\(\int \frac {\sqrt [4]{a+b x^4}}{x} \, dx\) [993]

Optimal result

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

[Out]

(b*x^4+a)^(1/4)-1/2*a^(1/4)*arctan((b*x^4+a)^(1/4)/a^(1/4))-1/2*a^(1/4)*arctanh((b*x^4+a)^(1/4)/a^(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} \[ \int \frac {\sqrt [4]{a+b x^4}}{x} \, dx=-\frac {1}{2} \sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {1}{2} \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )+\sqrt [4]{a+b x^4} \]

[In]

Int[(a + b*x^4)^(1/4)/x,x]

[Out]

(a + b*x^4)^(1/4) - (a^(1/4)*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/2 - (a^(1/4)*ArcTanh[(a + b*x^4)^(1/4)/a^(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]{a+b x}}{x} \, dx,x,x^4\right ) \\ & = \sqrt [4]{a+b x^4}+\frac {1}{4} a \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/4}} \, dx,x,x^4\right ) \\ & = \sqrt [4]{a+b x^4}+\frac {a \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right )}{b} \\ & = \sqrt [4]{a+b x^4}-\frac {1}{2} \sqrt {a} \text {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )-\frac {1}{2} \sqrt {a} \text {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right ) \\ & = \sqrt [4]{a+b x^4}-\frac {1}{2} \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {1}{2} \sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{a+b x^4}}{x} \, dx=\sqrt [4]{a+b x^4}-\frac {1}{2} \sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac {1}{2} \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right ) \]

[In]

Integrate[(a + b*x^4)^(1/4)/x,x]

[Out]

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

Maple [A] (verified)

Time = 4.77 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\left (b \,x^{4}+a \right )^{\frac {1}{4}}-\frac {\ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) a^{\frac {1}{4}}}{4}-\frac {a^{\frac {1}{4}} \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{2}\)	\(71\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt [4]{a+b x^4}}{x} \, dx=-\frac {1}{4} \, a^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}\right ) - \frac {1}{4} i \, a^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} + i \, a^{\frac {1}{4}}\right ) + \frac {1}{4} i \, a^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} - i \, a^{\frac {1}{4}}\right ) + \frac {1}{4} \, a^{\frac {1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}\right ) + {\left (b x^{4} + a\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt [4]{a+b x^4}}{x} \, dx=- \frac {\sqrt [4]{b} x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{a+b x^4}}{x} \, dx=-\frac {1}{2} \, a^{\frac {1}{4}} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) + \frac {1}{4} \, a^{\frac {1}{4}} \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right ) + {\left (b x^{4} + a\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

-1/2*a^(1/4)*arctan((b*x^4 + a)^(1/4)/a^(1/4)) + 1/4*a^(1/4)*log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)^(1
/4) + a^(1/4))) + (b*x^4 + a)^(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 = 0.29 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.77 \[ \int \frac {\sqrt [4]{a+b x^4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - \frac {1}{8} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right ) + \frac {1}{8} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right ) + {\left (b x^{4} + a\right )}^{\frac {1}{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt [4]{a+b x^4}}{x} \, dx={\left (b\,x^4+a\right )}^{1/4}-\frac {a^{1/4}\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2}-\frac {a^{1/4}\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{2} \]

[In]

int((a + b*x^4)^(1/4)/x,x)

[Out]

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