3.858 \(\int \frac {1}{(a+b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=78 \[ \frac {6 \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^2}}+\frac {2 x}{5 a \left (a+b x^2\right )^{5/4}} \]

[Out]

2/5*x/a/(b*x^2+a)^(5/4)+6/5*(1+b*x^2/a)^(1/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^
(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^2+a)^(1/4)/b^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {199, 197, 196} \[ \frac {6 \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^2}}+\frac {2 x}{5 a \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(-9/4),x]

[Out]

(2*x)/(5*a*(a + b*x^2)^(5/4)) + (6*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*a^(3/
2)*Sqrt[b]*(a + b*x^2)^(1/4))

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 197

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + (b
*x^2)/a)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^2\right )^{9/4}} \, dx &=\frac {2 x}{5 a \left (a+b x^2\right )^{5/4}}+\frac {3 \int \frac {1}{\left (a+b x^2\right )^{5/4}} \, dx}{5 a}\\ &=\frac {2 x}{5 a \left (a+b x^2\right )^{5/4}}+\frac {\left (3 \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{5 a^2 \sqrt [4]{a+b x^2}}\\ &=\frac {2 x}{5 a \left (a+b x^2\right )^{5/4}}+\frac {6 \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 72, normalized size = 0.92 \[ \frac {-3 x \left (a+b x^2\right ) \sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )+8 a x+6 b x^3}{5 a^2 \left (a+b x^2\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(-9/4),x]

[Out]

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

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(-9/4), x)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(9/4),x)

[Out]

int(1/(b*x^2+a)^(9/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(-9/4), x)

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mupad [B]  time = 4.88, size = 37, normalized size = 0.47 \[ \frac {x\,{\left (\frac {b\,x^2}{a}+1\right )}^{9/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (b\,x^2+a\right )}^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^2)^(9/4),x)

[Out]

(x*((b*x^2)/a + 1)^(9/4)*hypergeom([1/2, 9/4], 3/2, -(b*x^2)/a))/(a + b*x^2)^(9/4)

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sympy [C]  time = 1.36, size = 24, normalized size = 0.31 \[ \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {9}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(9/4),x)

[Out]

x*hyper((1/2, 9/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(9/4)

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