2\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a - b*x^2)^(3/4),x)

[Out]

int(x^4/(a - b*x^2)^(3/4), x)

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sympy [C]  time = 0.90, size = 29, normalized size = 0.28 \[ \frac {x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{5 a^{\frac {3}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(-b*x**2+a)**(3/4),x)

[Out]

x**5*hyper((3/4, 5/2), (7/2,), b*x**2*exp_polar(2*I*pi)/a)/(5*a**(3/4))

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