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

Optimal. Leaf size=121 \[ \frac{8 a^{7/2} \left (\frac{b x^2}{a}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ),2\right )}{77 b^{5/2} \left (a+b x^2\right )^{3/4}}-\frac{4 a^2 x \sqrt [4]{a+b x^2}}{77 b^2}+\frac{2}{11} x^5 \sqrt [4]{a+b x^2}+\frac{2 a x^3 \sqrt [4]{a+b x^2}}{77 b} \]

[Out]

(-4*a^2*x*(a + b*x^2)^(1/4))/(77*b^2) + (2*a*x^3*(a + b*x^2)^(1/4))/(77*b) + (2*x^5*(a + b*x^2)^(1/4))/11 + (8
*a^(7/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(77*b^(5/2)*(a + b*x^2)^(3/4))

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Rubi [A]  time = 0.0487949, antiderivative size = 121, 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 = {279, 321, 233, 231} \[ -\frac{4 a^2 x \sqrt [4]{a+b x^2}}{77 b^2}+\frac{8 a^{7/2} \left (\frac{b x^2}{a}+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{77 b^{5/2} \left (a+b x^2\right )^{3/4}}+\frac{2}{11} x^5 \sqrt [4]{a+b x^2}+\frac{2 a x^3 \sqrt [4]{a+b x^2}}{77 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^2)^(1/4),x]

[Out]

(-4*a^2*x*(a + b*x^2)^(1/4))/(77*b^2) + (2*a*x^3*(a + b*x^2)^(1/4))/(77*b) + (2*x^5*(a + b*x^2)^(1/4))/11 + (8
*a^(7/2)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(77*b^(5/2)*(a + b*x^2)^(3/4))

Rule 279

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 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 233

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 231

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

Rubi steps

\begin{align*} \int x^4 \sqrt [4]{a+b x^2} \, dx &=\frac{2}{11} x^5 \sqrt [4]{a+b x^2}+\frac{1}{11} a \int \frac{x^4}{\left (a+b x^2\right )^{3/4}} \, dx\\ &=\frac{2 a x^3 \sqrt [4]{a+b x^2}}{77 b}+\frac{2}{11} x^5 \sqrt [4]{a+b x^2}-\frac{\left (6 a^2\right ) \int \frac{x^2}{\left (a+b x^2\right )^{3/4}} \, dx}{77 b}\\ &=-\frac{4 a^2 x \sqrt [4]{a+b x^2}}{77 b^2}+\frac{2 a x^3 \sqrt [4]{a+b x^2}}{77 b}+\frac{2}{11} x^5 \sqrt [4]{a+b x^2}+\frac{\left (4 a^3\right ) \int \frac{1}{\left (a+b x^2\right )^{3/4}} \, dx}{77 b^2}\\ &=-\frac{4 a^2 x \sqrt [4]{a+b x^2}}{77 b^2}+\frac{2 a x^3 \sqrt [4]{a+b x^2}}{77 b}+\frac{2}{11} x^5 \sqrt [4]{a+b x^2}+\frac{\left (4 a^3 \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}{77 b^2 \left (a+b x^2\right )^{3/4}}\\ &=-\frac{4 a^2 x \sqrt [4]{a+b x^2}}{77 b^2}+\frac{2 a x^3 \sqrt [4]{a+b x^2}}{77 b}+\frac{2}{11} x^5 \sqrt [4]{a+b x^2}+\frac{8 a^{7/2} \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 )}{77 b^{5/2} \left (a+b x^2\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0509471, size = 93, normalized size = 0.77 \[ \frac{2 x \sqrt [4]{a+b x^2} \left (\sqrt [4]{\frac{b x^2}{a}+1} \left (-6 a^2+a b x^2+7 b^2 x^4\right )+6 a^2 \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^2}{a}\right )\right )}{77 b^2 \sqrt [4]{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^2)^(1/4),x]

[Out]

(2*x*(a + b*x^2)^(1/4)*((1 + (b*x^2)/a)^(1/4)*(-6*a^2 + a*b*x^2 + 7*b^2*x^4) + 6*a^2*Hypergeometric2F1[-1/4, 1
/2, 3/2, -((b*x^2)/a)]))/(77*b^2*(1 + (b*x^2)/a)^(1/4))

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{x}^{4}\sqrt [4]{b{x}^{2}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{1}{4}} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(1/4)*x^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac{1}{4}} x^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(1/4)*x^4, x)

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Sympy [C]  time = 0.888968, size = 29, normalized size = 0.24 \begin{align*} \frac{\sqrt [4]{a} x^{5}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(1/4)*x**5*hyper((-1/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{1}{4}} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^2 + a)^(1/4)*x^4, x)