3.865 \(\int \frac{1}{x^4 (a+b x^4)^{3/2}} \, dx\)

Optimal. Leaf size=131 \[ -\frac{5 b^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{12 a^{9/4} \sqrt{a+b x^4}}-\frac{5 \sqrt{a+b x^4}}{6 a^2 x^3}+\frac{1}{2 a x^3 \sqrt{a+b x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0353922, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {290, 325, 220} \[ -\frac{5 b^{3/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{12 a^{9/4} \sqrt{a+b x^4}}-\frac{5 \sqrt{a+b x^4}}{6 a^2 x^3}+\frac{1}{2 a x^3 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^4)^(3/2)),x]

[Out]

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

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C]  time = 0.0090256, size = 54, normalized size = 0.41 \[ -\frac{\sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{3}{4},\frac{3}{2};\frac{1}{4};-\frac{b x^4}{a}\right )}{3 a x^3 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.014, size = 113, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,{x}^{3}{a}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{bx}{2\,{a}^{2}}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}-{\frac{5\,b}{6\,{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/x^4/(b*x^4+a)^(3/2),x)

[Out]

-1/3*(b*x^4+a)^(1/2)/x^3/a^2-1/2*b*x/a^2/((x^4+1/b*a)*b)^(1/2)-5/6*b/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{3}{2}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^4), 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^{2} x^{12} + 2 \, a b x^{8} + a^{2} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 1.40557, size = 41, normalized size = 0.31 \begin{align*} \frac{\Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{3}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} x^{3} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**4+a)**(3/2),x)

[Out]

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

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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{3}{2}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

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