Integrand size = 27, antiderivative size = 136 \[ \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx=\frac {1}{2 a^2 x^3 \sqrt {a-b x^2} \sqrt {a+b x^2}}-\frac {5 \sqrt {a-b x^2} \sqrt {a+b x^2}}{6 a^4 x^3}+\frac {5 b^{3/2} \sqrt {1-\frac {b^2 x^4}{a^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),-1\right )}{6 a^{7/2} \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Output:
1/2/a^2/x^3/(-b*x^2+a)^(1/2)/(b*x^2+a)^(1/2)-5/6*(-b*x^2+a)^(1/2)*(b*x^2+a )^(1/2)/a^4/x^3+5/6*b^(3/2)*(1-b^2*x^4/a^2)^(1/2)*EllipticF(b^(1/2)*x/a^(1 /2),I)/a^(7/2)/(-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 = 4.41 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1-\frac {b^2 x^4}{a^2}} \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};\frac {b^2 x^4}{a^2}\right )}{3 a^2 x^3 \sqrt {a-b x^2} \sqrt {a+b x^2}} \] Input:
Integrate[1/(x^4*(a - b*x^2)^(3/2)*(a + b*x^2)^(3/2)),x]
Output:
-1/3*(Sqrt[1 - (b^2*x^4)/a^2]*HypergeometricPFQ[{-3/4, 3/2}, {1/4}, (b^2*x ^4)/a^2])/(a^2*x^3*Sqrt[a - b*x^2]*Sqrt[a + b*x^2])
Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {342, 343, 289, 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.
\(\displaystyle \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 342 |
\(\displaystyle \frac {5 \int \frac {1}{x^4 \sqrt {a-b x^2} \sqrt {b x^2+a}}dx}{2 a^2}+\frac {1}{2 a^2 x^3 \sqrt {a-b x^2} \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 343 |
\(\displaystyle \frac {5 \left (\frac {b^2 \int \frac {1}{\sqrt {a-b x^2} \sqrt {b x^2+a}}dx}{3 a^2}-\frac {\sqrt {a-b x^2} \sqrt {a+b x^2}}{3 a^2 x^3}\right )}{2 a^2}+\frac {1}{2 a^2 x^3 \sqrt {a-b x^2} \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 289 |
\(\displaystyle \frac {5 \left (\frac {b^2 \sqrt {a^2-b^2 x^4} \int \frac {1}{\sqrt {a^2-b^2 x^4}}dx}{3 a^2 \sqrt {a-b x^2} \sqrt {a+b x^2}}-\frac {\sqrt {a-b x^2} \sqrt {a+b x^2}}{3 a^2 x^3}\right )}{2 a^2}+\frac {1}{2 a^2 x^3 \sqrt {a-b x^2} \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {5 \left (\frac {b^2 \sqrt {1-\frac {b^2 x^4}{a^2}} \int \frac {1}{\sqrt {1-\frac {b^2 x^4}{a^2}}}dx}{3 a^2 \sqrt {a-b x^2} \sqrt {a+b x^2}}-\frac {\sqrt {a-b x^2} \sqrt {a+b x^2}}{3 a^2 x^3}\right )}{2 a^2}+\frac {1}{2 a^2 x^3 \sqrt {a-b x^2} \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {1}{2 a^2 x^3 \sqrt {a-b x^2} \sqrt {a+b x^2}}+\frac {5 \left (\frac {b^{3/2} \sqrt {1-\frac {b^2 x^4}{a^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),-1\right )}{3 a^{3/2} \sqrt {a-b x^2} \sqrt {a+b x^2}}-\frac {\sqrt {a-b x^2} \sqrt {a+b x^2}}{3 a^2 x^3}\right )}{2 a^2}\) |
Input:
Int[1/(x^4*(a - b*x^2)^(3/2)*(a + b*x^2)^(3/2)),x]
Output:
1/(2*a^2*x^3*Sqrt[a - b*x^2]*Sqrt[a + b*x^2]) + (5*(-1/3*(Sqrt[a - b*x^2]* Sqrt[a + b*x^2])/(a^2*x^3) + (b^(3/2)*Sqrt[1 - (b^2*x^4)/a^2]*EllipticF[Ar cSin[(Sqrt[b]*x)/Sqrt[a]], -1])/(3*a^(3/2)*Sqrt[a - b*x^2]*Sqrt[a + b*x^2] )))/(2*a^2)
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Sim p[(a + b*x^2)^FracPart[p]*((c + d*x^2)^FracPart[p]/(a*c + b*d*x^4)^FracPart [p]) Int[(a*c + b*d*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b* c + a*d, 0] && !IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(-(e*x)^(m + 1))*(a + b*x^2)^(p + 1)*((c + d*x^2)^(p + 1)/(4*a*c*e*(p + 1))), x] + Simp[(m + 4*p + 5)/(4*a*c*(p + 1)) Int[(e*x) ^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c + a*d, 0] && LtQ[p, -1] && IntegerQ[2*m]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(p_) , x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(p + 1)/ (a*c*e*(m + 1))), x] - Simp[b*d*((m + 4*p + 5)/(a*c*e^4*(m + 1))) Int[(e* x)^(m + 4)*(a + b*x^2)^p*(c + d*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b*c + a*d, 0] && LtQ[m, -1]
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]
Time = 7.17 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91
method | result | size |
