\(\int \frac {1}{x (b+a x^4)^{3/4}} \, dx\) [711]

Optimal result

Integrand size = 15, antiderivative size = 55 \[ \int \frac {1}{x \left (b+a x^4\right )^{3/4}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{2 b^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{2 b^{3/4}} \]

[Out]

-1/2*arctan((a*x^4+b)^(1/4)/b^(1/4))/b^(3/4)-1/2*arctanh((a*x^4+b)^(1/4)/b^(1/4))/b^(3/4)

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, 218, 212, 209} \[ \int \frac {1}{x \left (b+a x^4\right )^{3/4}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{2 b^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{a x^4+b}}{\sqrt [4]{b}}\right )}{2 b^{3/4}} \]

[In]

Int[1/(x*(b + a*x^4)^(3/4)),x]

[Out]

-1/2*ArcTan[(b + a*x^4)^(1/4)/b^(1/4)]/b^(3/4) - ArcTanh[(b + a*x^4)^(1/4)/b^(1/4)]/(2*b^(3/4))

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x \left (b+a x^4\right )^{3/4}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{b+a x^4}}{\sqrt [4]{b}}\right )}{2 b^{3/4}} \]

[In]

Integrate[1/(x*(b + a*x^4)^(3/4)),x]

[Out]

-1/2*(ArcTan[(b + a*x^4)^(1/4)/b^(1/4)] + ArcTanh[(b + a*x^4)^(1/4)/b^(1/4)])/b^(3/4)

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(-\frac {\ln \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{\left (a \,x^{4}+b \right )^{\frac {1}{4}}-b^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{4 b^{\frac {3}{4}}}\)	\(55\)

[In]

int(1/x/(a*x^4+b)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-1/4/b^(3/4)*(ln(((a*x^4+b)^(1/4)+b^(1/4))/((a*x^4+b)^(1/4)-b^(1/4)))+2*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 = 104, normalized size of antiderivative = 1.89 \[ \int \frac {1}{x \left (b+a x^4\right )^{3/4}} \, dx=-\frac {1}{4} \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (b \frac {1}{b^{3}}^{\frac {1}{4}} + {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} i \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (i \, b \frac {1}{b^{3}}^{\frac {1}{4}} + {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} i \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (-i \, b \frac {1}{b^{3}}^{\frac {1}{4}} + {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \frac {1}{b^{3}}^{\frac {1}{4}} \log \left (-b \frac {1}{b^{3}}^{\frac {1}{4}} + {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right ) \]

[In]

integrate(1/x/(a*x^4+b)^(3/4),x, algorithm="fricas")

[Out]

-1/4*(b^(-3))^(1/4)*log(b*(b^(-3))^(1/4) + (a*x^4 + b)^(1/4)) - 1/4*I*(b^(-3))^(1/4)*log(I*b*(b^(-3))^(1/4) +
(a*x^4 + b)^(1/4)) + 1/4*I*(b^(-3))^(1/4)*log(-I*b*(b^(-3))^(1/4) + (a*x^4 + b)^(1/4)) + 1/4*(b^(-3))^(1/4)*lo
g(-b*(b^(-3))^(1/4) + (a*x^4 + b)^(1/4))

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x \left (b+a x^4\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{4}} x^{3} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate(1/x/(a*x**4+b)**(3/4),x)

[Out]

-gamma(3/4)*hyper((3/4, 3/4), (7/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(3/4)*x**3*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x \left (b+a x^4\right )^{3/4}} \, dx=-\frac {\arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{2 \, b^{\frac {3}{4}}} + \frac {\log \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{4} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{4 \, b^{\frac {3}{4}}} \]

[In]

integrate(1/x/(a*x^4+b)^(3/4),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

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

Time = 0.26 (sec) , antiderivative size = 186, normalized size of antiderivative = 3.38 \[ \int \frac {1}{x \left (b+a x^4\right )^{3/4}} \, dx=-\frac {\sqrt {2} \left (-b\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{4 \, b} - \frac {\sqrt {2} \left (-b\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{4 \, b} - \frac {\sqrt {2} \left (-b\right )^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right )}{8 \, b} + \frac {\sqrt {2} \left (-b\right )^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} + \sqrt {-b}\right )}{8 \, b} \]

[In]

integrate(1/x/(a*x^4+b)^(3/4),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))/b - 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))/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))/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))/b

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x \left (b+a x^4\right )^{3/4}} \, dx=-\frac {\mathrm {atan}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )+\mathrm {atanh}\left (\frac {{\left (a\,x^4+b\right )}^{1/4}}{b^{1/4}}\right )}{2\,b^{3/4}} \]

[In]

int(1/(x*(b + a*x^4)^(3/4)),x)

[Out]

-(atan((b + a*x^4)^(1/4)/b^(1/4)) + atanh((b + a*x^4)^(1/4)/b^(1/4)))/(2*b^(3/4))