\(\int \frac {-b+a x^4}{x^4 \sqrt [4]{-b+2 a x^4}} \, dx\) [1482]

Optimal result

Integrand size = 27, antiderivative size = 104 \[ \int \frac {-b+a x^4}{x^4 \sqrt [4]{-b+2 a x^4}} \, dx=-\frac {\left (-b+2 a x^4\right )^{3/4}}{3 x^3}+\frac {a^{3/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+2 a x^4}}\right )}{2 \sqrt [4]{2}}+\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+2 a x^4}}\right )}{2 \sqrt [4]{2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {462, 246, 218, 212, 209} \[ \int \frac {-b+a x^4}{x^4 \sqrt [4]{-b+2 a x^4}} \, dx=\frac {a^{3/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{2 a x^4-b}}\right )}{2 \sqrt [4]{2}}+\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{2 a x^4-b}}\right )}{2 \sqrt [4]{2}}-\frac {\left (2 a x^4-b\right )^{3/4}}{3 x^3} \]

[In]

Int[(-b + a*x^4)/(x^4*(-b + 2*a*x^4)^(1/4)),x]

[Out]

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

Rule 462

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {-b+a x^4}{x^4 \sqrt [4]{-b+2 a x^4}} \, dx=-\frac {\left (-b+2 a x^4\right )^{3/4}}{3 x^3}+\frac {a^{3/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+2 a x^4}}\right )}{2 \sqrt [4]{2}}+\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+2 a x^4}}\right )}{2 \sqrt [4]{2}} \]

[In]

Integrate[(-b + a*x^4)/(x^4*(-b + 2*a*x^4)^(1/4)),x]

[Out]

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {3}{4}} \left (\arctan \left (\frac {2^{\frac {3}{4}} \left (2 a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right ) a \,x^{3}-\frac {\ln \left (\frac {-x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (2 a \,x^{4}-b \right )^{\frac {1}{4}}}{x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (2 a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) a \,x^{3}}{2}+\frac {2 \left (2 a \,x^{4}-b \right )^{\frac {3}{4}} 2^{\frac {1}{4}} a^{\frac {1}{4}}}{3}\right )}{4 a^{\frac {1}{4}} x^{3}}\)	\(119\)

[In]

int((a*x^4-b)/x^4/(2*a*x^4-b)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {-b+a x^4}{x^4 \sqrt [4]{-b+2 a x^4}} \, dx=\text {Timed out} \]

[In]

integrate((a*x^4-b)/x^4/(2*a*x^4-b)^(1/4),x, algorithm="fricas")

[Out]

Timed out

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.14 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.34 \[ \int \frac {-b+a x^4}{x^4 \sqrt [4]{-b+2 a x^4}} \, dx=\frac {a x e^{- \frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {2 a x^{4}}{b}} \right )}}{4 \sqrt [4]{b} \Gamma \left (\frac {5}{4}\right )} - b \left (\begin {cases} - \frac {2^{\frac {3}{4}} a^{\frac {3}{4}} \left (-1 + \frac {b}{2 a x^{4}}\right )^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} \Gamma \left (- \frac {3}{4}\right )}{4 b \Gamma \left (\frac {1}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 2 \\- \frac {2^{\frac {3}{4}} a^{\frac {3}{4}} \left (1 - \frac {b}{2 a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{4 b \Gamma \left (\frac {1}{4}\right )} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((a*x**4-b)/x**4/(2*a*x**4-b)**(1/4),x)

[Out]

a*x*exp(-I*pi/4)*gamma(1/4)*hyper((1/4, 1/4), (5/4,), 2*a*x**4/b)/(4*b**(1/4)*gamma(5/4)) - b*Piecewise((-2**(
3/4)*a**(3/4)*(-1 + b/(2*a*x**4))**(3/4)*exp(3*I*pi/4)*gamma(-3/4)/(4*b*gamma(1/4)), Abs(b/(a*x**4)) > 2), (-2
**(3/4)*a**(3/4)*(1 - b/(2*a*x**4))**(3/4)*gamma(-3/4)/(4*b*gamma(1/4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11 \[ \int \frac {-b+a x^4}{x^4 \sqrt [4]{-b+2 a x^4}} \, dx=-\frac {1}{8} \, {\left (\frac {2 \cdot 2^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2 \, a x^{4} - b\right )}^{\frac {1}{4}}}{2 \, a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {1}{4}} a^{\frac {1}{4}} - \frac {{\left (2 \, a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{2^{\frac {1}{4}} a^{\frac {1}{4}} + \frac {{\left (2 \, a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )} a - \frac {{\left (2 \, a x^{4} - b\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

integrate((a*x^4-b)/x^4/(2*a*x^4-b)^(1/4),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((a*x^4-b)/x^4/(2*a*x^4-b)^(1/4),x, algorithm="giac")

[Out]

integrate((a*x^4 - b)/((2*a*x^4 - b)^(1/4)*x^4), x)

Mupad [B] (verification not implemented)

Time = 6.84 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.58 \[ \int \frac {-b+a x^4}{x^4 \sqrt [4]{-b+2 a x^4}} \, dx=\frac {a\,x\,{\left (1-\frac {2\,a\,x^4}{b}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ \frac {2\,a\,x^4}{b}\right )}{{\left (2\,a\,x^4-b\right )}^{1/4}}-\frac {{\left (2\,a\,x^4-b\right )}^{3/4}}{3\,x^3} \]

[In]

int(-(b - a*x^4)/(x^4*(2*a*x^4 - b)^(1/4)),x)

[Out]

(a*x*(1 - (2*a*x^4)/b)^(1/4)*hypergeom([1/4, 1/4], 5/4, (2*a*x^4)/b))/(2*a*x^4 - b)^(1/4) - (2*a*x^4 - b)^(3/4
)/(3*x^3)