Integrand size = 24, antiderivative size = 58 \[ \int \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\frac {2}{7} x \sqrt {1-b^2 x^4}+\frac {1}{7} x \left (1-b^2 x^4\right )^{3/2}+\frac {4 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{7 \sqrt {b}} \] Output:
2/7*x*(-b^2*x^4+1)^(1/2)+1/7*x*(-b^2*x^4+1)^(3/2)+4/7*EllipticF(b^(1/2)*x, I)/b^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.87 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.33 \[ \int \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},b^2 x^4\right ) \] Input:
Integrate[(1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2),x]
Output:
x*Hypergeometric2F1[-3/2, 1/4, 5/4, b^2*x^4]
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {284, 748, 748, 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.
\(\displaystyle \int \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 284 |
\(\displaystyle \int \left (1-b^2 x^4\right )^{3/2}dx\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {6}{7} \int \sqrt {1-b^2 x^4}dx+\frac {1}{7} x \left (1-b^2 x^4\right )^{3/2}\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \frac {1}{\sqrt {1-b^2 x^4}}dx+\frac {1}{3} x \sqrt {1-b^2 x^4}\right )+\frac {1}{7} x \left (1-b^2 x^4\right )^{3/2}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {6}{7} \left (\frac {2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{3 \sqrt {b}}+\frac {1}{3} x \sqrt {1-b^2 x^4}\right )+\frac {1}{7} x \left (1-b^2 x^4\right )^{3/2}\) |
Input:
Int[(1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2),x]
Output:
(x*(1 - b^2*x^4)^(3/2))/7 + (6*((x*Sqrt[1 - b^2*x^4])/3 + (2*EllipticF[Arc Sin[Sqrt[b]*x], -1])/(3*Sqrt[b])))/7
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> I nt[(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0]))
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Simp[a*n*(p/(n*p + 1)) Int[(a + b*x^n)^(p - 1), x], x] /; Fre eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat or[p + 1/n], Denominator[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]
Time = 1.84 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.50
method | result | size |
default | \(-\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (b^{\frac {9}{2}} x^{9}-4 b^{\frac {5}{2}} x^{5}+4 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )+3 \sqrt {b}\, x \right )}{7 \left (b^{2} x^{4}-1\right ) \sqrt {b}}\) | \(87\) |
elliptic | \(\frac {\sqrt {-b^{2} x^{4}+1}\, \left (-\frac {b^{2} x^{5} \sqrt {-b^{2} x^{4}+1}}{7}+\frac {3 x \sqrt {-b^{2} x^{4}+1}}{7}+\frac {4 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{7 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\right )}{\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}}\) | \(113\) |
risch | \(\frac {x \left (b^{2} x^{4}-3\right ) \left (b \,x^{2}-1\right ) \sqrt {b \,x^{2}+1}\, \sqrt {\left (-b \,x^{2}+1\right ) \left (b \,x^{2}+1\right )}}{7 \sqrt {-\left (b \,x^{2}-1\right ) \left (b \,x^{2}+1\right )}\, \sqrt {-b \,x^{2}+1}}+\frac {4 \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right ) \sqrt {\left (-b \,x^{2}+1\right ) \left (b \,x^{2}+1\right )}}{7 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\) | \(119\) |
Input:
int((-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/7*(-b*x^2+1)^(1/2)*(b*x^2+1)^(1/2)*(b^(9/2)*x^9-4*b^(5/2)*x^5+4*(-b*x^2 +1)^(1/2)*(b*x^2+1)^(1/2)*EllipticF(b^(1/2)*x,I)+3*b^(1/2)*x)/(b^2*x^4-1)/ b^(1/2)
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=-\frac {{\left (b^{3} x^{5} - 3 \, b x\right )} \sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1} - \frac {4 \, \sqrt {-b^{2}} F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}}}{7 \, b} \] Input:
integrate((-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x, algorithm="fricas")
Output:
-1/7*((b^3*x^5 - 3*b*x)*sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1) - 4*sqrt(-b^2)*el liptic_f(arcsin(1/(sqrt(b)*x)), -1)/sqrt(b))/b
\[ \int \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int \left (- b x^{2} + 1\right )^{\frac {3}{2}} \left (b x^{2} + 1\right )^{\frac {3}{2}}\, dx \] Input:
integrate((-b*x**2+1)**(3/2)*(b*x**2+1)**(3/2),x)
Output:
Integral((-b*x**2 + 1)**(3/2)*(b*x**2 + 1)**(3/2), x)
\[ \int \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int { {\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x, algorithm="maxima")
Output:
integrate((b*x^2 + 1)^(3/2)*(-b*x^2 + 1)^(3/2), x)
\[ \int \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int { {\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x, algorithm="giac")
Output:
integrate((b*x^2 + 1)^(3/2)*(-b*x^2 + 1)^(3/2), x)
Timed out. \[ \int \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int {\left (1-b\,x^2\right )}^{3/2}\,{\left (b\,x^2+1\right )}^{3/2} \,d x \] Input:
int((1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2),x)
Output:
int((1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2), x)
\[ \int \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, b^{2} x^{5}}{7}+\frac {3 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x}{7}-\frac {4 \left (\int \frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} x^{4}-1}d x \right )}{7} \] Input:
int((-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x)
Output:
( - sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*b**2*x**5 + 3*sqrt(b*x**2 + 1)*sq rt( - b*x**2 + 1)*x - 4*int((sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1))/(b**2*x **4 - 1),x))/7