\(\int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx\) [1368]

Optimal result

Integrand size = 31, antiderivative size = 31 \[ \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {2 (e x)^{9/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{8},\frac {17}{8},b^2 x^4\right )}{9 e} \] Output:

2/9*(e*x)^(9/2)*hypergeom([1/2, 9/8],[17/8],b^2*x^4)/e
 

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {2 e^3 \sqrt {e x} \left (-\sqrt {1-b^2 x^4}+\operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},b^2 x^4\right )\right )}{5 b^2} \] Input:

Integrate[(e*x)^(7/2)/(Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
 

Output:

(2*e^3*Sqrt[e*x]*(-Sqrt[1 - b^2*x^4] + Hypergeometric2F1[1/8, 1/2, 9/8, b^ 
2*x^4]))/(5*b^2)
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(479\) vs. \(2(31)=62\).

Time = 0.55 (sec) , antiderivative size = 479, normalized size of antiderivative = 15.45, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {335, 843, 851, 767, 27, 2422}

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.

\(\displaystyle \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {b x^2+1}} \, dx\)

\(\Big \downarrow \) 335

\(\displaystyle \int \frac {(e x)^{7/2}}{\sqrt {1-b^2 x^4}}dx\)

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 767

\(\displaystyle \frac {2 e^3 \left (\frac {1}{2} \int \frac {e-\sqrt [4]{-b^2} e x}{e \sqrt {1-b^2 x^4}}d\sqrt {e x}+\frac {1}{2} \int \frac {\sqrt [4]{-b^2} x e+e}{e \sqrt {1-b^2 x^4}}d\sqrt {e x}\right )}{5 b^2}-\frac {2 e^3 \sqrt {1-b^2 x^4} \sqrt {e x}}{5 b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 e^3 \left (\frac {\int \frac {e-\sqrt [4]{-b^2} e x}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{2 e}+\frac {\int \frac {\sqrt [4]{-b^2} x e+e}{\sqrt {1-b^2 x^4}}d\sqrt {e x}}{2 e}\right )}{5 b^2}-\frac {2 e^3 \sqrt {1-b^2 x^4} \sqrt {e x}}{5 b^2}\)

\(\Big \downarrow \) 2422

\(\displaystyle \frac {2 e^3 \left (\frac {\sqrt [4]{-b^2} (e x)^{3/2} \sqrt {\frac {\left (\sqrt [4]{-b^2} e x+e\right )^2}{\sqrt [4]{-b^2} e^2 x}} \sqrt {-\frac {e^4-b^2 e^4 x^4}{\sqrt {-b^2} e^4 x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt {2} \sqrt {-b^2} x^2 e^2-2 \sqrt [4]{-b^2} x e^2+\sqrt {2} e^2}{\sqrt [4]{-b^2} e^2 x}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \sqrt {1-b^2 x^4} \left (\sqrt [4]{-b^2} e x+e\right )}-\frac {\sqrt [4]{-b^2} (e x)^{3/2} \sqrt {-\frac {\left (e-\sqrt [4]{-b^2} e x\right )^2}{\sqrt [4]{-b^2} e^2 x}} \sqrt {-\frac {e^4-b^2 e^4 x^4}{\sqrt {-b^2} e^4 x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt {2} \sqrt {-b^2} x^2 e^2+2 \sqrt [4]{-b^2} x e^2+\sqrt {2} e^2}{\sqrt [4]{-b^2} e^2 x}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \sqrt {1-b^2 x^4} \left (e-\sqrt [4]{-b^2} e x\right )}\right )}{5 b^2}-\frac {2 e^3 \sqrt {1-b^2 x^4} \sqrt {e x}}{5 b^2}\)

Input:

Int[(e*x)^(7/2)/(Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2]),x]
 

Output:

