\(\int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx\) [1208]

Optimal result

Integrand size = 27, antiderivative size = 88 \[ \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx=\frac {\left (-3 b-4 a x^4\right ) \left (-b+a x^4\right )^{3/4}}{7 b x^7}+a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \]

[Out]

1/7*(-4*a*x^4-3*b)*(a*x^4-b)^(3/4)/b/x^7+a^(3/4)*arctan(a^(1/4)*x/(a*x^4-b)^(1/4))+a^(3/4)*arctanh(a^(1/4)*x/(
a*x^4-b)^(1/4))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1503, 246, 218, 212, 209, 277, 270} \[ \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx=a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )-\frac {3 \left (a x^4-b\right )^{3/4}}{7 x^7}-\frac {4 a \left (a x^4-b\right )^{3/4}}{7 b x^3} \]

[In]

Int[(-3*b + 2*a*x^8)/(x^8*(-b + a*x^4)^(1/4)),x]

[Out]

(-3*(-b + a*x^4)^(3/4))/(7*x^7) - (4*a*(-b + a*x^4)^(3/4))/(7*b*x^3) + a^(3/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)
^(1/4)] + a^(3/4)*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(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 270

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

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 1503

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[Expan
dIntegrand[(f*x)^m*(d + e*x^n)^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2, 2*n]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 a}{\sqrt [4]{-b+a x^4}}-\frac {3 b}{x^8 \sqrt [4]{-b+a x^4}}\right ) \, dx \\ & = (2 a) \int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx-(3 b) \int \frac {1}{x^8 \sqrt [4]{-b+a x^4}} \, dx \\ & = -\frac {3 \left (-b+a x^4\right )^{3/4}}{7 x^7}-\frac {1}{7} (12 a) \int \frac {1}{x^4 \sqrt [4]{-b+a x^4}} \, dx+(2 a) \text {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = -\frac {3 \left (-b+a x^4\right )^{3/4}}{7 x^7}-\frac {4 a \left (-b+a x^4\right )^{3/4}}{7 b x^3}+a \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+a \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = -\frac {3 \left (-b+a x^4\right )^{3/4}}{7 x^7}-\frac {4 a \left (-b+a x^4\right )^{3/4}}{7 b x^3}+a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx=\frac {\left (-3 b-4 a x^4\right ) \left (-b+a x^4\right )^{3/4}}{7 b x^7}+a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \]

[In]

Integrate[(-3*b + 2*a*x^8)/(x^8*(-b + a*x^4)^(1/4)),x]

[Out]

((-3*b - 4*a*x^4)*(-b + a*x^4)^(3/4))/(7*b*x^7) + a^(3/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + a^(3/4)*Arc
Tanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.38

method	result	size
pseudoelliptic	\(\frac {7 a^{\frac {3}{4}} \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) b \,x^{7}-14 a^{\frac {3}{4}} \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b \,x^{7}-8 \left (a \,x^{4}-b \right )^{\frac {3}{4}} a \,x^{4}-6 b \left (a \,x^{4}-b \right )^{\frac {3}{4}}}{14 b \,x^{7}}\)	\(121\)

[In]

int((2*a*x^8-3*b)/x^8/(a*x^4-b)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/14*(7*a^(3/4)*ln((-a^(1/4)*x-(a*x^4-b)^(1/4))/(a^(1/4)*x-(a*x^4-b)^(1/4)))*b*x^7-14*a^(3/4)*arctan(1/a^(1/4)
/x*(a*x^4-b)^(1/4))*b*x^7-8*(a*x^4-b)^(3/4)*a*x^4-6*b*(a*x^4-b)^(3/4))/b/x^7

Fricas [F(-1)]

Timed out. \[ \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx=\text {Timed out} \]

