3.1025 \(\int \frac {x^6}{(a+b x^2)^{5/6}} \, dx\)

Optimal. Leaf size=324 \[ -\frac {81\ 3^{3/4} \sqrt {2-\sqrt {3}} a^3 \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {\left (\frac {a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac {a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {-\sqrt [3]{\frac {a}{a+b x^2}}+\sqrt {3}+1}{-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1}\right ),4 \sqrt {3}-7\right )}{128 b^4 x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}}}+\frac {81 a^2 x \sqrt [6]{a+b x^2}}{128 b^3}-\frac {9 a x^3 \sqrt [6]{a+b x^2}}{32 b^2}+\frac {3 x^5 \sqrt [6]{a+b x^2}}{16 b} \]

[Out]

81/128*a^2*x*(b*x^2+a)^(1/6)/b^3-9/32*a*x^3*(b*x^2+a)^(1/6)/b^2+3/16*x^5*(b*x^2+a)^(1/6)/b-81/128*3^(3/4)*a^3*
(b*x^2+a)^(1/6)*(1-(a/(b*x^2+a))^(1/3))*EllipticF((1-(a/(b*x^2+a))^(1/3)+3^(1/2))/(1-(a/(b*x^2+a))^(1/3)-3^(1/
2)),2*I-I*3^(1/2))*(1/2*6^(1/2)-1/2*2^(1/2))*((1+(a/(b*x^2+a))^(1/3)+(a/(b*x^2+a))^(2/3))/(1-(a/(b*x^2+a))^(1/
3)-3^(1/2))^2)^(1/2)/b^4/x/(a/(b*x^2+a))^(1/3)/((-1+(a/(b*x^2+a))^(1/3))/(1-(a/(b*x^2+a))^(1/3)-3^(1/2))^2)^(1
/2)

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Rubi [A]  time = 0.29, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {321, 241, 236, 219} \[ \frac {81 a^2 x \sqrt [6]{a+b x^2}}{128 b^3}-\frac {81\ 3^{3/4} \sqrt {2-\sqrt {3}} a^3 \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {\left (\frac {a}{a+b x^2}\right )^{2/3}+\sqrt [3]{\frac {a}{a+b x^2}}+1}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-\sqrt [3]{\frac {a}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{\frac {a}{b x^2+a}}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{128 b^4 x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (-\sqrt [3]{\frac {a}{a+b x^2}}-\sqrt {3}+1\right )^2}}}-\frac {9 a x^3 \sqrt [6]{a+b x^2}}{32 b^2}+\frac {3 x^5 \sqrt [6]{a+b x^2}}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^2)^(5/6),x]

[Out]

(81*a^2*x*(a + b*x^2)^(1/6))/(128*b^3) - (9*a*x^3*(a + b*x^2)^(1/6))/(32*b^2) + (3*x^5*(a + b*x^2)^(1/6))/(16*
b) - (81*3^(3/4)*Sqrt[2 - Sqrt[3]]*a^3*(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]])/(128*b^4*x*(a/(a + b*x^2))^(1/3)*Sq
rt[-((1 - (a/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (a/(a + b*x^2))^(1/3))^2)])

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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {x^6}{\left (a+b x^2\right )^{5/6}} \, dx &=\frac {3 x^5 \sqrt [6]{a+b x^2}}{16 b}-\frac {(15 a) \int \frac {x^4}{\left (a+b x^2\right )^{5/6}} \, dx}{16 b}\\ &=-\frac {9 a x^3 \sqrt [6]{a+b x^2}}{32 b^2}+\frac {3 x^5 \sqrt [6]{a+b x^2}}{16 b}+\frac {\left (27 a^2\right ) \int \frac {x^2}{\left (a+b x^2\right )^{5/6}} \, dx}{32 b^2}\\ &=\frac {81 a^2 x \sqrt [6]{a+b x^2}}{128 b^3}-\frac {9 a x^3 \sqrt [6]{a+b x^2}}{32 b^2}+\frac {3 x^5 \sqrt [6]{a+b x^2}}{16 b}-\frac {\left (81 a^3\right ) \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx}{128 b^3}\\ &=\frac {81 a^2 x \sqrt [6]{a+b x^2}}{128 b^3}-\frac {9 a x^3 \sqrt [6]{a+b x^2}}{32 b^2}+\frac {3 x^5 \sqrt [6]{a+b x^2}}{16 b}-\frac {\left (81 a^3\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 )}{128 b^3 \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}\\ &=\frac {81 a^2 x \sqrt [6]{a+b x^2}}{128 b^3}-\frac {9 a x^3 \sqrt [6]{a+b x^2}}{32 b^2}+\frac {3 x^5 \sqrt [6]{a+b x^2}}{16 b}+\frac {\left (243 a^3 \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 )}{256 b^4 x \sqrt [3]{\frac {a}{a+b x^2}}}\\ &=\frac {81 a^2 x \sqrt [6]{a+b x^2}}{128 b^3}-\frac {9 a x^3 \sqrt [6]{a+b x^2}}{32 b^2}+\frac {3 x^5 \sqrt [6]{a+b x^2}}{16 b}-\frac {81\ 3^{3/4} \sqrt {2-\sqrt {3}} a^3 \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 )}{128 b^4 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*}

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Mathematica [C]  time = 0.03, size = 89, normalized size = 0.27 \[ \frac {3 x \left (-27 a^3 \left (\frac {b x^2}{a}+1\right )^{5/6} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};-\frac {b x^2}{a}\right )+27 a^3+15 a^2 b x^2-4 a b^2 x^4+8 b^3 x^6\right )}{128 b^3 \left (a+b x^2\right )^{5/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^2)^(5/6),x]

[Out]

(3*x*(27*a^3 + 15*a^2*b*x^2 - 4*a*b^2*x^4 + 8*b^3*x^6 - 27*a^3*(1 + (b*x^2)/a)^(5/6)*Hypergeometric2F1[1/2, 5/
6, 3/2, -((b*x^2)/a)]))/(128*b^3*(a + b*x^2)^(5/6))

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {5}{6}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^6/(b*x^2 + a)^(5/6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {5}{6}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(b*x^2 + a)^(5/6), x)

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {5}{6}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (b x^{2} + a\right )}^{\frac {5}{6}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^6/(b*x^2 + a)^(5/6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^6}{{\left (b\,x^2+a\right )}^{5/6}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a + b*x^2)^(5/6),x)

[Out]

int(x^6/(a + b*x^2)^(5/6), x)

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sympy [A]  time = 0.99, size = 27, normalized size = 0.08 \[ \frac {x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{6}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac {5}{6}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**7*hyper((5/6, 7/2), (9/2,), b*x**2*exp_polar(I*pi)/a)/(7*a**(5/6))

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