(-2*e^3*Sqrt[e*x]*Sqrt[1 - b^2*x^4])/(5*b^2) + (2*e^3*(((-b^2)^(1/4)*(e*x) 
^(3/2)*Sqrt[(e + (-b^2)^(1/4)*e*x)^2/((-b^2)^(1/4)*e^2*x)]*Sqrt[-((e^4 - b 
^2*e^4*x^4)/(Sqrt[-b^2]*e^4*x^2))]*EllipticF[ArcSin[Sqrt[-((Sqrt[2]*e^2 - 
2*(-b^2)^(1/4)*e^2*x + Sqrt[2]*Sqrt[-b^2]*e^2*x^2)/((-b^2)^(1/4)*e^2*x))]/ 
2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]*(e + (-b^2)^(1/4)*e*x)*Sqrt[1 
- b^2*x^4]) - ((-b^2)^(1/4)*(e*x)^(3/2)*Sqrt[-((e - (-b^2)^(1/4)*e*x)^2/(( 
-b^2)^(1/4)*e^2*x))]*Sqrt[-((e^4 - b^2*e^4*x^4)/(Sqrt[-b^2]*e^4*x^2))]*Ell 
ipticF[ArcSin[Sqrt[(Sqrt[2]*e^2 + 2*(-b^2)^(1/4)*e^2*x + Sqrt[2]*Sqrt[-b^2 
]*e^2*x^2)/((-b^2)^(1/4)*e^2*x)]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2 
]]*(e - (-b^2)^(1/4)*e*x)*Sqrt[1 - b^2*x^4])))/(5*b^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 767

Int[1/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[1/2   Int[(1 - Rt[b/a, 4 
]*x^2)/Sqrt[a + b*x^8], x], x] + Simp[1/2   Int[(1 + Rt[b/a, 4]*x^2)/Sqrt[a 
 + b*x^8], x], x] /; FreeQ[{a, b}, x]
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && 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 2422

Int[((c_) + (d_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^8], x_Symbol] :> Simp[(-c) 
*d*x^3*Sqrt[-(c - d*x^2)^2/(c*d*x^2)]*(Sqrt[(-d^2)*((a + b*x^8)/(b*c^2*x^4) 
)]/(Sqrt[2 + Sqrt[2]]*(c - d*x^2)*Sqrt[a + b*x^8]))*EllipticF[ArcSin[(1/2)* 
Sqrt[(Sqrt[2]*c^2 + 2*c*d*x^2 + Sqrt[2]*d^2*x^4)/(c*d*x^2)]], -2*(1 - Sqrt[ 
2])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^4 - a*d^4, 0]
 

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {7}{2}}}{\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}}d x\]

Input:

int((e*x)^(7/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
 

Output:

int((e*x)^(7/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
 

Fricas [F]

\[ \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {\left (e x\right )^{\frac {7}{2}}}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}} \,d x } \] Input:

integrate((e*x)^(7/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="frica 
s")
 

Output:

integral(-sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)*sqrt(e*x)*e^3*x^3/(b^2*x^4 - 1) 
, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\text {Timed out} \] Input:

integrate((e*x)**(7/2)/(-b*x**2+1)**(1/2)/(b*x**2+1)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int { \frac {\left (e x\right )^{\frac {7}{2}}}{\sqrt {b x^{2} + 1} \sqrt {-b x^{2} + 1}} \,d x } \] Input:

integrate((e*x)^(7/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x, algorithm="maxim 
a")
 

Output:

integrate((e*x)^(7/2)/(sqrt(b*x^2 + 1)*sqrt(-b*x^2 + 1)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*x)^(7/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),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
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\int \frac {{\left (e\,x\right )}^{7/2}}{\sqrt {1-b\,x^2}\,\sqrt {b\,x^2+1}} \,d x \] Input:

int((e*x)^(7/2)/((1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2)),x)
 

Output:

int((e*x)^(7/2)/((1 - b*x^2)^(1/2)*(b*x^2 + 1)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {(e x)^{7/2}}{\sqrt {1-b x^2} \sqrt {1+b x^2}} \, dx=\frac {\sqrt {e}\, e^{3} \left (-2 \sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}-\left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}}{b^{2} x^{5}-x}d x \right )\right )}{5 b^{2}} \] Input:

int((e*x)^(7/2)/(-b*x^2+1)^(1/2)/(b*x^2+1)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(e)*e**3*( - 2*sqrt(x)*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1) - int((sq 
rt(x)*sqrt(b*x**2 + 1)*sqrt( - b*x**2 + 1))/(b**2*x**5 - x),x)))/(5*b**2)