3.1010 \(\int x^8 \sqrt [4]{a+b x^4} \, dx\)

Optimal. Leaf size=126 \[ -\frac{a^{5/2} x^3 \left (\frac{a}{b x^4}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{24 b^{3/2} \left (a+b x^4\right )^{3/4}}-\frac{a^2 x \sqrt [4]{a+b x^4}}{24 b^2}+\frac{1}{10} x^9 \sqrt [4]{a+b x^4}+\frac{a x^5 \sqrt [4]{a+b x^4}}{60 b} \]

[Out]

-(a^2*x*(a + b*x^4)^(1/4))/(24*b^2) + (a*x^5*(a + b*x^4)^(1/4))/(60*b) + (x^9*(a + b*x^4)^(1/4))/10 - (a^(5/2)
*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(24*b^(3/2)*(a + b*x^4)^(3/4))

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Rubi [A]  time = 0.0561152, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {279, 321, 237, 335, 275, 231} \[ -\frac{a^2 x \sqrt [4]{a+b x^4}}{24 b^2}-\frac{a^{5/2} x^3 \left (\frac{a}{b x^4}+1\right )^{3/4} F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{24 b^{3/2} \left (a+b x^4\right )^{3/4}}+\frac{1}{10} x^9 \sqrt [4]{a+b x^4}+\frac{a x^5 \sqrt [4]{a+b x^4}}{60 b} \]

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*x^4)^(1/4),x]

[Out]

-(a^2*x*(a + b*x^4)^(1/4))/(24*b^2) + (a*x^5*(a + b*x^4)^(1/4))/(60*b) + (x^9*(a + b*x^4)^(1/4))/10 - (a^(5/2)
*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(24*b^(3/2)*(a + b*x^4)^(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 237

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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^8 \sqrt [4]{a+b x^4} \, dx &=\frac{1}{10} x^9 \sqrt [4]{a+b x^4}+\frac{1}{10} a \int \frac{x^8}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac{a x^5 \sqrt [4]{a+b x^4}}{60 b}+\frac{1}{10} x^9 \sqrt [4]{a+b x^4}-\frac{a^2 \int \frac{x^4}{\left (a+b x^4\right )^{3/4}} \, dx}{12 b}\\ &=-\frac{a^2 x \sqrt [4]{a+b x^4}}{24 b^2}+\frac{a x^5 \sqrt [4]{a+b x^4}}{60 b}+\frac{1}{10} x^9 \sqrt [4]{a+b x^4}+\frac{a^3 \int \frac{1}{\left (a+b x^4\right )^{3/4}} \, dx}{24 b^2}\\ &=-\frac{a^2 x \sqrt [4]{a+b x^4}}{24 b^2}+\frac{a x^5 \sqrt [4]{a+b x^4}}{60 b}+\frac{1}{10} x^9 \sqrt [4]{a+b x^4}+\frac{\left (a^3 \left (1+\frac{a}{b x^4}\right )^{3/4} x^3\right ) \int \frac{1}{\left (1+\frac{a}{b x^4}\right )^{3/4} x^3} \, dx}{24 b^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{a^2 x \sqrt [4]{a+b x^4}}{24 b^2}+\frac{a x^5 \sqrt [4]{a+b x^4}}{60 b}+\frac{1}{10} x^9 \sqrt [4]{a+b x^4}-\frac{\left (a^3 \left (1+\frac{a}{b x^4}\right )^{3/4} x^3\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1+\frac{a x^4}{b}\right )^{3/4}} \, dx,x,\frac{1}{x}\right )}{24 b^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{a^2 x \sqrt [4]{a+b x^4}}{24 b^2}+\frac{a x^5 \sqrt [4]{a+b x^4}}{60 b}+\frac{1}{10} x^9 \sqrt [4]{a+b x^4}-\frac{\left (a^3 \left (1+\frac{a}{b x^4}\right )^{3/4} x^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{a x^2}{b}\right )^{3/4}} \, dx,x,\frac{1}{x^2}\right )}{48 b^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{a^2 x \sqrt [4]{a+b x^4}}{24 b^2}+\frac{a x^5 \sqrt [4]{a+b x^4}}{60 b}+\frac{1}{10} x^9 \sqrt [4]{a+b x^4}-\frac{a^{5/2} \left (1+\frac{a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{24 b^{3/2} \left (a+b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0480913, size = 93, normalized size = 0.74 \[ \frac{x \sqrt [4]{a+b x^4} \left (\sqrt [4]{\frac{b x^4}{a}+1} \left (-5 a^2+a b x^4+6 b^2 x^8\right )+5 a^2 \, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{5}{4};-\frac{b x^4}{a}\right )\right )}{60 b^2 \sqrt [4]{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a + b*x^4)^(1/4),x]

[Out]

(x*(a + b*x^4)^(1/4)*((1 + (b*x^4)/a)^(1/4)*(-5*a^2 + a*b*x^4 + 6*b^2*x^8) + 5*a^2*Hypergeometric2F1[-1/4, 1/4
, 5/4, -((b*x^4)/a)]))/(60*b^2*(1 + (b*x^4)/a)^(1/4))

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{x}^{8}\sqrt [4]{b{x}^{4}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(b*x^4+a)^(1/4),x)

[Out]

int(x^8*(b*x^4+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{4} + a\right )}^{\frac{1}{4}} x^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(1/4)*x^8, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{4} + a\right )}^{\frac{1}{4}} x^{8}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(1/4)*x^8, x)

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Sympy [C]  time = 1.42892, size = 39, normalized size = 0.31 \begin{align*} \frac{\sqrt [4]{a} x^{9} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(b*x**4+a)**(1/4),x)

[Out]

a**(1/4)*x**9*gamma(9/4)*hyper((-1/4, 9/4), (13/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(13/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{4} + a\right )}^{\frac{1}{4}} x^{8}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(1/4)*x^8, x)