3.17.87 \(\int \frac {-b+a x^2}{x^2 (b+a x^2) \sqrt [4]{-b x^2+a x^4}} \, dx\) [1687]

3.17.87.1 Optimal result
3.17.87.2 Mathematica [A] (verified)
3.17.87.3 Rubi [C] (verified)
3.17.87.4 Maple [A] (verified)
3.17.87.5 Fricas [C] (verification not implemented)
3.17.87.6 Sympy [F]
3.17.87.7 Maxima [F]
3.17.87.8 Giac [B] (verification not implemented)
3.17.87.9 Mupad [F(-1)]

3.17.87.1 Optimal result

Integrand size = 38, antiderivative size = 113 \[ \int \frac {-b+a x^2}{x^2 \left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=-\frac {2 \left (-b x^2+a x^4\right )^{3/4}}{3 b x^3}+\frac {2^{3/4} a^{3/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{b}+\frac {2^{3/4} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )}{b} \]

output
-2/3*(a*x^4-b*x^2)^(3/4)/b/x^3+2^(3/4)*a^(3/4)*arctan(2^(1/4)*a^(1/4)*x/(a 
*x^4-b*x^2)^(1/4))/b+2^(3/4)*a^(3/4)*arctanh(2^(1/4)*a^(1/4)*x/(a*x^4-b*x^ 
2)^(1/4))/b
 
3.17.87.2 Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.32 \[ \int \frac {-b+a x^2}{x^2 \left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\frac {\left (-b x^2+a x^4\right )^{3/4} \left (-2 \left (-b+a x^2\right )^{3/4}+3\ 2^{3/4} a^{3/4} x^{3/2} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+3\ 2^{3/4} a^{3/4} x^{3/2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )}{3 b x^3 \left (-b+a x^2\right )^{3/4}} \]

input
Integrate[(-b + a*x^2)/(x^2*(b + a*x^2)*(-(b*x^2) + a*x^4)^(1/4)),x]
 
output
((-(b*x^2) + a*x^4)^(3/4)*(-2*(-b + a*x^2)^(3/4) + 3*2^(3/4)*a^(3/4)*x^(3/ 
2)*ArcTan[(2^(1/4)*a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)] + 3*2^(3/4)*a^(3/4 
)*x^(3/2)*ArcTanh[(2^(1/4)*a^(1/4)*Sqrt[x])/(-b + a*x^2)^(1/4)]))/(3*b*x^3 
*(-b + a*x^2)^(3/4))
 
3.17.87.3 Rubi [C] (verified)

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

Time = 0.51 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.52, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2467, 368, 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 368

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

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

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

3.17.87.3.1 Defintions of rubi rules used

rule 368
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) 
 - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], 
 x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 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]
 
3.17.87.4 Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.15

method result size
pseudoelliptic \(-\frac {2^{\frac {3}{4}} \left (\arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x \,a^{\frac {1}{4}}}\right ) a \,x^{3}-\frac {\ln \left (\frac {x 2^{\frac {1}{4}} a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{-x 2^{\frac {1}{4}} a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right ) a \,x^{3}}{2}+\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {3}{4}} 2^{\frac {1}{4}} a^{\frac {1}{4}}}{3}\right )}{a^{\frac {1}{4}} x^{3} b}\) \(130\)

input
int((a*x^2-b)/x^2/(a*x^2+b)/(a*x^4-b*x^2)^(1/4),x,method=_RETURNVERBOSE)
 
output
-2^(3/4)*(arctan(1/2*(x^2*(a*x^2-b))^(1/4)/x*2^(3/4)/a^(1/4))*a*x^3-1/2*ln 
((x*2^(1/4)*a^(1/4)+(x^2*(a*x^2-b))^(1/4))/(-x*2^(1/4)*a^(1/4)+(x^2*(a*x^2 
-b))^(1/4)))*a*x^3+1/3*(x^2*(a*x^2-b))^(3/4)*2^(1/4)*a^(1/4))/a^(1/4)/x^3/ 
b
 
3.17.87.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 42.72 (sec) , antiderivative size = 622, normalized size of antiderivative = 5.50 \[ \int \frac {-b+a x^2}{x^2 \left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\frac {3 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \sqrt {\frac {1}{2}} {\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} + 4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {a x^{4} - b x^{2}} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 2 \, {\left (a x^{4} - b x^{2}\right )}^{\frac {3}{4}} a^{2} + \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (3 \, a^{2} b x^{3} - a b^{2} x\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} + b x}\right ) + 3 i \, \left (\frac {1}{2}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {4 \, \sqrt {\frac {1}{2}} {\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} + 4 i \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {a x^{4} - b x^{2}} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} - 2 \, {\left (a x^{4} - b x^{2}\right )}^{\frac {3}{4}} a^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (3 i \, a^{2} b x^{3} - i \, a b^{2} x\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} + b x}\right ) - 3 i \, \left (\frac {1}{2}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {4 \, \sqrt {\frac {1}{2}} {\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} - 4 i \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {a x^{4} - b x^{2}} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} - 2 \, {\left (a x^{4} - b x^{2}\right )}^{\frac {3}{4}} a^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (-3 i \, a^{2} b x^{3} + i \, a b^{2} x\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} + b x}\right ) - 3 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \sqrt {\frac {1}{2}} {\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} - 4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {a x^{4} - b x^{2}} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 2 \, {\left (a x^{4} - b x^{2}\right )}^{\frac {3}{4}} a^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (3 \, a^{2} b x^{3} - a b^{2} x\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} + b x}\right ) - 4 \, {\left (a x^{4} - b x^{2}\right )}^{\frac {3}{4}}}{6 \, b x^{3}} \]

