3.9.53 \(\int \frac {1}{(a-b x^2)^{5/4}} \, dx\) [853]

Optimal. Leaf size=77 \[ \frac {2 x}{a \sqrt [4]{a-b x^2}}-\frac {2 \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt {b} \sqrt [4]{a-b x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {205, 235, 234} \begin {gather*} \frac {2 x}{a \sqrt [4]{a-b x^2}}-\frac {2 \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \text {ArcSin}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt {b} \sqrt [4]{a-b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x^2)^(-5/4),x]

[Out]

(2*x)/(a*(a - b*x^2)^(1/4)) - (2*(1 - (b*x^2)/a)^(1/4)*EllipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*S
qrt[b]*(a - b*x^2)^(1/4))

Rule 205

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

Rule 234

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

Rule 235

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.03, size = 56, normalized size = 0.73 \begin {gather*} \frac {2 x-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 )}{a \sqrt [4]{a-b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x^2)^(-5/4),x]

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2+a)^(5/4),x)

[Out]

int(1/(-b*x^2+a)^(5/4),x)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((-b*x^2 + a)^(-5/4), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 0.46, size = 26, normalized size = 0.34 \begin {gather*} \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{a^{\frac {5}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2+a)**(5/4),x)

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((-b*x^2 + a)^(-5/4), x)

________________________________________________________________________________________

Mupad [B]
time = 4.91, size = 38, normalized size = 0.49 \begin {gather*} \frac {x\,{\left (1-\frac {b\,x^2}{a}\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {5}{4};\ \frac {3}{2};\ \frac {b\,x^2}{a}\right )}{{\left (a-b\,x^2\right )}^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - b*x^2)^(5/4),x)

[Out]

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

________________________________________________________________________________________