\(\int \frac {1}{\sqrt [4]{a+b x^4}} \, dx\) [1097]

Optimal result

Integrand size = 11, antiderivative size = 57 \[ \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}} \]

[Out]

1/2*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/4)+1/2*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {246, 218, 212, 209} \[ \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}} \]

[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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 246

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]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )+\frac {1}{2} \text {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] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\log \left (1-\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\log \left (1+\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4 \sqrt [4]{b}} \]

[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))

Maple [A] (verified)

Time = 4.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(\frac {-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )+\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )}{4 b^{\frac {1}{4}}}\)	\(65\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.86 \[ \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx=\frac {\log \left (\frac {b^{\frac {1}{4}} x + {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}{4 \, b^{\frac {1}{4}}} - \frac {\log \left (-\frac {b^{\frac {1}{4}} x - {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}{4 \, b^{\frac {1}{4}}} - \frac {i \, \log \left (\frac {i \, b^{\frac {1}{4}} x + {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}{4 \, b^{\frac {1}{4}}} + \frac {i \, \log \left (\frac {-i \, b^{\frac {1}{4}} x + {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}{4 \, b^{\frac {1}{4}}} \]

[In]

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

[Out]

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) - 1/4
*I*log((I*b^(1/4)*x + (b*x^4 + a)^(1/4))/x)/b^(1/4) + 1/4*I*log((-I*b^(1/4)*x + (b*x^4 + a)^(1/4))/x)/b^(1/4)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {5}{4}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx=-\frac {\arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{2 \, b^{\frac {1}{4}}} - \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{4 \, b^{\frac {1}{4}}} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

integrate((b*x^4 + a)^(-1/4), x)

Mupad [B] (verification not implemented)

Time = 5.70 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx=\frac {x\,{\left (\frac {b\,x^4}{a}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (b\,x^4+a\right )}^{1/4}} \]

[In]

int(1/(a + b*x^4)^(1/4),x)

[Out]

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