3.851 \(\int \frac {x^4}{(a-b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {24 a^{3/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a-b x^2}}+\frac {12 x \left (a-b x^2\right )^{3/4}}{5 b^2}+\frac {2 x^3}{b \sqrt [4]{a-b x^2}} \]

[Out]

2*x^3/b/(-b*x^2+a)^(1/4)+12/5*x*(-b*x^2+a)^(3/4)/b^2-24/5*a^(3/2)*(1-b*x^2/a)^(1/4)*(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)))*EllipticE(sin(1/2*arcsin(x*b^(1/2)/a^(1/2))),2^(1/2))/b^
(5/2)/(-b*x^2+a)^(1/4)

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Rubi [A]  time = 0.03, antiderivative size = 101, 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 = {288, 321, 229, 228} \[ -\frac {24 a^{3/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a-b x^2}}+\frac {12 x \left (a-b x^2\right )^{3/4}}{5 b^2}+\frac {2 x^3}{b \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a - b*x^2)^(5/4),x]

[Out]

(2*x^3)/(b*(a - b*x^2)^(1/4)) + (12*x*(a - b*x^2)^(3/4))/(5*b^2) - (24*a^(3/2)*(1 - (b*x^2)/a)^(1/4)*EllipticE
[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^(5/2)*(a - b*x^2)^(1/4))

Rule 228

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

Rule 229

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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

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Mathematica [C]  time = 0.02, size = 66, normalized size = 0.65 \[ -\frac {2 \left (6 a x \sqrt [4]{1-\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {b x^2}{a}\right )-6 a x+b x^3\right )}{5 b^2 \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a - b*x^2)^(5/4),x]

[Out]

(-2*(-6*a*x + b*x^3 + 6*a*x*(1 - (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (b*x^2)/a]))/(5*b^2*(a - b*
x^2)^(1/4))

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fricas [F]  time = 1.20, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}} x^{4}}{b^{2} x^{4} - 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^4/(-b*x^2 + a)^(5/4), x)

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (-b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(-b*x^2+a)^(5/4),x)

[Out]

int(x^4/(-b*x^2+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^4/(-b*x^2 + a)^(5/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (a-b\,x^2\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a - b*x^2)^(5/4),x)

[Out]

int(x^4/(a - b*x^2)^(5/4), x)

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sympy [C]  time = 0.95, size = 29, normalized size = 0.29 \[ \frac {x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{5 a^{\frac {5}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(-b*x**2+a)**(5/4),x)

[Out]

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

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