3.999 \(\int \frac{\sqrt [4]{a+b x^4}}{x^3} \, dx\)

Optimal. Leaf size=79 \[ \frac{\sqrt{a} \sqrt{b} \left (\frac{b x^4}{a}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{2 \left (a+b x^4\right )^{3/4}}-\frac{\sqrt [4]{a+b x^4}}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0466232, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {275, 277, 233, 231} \[ \frac{\sqrt{a} \sqrt{b} \left (\frac{b x^4}{a}+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{2 \left (a+b x^4\right )^{3/4}}-\frac{\sqrt [4]{a+b x^4}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(1/4)/x^3,x]

[Out]

-(a + b*x^4)^(1/4)/(2*x^2) + (Sqrt[a]*Sqrt[b]*(1 + (b*x^4)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2,
2])/(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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 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{\sqrt [4]{a+b x^4}}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt [4]{a+b x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\sqrt [4]{a+b x^4}}{2 x^2}+\frac{1}{4} b \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/4}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt [4]{a+b x^4}}{2 x^2}+\frac{\left (b \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 )}{4 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a+b x^4}}{2 x^2}+\frac{\sqrt{a} \sqrt{b} \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 )}{2 \left (a+b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0112106, size = 51, normalized size = 0.65 \[ -\frac{\sqrt [4]{a+b x^4} \, _2F_1\left (-\frac{1}{2},-\frac{1}{4};\frac{1}{2};-\frac{b x^4}{a}\right )}{2 x^2 \sqrt [4]{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(1/4)/x^3,x]

[Out]

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

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}}\sqrt [4]{b{x}^{4}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(1/4)/x^3,x)

[Out]

int((b*x^4+a)^(1/4)/x^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(1/4)/x^3,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(1/4)/x^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(1/4)/x^3,x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(1/4)/x^3, x)

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Sympy [C]  time = 1.09684, size = 32, normalized size = 0.41 \begin{align*} - \frac{\sqrt [4]{a}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(1/4)/x**3,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(1/4)/x^3,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(1/4)/x^3, x)