\(\int \frac {(1-b x^2)^{3/2} (1+b x^2)^{3/2}}{(e x)^{3/2}} \, dx\) [1356]

Optimal result

Integrand size = 31, antiderivative size = 92 \[ \int \frac {\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}}{(e x)^{3/2}} \, dx=-\frac {2 \sqrt {1-b^2 x^4}}{e \sqrt {e x}}-\frac {2 b^2 (e x)^{7/2} \sqrt {1-b^2 x^4}}{11 e^5}-\frac {96 b^2 (e x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},b^2 x^4\right )}{77 e^5} \] Output:

-2*(-b^2*x^4+1)^(1/2)/e/(e*x)^(1/2)-2/11*b^2*(e*x)^(7/2)*(-b^2*x^4+1)^(1/2 
)/e^5-96/77*b^2*(e*x)^(7/2)*hypergeom([1/2, 7/8],[15/8],b^2*x^4)/e^5
 

Mathematica [A] (verified)

Time = 10.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.29 \[ \int \frac {\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}}{(e x)^{3/2}} \, dx=-\frac {2 x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{8},\frac {7}{8},b^2 x^4\right )}{(e x)^{3/2}} \] Input:

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

Output:

(-2*x*Hypergeometric2F1[-3/2, -1/8, 7/8, b^2*x^4])/(e*x)^(3/2)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {335, 809, 811, 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.

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

\(\Big \downarrow \) 335

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

\(\Big \downarrow \) 809

\(\displaystyle -\frac {12 b^2 \int (e x)^{5/2} \sqrt {1-b^2 x^4}dx}{e^4}-\frac {2 \left (1-b^2 x^4\right )^{3/2}}{e \sqrt {e x}}\)

\(\Big \downarrow \) 811

\(\displaystyle -\frac {12 b^2 \left (\frac {4}{11} \int \frac {(e x)^{5/2}}{\sqrt {1-b^2 x^4}}dx+\frac {2 \sqrt {1-b^2 x^4} (e x)^{7/2}}{11 e}\right )}{e^4}-\frac {2 \left (1-b^2 x^4\right )^{3/2}}{e \sqrt {e x}}\)

\(\Big \downarrow \) 851

\(\displaystyle -\frac {12 b^2 \left (\frac {8 \int \frac {e^3 x^3}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{11 e}+\frac {2 \sqrt {1-b^2 x^4} (e x)^{7/2}}{11 e}\right )}{e^4}-\frac {2 \left (1-b^2 x^4\right )^{3/2}}{e \sqrt {e x}}\)

\(\Big \downarrow \) 888

\(\displaystyle -\frac {12 b^2 \left (\frac {8 (e x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},b^2 x^4\right )}{77 e}+\frac {2 \sqrt {1-b^2 x^4} (e x)^{7/2}}{11 e}\right )}{e^4}-\frac {2 \left (1-b^2 x^4\right )^{3/2}}{e \sqrt {e x}}\)

Input:

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

Output:

(-2*(1 - b^2*x^4)^(3/2))/(e*Sqrt[e*x]) - (12*b^2*((2*(e*x)^(7/2)*Sqrt[1 - 
b^2*x^4])/(11*e) + (8*(e*x)^(7/2)*Hypergeometric2F1[1/2, 7/8, 15/8, b^2*x^ 
4])/(77*e)))/e^4
 

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 809

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1)))   I 
nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ 
[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntB 
inomialQ[a, b, c, n, m, p, x]
 

rule 811

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]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (1-b x^2\right )^{3/2} \left (1+b x^2\right )^{3/2}}{(e x)^{3/2}} \, dx=\frac {2 \sqrt {e}\, \left (-\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, b^{2} x^{4}+5 \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}-8 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} x^{6}-x^{2}}d x \right )\right )}{11 \sqrt {x}\, e^{2}} \] Input:

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

Output:

(2*sqrt(e)*( - sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1)*b**2*x**4 + 5*sqrt(b*x 
**2 + 1)*sqrt( - b*x**2 + 1) - 8*sqrt(x)*int((sqrt(x)*sqrt(b*x**2 + 1)*sqr 
t( - b*x**2 + 1))/(b**2*x**6 - x**2),x)))/(11*sqrt(x)*e**2)