\(\int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx\) [712]

Optimal result

Integrand size = 15, antiderivative size = 55 \[ \int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 65, 304, 209, 212} \[ \int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}} \]

[In]

Int[1/(x*(b + a*x^4)^(1/4)),x]

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 272

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

Rule 304

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}} \]

[In]

Integrate[1/(x*(b + a*x^4)^(1/4)),x]

[Out]

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

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(\frac {2 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )-\ln \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{\left (a \,x^{4}+b \right )^{\frac {1}{4}}-b^{\frac {1}{4}}}\right )}{4 b^{\frac {1}{4}}}\)	\(57\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.51 \[ \int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx=-\frac {\log \left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}\right )}{4 \, b^{\frac {1}{4}}} + \frac {i \, \log \left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} + i \, b^{\frac {1}{4}}\right )}{4 \, b^{\frac {1}{4}}} - \frac {i \, \log \left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} - i \, b^{\frac {1}{4}}\right )}{4 \, b^{\frac {1}{4}}} + \frac {\log \left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}\right )}{4 \, b^{\frac {1}{4}}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx=- \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \sqrt [4]{a} x \Gamma \left (\frac {5}{4}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx=\frac {\arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{2 \, b^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{4} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{4 \, b^{\frac {1}{4}}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (39) = 78\).

Time = 0.29 (sec) , antiderivative size = 186, normalized size of antiderivative = 3.38 \[ \int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx=-\frac {\sqrt {2} \left (-b\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{4 \, b} - \frac {\sqrt {2} \left (-b\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{4 \, b} + \frac {\sqrt {2} \left (-b\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right )}{8 \, b} - \frac {\sqrt {2} \left (-b\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right )}{8 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.85 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x \sqrt [4]{b+a x^4}} \, dx=\frac {\mathrm {atan}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )-\mathrm {atanh}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )}{2\,b^{1/4}} \]

[In]

int(1/(x*(b + a*x^4)^(1/4)),x)

[Out]

(atan((b + a*x^4)^(1/4)/b^(1/4)) - atanh((b + a*x^4)^(1/4)/b^(1/4)))/(2*b^(1/4))