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