3.873 \(\int \frac{1}{(a+b x^4)^{5/2}} \, dx\)

Optimal. Leaf size=127 \[ \frac{5 \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{24 a^{9/4} \sqrt [4]{b} \sqrt{a+b x^4}}+\frac{5 x}{12 a^2 \sqrt{a+b x^4}}+\frac{x}{6 a \left (a+b x^4\right )^{3/2}} \]

[Out]

x/(6*a*(a + b*x^4)^(3/2)) + (5*x)/(12*a^2*Sqrt[a + b*x^4]) + (5*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt
[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(24*a^(9/4)*b^(1/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.0283282, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {199, 220} \[ \frac{5 x}{12 a^2 \sqrt{a+b x^4}}+\frac{5 \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{24 a^{9/4} \sqrt [4]{b} \sqrt{a+b x^4}}+\frac{x}{6 a \left (a+b x^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(-5/2),x]

[Out]

x/(6*a*(a + b*x^4)^(3/2)) + (5*x)/(12*a^2*Sqrt[a + b*x^4]) + (5*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt
[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(24*a^(9/4)*b^(1/4)*Sqrt[a + b*x^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 220

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^4\right )^{5/2}} \, dx &=\frac{x}{6 a \left (a+b x^4\right )^{3/2}}+\frac{5 \int \frac{1}{\left (a+b x^4\right )^{3/2}} \, dx}{6 a}\\ &=\frac{x}{6 a \left (a+b x^4\right )^{3/2}}+\frac{5 x}{12 a^2 \sqrt{a+b x^4}}+\frac{5 \int \frac{1}{\sqrt{a+b x^4}} \, dx}{12 a^2}\\ &=\frac{x}{6 a \left (a+b x^4\right )^{3/2}}+\frac{5 x}{12 a^2 \sqrt{a+b x^4}}+\frac{5 \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{24 a^{9/4} \sqrt [4]{b} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0312654, size = 72, normalized size = 0.57 \[ \frac{5 x \left (a+b x^4\right ) \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^4}{a}\right )+7 a x+5 b x^5}{12 a^2 \left (a+b x^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(-5/2),x]

[Out]

(7*a*x + 5*b*x^5 + 5*x*(a + b*x^4)*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(12*a^2
*(a + b*x^4)^(3/2))

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Maple [C]  time = 0.012, size = 123, normalized size = 1. \begin{align*}{\frac{x}{6\,{b}^{2}a}\sqrt{b{x}^{4}+a} \left ({x}^{4}+{\frac{a}{b}} \right ) ^{-2}}+{\frac{5\,x}{12\,{a}^{2}}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}+{\frac{5}{12\,{a}^{2}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x/a/b^2*(b*x^4+a)^(1/2)/(x^4+1/b*a)^2+5/12*x/a^2/((x^4+1/b*a)*b)^(1/2)+5/12/a^2/(I/a^(1/2)*b^(1/2))^(1/2)*
(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2)
)^(1/2),I)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(-5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{4} + a}}{b^{3} x^{12} + 3 \, a b^{2} x^{8} + 3 \, a^{2} b x^{4} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^4 + a)/(b^3*x^12 + 3*a*b^2*x^8 + 3*a^2*b*x^4 + a^3), x)

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Sympy [C]  time = 1.13579, size = 36, normalized size = 0.28 \begin{align*} \frac{x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{5}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(-5/2), x)