\(\int \frac {\sqrt [4]{b+a x^4}}{x} \, dx\) [869]

Optimal result

Integrand size = 15, antiderivative size = 66 \[ \int \frac {\sqrt [4]{b+a x^4}}{x} \, dx=\sqrt [4]{b+a x^4}-\frac {1}{2} \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )-\frac {1}{2} \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 52, 65, 218, 212, 209} \[ \int \frac {\sqrt [4]{b+a x^4}}{x} \, dx=-\frac {1}{2} \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )-\frac {1}{2} \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )+\sqrt [4]{a x^4+b} \]

[In]

Int[(b + a*x^4)^(1/4)/x,x]

[Out]

(b + a*x^4)^(1/4) - (b^(1/4)*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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 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 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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {\sqrt [4]{b+a x}}{x} \, dx,x,x^4\right ) \\ & = \sqrt [4]{b+a x^4}+\frac {1}{4} b \text {Subst}\left (\int \frac {1}{x (b+a x)^{3/4}} \, dx,x,x^4\right ) \\ & = \sqrt [4]{b+a x^4}+\frac {b \text {Subst}\left (\int \frac {1}{-\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{b+a x^4}\right )}{a} \\ & = \sqrt [4]{b+a x^4}-\frac {1}{2} \sqrt {b} \text {Subst}\left (\int \frac {1}{\sqrt {b}-x^2} \, dx,x,\sqrt [4]{b+a x^4}\right )-\frac {1}{2} \sqrt {b} \text {Subst}\left (\int \frac {1}{\sqrt {b}+x^2} \, dx,x,\sqrt [4]{b+a x^4}\right ) \\ & = \sqrt [4]{b+a x^4}-\frac {1}{2} \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )-\frac {1}{2} \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{b+a x^4}}{x} \, dx=\sqrt [4]{b+a x^4}-\frac {1}{2} \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )-\frac {1}{2} \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right ) \]

[In]

Integrate[(b + a*x^4)^(1/4)/x,x]

[Out]

(b + a*x^4)^(1/4) - (b^(1/4)*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

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(\left (a \,x^{4}+b \right )^{\frac {1}{4}}-\frac {\ln \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{\left (a \,x^{4}+b \right )^{\frac {1}{4}}-b^{\frac {1}{4}}}\right ) b^{\frac {1}{4}}}{4}-\frac {b^{\frac {1}{4}} \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{2}\)	\(67\)

[In]

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

[Out]

(a*x^4+b)^(1/4)-1/4*ln(((a*x^4+b)^(1/4)+b^(1/4))/((a*x^4+b)^(1/4)-b^(1/4)))*b^(1/4)-1/2*b^(1/4)*arctan((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 = 92, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt [4]{b+a x^4}}{x} \, dx=-\frac {1}{4} \, b^{\frac {1}{4}} \log \left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}\right ) - \frac {1}{4} i \, b^{\frac {1}{4}} \log \left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} + i \, b^{\frac {1}{4}}\right ) + \frac {1}{4} i \, b^{\frac {1}{4}} \log \left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} - i \, b^{\frac {1}{4}}\right ) + \frac {1}{4} \, b^{\frac {1}{4}} \log \left ({\left (a x^{4} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}\right ) + {\left (a x^{4} + b\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt [4]{b+a x^4}}{x} \, dx=- \frac {\sqrt [4]{a} x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

-a**(1/4)*x*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), b*exp_polar(I*pi)/(a*x**4))/(4*gamma(3/4))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{b+a x^4}}{x} \, dx=-\frac {1}{2} \, b^{\frac {1}{4}} \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) + \frac {1}{4} \, b^{\frac {1}{4}} \log \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{4} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right ) + {\left (a x^{4} + b\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (48) = 96\).

Time = 0.26 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.77 \[ \int \frac {\sqrt [4]{b+a x^4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} \left (-b\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-b\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right ) - \frac {1}{8} \, \sqrt {2} \left (-b\right )^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right ) + \frac {1}{8} \, \sqrt {2} \left (-b\right )^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right ) + {\left (a x^{4} + b\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt [4]{b+a x^4}}{x} \, dx={\left (a\,x^4+b\right )}^{1/4}-\frac {b^{1/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )}{2}-\frac {b^{1/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )}{2} \]

[In]

int((b + a*x^4)^(1/4)/x,x)

[Out]

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