3.1097 \(\int \frac{1}{\sqrt [4]{a+b x^4}} \, dx\)

Optimal. Leaf size=57 \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}} \]

[Out]

ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4)) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4))

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Rubi [A]  time = 0.013011, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {240, 212, 206, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(-1/4),x]

[Out]

ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4)) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4))

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [4]{a+b x^4}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{1-b x^4} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{b} x^2} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{b} x^2} \, dx,x,\frac{x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\\ \end{align*}

Mathematica [A]  time = 0.0293327, size = 76, normalized size = 1.33 \[ \frac{-\log \left (1-\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\log \left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}+1\right )+2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4 \sqrt [4]{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(-1/4),x]

[Out]

(2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] - Log[1 - (b^(1/4)*x)/(a + b*x^4)^(1/4)] + Log[1 + (b^(1/4)*x)/(a + b
*x^4)^(1/4)])/(4*b^(1/4))

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt [4]{b{x}^{4}+a}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.6457, size = 281, normalized size = 4.93 \begin{align*} \frac{\arctan \left (\frac{\frac{x \sqrt{\frac{\sqrt{b} x^{2} + \sqrt{b x^{4} + a}}{x^{2}}}}{b^{\frac{1}{4}}} - \frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{b^{\frac{1}{4}}}}{x}\right )}{b^{\frac{1}{4}}} + \frac{\log \left (\frac{b^{\frac{1}{4}} x +{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}{4 \, b^{\frac{1}{4}}} - \frac{\log \left (-\frac{b^{\frac{1}{4}} x -{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}{4 \, b^{\frac{1}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan((x*sqrt((sqrt(b)*x^2 + sqrt(b*x^4 + a))/x^2)/b^(1/4) - (b*x^4 + a)^(1/4)/b^(1/4))/x)/b^(1/4) + 1/4*log(
(b^(1/4)*x + (b*x^4 + a)^(1/4))/x)/b^(1/4) - 1/4*log(-(b^(1/4)*x - (b*x^4 + a)^(1/4))/x)/b^(1/4)

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Sympy [C]  time = 1.27058, size = 36, normalized size = 0.63 \begin{align*} \frac{x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*gamma(1/4)*hyper((1/4, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(1/4)*gamma(5/4))

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Giac [B]  time = 1.14823, size = 278, normalized size = 4.88 \begin{align*} \frac{\sqrt{2} \left (-b\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} + \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{4 \, b} + \frac{\sqrt{2} \left (-b\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} - \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{4 \, b} - \frac{\sqrt{2} \left (-b\right )^{\frac{3}{4}} \log \left (\sqrt{-b} + \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{8 \, b} + \frac{\sqrt{2} \left (-b\right )^{\frac{3}{4}} \log \left (\sqrt{-b} - \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*(-b)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4))/b + 1/4*sqr
t(2)*(-b)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4))/b - 1/8*sqrt(2)*(
-b)^(3/4)*log(sqrt(-b) + sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2)/b + 1/8*sqrt(2)*(-b)^(3
/4)*log(sqrt(-b) - sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2)/b