3.1092 \(\int \frac{1}{x^3 \sqrt [4]{a+b x^4}} \, dx\)

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

[Out]

(b*x^2)/(2*a*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(2*a*x^2) - (Sqrt[b]*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTa
n[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(2*Sqrt[a]*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0537696, 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, 325, 229, 227, 196} \[ \frac{b x^2}{2 a \sqrt [4]{a+b x^4}}-\frac{\left (a+b x^4\right )^{3/4}}{2 a x^2}-\frac{\sqrt{b} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{2 \sqrt{a} \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^4)^(1/4)),x]

[Out]

(b*x^2)/(2*a*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(2*a*x^2) - (Sqrt[b]*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTa
n[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(2*Sqrt[a]*(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 325

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

Mathematica [C]  time = 0.0113872, size = 51, normalized size = 0.49 \[ -\frac{\sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{1}{2};-\frac{b x^4}{a}\right )}{2 x^2 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}}{\frac{1}{\sqrt [4]{b{x}^{4}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/x^3/(b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((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{3}{4}}}{b x^{7} + a x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^4 + a)^(3/4)/(b*x^7 + a*x^3), x)

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Sympy [C]  time = 1.07404, size = 31, normalized size = 0.3 \begin{align*} - \frac{{{}_{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 \sqrt [4]{a} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/x^3/(b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

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