Integrand size = 27, antiderivative size = 60 \[ \int x^2 \sqrt {1-b x^2} \sqrt {1+b x^2} \, dx=\frac {1}{5} x^3 \sqrt {1-b^2 x^4}+\frac {2 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{5 b^{3/2}}-\frac {2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{5 b^{3/2}} \] Output:
1/5*x^3*(-b^2*x^4+1)^(1/2)+2/5*EllipticE(b^(1/2)*x,I)/b^(3/2)-2/5*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 = 3.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.40 \[ \int x^2 \sqrt {1-b x^2} \sqrt {1+b x^2} \, dx=\frac {1}{3} x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right ) \] Input:
Integrate[x^2*Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2],x]
Output:
(x^3*Hypergeometric2F1[-1/2, 3/4, 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, 811, 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.
\(\displaystyle \int x^2 \sqrt {1-b x^2} \sqrt {b x^2+1} \, dx\) |
\(\Big \downarrow \) 335 |
\(\displaystyle \int x^2 \sqrt {1-b^2 x^4}dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {2}{5} \int \frac {x^2}{\sqrt {1-b^2 x^4}}dx+\frac {1}{5} x^3 \sqrt {1-b^2 x^4}\) |
\(\Big \downarrow \) 836 |
\(\displaystyle \frac {2}{5} \left (\frac {\int \frac {b x^2+1}{\sqrt {1-b^2 x^4}}dx}{b}-\frac {\int \frac {1}{\sqrt {1-b^2 x^4}}dx}{b}\right )+\frac {1}{5} x^3 \sqrt {1-b^2 x^4}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {2}{5} \left (\frac {\int \frac {b x^2+1}{\sqrt {1-b^2 x^4}}dx}{b}-\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{b^{3/2}}\right )+\frac {1}{5} x^3 \sqrt {1-b^2 x^4}\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \frac {2}{5} \left (\frac {\int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx}{b}-\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{b^{3/2}}\right )+\frac {1}{5} x^3 \sqrt {1-b^2 x^4}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {2}{5} \left (\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{b^{3/2}}-\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{b^{3/2}}\right )+\frac {1}{5} x^3 \sqrt {1-b^2 x^4}\) |
Input:
Int[x^2*Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2],x]
Output:
(x^3*Sqrt[1 - b^2*x^4])/5 + (2*(EllipticE[ArcSin[Sqrt[b]*x], -1]/b^(3/2) - EllipticF[ArcSin[Sqrt[b]*x], -1]/b^(3/2)))/5
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/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && 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. 105 vs. \(2 (44 ) = 88\).
Time = 0.74 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.77
method | result | size |
elliptic | \(\frac {\sqrt {-b^{2} x^{4}+1}\, \left (\frac {x^{3} \sqrt {-b^{2} x^{4}+1}}{5}-\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )}{\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}}\) | \(106\) |
default | \(\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (x^{7} b^{\frac {7}{2}}-b^{\frac {3}{2}} x^{3}+2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{5 \left (b^{2} x^{4}-1\right ) b^{\frac {3}{2}}}\) | \(110\) |
risch | \(-\frac {x^{3} \left (b \,x^{2}-1\right ) \sqrt {b \,x^{2}+1}\, \sqrt {\left (-b \,x^{2}+1\right ) \left (b \,x^{2}+1\right )}}{5 \sqrt {-\left (b \,x^{2}-1\right ) \left (b \,x^{2}+1\right )}\, \sqrt {-b \,x^{2}+1}}-\frac {2 \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right ) \sqrt {\left (-b \,x^{2}+1\right ) \left (b \,x^{2}+1\right )}}{5 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\) | \(123\) |
Input:
int(x^2*(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
(-b^2*x^4+1)^(1/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2)*(1/5*x^3*(-b^2*x^4+1)^ (1/2)-2/5/b^(3/2)*(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)/(-b^2*x^4+1)^(1/2)*(Ell ipticF(b^(1/2)*x,I)-EllipticE(b^(1/2)*x,I)))
Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (42) = 84\).
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.43 \[ \int x^2 \sqrt {1-b x^2} \sqrt {1+b x^2} \, dx=\frac {{\left (b^{3} x^{4} - 2 \, b\right )} \sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1} - \frac {2 \, \sqrt {-b^{2}} x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + \frac {2 \, \sqrt {-b^{2}} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}}}{5 \, b^{3} x} \] Input:
integrate(x^2*(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2),x, algorithm="fricas")
Output:
1/5*((b^3*x^4 - 2*b)*sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1) - 2*sqrt(-b^2)*x*ell iptic_e(arcsin(1/(sqrt(b)*x)), -1)/sqrt(b) + 2*sqrt(-b^2)*x*elliptic_f(arc sin(1/(sqrt(b)*x)), -1)/sqrt(b))/(b^3*x)
\[ \int x^2 \sqrt {1-b x^2} \sqrt {1+b x^2} \, dx=\int x^{2} \sqrt {- b x^{2} + 1} \sqrt {b x^{2} + 1}\, dx \] Input:
integrate(x**2*(-b*x**2+1)**(1/2)*(b*x**2+1)**(1/2),x)
Output:
Integral(x**2*sqrt(-b*x**2 + 1)*sqrt(b*x**2 + 1), x)
\[ \int x^2 \sqrt {1-b x^2} \sqrt {1+b x^2} \, dx=\int { \sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1} x^{2} \,d x } \] Input:
integrate(x^2*(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*x^2, x)
\[ \int x^2 \sqrt {1-b x^2} \sqrt {1+b x^2} \, dx=\int { \sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1} x^{2} \,d x } \] Input:
integrate(x^2*(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*x^2, x)
Timed out. \[ \int x^2 \sqrt {1-b x^2} \sqrt {1+b x^2} \, dx=\int x^2\,\sqrt {1-b\,x^2}\,\sqrt {b\,x^2+1} \,d x \] Input:
int(x^2*(1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2),x)
Output:
int(x^2*(1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2), x)
\[ \int x^2 \sqrt {1-b x^2} \sqrt {1+b x^2} \, dx=\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x^{3}}{5}-\frac {2 \left (\int \frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x^{2}}{b^{2} x^{4}-1}d x \right )}{5} \] Input:
int(x^2*(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2),x)
Output:
(sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x**3 - 2*int((sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x**2)/(b**2*x**4 - 1),x))/5