3.1090 \(\int \frac{x^5}{\sqrt [4]{a+b x^4}} \, dx\)

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

[Out]

(-2*a*x^2)/(5*b*(a + b*x^4)^(1/4)) + (x^2*(a + b*x^4)^(3/4))/(5*b) + (2*a^(3/2)*(1 + (b*x^4)/a)^(1/4)*Elliptic
E[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(5*b^(3/2)*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0562071, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{2 a^{3/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a+b x^4}}+\frac{x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac{2 a x^2}{5 b \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^4)^(1/4),x]

[Out]

(-2*a*x^2)/(5*b*(a + b*x^4)^(1/4)) + (x^2*(a + b*x^4)^(3/4))/(5*b) + (2*a^(3/2)*(1 + (b*x^4)/a)^(1/4)*Elliptic
E[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(5*b^(3/2)*(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^5}{\sqrt [4]{a+b x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac{x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{5 b}\\ &=\frac{x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac{\left (a \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 )}{5 b \sqrt [4]{a+b x^4}}\\ &=-\frac{2 a x^2}{5 b \sqrt [4]{a+b x^4}}+\frac{x^2 \left (a+b x^4\right )^{3/4}}{5 b}+\frac{\left (a \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 )}{5 b \sqrt [4]{a+b x^4}}\\ &=-\frac{2 a x^2}{5 b \sqrt [4]{a+b x^4}}+\frac{x^2 \left (a+b x^4\right )^{3/4}}{5 b}+\frac{2 a^{3/2} \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 )}{5 b^{3/2} \sqrt [4]{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0187022, size = 64, normalized size = 0.62 \[ \frac{x^2 \left (-a \sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^4}{a}\right )+a+b x^4\right )}{5 b \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^4)^(1/4),x]

[Out]

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

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{{x}^{5}{\frac{1}{\sqrt [4]{b{x}^{4}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^4+a)^(1/4),x)

[Out]

int(x^5/(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^{5}}{{\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^5/(b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(x^5/(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^{5}}{{\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^5/(b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral(x^5/(b*x^4 + a)^(1/4), x)

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Sympy [C]  time = 1.18742, size = 27, normalized size = 0.26 \begin{align*} \frac{x^{6}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{6 \sqrt [4]{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**4+a)**(1/4),x)

[Out]

x**6*hyper((1/4, 3/2), (5/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{{\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^5/(b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate(x^5/(b*x^4 + a)^(1/4), x)