default | \(\frac {\sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}\, \left (5 b^{2} \sqrt {\frac {-b \,x^{2}+a}{a}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right ) x^{3}+5 \sqrt {\frac {b}{a}}\, b^{2} x^{4}-2 \sqrt {\frac {b}{a}}\, a^{2}\right )}{6 a^{4} \left (-b^{2} x^{4}+a^{2}\right ) x^{3} \sqrt {\frac {b}{a}}}\) | \(124\) |
elliptic | \(\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \left (-\frac {\sqrt {-b^{2} x^{4}+a^{2}}}{3 a^{4} x^{3}}+\frac {b^{2} x}{2 a^{4} \sqrt {-\left (x^{4}-\frac {a^{2}}{b^{2}}\right ) b^{2}}}+\frac {5 b^{2} \sqrt {1-\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right )}{6 a^{4} \sqrt {\frac {b}{a}}\, \sqrt {-b^{2} x^{4}+a^{2}}}\right )}{\sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}}\) | \(152\) |
risch | \(-\frac {\sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}}{3 a^{4} x^{3}}+\frac {b^{2} \left (\frac {\sqrt {1-\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right )}{\sqrt {\frac {b}{a}}\, \sqrt {-b^{2} x^{4}+a^{2}}}-\frac {3 a \left (\frac {\left (-b^{2} x^{2}-a b \right ) x}{2 a^{2} b \sqrt {\left (x^{2}-\frac {a}{b}\right ) \left (-b^{2} x^{2}-a b \right )}}-\frac {\sqrt {1-\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right )}{2 a \sqrt {\frac {b}{a}}\, \sqrt {-b^{2} x^{4}+a^{2}}}-\frac {\sqrt {1-\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {b}{a}}, i\right )\right )}{2 a \sqrt {\frac {b}{a}}\, \sqrt {-b^{2} x^{4}+a^{2}}}\right )}{2}+\frac {3 a \left (\frac {\left (-b^{2} x^{2}+a b \right ) x}{2 a^{2} b \sqrt {\left (x^{2}+\frac {a}{b}\right ) \left (-b^{2} x^{2}+a b \right )}}+\frac {\sqrt {1-\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right )}{2 a \sqrt {\frac {b}{a}}\, \sqrt {-b^{2} x^{4}+a^{2}}}-\frac {\sqrt {1-\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {b}{a}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {b}{a}}, i\right )\right )}{2 a \sqrt {\frac {b}{a}}\, \sqrt {-b^{2} x^{4}+a^{2}}}\right )}{2}\right ) \sqrt {\left (-b \,x^{2}+a \right ) \left (b \,x^{2}+a \right )}}{3 a^{4} \sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}}\) | \(517\) |
Input:
int(1/x^4/(-b*x^2+a)^(3/2)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/6*(-b*x^2+a)^(1/2)*(b*x^2+a)^(1/2)*(5*b^2*((-b*x^2+a)/a)^(1/2)*((b*x^2+a )/a)^(1/2)*EllipticF(x*(b/a)^(1/2),I)*x^3+5*(b/a)^(1/2)*b^2*x^4-2*(b/a)^(1 /2)*a^2)/a^4/(-b^2*x^4+a^2)/x^3/(b/a)^(1/2)
Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx=\frac {5 \, {\left (b^{3} x^{7} - a^{2} b x^{3}\right )} \sqrt {\frac {b}{a}} F(\arcsin \left (x \sqrt {\frac {b}{a}}\right )\,|\,-1) - {\left (5 \, b^{2} x^{4} - 2 \, a^{2}\right )} \sqrt {b x^{2} + a} \sqrt {-b x^{2} + a}}{6 \, {\left (a^{4} b^{2} x^{7} - a^{6} x^{3}\right )}} \] Input:
integrate(1/x^4/(-b*x^2+a)^(3/2)/(b*x^2+a)^(3/2),x, algorithm="fricas")
Output:
1/6*(5*(b^3*x^7 - a^2*b*x^3)*sqrt(b/a)*elliptic_f(arcsin(x*sqrt(b/a)), -1) - (5*b^2*x^4 - 2*a^2)*sqrt(b*x^2 + a)*sqrt(-b*x^2 + a))/(a^4*b^2*x^7 - a^ 6*x^3)
Timed out. \[ \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/x**4/(-b*x**2+a)**(3/2)/(b*x**2+a)**(3/2),x)
Output:
Timed out
\[ \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (-b x^{2} + a\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(-b*x^2+a)^(3/2)/(b*x^2+a)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(3/2)*(-b*x^2 + a)^(3/2)*x^4), x)
\[ \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (-b x^{2} + a\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(-b*x^2+a)^(3/2)/(b*x^2+a)^(3/2),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(3/2)*(-b*x^2 + a)^(3/2)*x^4), x)
Timed out. \[ \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^2+a\right )}^{3/2}\,{\left (a-b\,x^2\right )}^{3/2}} \,d x \] Input:
int(1/(x^4*(a + b*x^2)^(3/2)*(a - b*x^2)^(3/2)),x)
Output:
int(1/(x^4*(a + b*x^2)^(3/2)*(a - b*x^2)^(3/2)), x)
\[ \int \frac {1}{x^4 \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {b \,x^{2}+a}\, \sqrt {-b \,x^{2}+a}}{b^{4} x^{12}-2 a^{2} b^{2} x^{8}+a^{4} x^{4}}d x \] Input:
int(1/x^4/(-b*x^2+a)^(3/2)/(b*x^2+a)^(3/2),x)
Output:
int((sqrt(a + b*x**2)*sqrt(a - b*x**2))/(a**4*x**4 - 2*a**2*b**2*x**8 + b* *4*x**12),x)