3.8.88 \(\int x^2 \sqrt [4]{a+b x^2} \, dx\) [788]

Optimal. Leaf size=97 \[ \frac {2 a x \sqrt [4]{a+b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a+b x^2}-\frac {4 a^{5/2} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 b^{3/2} \left (a+b x^2\right )^{3/4}} \]

[Out]

2/21*a*x*(b*x^2+a)^(1/4)/b+2/7*x^3*(b*x^2+a)^(1/4)-4/21*a^(5/2)*(1+b*x^2/a)^(3/4)*(cos(1/2*arctan(x*b^(1/2)/a^
(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/b^(3
/2)/(b*x^2+a)^(3/4)

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Rubi [A]
time = 0.02, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {285, 327, 239, 237} \begin {gather*} -\frac {4 a^{5/2} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac {2 a x \sqrt [4]{a+b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2)^(1/4),x]

[Out]

(2*a*x*(a + b*x^2)^(1/4))/(21*b) + (2*x^3*(a + b*x^2)^(1/4))/7 - (4*a^(5/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[Ar
cTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(21*b^(3/2)*(a + b*x^2)^(3/4))

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 239

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + b*(x^2
/a))^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

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 [4]{a+b x^2} \, dx &=\frac {2}{7} x^3 \sqrt [4]{a+b x^2}+\frac {1}{7} a \int \frac {x^2}{\left (a+b x^2\right )^{3/4}} \, dx\\ &=\frac {2 a x \sqrt [4]{a+b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a+b x^2}-\frac {\left (2 a^2\right ) \int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx}{21 b}\\ &=\frac {2 a x \sqrt [4]{a+b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a+b x^2}-\frac {\left (2 a^2 \left (1+\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx}{21 b \left (a+b x^2\right )^{3/4}}\\ &=\frac {2 a x \sqrt [4]{a+b x^2}}{21 b}+\frac {2}{7} x^3 \sqrt [4]{a+b x^2}-\frac {4 a^{5/2} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 b^{3/2} \left (a+b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.98, size = 62, normalized size = 0.64 \begin {gather*} \frac {2 x \sqrt [4]{a+b x^2} \left (a+b x^2-\frac {a \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^2}{a}\right )}{\sqrt [4]{1+\frac {b x^2}{a}}}\right )}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^2)^(1/4),x]

[Out]

(2*x*(a + b*x^2)^(1/4)*(a + b*x^2 - (a*Hypergeometric2F1[-1/4, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/4))
)/(7*b)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{2} \left (b \,x^{2}+a \right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((b*x^2 + a)^(1/4)*x^2, x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.46, size = 29, normalized size = 0.30 \begin {gather*} \frac {\sqrt [4]{a} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \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)**(1/4),x)

[Out]

a**(1/4)*x**3*hyper((-1/4, 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)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/4)*x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\left (b\,x^2+a\right )}^{1/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^2)^(1/4),x)

[Out]

int(x^2*(a + b*x^2)^(1/4), x)

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