3.1013 \(\int \frac{\sqrt [4]{a+b x^4}}{x^4} \, dx\)

Optimal. Leaf size=82 \[ -\frac{b^{3/2} x^3 \left (\frac{a}{b x^4}+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{3 \sqrt{a} \left (a+b x^4\right )^{3/4}}-\frac{\sqrt [4]{a+b x^4}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0340621, antiderivative size = 82, 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 = {277, 237, 335, 275, 231} \[ -\frac{b^{3/2} x^3 \left (\frac{a}{b x^4}+1\right )^{3/4} F\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{3 \sqrt{a} \left (a+b x^4\right )^{3/4}}-\frac{\sqrt [4]{a+b x^4}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(1/4)/x^4,x]

[Out]

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

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 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 \frac{\sqrt [4]{a+b x^4}}{x^4} \, dx &=-\frac{\sqrt [4]{a+b x^4}}{3 x^3}+\frac{1}{3} b \int \frac{1}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=-\frac{\sqrt [4]{a+b x^4}}{3 x^3}+\frac{\left (b \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}{3 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a+b x^4}}{3 x^3}-\frac{\left (b \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 )}{3 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a+b x^4}}{3 x^3}-\frac{\left (b \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 )}{6 \left (a+b x^4\right )^{3/4}}\\ &=-\frac{\sqrt [4]{a+b x^4}}{3 x^3}-\frac{b^{3/2} \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 )}{3 \sqrt{a} \left (a+b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0089323, size = 51, normalized size = 0.62 \[ -\frac{\sqrt [4]{a+b x^4} \, _2F_1\left (-\frac{3}{4},-\frac{1}{4};\frac{1}{4};-\frac{b x^4}{a}\right )}{3 x^3 \sqrt [4]{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(1/4)/x^4,x]

[Out]

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

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}}\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^4,x)

[Out]

int((b*x^4+a)^(1/4)/x^4,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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(1/4)/x^4, 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^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^4 + a)^(1/4)/x^4, x)

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Sympy [C]  time = 1.20434, size = 31, normalized size = 0.38 \begin{align*} - \frac{\sqrt [4]{b}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{4}}} \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**4,x)

[Out]

-b**(1/4)*hyper((-1/4, 1/2), (3/2,), a*exp_polar(I*pi)/(b*x**4))/(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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(1/4)/x^4, x)