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

Optimal. Leaf size=96 \[ \frac {10 a^{5/2} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 \sqrt {b} \left (a-b x^2\right )^{3/4}}+\frac {10}{21} a x \sqrt [4]{a-b x^2}+\frac {2}{7} x \left (a-b x^2\right )^{5/4} \]

[Out]

10/21*a*x*(-b*x^2+a)^(1/4)+2/7*x*(-b*x^2+a)^(5/4)+10/21*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*
x^2+a)^(3/4)/b^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {195, 233, 232} \[ \frac {10 a^{5/2} \left (1-\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 \sqrt {b} \left (a-b x^2\right )^{3/4}}+\frac {10}{21} a x \sqrt [4]{a-b x^2}+\frac {2}{7} x \left (a-b x^2\right )^{5/4} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(5/4),x]

[Out]

(10*a*x*(a - b*x^2)^(1/4))/21 + (2*x*(a - b*x^2)^(5/4))/7 + (10*a^(5/2)*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin
[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(21*Sqrt[b]*(a - b*x^2)^(3/4))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]

Rubi steps

\begin {align*} \int \left (a-b x^2\right )^{5/4} \, dx &=\frac {2}{7} x \left (a-b x^2\right )^{5/4}+\frac {1}{7} (5 a) \int \sqrt [4]{a-b x^2} \, dx\\ &=\frac {10}{21} a x \sqrt [4]{a-b x^2}+\frac {2}{7} x \left (a-b x^2\right )^{5/4}+\frac {1}{21} \left (5 a^2\right ) \int \frac {1}{\left (a-b x^2\right )^{3/4}} \, dx\\ &=\frac {10}{21} a x \sqrt [4]{a-b x^2}+\frac {2}{7} x \left (a-b x^2\right )^{5/4}+\frac {\left (5 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}{21 \left (a-b x^2\right )^{3/4}}\\ &=\frac {10}{21} a x \sqrt [4]{a-b x^2}+\frac {2}{7} x \left (a-b x^2\right )^{5/4}+\frac {10 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 )}{21 \sqrt {b} \left (a-b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 48, normalized size = 0.50 \[ \frac {a x \sqrt [4]{a-b x^2} \, _2F_1\left (-\frac {5}{4},\frac {1}{2};\frac {3}{2};\frac {b x^2}{a}\right )}{\sqrt [4]{1-\frac {b x^2}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(5/4),x]

[Out]

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

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-b x^{2} + a\right )}^{\frac {5}{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-b*x^2 + a)^(5/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-b*x^2 + a)^(5/4), x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \left (-b \,x^{2}+a \right )^{\frac {5}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-b*x^2 + a)^(5/4), x)

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mupad [B]  time = 4.84, size = 38, normalized size = 0.40 \[ \frac {x\,{\left (a-b\,x^2\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ \frac {3}{2};\ \frac {b\,x^2}{a}\right )}{{\left (1-\frac {b\,x^2}{a}\right )}^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x^2)^(5/4),x)

[Out]

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

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sympy [C]  time = 1.19, size = 27, normalized size = 0.28 \[ a^{\frac {5}{4}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/4)*x*hyper((-5/4, 1/2), (3/2,), b*x**2*exp_polar(2*I*pi)/a)

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