\(\int \frac {-b+a x^8}{x^2 (-b+a x^4)^{3/4}} \, dx\) [1207]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {\left (-4+x^4\right ) \sqrt [4]{-b+a x^4}}{4 x}-\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {\left (-4+x^4\right ) \sqrt [4]{-b+a x^4}}{4 x}-\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}} \] Input:

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

Output:

((-4 + x^4)*(-b + a*x^4)^(1/4))/(4*x) - (3*b*ArcTan[(a^(1/4)*x)/(-b + a*x^ 
4)^(1/4)])/(8*a^(3/4)) + (3*b*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/(8* 
a^(3/4))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.26, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1811, 25, 27, 953, 854, 827, 216, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {a x^8-b}{x^2 \left (a x^4-b\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 1811

\(\displaystyle \frac {\int -\frac {a b \left (4-3 x^4\right )}{x^2 \left (a x^4-b\right )^{3/4}}dx}{4 a}+\frac {1}{4} x^3 \sqrt [4]{a x^4-b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4-b}-\frac {\int \frac {a b \left (4-3 x^4\right )}{x^2 \left (a x^4-b\right )^{3/4}}dx}{4 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4-b}-\frac {1}{4} b \int \frac {4-3 x^4}{x^2 \left (a x^4-b\right )^{3/4}}dx\)

\(\Big \downarrow \) 953

\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4-b}-\frac {1}{4} b \left (\frac {4 \sqrt [4]{a x^4-b}}{b x}-3 \int \frac {x^2}{\left (a x^4-b\right )^{3/4}}dx\right )\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4-b}-\frac {1}{4} b \left (\frac {4 \sqrt [4]{a x^4-b}}{b x}-3 \int \frac {x^2}{\sqrt {a x^4-b} \left (1-\frac {a x^4}{a x^4-b}\right )}d\frac {x}{\sqrt [4]{a x^4-b}}\right )\)

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4-b}-\frac {1}{4} b \left (\frac {4 \sqrt [4]{a x^4-b}}{b x}-3 \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 a^{3/4}}\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4-b}-\frac {1}{4} b \left (\frac {4 \sqrt [4]{a x^4-b}}{b x}-3 \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 a^{3/4}}\right )\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 827

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

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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 953

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] + Simp[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] && G 
tQ[m + n, -1]))
 

rule 1811

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ 
(n_))^(q_.), x_Symbol] :> Simp[c^p*(f*x)^(m + 2*n*p - n + 1)*((d + e*x^n)^( 
q + 1)/(e*f^(2*n*p - n + 1)*(m + 2*n*p + n*q + 1))), x] + Simp[1/(e*(m + 2* 
n*p + n*q + 1))   Int[(f*x)^m*(d + e*x^n)^q*ExpandToSum[e*(m + 2*n*p + n*q 
+ 1)*((a + c*x^(2*n))^p - c^p*x^(2*n*p)) - d*c^p*(m + 2*n*p - n + 1)*x^(2*n 
*p - n), x], x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2, 2*n] && I 
GtQ[n, 0] && IGtQ[p, 0] && GtQ[2*n*p, n - 1] &&  !IntegerQ[q] && NeQ[m + 2* 
n*p + n*q + 1, 0]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.23

method	result	size
pseudoelliptic	\(-\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x}+\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}} x^{3}}{4}+\frac {3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) b}{16 a^{\frac {3}{4}}}+\frac {3 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b}{8 a^{\frac {3}{4}}}\)	\(108\)

Input:

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

Output:

-(a*x^4-b)^(1/4)/x+1/4*(a*x^4-b)^(1/4)*x^3+3/16*ln((-a^(1/4)*x-(a*x^4-b)^( 
1/4))/(a^(1/4)*x-(a*x^4-b)^(1/4)))/a^(3/4)*b+3/8*arctan(1/a^(1/4)/x*(a*x^4 
-b)^(1/4))/a^(3/4)*b
 

Fricas [F(-1)]

Timed out. \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.39 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.43 \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=- \frac {a x^{7} e^{\frac {i \pi }{4}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} - b \left (\begin {cases} - \frac {\sqrt [4]{a} \sqrt [4]{-1 + \frac {b}{a x^{4}}} e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\- \frac {\sqrt [4]{a} \sqrt [4]{1 - \frac {b}{a x^{4}}} \Gamma \left (- \frac {1}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-a*x**7*exp(I*pi/4)*gamma(7/4)*hyper((3/4, 7/4), (11/4,), a*x**4/b)/(4*b** 
(3/4)*gamma(11/4)) - b*Piecewise((-a**(1/4)*(-1 + b/(a*x**4))**(1/4)*exp(I 
*pi/4)*gamma(-1/4)/(4*b*gamma(3/4)), Abs(b/(a*x**4)) > 1), (-a**(1/4)*(1 - 
 b/(a*x**4))**(1/4)*gamma(-1/4)/(4*b*gamma(3/4)), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (68) = 136\).

Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.59 \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {1}{16} \, a {\left (\frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )}}{a} + \frac {4 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} b}{{\left (a^{2} - \frac {{\left (a x^{4} - b\right )} a}{x^{4}}\right )} x}\right )} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x} \] Input:

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

Output:

1/16*a*(3*(2*b*arctan((a*x^4 - b)^(1/4)/(a^(1/4)*x))/a^(3/4) - b*log(-(a^( 
1/4) - (a*x^4 - b)^(1/4)/x)/(a^(1/4) + (a*x^4 - b)^(1/4)/x))/a^(3/4))/a + 
4*(a*x^4 - b)^(1/4)*b/((a^2 - (a*x^4 - b)*a/x^4)*x)) - (a*x^4 - b)^(1/4)/x
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=-\int \frac {b-a\,x^8}{x^2\,{\left (a\,x^4-b\right )}^{3/4}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\left (\int \frac {x^{6}}{\left (a \,x^{4}-b \right )^{\frac {3}{4}}}d x \right ) a -\left (\int \frac {1}{\left (a \,x^{4}-b \right )^{\frac {3}{4}} x^{2}}d x \right ) b \] Input:

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

Output:

int(x**6/(a*x**4 - b)**(3/4),x)*a - int(1/((a*x**4 - b)**(3/4)*x**2),x)*b