\(\int \frac {-b+a x^8}{x^6 (b+a x^4)^{3/4}} \, dx\) [1175]

Optimal result

Integrand size = 24, antiderivative size = 86 \[ \int \frac {-b+a x^8}{x^6 \left (b+a x^4\right )^{3/4}} \, dx=\frac {\left (b-4 a x^4\right ) \sqrt [4]{b+a x^4}}{5 b x^5}-\frac {1}{2} \sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )+\frac {1}{2} \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right ) \]

[Out]

1/5*(-4*a*x^4+b)*(a*x^4+b)^(1/4)/b/x^5-1/2*a^(1/4)*arctan(a^(1/4)*x/(a*x^4+b)^(1/4))+1/2*a^(1/4)*arctanh(a^(1/
4)*x/(a*x^4+b)^(1/4))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1503, 277, 270, 338, 304, 209, 212} \[ \int \frac {-b+a x^8}{x^6 \left (b+a x^4\right )^{3/4}} \, dx=-\frac {1}{2} \sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )+\frac {1}{2} \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )-\frac {4 a \sqrt [4]{a x^4+b}}{5 b x}+\frac {\sqrt [4]{a x^4+b}}{5 x^5} \]

[In]

Int[(-b + a*x^8)/(x^6*(b + a*x^4)^(3/4)),x]

[Out]

(b + a*x^4)^(1/4)/(5*x^5) - (4*a*(b + a*x^4)^(1/4))/(5*b*x) - (a^(1/4)*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/
2 + (a^(1/4)*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/2

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 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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

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 {b}{x^6 \left (b+a x^4\right )^{3/4}}+\frac {a x^2}{\left (b+a x^4\right )^{3/4}}\right ) \, dx \\ & = a \int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx-b \int \frac {1}{x^6 \left (b+a x^4\right )^{3/4}} \, dx \\ & = \frac {\sqrt [4]{b+a x^4}}{5 x^5}+\frac {1}{5} (4 a) \int \frac {1}{x^2 \left (b+a x^4\right )^{3/4}} \, dx+a \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right ) \\ & = \frac {\sqrt [4]{b+a x^4}}{5 x^5}-\frac {4 a \sqrt [4]{b+a x^4}}{5 b x}+\frac {1}{2} \sqrt {a} \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )-\frac {1}{2} \sqrt {a} \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right ) \\ & = \frac {\sqrt [4]{b+a x^4}}{5 x^5}-\frac {4 a \sqrt [4]{b+a x^4}}{5 b x}-\frac {1}{2} \sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )+\frac {1}{2} \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {-b+a x^8}{x^6 \left (b+a x^4\right )^{3/4}} \, dx=\frac {\left (b-4 a x^4\right ) \sqrt [4]{b+a x^4}}{5 b x^5}-\frac {1}{2} \sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )+\frac {1}{2} \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right ) \]

[In]

Integrate[(-b + a*x^8)/(x^6*(b + a*x^4)^(3/4)),x]

[Out]

((b - 4*a*x^4)*(b + a*x^4)^(1/4))/(5*b*x^5) - (a^(1/4)*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/2 + (a^(1/4)*Arc
Tanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/2

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.29

method	result	size
pseudoelliptic	\(\frac {5 b \,a^{\frac {1}{4}} \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) x^{5}+10 b \,a^{\frac {1}{4}} \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) x^{5}-16 a \left (a \,x^{4}+b \right )^{\frac {1}{4}} x^{4}+4 b \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{20 b \,x^{5}}\)	\(111\)

[In]

int((a*x^8-b)/x^6/(a*x^4+b)^(3/4),x,method=_RETURNVERBOSE)

[Out]

1/20*(5*b*a^(1/4)*ln((-a^(1/4)*x-(a*x^4+b)^(1/4))/(a^(1/4)*x-(a*x^4+b)^(1/4)))*x^5+10*b*a^(1/4)*arctan(1/a^(1/
4)/x*(a*x^4+b)^(1/4))*x^5-16*a*(a*x^4+b)^(1/4)*x^4+4*b*(a*x^4+b)^(1/4))/b/x^5

Fricas [F(-1)]

Timed out. \[ \int \frac {-b+a x^8}{x^6 \left (b+a x^4\right )^{3/4}} \, dx=\text {Timed out} \]

[In]

integrate((a*x^8-b)/x^6/(a*x^4+b)^(3/4),x, algorithm="fricas")

[Out]

Timed out

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \frac {-b+a x^8}{x^6 \left (b+a x^4\right )^{3/4}} \, dx=- \frac {a^{\frac {5}{4}} \sqrt [4]{1 + \frac {b}{a x^{4}}} \Gamma \left (- \frac {5}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} + \frac {\sqrt [4]{a} \sqrt [4]{1 + \frac {b}{a x^{4}}} \Gamma \left (- \frac {5}{4}\right )}{16 x^{4} \Gamma \left (\frac {3}{4}\right )} + \frac {a x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate((a*x**8-b)/x**6/(a*x**4+b)**(3/4),x)

[Out]

-a**(5/4)*(1 + b/(a*x**4))**(1/4)*gamma(-5/4)/(4*b*gamma(3/4)) + a**(1/4)*(1 + b/(a*x**4))**(1/4)*gamma(-5/4)/
(16*x**4*gamma(3/4)) + a*x**3*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*x**4*exp_polar(I*pi)/b)/(4*b**(3/4)*gamma
(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24 \[ \int \frac {-b+a x^8}{x^6 \left (b+a x^4\right )^{3/4}} \, dx=\frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )} - \frac {\frac {5 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} a}{x} - \frac {{\left (a x^{4} + b\right )}^{\frac {5}{4}}}{x^{5}}}{5 \, b} \]

[In]

integrate((a*x^8-b)/x^6/(a*x^4+b)^(3/4),x, algorithm="maxima")

[Out]

1/4*a*(2*arctan((a*x^4 + b)^(1/4)/(a^(1/4)*x))/a^(3/4) - log(-(a^(1/4) - (a*x^4 + b)^(1/4)/x)/(a^(1/4) + (a*x^
4 + b)^(1/4)/x))/a^(3/4)) - 1/5*(5*(a*x^4 + b)^(1/4)*a/x - (a*x^4 + b)^(5/4)/x^5)/b

Giac [F]

\[ \int \frac {-b+a x^8}{x^6 \left (b+a x^4\right )^{3/4}} \, dx=\int { \frac {a x^{8} - b}{{\left (a x^{4} + b\right )}^{\frac {3}{4}} x^{6}} \,d x } \]

[In]

integrate((a*x^8-b)/x^6/(a*x^4+b)^(3/4),x, algorithm="giac")

[Out]

integrate((a*x^8 - b)/((a*x^4 + b)^(3/4)*x^6), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-b+a x^8}{x^6 \left (b+a x^4\right )^{3/4}} \, dx=-\int \frac {b-a\,x^8}{x^6\,{\left (a\,x^4+b\right )}^{3/4}} \,d x \]

[In]

int(-(b - a*x^8)/(x^6*(b + a*x^4)^(3/4)),x)

[Out]

-int((b - a*x^8)/(x^6*(b + a*x^4)^(3/4)), x)