[In]

integrate((2*a*x^8-3*b)/x^8/(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 = 1.26 (sec) , antiderivative size = 345, normalized size of antiderivative = 3.92 \[ \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx=\frac {a x e^{- \frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{2 \sqrt [4]{b} \Gamma \left (\frac {5}{4}\right )} - 3 b \left (\begin {cases} - \frac {4 a^{\frac {15}{4}} x^{8} \left (-1 + \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 a^{2} b^{2} x^{8} e^{\frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) - 16 a b^{3} x^{4} e^{\frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right )} + \frac {a^{\frac {11}{4}} b x^{4} \left (-1 + \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 a^{2} b^{2} x^{8} e^{\frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) - 16 a b^{3} x^{4} e^{\frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right )} + \frac {3 a^{\frac {7}{4}} b^{2} \left (-1 + \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 a^{2} b^{2} x^{8} e^{\frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) - 16 a b^{3} x^{4} e^{\frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\\frac {a^{\frac {7}{4}} \left (1 - \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{4 b^{2} \Gamma \left (\frac {1}{4}\right )} + \frac {3 a^{\frac {3}{4}} \left (1 - \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 b x^{4} \Gamma \left (\frac {1}{4}\right )} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((2*a*x**8-3*b)/x**8/(a*x**4-b)**(1/4),x)

[Out]

a*x*exp(-I*pi/4)*gamma(1/4)*hyper((1/4, 1/4), (5/4,), a*x**4/b)/(2*b**(1/4)*gamma(5/4)) - 3*b*Piecewise((-4*a*
*(15/4)*x**8*(-1 + b/(a*x**4))**(3/4)*gamma(-7/4)/(16*a**2*b**2*x**8*exp(I*pi/4)*gamma(1/4) - 16*a*b**3*x**4*e
xp(I*pi/4)*gamma(1/4)) + a**(11/4)*b*x**4*(-1 + b/(a*x**4))**(3/4)*gamma(-7/4)/(16*a**2*b**2*x**8*exp(I*pi/4)*
gamma(1/4) - 16*a*b**3*x**4*exp(I*pi/4)*gamma(1/4)) + 3*a**(7/4)*b**2*(-1 + b/(a*x**4))**(3/4)*gamma(-7/4)/(16
*a**2*b**2*x**8*exp(I*pi/4)*gamma(1/4) - 16*a*b**3*x**4*exp(I*pi/4)*gamma(1/4)), Abs(b/(a*x**4)) > 1), (a**(7/
4)*(1 - b/(a*x**4))**(3/4)*gamma(-7/4)/(4*b**2*gamma(1/4)) + 3*a**(3/4)*(1 - b/(a*x**4))**(3/4)*gamma(-7/4)/(1
6*b*x**4*gamma(1/4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.32 \[ \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx=-\frac {1}{2} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )} - \frac {\frac {7 \, {\left (a x^{4} - b\right )}^{\frac {3}{4}} a}{x^{3}} - \frac {3 \, {\left (a x^{4} - b\right )}^{\frac {7}{4}}}{x^{7}}}{7 \, b} \]

[In]

integrate((2*a*x^8-3*b)/x^8/(a*x^4-b)^(1/4),x, algorithm="maxima")

[Out]

-1/2*a*(2*arctan((a*x^4 - b)^(1/4)/(a^(1/4)*x))/a^(1/4) + log(-(a^(1/4) - (a*x^4 - b)^(1/4)/x)/(a^(1/4) + (a*x
^4 - b)^(1/4)/x))/a^(1/4)) - 1/7*(7*(a*x^4 - b)^(3/4)*a/x^3 - 3*(a*x^4 - b)^(7/4)/x^7)/b

Giac [F]

\[ \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx=\int { \frac {2 \, a x^{8} - 3 \, b}{{\left (a x^{4} - b\right )}^{\frac {1}{4}} x^{8}} \,d x } \]

[In]

integrate((2*a*x^8-3*b)/x^8/(a*x^4-b)^(1/4),x, algorithm="giac")

[Out]

integrate((2*a*x^8 - 3*b)/((a*x^4 - b)^(1/4)*x^8), x)

Mupad [B] (verification not implemented)

Time = 6.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81 \[ \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx=\frac {2\,a\,x\,{\left (1-\frac {a\,x^4}{b}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ \frac {a\,x^4}{b}\right )}{{\left (a\,x^4-b\right )}^{1/4}}-\frac {{\left (a\,x^4-b\right )}^{3/4}\,\left (4\,a\,x^4+3\,b\right )}{7\,b\,x^7} \]

[In]

int(-(3*b - 2*a*x^8)/(x^8*(a*x^4 - b)^(1/4)),x)

[Out]

(2*a*x*(1 - (a*x^4)/b)^(1/4)*hypergeom([1/4, 1/4], 5/4, (a*x^4)/b))/(a*x^4 - b)^(1/4) - ((a*x^4 - b)^(3/4)*(3*
b + 4*a*x^4))/(7*b*x^7)