Integrand size = 29, antiderivative size = 41 \[ \int (e x)^m \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{4},\frac {5+m}{4},b^2 x^4\right )}{e (1+m)} \] Output:
(e*x)^(1+m)*hypergeom([-3/2, 1/4+1/4*m],[5/4+1/4*m],b^2*x^4)/e/(1+m)
Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int (e x)^m \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\frac {x (e x)^m \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{4},1+\frac {1+m}{4},b^2 x^4\right )}{1+m} \] Input:
Integrate[(e*x)^m*(1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2),x]
Output:
(x*(e*x)^m*Hypergeometric2F1[-3/2, (1 + m)/4, 1 + (1 + m)/4, b^2*x^4])/(1 + m)
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {335, 888}
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.
\(\displaystyle \int \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2} (e x)^m \, dx\) |
\(\Big \downarrow \) 335 |
\(\displaystyle \int \left (1-b^2 x^4\right )^{3/2} (e x)^mdx\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{4},\frac {m+5}{4},b^2 x^4\right )}{e (m+1)}\) |
Input:
Int[(e*x)^m*(1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2),x]
Output:
((e*x)^(1 + m)*Hypergeometric2F1[-3/2, (1 + m)/4, (5 + m)/4, b^2*x^4])/(e* (1 + m))
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
\[\int \left (e x \right )^{m} \left (-b \,x^{2}+1\right )^{\frac {3}{2}} \left (b \,x^{2}+1\right )^{\frac {3}{2}}d x\]
Input:
int((e*x)^m*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x)
Output:
int((e*x)^m*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x)
\[ \int (e x)^m \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int { {\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x, algorithm="fricas")
Output:
integral(-(b^2*x^4 - 1)*sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*(e*x)^m, x)
\[ \int (e x)^m \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int \left (e x\right )^{m} \left (- b x^{2} + 1\right )^{\frac {3}{2}} \left (b x^{2} + 1\right )^{\frac {3}{2}}\, dx \] Input:
integrate((e*x)**m*(-b*x**2+1)**(3/2)*(b*x**2+1)**(3/2),x)
Output:
Integral((e*x)**m*(-b*x**2 + 1)**(3/2)*(b*x**2 + 1)**(3/2), x)
\[ \int (e x)^m \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int { {\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(-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)*(e*x)^m, x)
Exception generated. \[ \int (e x)^m \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((e*x)^m*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int (e x)^m \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\int {\left (e\,x\right )}^m\,{\left (1-b\,x^2\right )}^{3/2}\,{\left (b\,x^2+1\right )}^{3/2} \,d x \] Input:
int((e*x)^m*(1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2),x)
Output:
int((e*x)^m*(1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2), x)
\[ \int (e x)^m \left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2} \, dx=\frac {e^{m} \left (-x^{m} \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, b^{2} m \,x^{5}-3 x^{m} \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, b^{2} x^{5}+x^{m} \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, m x +9 x^{m} \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x -12 \left (\int \frac {x^{m} \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} m^{2} x^{4}+10 b^{2} m \,x^{4}+21 b^{2} x^{4}-m^{2}-10 m -21}d x \right ) m^{2}-120 \left (\int \frac {x^{m} \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} m^{2} x^{4}+10 b^{2} m \,x^{4}+21 b^{2} x^{4}-m^{2}-10 m -21}d x \right ) m -252 \left (\int \frac {x^{m} \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} m^{2} x^{4}+10 b^{2} m \,x^{4}+21 b^{2} x^{4}-m^{2}-10 m -21}d x \right )\right )}{m^{2}+10 m +21} \] Input:
int((e*x)^m*(-b*x^2+1)^(3/2)*(b*x^2+1)^(3/2),x)
Output:
(e**m*( - x**m*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*b**2*m*x**5 - 3*x**m*s qrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*b**2*x**5 + x**m*sqrt(b*x**2 + 1)*sqrt ( - b*x**2 + 1)*m*x + 9*x**m*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x - 12*i nt((x**m*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1))/(b**2*m**2*x**4 + 10*b**2*m *x**4 + 21*b**2*x**4 - m**2 - 10*m - 21),x)*m**2 - 120*int((x**m*sqrt(b*x* *2 + 1)*sqrt( - b*x**2 + 1))/(b**2*m**2*x**4 + 10*b**2*m*x**4 + 21*b**2*x* *4 - m**2 - 10*m - 21),x)*m - 252*int((x**m*sqrt(b*x**2 + 1)*sqrt( - b*x** 2 + 1))/(b**2*m**2*x**4 + 10*b**2*m*x**4 + 21*b**2*x**4 - m**2 - 10*m - 21 ),x)))/(m**2 + 10*m + 21)