3.1030 \(\int \frac{1}{x^4 (a+b x^2)^{5/6}} \, dx\)

Optimal. Leaf size=300 \[ \frac{16 \sqrt{2-\sqrt{3}} b \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac{a}{a+b x^2}}\right ) \sqrt{\frac{\left (\frac{a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac{a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{-\sqrt [3]{\frac{a}{a+b x^2}}+\sqrt{3}+1}{-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1}\right ),4 \sqrt{3}-7\right )}{9 \sqrt [4]{3} a^2 x \sqrt [3]{\frac{a}{a+b x^2}} \sqrt{-\frac{1-\sqrt [3]{\frac{a}{a+b x^2}}}{\left (-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1\right )^2}}}+\frac{8 b \sqrt [6]{a+b x^2}}{9 a^2 x}-\frac{\sqrt [6]{a+b x^2}}{3 a x^3} \]

[Out]

-(a + b*x^2)^(1/6)/(3*a*x^3) + (8*b*(a + b*x^2)^(1/6))/(9*a^2*x) + (16*Sqrt[2 - Sqrt[3]]*b*(a + b*x^2)^(1/6)*(
1 - (a/(a + b*x^2))^(1/3))*Sqrt[(1 + (a/(a + b*x^2))^(1/3) + (a/(a + b*x^2))^(2/3))/(1 - Sqrt[3] - (a/(a + b*x
^2))^(1/3))^2]*EllipticF[ArcSin[(1 + Sqrt[3] - (a/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (a/(a + b*x^2))^(1/3))],
-7 + 4*Sqrt[3]])/(9*3^(1/4)*a^2*x*(a/(a + b*x^2))^(1/3)*Sqrt[-((1 - (a/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (a/(
a + b*x^2))^(1/3))^2)])

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Rubi [A]  time = 0.242957, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {325, 241, 236, 219} \[ \frac{8 b \sqrt [6]{a+b x^2}}{9 a^2 x}+\frac{16 \sqrt{2-\sqrt{3}} b \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac{a}{a+b x^2}}\right ) \sqrt{\frac{\left (\frac{a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac{a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-\sqrt [3]{\frac{a}{b x^2+a}}+\sqrt{3}+1}{-\sqrt [3]{\frac{a}{b x^2+a}}-\sqrt{3}+1}\right )|-7+4 \sqrt{3}\right )}{9 \sqrt [4]{3} a^2 x \sqrt [3]{\frac{a}{a+b x^2}} \sqrt{-\frac{1-\sqrt [3]{\frac{a}{a+b x^2}}}{\left (-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1\right )^2}}}-\frac{\sqrt [6]{a+b x^2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)^(5/6)),x]

[Out]

-(a + b*x^2)^(1/6)/(3*a*x^3) + (8*b*(a + b*x^2)^(1/6))/(9*a^2*x) + (16*Sqrt[2 - Sqrt[3]]*b*(a + b*x^2)^(1/6)*(
1 - (a/(a + b*x^2))^(1/3))*Sqrt[(1 + (a/(a + b*x^2))^(1/3) + (a/(a + b*x^2))^(2/3))/(1 - Sqrt[3] - (a/(a + b*x
^2))^(1/3))^2]*EllipticF[ArcSin[(1 + Sqrt[3] - (a/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (a/(a + b*x^2))^(1/3))],
-7 + 4*Sqrt[3]])/(9*3^(1/4)*a^2*x*(a/(a + b*x^2))^(1/3)*Sqrt[-((1 - (a/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (a/(
a + b*x^2))^(1/3))^2)])

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 241

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a/(a + b*x^n))^(p + 1/n)*(a + b*x^n)^(p + 1/n), Subst[In
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p,
0] && NeQ[p, -2^(-1)] && LtQ[Denominator[p + 1/n], Denominator[p]]

Rule 236

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[(3*Sqrt[b*x^2])/(2*b*x), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx &=-\frac{\sqrt [6]{a+b x^2}}{3 a x^3}-\frac{(8 b) \int \frac{1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx}{9 a}\\ &=-\frac{\sqrt [6]{a+b x^2}}{3 a x^3}+\frac{8 b \sqrt [6]{a+b x^2}}{9 a^2 x}+\frac{\left (16 b^2\right ) \int \frac{1}{\left (a+b x^2\right )^{5/6}} \, dx}{27 a^2}\\ &=-\frac{\sqrt [6]{a+b x^2}}{3 a x^3}+\frac{8 b \sqrt [6]{a+b x^2}}{9 a^2 x}+\frac{\left (16 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-b x^2\right )^{2/3}} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{27 a^2 \sqrt [3]{\frac{a}{a+b x^2}} \sqrt [3]{a+b x^2}}\\ &=-\frac{\sqrt [6]{a+b x^2}}{3 a x^3}+\frac{8 b \sqrt [6]{a+b x^2}}{9 a^2 x}-\frac{\left (8 b \sqrt{-\frac{b x^2}{a+b x^2}} \sqrt [6]{a+b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x^3}} \, dx,x,\sqrt [3]{\frac{a}{a+b x^2}}\right )}{9 a^2 x \sqrt [3]{\frac{a}{a+b x^2}}}\\ &=-\frac{\sqrt [6]{a+b x^2}}{3 a x^3}+\frac{8 b \sqrt [6]{a+b x^2}}{9 a^2 x}+\frac{16 \sqrt{2-\sqrt{3}} b \sqrt{-\frac{b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac{a}{a+b x^2}}\right ) \sqrt{\frac{1+\sqrt [3]{\frac{a}{a+b x^2}}+\left (\frac{a}{a+b x^2}\right )^{2/3}}{\left (1-\sqrt{3}-\sqrt [3]{\frac{a}{a+b x^2}}\right )^2}} F\left (\sin ^{-1}\left (\frac{1+\sqrt{3}-\sqrt [3]{\frac{a}{a+b x^2}}}{1-\sqrt{3}-\sqrt [3]{\frac{a}{a+b x^2}}}\right )|-7+4 \sqrt{3}\right )}{9 \sqrt [4]{3} a^2 x \sqrt [3]{\frac{a}{a+b x^2}} \sqrt{-\frac{1-\sqrt [3]{\frac{a}{a+b x^2}}}{\left (1-\sqrt{3}-\sqrt [3]{\frac{a}{a+b x^2}}\right )^2}} \sqrt{-1+\frac{a}{a+b x^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0089693, size = 51, normalized size = 0.17 \[ -\frac{\left (\frac{b x^2}{a}+1\right )^{5/6} \, _2F_1\left (-\frac{3}{2},\frac{5}{6};-\frac{1}{2};-\frac{b x^2}{a}\right )}{3 x^3 \left (a+b x^2\right )^{5/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)^(5/6)),x]

[Out]

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

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{6}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^2+a)^(5/6),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^(5/6),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^(5/6),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.49648, size = 32, normalized size = 0.11 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{5}{6} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac{5}{6}} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**2+a)**(5/6),x)

[Out]

-hyper((-3/2, 5/6), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(5/6)*x**3)

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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{5}{6}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^(5/6),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/6)*x^4), x)