3.1089 \(\int \frac{x^9}{\sqrt [4]{a+b x^4}} \, dx\)

Optimal. Leaf size=128 \[ \frac{4 a^2 x^2}{15 b^2 \sqrt [4]{a+b x^4}}-\frac{4 a^{5/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a+b x^4}}-\frac{2 a x^2 \left (a+b x^4\right )^{3/4}}{15 b^2}+\frac{x^6 \left (a+b x^4\right )^{3/4}}{9 b} \]

[Out]

(4*a^2*x^2)/(15*b^2*(a + b*x^4)^(1/4)) - (2*a*x^2*(a + b*x^4)^(3/4))/(15*b^2) + (x^6*(a + b*x^4)^(3/4))/(9*b)
- (4*a^(5/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(15*b^(5/2)*(a + b*x^4)^(1/4
))

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Rubi [A]  time = 0.0741468, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {275, 321, 229, 227, 196} \[ \frac{4 a^2 x^2}{15 b^2 \sqrt [4]{a+b x^4}}-\frac{4 a^{5/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a+b x^4}}-\frac{2 a x^2 \left (a+b x^4\right )^{3/4}}{15 b^2}+\frac{x^6 \left (a+b x^4\right )^{3/4}}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(a + b*x^4)^(1/4),x]

[Out]

(4*a^2*x^2)/(15*b^2*(a + b*x^4)^(1/4)) - (2*a*x^2*(a + b*x^4)^(3/4))/(15*b^2) + (x^6*(a + b*x^4)^(3/4))/(9*b)
- (4*a^(5/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(15*b^(5/2)*(a + b*x^4)^(1/4
))

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 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 229

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

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{x^9}{\sqrt [4]{a+b x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac{x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac{a \operatorname{Subst}\left (\int \frac{x^2}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{3 b}\\ &=-\frac{2 a x^2 \left (a+b x^4\right )^{3/4}}{15 b^2}+\frac{x^6 \left (a+b x^4\right )^{3/4}}{9 b}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{15 b^2}\\ &=-\frac{2 a x^2 \left (a+b x^4\right )^{3/4}}{15 b^2}+\frac{x^6 \left (a+b x^4\right )^{3/4}}{9 b}+\frac{\left (2 a^2 \sqrt [4]{1+\frac{b x^4}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{1+\frac{b x^2}{a}}} \, dx,x,x^2\right )}{15 b^2 \sqrt [4]{a+b x^4}}\\ &=\frac{4 a^2 x^2}{15 b^2 \sqrt [4]{a+b x^4}}-\frac{2 a x^2 \left (a+b x^4\right )^{3/4}}{15 b^2}+\frac{x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac{\left (2 a^2 \sqrt [4]{1+\frac{b x^4}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{15 b^2 \sqrt [4]{a+b x^4}}\\ &=\frac{4 a^2 x^2}{15 b^2 \sqrt [4]{a+b x^4}}-\frac{2 a x^2 \left (a+b x^4\right )^{3/4}}{15 b^2}+\frac{x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac{4 a^{5/2} \sqrt [4]{1+\frac{b x^4}{a}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0262749, size = 80, normalized size = 0.62 \[ \frac{x^2 \left (6 a^2 \sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^4}{a}\right )-6 a^2-a b x^4+5 b^2 x^8\right )}{45 b^2 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a + b*x^4)^(1/4),x]

[Out]

(x^2*(-6*a^2 - a*b*x^4 + 5*b^2*x^8 + 6*a^2*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^4)/a)
]))/(45*b^2*(a + b*x^4)^(1/4))

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{{x}^{9}{\frac{1}{\sqrt [4]{b{x}^{4}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^4+a)^(1/4),x)

[Out]

int(x^9/(b*x^4+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{9}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(x^9/(b*x^4 + a)^(1/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{9}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral(x^9/(b*x^4 + a)^(1/4), x)

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Sympy [C]  time = 1.47964, size = 27, normalized size = 0.21 \begin{align*} \frac{x^{10}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{10 \sqrt [4]{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**4+a)**(1/4),x)

[Out]

x**10*hyper((1/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{9}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate(x^9/(b*x^4 + a)^(1/4), x)