\(\int \frac {-b+a x^3}{x^3 (b+a x^3) \sqrt [4]{-b x+a x^4}} \, dx\) [1913]

Optimal result

Integrand size = 36, antiderivative size = 133 \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\frac {4 \left (-b x+a x^4\right )^{3/4}}{9 b x^3}+\frac {2\ 2^{3/4} a^{3/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 b}+\frac {2\ 2^{3/4} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 b} \] Output:

-4/9*(a*x^4-b*x)^(3/4)/b/x^3+2/3*2^(3/4)*a^(3/4)*arctan(2^(1/4)*a^(1/4)*(a 
*x^4-b*x)^(3/4)/(a*x^3-b))/b+2/3*2^(3/4)*a^(3/4)*arctanh(2^(1/4)*a^(1/4)*( 
a*x^4-b*x)^(3/4)/(a*x^3-b))/b
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.59 \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\frac {4 \left (1+\frac {a x^3}{b}\right )^{3/4} \left (-b x+a x^4\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {3}{4},\frac {1}{4},\frac {2 a x^3}{b+a x^3}\right )}{9 b x^3 \left (1-\frac {a x^3}{b}\right )^{3/4}} \] Input:

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

Output:

(-4*(1 + (a*x^3)/b)^(3/4)*(-(b*x) + a*x^4)^(3/4)*Hypergeometric2F1[-3/4, - 
3/4, 1/4, (2*a*x^3)/(b + a*x^3)])/(9*b*x^3*(1 - (a*x^3)/b)^(3/4))
 

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.51 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.53, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {2467, 966, 965, 1013, 1012}

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.

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

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^3-b} \int \frac {\left (a x^3-b\right )^{3/4}}{x^{13/4} \left (a x^3+b\right )}dx}{\sqrt [4]{a x^4-b x}}\)

\(\Big \downarrow \) 966

\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3-b} \int \frac {\left (a x^3-b\right )^{3/4}}{x^{5/2} \left (a x^3+b\right )}d\sqrt [4]{x}}{\sqrt [4]{a x^4-b x}}\)

\(\Big \downarrow \) 965

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle -\frac {4 (a x-b)^{3/4} \sqrt [4]{a x^3-b} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\frac {2 a x}{b-a x}\right )}{9 b \sqrt {x} \sqrt [4]{a x^4-b x}}\)

Input:

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

Output:

(-4*(-b + a*x)^(3/4)*(-b + a*x^3)^(1/4)*Hypergeometric2F1[-3/4, 1, 1/4, (- 
2*a*x)/(b - a*x)])/(9*b*Sqrt[x]*(-(b*x) + a*x^4)^(1/4))
 

Defintions of rubi rules used

rule 965

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), 
 x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 
 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free 
Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
 

rule 966

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) 
)^(q_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*( 
m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*x)^( 
1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[ 
n, 0] && FractionQ[m] && IntegerQ[p]
 

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

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

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(-\frac {2 \left (\arctan \left (\frac {{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x \,a^{\frac {1}{4}}}\right ) a \,x^{3}-\frac {\ln \left (\frac {-2^{\frac {1}{4}} a^{\frac {1}{4}} x -{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{2^{\frac {1}{4}} a^{\frac {1}{4}} x -{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}\right ) a \,x^{3}}{2}+\frac {{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {3}{4}} 2^{\frac {1}{4}} a^{\frac {1}{4}}}{3}\right ) 2^{\frac {3}{4}}}{3 a^{\frac {1}{4}} x^{3} b}\)	\(126\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 51.81 (sec) , antiderivative size = 581, normalized size of antiderivative = 4.37 \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/18*(3*8^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log((4*sqrt(2)*(a*x^4 - b*x)^(1/4)*a 
*b^2*x^2*sqrt(a^3/b^4) + 8^(3/4)*sqrt(a*x^4 - b*x)*b^3*x*(a^3/b^4)^(3/4) + 
 4*(a*x^4 - b*x)^(3/4)*a^2 + 8^(1/4)*(3*a^2*b*x^3 - a*b^2)*(a^3/b^4)^(1/4) 
)/(a*x^3 + b)) + 3*I*8^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(-(4*sqrt(2)*(a*x^4 
- b*x)^(1/4)*a*b^2*x^2*sqrt(a^3/b^4) + I*8^(3/4)*sqrt(a*x^4 - b*x)*b^3*x*( 
a^3/b^4)^(3/4) - 4*(a*x^4 - b*x)^(3/4)*a^2 + 8^(1/4)*(-3*I*a^2*b*x^3 + I*a 
*b^2)*(a^3/b^4)^(1/4))/(a*x^3 + b)) - 3*I*8^(1/4)*b*x^3*(a^3/b^4)^(1/4)*lo 
g(-(4*sqrt(2)*(a*x^4 - b*x)^(1/4)*a*b^2*x^2*sqrt(a^3/b^4) - I*8^(3/4)*sqrt 
(a*x^4 - b*x)*b^3*x*(a^3/b^4)^(3/4) - 4*(a*x^4 - b*x)^(3/4)*a^2 + 8^(1/4)* 
(3*I*a^2*b*x^3 - I*a*b^2)*(a^3/b^4)^(1/4))/(a*x^3 + b)) - 3*8^(1/4)*b*x^3* 
(a^3/b^4)^(1/4)*log((4*sqrt(2)*(a*x^4 - b*x)^(1/4)*a*b^2*x^2*sqrt(a^3/b^4) 
 - 8^(3/4)*sqrt(a*x^4 - b*x)*b^3*x*(a^3/b^4)^(3/4) + 4*(a*x^4 - b*x)^(3/4) 
*a^2 - 8^(1/4)*(3*a^2*b*x^3 - a*b^2)*(a^3/b^4)^(1/4))/(a*x^3 + b)) - 8*(a* 
x^4 - b*x)^(3/4))/(b*x^3)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (105) = 210\).

Time = 0.26 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.62 \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\frac {2 \cdot 2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, b} + \frac {2 \cdot 2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, b} - \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right )}{3 \, b} + \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right )}{3 \, b} - \frac {4 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {3}{4}}}{9 \, b} \] Input:

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

Output:

2/3*2^(1/4)*(-a)^(3/4)*arctan(1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) + 2*(a - b/x 
^3)^(1/4))/(-a)^(1/4))/b + 2/3*2^(1/4)*(-a)^(3/4)*arctan(-1/2*2^(1/4)*(2^( 
3/4)*(-a)^(1/4) - 2*(a - b/x^3)^(1/4))/(-a)^(1/4))/b - 1/3*2^(1/4)*(-a)^(3 
/4)*log(2^(3/4)*(-a)^(1/4)*(a - b/x^3)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a - 
 b/x^3))/b + 1/3*2^(1/4)*(-a)^(3/4)*log(-2^(3/4)*(-a)^(1/4)*(a - b/x^3)^(1 
/4) + sqrt(2)*sqrt(-a) + sqrt(a - b/x^3))/b - 4/9*(a - b/x^3)^(3/4)/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\int -\frac {b-a\,x^3}{x^3\,{\left (a\,x^4-b\,x\right )}^{1/4}\,\left (a\,x^3+b\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {-b+a x^3}{x^3 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\left (\int \frac {1}{x^{\frac {25}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} a +x^{\frac {13}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} b}d x \right ) b +\left (\int \frac {1}{x^{\frac {13}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} a +x^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} b}d x \right ) a \] Input:

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

Output:

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