\(\int \frac {x^4}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx\) [1325]

Optimal result

Integrand size = 27, antiderivative size = 101 \[ \int \frac {x^4}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=-\frac {x \sqrt {a-b x^2} \sqrt {a+b x^2}}{3 b^2}+\frac {a^{5/2} \sqrt {1-\frac {b^2 x^4}{a^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),-1\right )}{3 b^{5/2} \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Output:

-1/3*x*(-b*x^2+a)^(1/2)*(b*x^2+a)^(1/2)/b^2+1/3*a^(5/2)*(1-b^2*x^4/a^2)^(1 
/2)*EllipticF(b^(1/2)*x/a^(1/2),I)/b^(5/2)/(-b*x^2+a)^(1/2)/(b*x^2+a)^(1/2 
)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.71 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\frac {x^5 \sqrt {1-\frac {b^2 x^4}{a^2}} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};\frac {b^2 x^4}{a^2}\right )}{5 \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Input:

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

Output:

(x^5*Sqrt[1 - (b^2*x^4)/a^2]*HypergeometricPFQ[{1/2, 5/4}, {9/4}, (b^2*x^4 
)/a^2])/(5*Sqrt[a - b*x^2]*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.26, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {344, 843, 765, 762}

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.

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

\(\Big \downarrow \) 344

\(\displaystyle \frac {\sqrt {a^2-b^2 x^4} \int \frac {x^4}{\sqrt {a^2-b^2 x^4}}dx}{\sqrt {a-b x^2} \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 765

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

\(\Big \downarrow \) 762

\(\displaystyle \frac {\sqrt {a^2-b^2 x^4} \left (\frac {a^{5/2} \sqrt {1-\frac {b^2 x^4}{a^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),-1\right )}{3 b^{5/2} \sqrt {a^2-b^2 x^4}}-\frac {x \sqrt {a^2-b^2 x^4}}{3 b^2}\right )}{\sqrt {a-b x^2} \sqrt {a+b x^2}}\)

Input:

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

Output:

(Sqrt[a^2 - b^2*x^4]*(-1/3*(x*Sqrt[a^2 - b^2*x^4])/b^2 + (a^(5/2)*Sqrt[1 - 
 (b^2*x^4)/a^2]*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]], -1])/(3*b^(5/2)*Sqr 
t[a^2 - b^2*x^4])))/(Sqrt[a - b*x^2]*Sqrt[a + b*x^2])
 

Defintions of rubi rules used

rule 344

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(a + b*x^2)^FracPart[p]*((c + d*x^2)^FracPart[p]/(a*c 
+ b*d*x^4)^FracPart[p])   Int[(e*x)^m*(a*c + b*d*x^4)^p, x], x] /; FreeQ[{a 
, b, c, d, e, m, p}, x] && EqQ[b*c + a*d, 0] &&  !IntegerQ[p]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

rule 843

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

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {\sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}\, \left (\sqrt {\frac {b}{a}}\, b^{2} x^{5}+a^{2} \sqrt {\frac {-b \,x^{2}+a}{a}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right )-\sqrt {\frac {b}{a}}\, a^{2} x \right )}{3 \left (-b^{2} x^{4}+a^{2}\right ) b^{2} \sqrt {\frac {b}{a}}}\)	\(117\)
elliptic	\(\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \left (-\frac {x \sqrt {-b^{2} x^{4}+a^{2}}}{3 b^{2}}+\frac {a^{2} \sqrt {1-\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right )}{3 b^{2} \sqrt {\frac {b}{a}}\, \sqrt {-b^{2} x^{4}+a^{2}}}\right )}{\sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}}\)	\(122\)
risch	\(-\frac {x \sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}}{3 b^{2}}+\frac {a^{2} \sqrt {1-\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right ) \sqrt {\left (-b \,x^{2}+a \right ) \left (b \,x^{2}+a \right )}}{3 b^{2} \sqrt {\frac {b}{a}}\, \sqrt {-b^{2} x^{4}+a^{2}}\, \sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}}\)	\(130\)

Input:

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

Output:

1/3*(-b*x^2+a)^(1/2)*(b*x^2+a)^(1/2)*((b/a)^(1/2)*b^2*x^5+a^2*((-b*x^2+a)/ 
a)^(1/2)*((b*x^2+a)/a)^(1/2)*EllipticF(x*(b/a)^(1/2),I)-(b/a)^(1/2)*a^2*x) 
/(-b^2*x^4+a^2)/b^2/(b/a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58 \[ \int \frac {x^4}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=-\frac {\sqrt {b x^{2} + a} \sqrt {-b x^{2} + a} b x - \sqrt {-b^{2}} a \sqrt {\frac {a}{b}} F(\arcsin \left (\frac {\sqrt {\frac {a}{b}}}{x}\right )\,|\,-1)}{3 \, b^{3}} \] Input:

integrate(x^4/(-b*x^2+a)^(1/2)/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

-1/3*(sqrt(b*x^2 + a)*sqrt(-b*x^2 + a)*b*x - sqrt(-b^2)*a*sqrt(a/b)*ellipt 
ic_f(arcsin(sqrt(a/b)/x), -1))/b^3
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^4}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\int { \frac {x^{4}}{\sqrt {b x^{2} + a} \sqrt {-b x^{2} + a}} \,d x } \] Input:

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

Output:

integrate(x^4/(sqrt(b*x^2 + a)*sqrt(-b*x^2 + a)), x)
 

Giac [F]

\[ \int \frac {x^4}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\int { \frac {x^{4}}{\sqrt {b x^{2} + a} \sqrt {-b x^{2} + a}} \,d x } \] Input:

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

Output:

integrate(x^4/(sqrt(b*x^2 + a)*sqrt(-b*x^2 + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt {a-b x^2} \sqrt {a+b x^2}} \, dx=\int \frac {x^4}{\sqrt {b\,x^2+a}\,\sqrt {a-b\,x^2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

( - sqrt(a + b*x**2)*sqrt(a - b*x**2)*x + int((sqrt(a + b*x**2)*sqrt(a - b 
*x**2))/(a**2 - b**2*x**4),x)*a**2)/(3*b**2)