3.1026 \(\int \frac{x^4}{(a+b x^2)^{5/6}} \, dx\)

Optimal. Leaf size=300 \[ \frac{27\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac{a}{a+b x^2}}\right ) \sqrt{\frac{\left (\frac{a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac{a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{-\sqrt [3]{\frac{a}{a+b x^2}}+\sqrt{3}+1}{-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1}\right ),4 \sqrt{3}-7\right )}{40 b^3 x \sqrt [3]{\frac{a}{a+b x^2}} \sqrt{-\frac{1-\sqrt [3]{\frac{a}{a+b x^2}}}{\left (-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1\right )^2}}}-\frac{27 a x \sqrt [6]{a+b x^2}}{40 b^2}+\frac{3 x^3 \sqrt [6]{a+b x^2}}{10 b} \]

[Out]

(-27*a*x*(a + b*x^2)^(1/6))/(40*b^2) + (3*x^3*(a + b*x^2)^(1/6))/(10*b) + (27*3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(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]])/(40*b^3*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 = 0.251414, 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 = {321, 241, 236, 219} \[ \frac{27\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac{a}{a+b x^2}}\right ) \sqrt{\frac{\left (\frac{a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac{a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac{-\sqrt [3]{\frac{a}{b x^2+a}}+\sqrt{3}+1}{-\sqrt [3]{\frac{a}{b x^2+a}}-\sqrt{3}+1}\right )|-7+4 \sqrt{3}\right )}{40 b^3 x \sqrt [3]{\frac{a}{a+b x^2}} \sqrt{-\frac{1-\sqrt [3]{\frac{a}{a+b x^2}}}{\left (-\sqrt [3]{\frac{a}{a+b x^2}}-\sqrt{3}+1\right )^2}}}-\frac{27 a x \sqrt [6]{a+b x^2}}{40 b^2}+\frac{3 x^3 \sqrt [6]{a+b x^2}}{10 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^2)^(5/6),x]

[Out]

(-27*a*x*(a + b*x^2)^(1/6))/(40*b^2) + (3*x^3*(a + b*x^2)^(1/6))/(10*b) + (27*3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(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]])/(40*b^3*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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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{x^4}{\left (a+b x^2\right )^{5/6}} \, dx &=\frac{3 x^3 \sqrt [6]{a+b x^2}}{10 b}-\frac{(9 a) \int \frac{x^2}{\left (a+b x^2\right )^{5/6}} \, dx}{10 b}\\ &=-\frac{27 a x \sqrt [6]{a+b x^2}}{40 b^2}+\frac{3 x^3 \sqrt [6]{a+b x^2}}{10 b}+\frac{\left (27 a^2\right ) \int \frac{1}{\left (a+b x^2\right )^{5/6}} \, dx}{40 b^2}\\ &=-\frac{27 a x \sqrt [6]{a+b x^2}}{40 b^2}+\frac{3 x^3 \sqrt [6]{a+b x^2}}{10 b}+\frac{\left (27 a^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 )}{40 b^2 \sqrt [3]{\frac{a}{a+b x^2}} \sqrt [3]{a+b x^2}}\\ &=-\frac{27 a x \sqrt [6]{a+b x^2}}{40 b^2}+\frac{3 x^3 \sqrt [6]{a+b x^2}}{10 b}-\frac{\left (81 a^2 \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 )}{80 b^3 x \sqrt [3]{\frac{a}{a+b x^2}}}\\ &=-\frac{27 a x \sqrt [6]{a+b x^2}}{40 b^2}+\frac{3 x^3 \sqrt [6]{a+b x^2}}{10 b}+\frac{27\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \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 )}{40 b^3 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 = 0.0215768, size = 79, normalized size = 0.26 \[ \frac{3 \left (9 a^2 x \left (\frac{b x^2}{a}+1\right )^{5/6} \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};-\frac{b x^2}{a}\right )-9 a^2 x-5 a b x^3+4 b^2 x^5\right )}{40 b^2 \left (a+b x^2\right )^{5/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^2)^(5/6),x]

[Out]

(3*(-9*a^2*x - 5*a*b*x^3 + 4*b^2*x^5 + 9*a^2*x*(1 + (b*x^2)/a)^(5/6)*Hypergeometric2F1[1/2, 5/6, 3/2, -((b*x^2
)/a)]))/(40*b^2*(a + b*x^2)^(5/6))

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int{{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(x^4/(b*x^2+a)^(5/6),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(b*x^2 + a)^(5/6), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{4}}{{\left (b x^{2} + a\right )}^{\frac{5}{6}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^4/(b*x^2 + a)^(5/6), x)

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Sympy [A]  time = 0.925291, size = 27, normalized size = 0.09 \begin{align*} \frac{x^{5}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{6}, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac{5}{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**5*hyper((5/6, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(5/6))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(b*x^2 + a)^(5/6), x)