3.859 \(\int \frac{1}{(a+b x^2)^{11/4}} \, dx\)

Optimal. Leaf size=97 \[ \frac{10 \left (\frac{b x^2}{a}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{21 a^{3/2} \sqrt{b} \left (a+b x^2\right )^{3/4}}+\frac{10 x}{21 a^2 \left (a+b x^2\right )^{3/4}}+\frac{2 x}{7 a \left (a+b x^2\right )^{7/4}} \]

[Out]

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

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Rubi [A]  time = 0.0235896, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {199, 233, 231} \[ \frac{10 x}{21 a^2 \left (a+b x^2\right )^{3/4}}+\frac{10 \left (\frac{b x^2}{a}+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{21 a^{3/2} \sqrt{b} \left (a+b x^2\right )^{3/4}}+\frac{2 x}{7 a \left (a+b x^2\right )^{7/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(-11/4),x]

[Out]

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

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]
)

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 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^2\right )^{11/4}} \, dx &=\frac{2 x}{7 a \left (a+b x^2\right )^{7/4}}+\frac{5 \int \frac{1}{\left (a+b x^2\right )^{7/4}} \, dx}{7 a}\\ &=\frac{2 x}{7 a \left (a+b x^2\right )^{7/4}}+\frac{10 x}{21 a^2 \left (a+b x^2\right )^{3/4}}+\frac{5 \int \frac{1}{\left (a+b x^2\right )^{3/4}} \, dx}{21 a^2}\\ &=\frac{2 x}{7 a \left (a+b x^2\right )^{7/4}}+\frac{10 x}{21 a^2 \left (a+b x^2\right )^{3/4}}+\frac{\left (5 \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 a^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac{2 x}{7 a \left (a+b x^2\right )^{7/4}}+\frac{10 x}{21 a^2 \left (a+b x^2\right )^{3/4}}+\frac{10 \left (1+\frac{b x^2}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{21 a^{3/2} \sqrt{b} \left (a+b x^2\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0309899, size = 75, normalized size = 0.77 \[ \frac{5 x \left (a+b x^2\right ) \left (\frac{b x^2}{a}+1\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};-\frac{b x^2}{a}\right )+2 x \left (8 a+5 b x^2\right )}{21 a^2 \left (a+b x^2\right )^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(-11/4),x]

[Out]

(2*x*(8*a + 5*b*x^2) + 5*x*(a + b*x^2)*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -((b*x^2)/a)])/(
21*a^2*(a + b*x^2)^(7/4))

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int \left ( b{x}^{2}+a \right ) ^{-{\frac{11}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(11/4),x)

[Out]

int(1/(b*x^2+a)^(11/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{11}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(11/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(-11/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(11/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)

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Sympy [C]  time = 2.94785, size = 24, normalized size = 0.25 \begin{align*} \frac{x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{11}{4} \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac{11}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(11/4),x)

[Out]

x*hyper((1/2, 11/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(11/4)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{11}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(11/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(-11/4), x)