Integrand size = 27, antiderivative size = 60 \[ \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\frac {x^3}{2 \sqrt {1-b^2 x^4}}-\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{2 b^{3/2}}+\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{2 b^{3/2}} \] Output:
1/2*x^3/(-b^2*x^4+1)^(1/2)-1/2*EllipticE(b^(1/2)*x,I)/b^(3/2)+1/2*Elliptic F(b^(1/2)*x,I)/b^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.74 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.40 \[ \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},b^2 x^4\right ) \] Input:
Integrate[x^2/((1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2)),x]
Output:
(x^3*Hypergeometric2F1[3/4, 3/2, 7/4, b^2*x^4])/3
Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {335, 819, 836, 762, 1388, 327}
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.
\(\displaystyle \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 335 |
\(\displaystyle \int \frac {x^2}{\left (1-b^2 x^4\right )^{3/2}}dx\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {x^3}{2 \sqrt {1-b^2 x^4}}-\frac {1}{2} \int \frac {x^2}{\sqrt {1-b^2 x^4}}dx\) |
\(\Big \downarrow \) 836 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{\sqrt {1-b^2 x^4}}dx}{b}-\frac {\int \frac {b x^2+1}{\sqrt {1-b^2 x^4}}dx}{b}\right )+\frac {x^3}{2 \sqrt {1-b^2 x^4}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {1}{2} \left (\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{b^{3/2}}-\frac {\int \frac {b x^2+1}{\sqrt {1-b^2 x^4}}dx}{b}\right )+\frac {x^3}{2 \sqrt {1-b^2 x^4}}\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \frac {1}{2} \left (\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{b^{3/2}}-\frac {\int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx}{b}\right )+\frac {x^3}{2 \sqrt {1-b^2 x^4}}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {1}{2} \left (\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{b^{3/2}}-\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{b^{3/2}}\right )+\frac {x^3}{2 \sqrt {1-b^2 x^4}}\) |
Input:
Int[x^2/((1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2)),x]
Output:
x^3/(2*Sqrt[1 - b^2*x^4]) + (-(EllipticE[ArcSin[Sqrt[b]*x], -1]/b^(3/2)) + EllipticF[ArcSin[Sqrt[b]*x], -1]/b^(3/2))/2
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p _.), x_Symbol] :> Int[(e*x)^m*(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, e , m, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] ))
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Simp[-q^(-1) Int[1/Sqrt[a + b*x^4], x], x] + Simp[1/q Int[(1 + q*x^2)/S qrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (44 ) = 88\).
Time = 3.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.68
method | result | size |
default | \(-\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (b^{\frac {3}{2}} x^{3}+\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{2 b^{\frac {3}{2}} \left (b^{2} x^{4}-1\right )}\) | \(101\) |
elliptic | \(\frac {\sqrt {-b^{2} x^{4}+1}\, \left (\frac {x^{3}}{2 \sqrt {-\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )}{\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}}\) | \(110\) |
Input:
int(x^2/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2*(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)*(b^(3/2)*x^3+(-b*x^2+1)^(1/2)*(b*x^2 +1)^(1/2)*EllipticF(b^(1/2)*x,I)-(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)*Elliptic E(b^(1/2)*x,I))/b^(3/2)/(b^2*x^4-1)
Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=-\frac {\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1} b x^{3} + {\left (b^{2} x^{4} - 1\right )} \sqrt {b} E(\arcsin \left (\sqrt {b} x\right )\,|\,-1) - {\left (b^{2} x^{4} - 1\right )} \sqrt {b} F(\arcsin \left (\sqrt {b} x\right )\,|\,-1)}{2 \, {\left (b^{3} x^{4} - b\right )}} \] Input:
integrate(x^2/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="fricas")
Output:
-1/2*(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*b*x^3 + (b^2*x^4 - 1)*sqrt(b)*ellip tic_e(arcsin(sqrt(b)*x), -1) - (b^2*x^4 - 1)*sqrt(b)*elliptic_f(arcsin(sqr t(b)*x), -1))/(b^3*x^4 - b)
Time = 29.47 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.45 \[ \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=- \frac {i {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {1}{2}, 1, 1 & \frac {1}{4}, \frac {5}{4}, \frac {7}{4} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {7}{4} & 0 \end {matrix} \middle | {\frac {1}{b^{2} x^{4}}} \right )}}{4 \pi ^{\frac {3}{2}} b^{\frac {3}{2}}} + \frac {i {G_{6, 6}^{3, 5}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2} & 1 \\0, \frac {1}{2}, 0 & - \frac {3}{4}, - \frac {1}{4}, \frac {3}{4} \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{b^{2} x^{4}}} \right )}}{4 \pi ^{\frac {3}{2}} b^{\frac {3}{2}}} \] Input:
integrate(x**2/(-b*x**2+1)**(3/2)/(b*x**2+1)**(3/2),x)
Output:
-I*meijerg(((1/2, 1, 1), (1/4, 5/4, 7/4)), ((1/2, 3/4, 1, 5/4, 7/4), (0,)) , 1/(b**2*x**4))/(4*pi**(3/2)*b**(3/2)) + I*meijerg(((-3/4, -1/4, 0, 1/4, 1/2), (1,)), ((0, 1/2, 0), (-3/4, -1/4, 3/4)), exp_polar(-2*I*pi)/(b**2*x* *4))/(4*pi**(3/2)*b**(3/2))
\[ \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int { \frac {x^{2}}{{\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^2/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="maxima")
Output:
integrate(x^2/((b*x^2 + 1)^(3/2)*(-b*x^2 + 1)^(3/2)), x)
\[ \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int { \frac {x^{2}}{{\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^2/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="giac")
Output:
integrate(x^2/((b*x^2 + 1)^(3/2)*(-b*x^2 + 1)^(3/2)), x)
Timed out. \[ \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int \frac {x^2}{{\left (1-b\,x^2\right )}^{3/2}\,{\left (b\,x^2+1\right )}^{3/2}} \,d x \] Input:
int(x^2/((1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2)),x)
Output:
int(x^2/((1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2)), x)
\[ \int \frac {x^2}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x^{2}}{b^{4} x^{8}-2 b^{2} x^{4}+1}d x \] Input:
int(x^2/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x)
Output:
int((sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x**2)/(b**4*x**8 - 2*b**2*x**4 + 1),x)