input
integrate((a*x^2-b)/x^2/(a*x^2+b)/(a*x^4-b*x^2)^(1/4),x, algorithm="fricas 
")
 
output
1/6*(3*(1/2)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log((4*sqrt(1/2)*(a*x^4 - b*x^2)^ 
(1/4)*a*b^2*x^2*sqrt(a^3/b^4) + 4*(1/2)^(3/4)*sqrt(a*x^4 - b*x^2)*b^3*x*(a 
^3/b^4)^(3/4) + 2*(a*x^4 - b*x^2)^(3/4)*a^2 + (1/2)^(1/4)*(3*a^2*b*x^3 - a 
*b^2*x)*(a^3/b^4)^(1/4))/(a*x^3 + b*x)) + 3*I*(1/2)^(1/4)*b*x^3*(a^3/b^4)^ 
(1/4)*log(-(4*sqrt(1/2)*(a*x^4 - b*x^2)^(1/4)*a*b^2*x^2*sqrt(a^3/b^4) + 4* 
I*(1/2)^(3/4)*sqrt(a*x^4 - b*x^2)*b^3*x*(a^3/b^4)^(3/4) - 2*(a*x^4 - b*x^2 
)^(3/4)*a^2 - (1/2)^(1/4)*(3*I*a^2*b*x^3 - I*a*b^2*x)*(a^3/b^4)^(1/4))/(a* 
x^3 + b*x)) - 3*I*(1/2)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(-(4*sqrt(1/2)*(a*x 
^4 - b*x^2)^(1/4)*a*b^2*x^2*sqrt(a^3/b^4) - 4*I*(1/2)^(3/4)*sqrt(a*x^4 - b 
*x^2)*b^3*x*(a^3/b^4)^(3/4) - 2*(a*x^4 - b*x^2)^(3/4)*a^2 - (1/2)^(1/4)*(- 
3*I*a^2*b*x^3 + I*a*b^2*x)*(a^3/b^4)^(1/4))/(a*x^3 + b*x)) - 3*(1/2)^(1/4) 
*b*x^3*(a^3/b^4)^(1/4)*log((4*sqrt(1/2)*(a*x^4 - b*x^2)^(1/4)*a*b^2*x^2*sq 
rt(a^3/b^4) - 4*(1/2)^(3/4)*sqrt(a*x^4 - b*x^2)*b^3*x*(a^3/b^4)^(3/4) + 2* 
(a*x^4 - b*x^2)^(3/4)*a^2 - (1/2)^(1/4)*(3*a^2*b*x^3 - a*b^2*x)*(a^3/b^4)^ 
(1/4))/(a*x^3 + b*x)) - 4*(a*x^4 - b*x^2)^(3/4))/(b*x^3)
 
3.17.87.6 Sympy [F]

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

input
integrate((a*x**2-b)/x**2/(a*x**2+b)/(a*x**4-b*x**2)**(1/4),x)
 
output
Integral((a*x**2 - b)/(x**2*(x**2*(a*x**2 - b))**(1/4)*(a*x**2 + b)), x)
 
3.17.87.7 Maxima [F]

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

input
integrate((a*x^2-b)/x^2/(a*x^2+b)/(a*x^4-b*x^2)^(1/4),x, algorithm="maxima 
")
 
output
integrate((a*x^2 - b)/((a*x^4 - b*x^2)^(1/4)*(a*x^2 + b)*x^2), x)
 
3.17.87.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (89) = 178\).

Time = 0.30 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.88 \[ \int \frac {-b+a x^2}{x^2 \left (b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}} \, dx=\frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{b} + \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{b} - \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{2 \, b} + \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right )}{2 \, b} - \frac {2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {3}{4}}}{3 \, b} \]

input
integrate((a*x^2-b)/x^2/(a*x^2+b)/(a*x^4-b*x^2)^(1/4),x, algorithm="giac")
 
output
2^(1/4)*(-a)^(3/4)*arctan(1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) + 2*(a - b/x^2)^ 
(1/4))/(-a)^(1/4))/b + 2^(1/4)*(-a)^(3/4)*arctan(-1/2*2^(1/4)*(2^(3/4)*(-a 
)^(1/4) - 2*(a - b/x^2)^(1/4))/(-a)^(1/4))/b - 1/2*2^(1/4)*(-a)^(3/4)*log( 
2^(3/4)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a - b/x^2)) 
/b + 1/2*2^(1/4)*(-a)^(3/4)*log(-2^(3/4)*(-a)^(1/4)*(a - b/x^2)^(1/4) + sq 
rt(2)*sqrt(-a) + sqrt(a - b/x^2))/b - 2/3*(a - b/x^2)^(3/4)/b
 
3.17.87.9 Mupad [F(-1)]

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

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