3.1119 \(\int \frac{x^9}{(a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=104 \[ \frac{4 a^{5/2} \left (\frac{b x^4}{a}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{7 b^{5/2} \left (a+b x^4\right )^{3/4}}-\frac{2 a x^2 \sqrt [4]{a+b x^4}}{7 b^2}+\frac{x^6 \sqrt [4]{a+b x^4}}{7 b} \]

[Out]

(-2*a*x^2*(a + b*x^4)^(1/4))/(7*b^2) + (x^6*(a + b*x^4)^(1/4))/(7*b) + (4*a^(5/2)*(1 + (b*x^4)/a)^(3/4)*Ellipt
icF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(7*b^(5/2)*(a + b*x^4)^(3/4))

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Rubi [A]  time = 0.0637318, antiderivative size = 104, 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 = {275, 321, 233, 231} \[ \frac{4 a^{5/2} \left (\frac{b x^4}{a}+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{7 b^{5/2} \left (a+b x^4\right )^{3/4}}-\frac{2 a x^2 \sqrt [4]{a+b x^4}}{7 b^2}+\frac{x^6 \sqrt [4]{a+b x^4}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(a + b*x^4)^(3/4),x]

[Out]

(-2*a*x^2*(a + b*x^4)^(1/4))/(7*b^2) + (x^6*(a + b*x^4)^(1/4))/(7*b) + (4*a^(5/2)*(1 + (b*x^4)/a)^(3/4)*Ellipt
icF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(7*b^(5/2)*(a + b*x^4)^(3/4))

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 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 \frac{x^9}{\left (a+b x^4\right )^{3/4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )\\ &=\frac{x^6 \sqrt [4]{a+b x^4}}{7 b}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{7 b}\\ &=-\frac{2 a x^2 \sqrt [4]{a+b x^4}}{7 b^2}+\frac{x^6 \sqrt [4]{a+b x^4}}{7 b}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )}{7 b^2}\\ &=-\frac{2 a x^2 \sqrt [4]{a+b x^4}}{7 b^2}+\frac{x^6 \sqrt [4]{a+b x^4}}{7 b}+\frac{\left (2 a^2 \left (1+\frac{b x^4}{a}\right )^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{7 b^2 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{2 a x^2 \sqrt [4]{a+b x^4}}{7 b^2}+\frac{x^6 \sqrt [4]{a+b x^4}}{7 b}+\frac{4 a^{5/2} \left (1+\frac{b x^4}{a}\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{7 b^{5/2} \left (a+b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0271018, size = 79, normalized size = 0.76 \[ \frac{x^2 \left (2 a^2 \left (\frac{b x^4}{a}+1\right )^{3/4} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{3}{2};-\frac{b x^4}{a}\right )-2 a^2-a b x^4+b^2 x^8\right )}{7 b^2 \left (a+b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a + b*x^4)^(3/4),x]

[Out]

(x^2*(-2*a^2 - a*b*x^4 + b^2*x^8 + 2*a^2*(1 + (b*x^4)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -((b*x^4)/a)])
)/(7*b^2*(a + b*x^4)^(3/4))

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{{x}^{9} \left ( b{x}^{4}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^4+a)^(3/4),x)

[Out]

int(x^9/(b*x^4+a)^(3/4),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{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(x^9/(b*x^4 + a)^(3/4), 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{3}{4}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

integral(x^9/(b*x^4 + a)^(3/4), x)

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Sympy [C]  time = 1.17288, size = 27, normalized size = 0.26 \begin{align*} \frac{x^{10}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{10 a^{\frac{3}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**4+a)**(3/4),x)

[Out]

x**10*hyper((3/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(3/4))

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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{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^9/(b*x^4 + a)^(3/4), x)