\(\int \frac {(e x)^{3/2}}{(1-b x^2)^{3/2} (1+b x^2)^{3/2}} \, dx\) [1375]

Optimal result

Integrand size = 31, antiderivative size = 60 \[ \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\frac {(e x)^{5/2}}{2 e \sqrt {1-b^2 x^4}}-\frac {(e x)^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right )}{10 e} \] Output:

1/2*(e*x)^(5/2)/e/(-b^2*x^4+1)^(1/2)-1/10*(e*x)^(5/2)*hypergeom([1/2, 5/8] 
,[13/8],b^2*x^4)/e
 

Mathematica [A] (verified)

Time = 10.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.48 \[ \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\frac {2}{5} x (e x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},\frac {3}{2},\frac {13}{8},b^2 x^4\right ) \] Input:

Integrate[(e*x)^(3/2)/((1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2)),x]
 

Output:

(2*x*(e*x)^(3/2)*Hypergeometric2F1[5/8, 3/2, 13/8, b^2*x^4])/5
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {335, 819, 851, 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.

\(\displaystyle \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 335

\(\displaystyle \int \frac {(e x)^{3/2}}{\left (1-b^2 x^4\right )^{3/2}}dx\)

\(\Big \downarrow \) 819

\(\displaystyle \frac {(e x)^{5/2}}{2 e \sqrt {1-b^2 x^4}}-\frac {1}{4} \int \frac {(e x)^{3/2}}{\sqrt {1-b^2 x^4}}dx\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {(e x)^{5/2}}{2 e \sqrt {1-b^2 x^4}}-\frac {\int \frac {e^2 x^2}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{2 e}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {(e x)^{5/2}}{2 e \sqrt {1-b^2 x^4}}-\frac {(e x)^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},b^2 x^4\right )}{10 e}\)

Input:

Int[(e*x)^(3/2)/((1 - b*x^2)^(3/2)*(1 + b*x^2)^(3/2)),x]
 

Output:

(e*x)^(5/2)/(2*e*Sqrt[1 - b^2*x^4]) - ((e*x)^(5/2)*Hypergeometric2F1[1/2, 
5/8, 13/8, b^2*x^4])/(10*e)
 

Defintions of rubi rules used

rule 335

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] 
))
 

rule 819

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]
 

rule 851

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ 
n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && 
FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 888

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])
 

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {3}{2}}}{\left (-b \,x^{2}+1\right )^{\frac {3}{2}} \left (b \,x^{2}+1\right )^{\frac {3}{2}}}d x\]

Input:

int((e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x)
 

Output:

int((e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x)
 

Fricas [F]

\[ \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int { \frac {\left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="frica 
s")
 

Output:

integral(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x)*e*x/(b^4*x^8 - 2*b^2*x 
^4 + 1), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 45.43 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.82 \[ \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=- \frac {i e^{\frac {3}{2}} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{8}, \frac {9}{8}, 1 & \frac {3}{8}, \frac {11}{8}, \frac {15}{8} \\\frac {5}{8}, \frac {7}{8}, \frac {9}{8}, \frac {11}{8}, \frac {15}{8} & 0 \end {matrix} \middle | {\frac {1}{b^{2} x^{4}}} \right )}}{4 \pi ^{\frac {3}{2}} b^{\frac {5}{4}}} + \frac {e^{\frac {3}{2}} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {5}{8}, - \frac {1}{8}, \frac {1}{8}, \frac {3}{8}, \frac {5}{8}, 1 & \\\frac {1}{8}, \frac {5}{8} & - \frac {5}{8}, - \frac {1}{8}, \frac {7}{8}, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{b^{2} x^{4}}} \right )} e^{- \frac {i \pi }{4}}}{4 \pi ^{\frac {3}{2}} b^{\frac {5}{4}}} \] Input:

integrate((e*x)**(3/2)/(-b*x**2+1)**(3/2)/(b*x**2+1)**(3/2),x)
 

Output:

-I*e**(3/2)*meijerg(((5/8, 9/8, 1), (3/8, 11/8, 15/8)), ((5/8, 7/8, 9/8, 1 
1/8, 15/8), (0,)), 1/(b**2*x**4))/(4*pi**(3/2)*b**(5/4)) + e**(3/2)*meijer 
g(((-5/8, -1/8, 1/8, 3/8, 5/8, 1), ()), ((1/8, 5/8), (-5/8, -1/8, 7/8, 0)) 
, exp_polar(-2*I*pi)/(b**2*x**4))*exp(-I*pi/4)/(4*pi**(3/2)*b**(5/4))
 

Maxima [F]

\[ \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int { \frac {\left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="maxim 
a")
 

Output:

integrate((e*x)^(3/2)/((b*x^2 + 1)^(3/2)*(-b*x^2 + 1)^(3/2)), x)
 

Giac [F]

\[ \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int { \frac {\left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + 1\right )}^{\frac {3}{2}} {\left (-b x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x, algorithm="giac" 
)
 

Output:

integrate((e*x)^(3/2)/((b*x^2 + 1)^(3/2)*(-b*x^2 + 1)^(3/2)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\int \frac {{\left (e\,x\right )}^{3/2}}{{\left (1-b\,x^2\right )}^{3/2}\,{\left (b\,x^2+1\right )}^{3/2}} \,d x \] Input:

int((e*x)^(3/2)/((1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2)),x)
 

Output:

int((e*x)^(3/2)/((1 - b*x^2)^(3/2)*(b*x^2 + 1)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {(e x)^{3/2}}{\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}} \, dx=\sqrt {e}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, x}{b^{4} x^{8}-2 b^{2} x^{4}+1}d x \right ) e \] Input:

int((e*x)^(3/2)/(-b*x^2+1)^(3/2)/(b*x^2+1)^(3/2),x)
 

Output:

sqrt(e)*int((sqrt(x)*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*x)/(b**4*x**8 - 
2*b**2*x**4 + 1),x)*e