3.11.12 \(\int x^2 \sqrt [6]{a+b x^2} \, dx\) [1012]

Optimal. Leaf size=297 \[ \frac {3 a x \sqrt [6]{a+b x^2}}{40 b}+\frac {3}{10} x^3 \sqrt [6]{a+b x^2}-\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {1+\sqrt [3]{\frac {a}{a+b x^2}}+\left (\frac {a}{a+b x^2}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}}{1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}}\right )|-7+4 \sqrt {3}\right )}{40 b^2 x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}\right )^2}}} \]

[Out]

3/40*a*x*(b*x^2+a)^(1/6)/b+3/10*x^3*(b*x^2+a)^(1/6)-3/40*3^(3/4)*a^2*(b*x^2+a)^(1/6)*(1-(a/(b*x^2+a))^(1/3))*E
llipticF((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^2/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 = 0.18, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {285, 327, 247, 242, 225} \begin {gather*} -\frac {3\ 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 (\text {ArcSin}\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^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 {3 a x \sqrt [6]{a+b x^2}}{40 b}+\frac {3}{10} x^3 \sqrt [6]{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2)^(1/6),x]

[Out]

(3*a*x*(a + b*x^2)^(1/6))/(40*b) + (3*x^3*(a + b*x^2)^(1/6))/10 - (3*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^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 225

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

Rule 242

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 247

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 285

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

Rule 327

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 x^2 \sqrt [6]{a+b x^2} \, dx &=\frac {3}{10} x^3 \sqrt [6]{a+b x^2}+\frac {1}{10} a \int \frac {x^2}{\left (a+b x^2\right )^{5/6}} \, dx\\ &=\frac {3 a x \sqrt [6]{a+b x^2}}{40 b}+\frac {3}{10} x^3 \sqrt [6]{a+b x^2}-\frac {\left (3 a^2\right ) \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx}{40 b}\\ &=\frac {3 a x \sqrt [6]{a+b x^2}}{40 b}+\frac {3}{10} x^3 \sqrt [6]{a+b x^2}-\frac {\left (3 a^2\right ) \text {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 \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}\\ &=\frac {3 a x \sqrt [6]{a+b x^2}}{40 b}+\frac {3}{10} x^3 \sqrt [6]{a+b x^2}+\frac {\left (9 a^2 \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^3}} \, dx,x,\sqrt [3]{\frac {a}{a+b x^2}}\right )}{80 b^2 x \sqrt [3]{\frac {a}{a+b x^2}}}\\ &=\frac {3 a x \sqrt [6]{a+b x^2}}{40 b}+\frac {3}{10} x^3 \sqrt [6]{a+b x^2}-\frac {3\ 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^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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 8.56, size = 62, normalized size = 0.21 \begin {gather*} \frac {3 x \sqrt [6]{a+b x^2} \left (a+b x^2-\frac {a \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )}{\sqrt [6]{1+\frac {b x^2}{a}}}\right )}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^2)^(1/6),x]

[Out]

(3*x*(a + b*x^2)^(1/6)*(a + b*x^2 - (a*Hypergeometric2F1[-1/6, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/6))
)/(10*b)

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int x^{2} \left (b \,x^{2}+a \right )^{\frac {1}{6}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(1/6)*x^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(1/6)*x^2, x)

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Sympy [A]
time = 0.45, size = 29, normalized size = 0.10 \begin {gather*} \frac {\sqrt [6]{a} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{6}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(1/6)*x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,{\left (b\,x^2+a\right )}^{1/6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^2)^(1/6),x)

[Out]

int(x^2*(a + b*x^2)^(1/6), x)

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