\(\int \frac {1}{(-1+e x)^{3/2} (1+e x)^{3/2} (a+c x^2)} \, dx\) [23]

Optimal result

Integrand size = 28, antiderivative size = 116 \[ \int \frac {1}{(-1+e x)^{3/2} (1+e x)^{3/2} \left (a+c x^2\right )} \, dx=-\frac {e^2 x}{\left (c+a e^2\right ) \sqrt {-1+e x} \sqrt {1+e x}}-\frac {c \sqrt {-1+e^2 x^2} \text {arctanh}\left (\frac {\sqrt {c+a e^2} x}{\sqrt {a} \sqrt {-1+e^2 x^2}}\right )}{\sqrt {a} \left (c+a e^2\right )^{3/2} \sqrt {-1+e x} \sqrt {1+e x}} \] Output:

-e^2*x/(a*e^2+c)/(e*x-1)^(1/2)/(e*x+1)^(1/2)-c*(e^2*x^2-1)^(1/2)*arctanh(( 
a*e^2+c)^(1/2)*x/a^(1/2)/(e^2*x^2-1)^(1/2))/a^(1/2)/(a*e^2+c)^(3/2)/(e*x-1 
)^(1/2)/(e*x+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(-1+e x)^{3/2} (1+e x)^{3/2} \left (a+c x^2\right )} \, dx=-\frac {e^2 x}{\left (c+a e^2\right ) \sqrt {-1+e x} \sqrt {1+e x}}-\frac {c \text {arctanh}\left (\frac {\sqrt {c+a e^2} x}{\sqrt {a} \sqrt {-1+e x} \sqrt {1+e x}}\right )}{\sqrt {a} \left (c+a e^2\right )^{3/2}} \] Input:

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

Output:

-((e^2*x)/((c + a*e^2)*Sqrt[-1 + e*x]*Sqrt[1 + e*x])) - (c*ArcTanh[(Sqrt[c 
 + a*e^2]*x)/(Sqrt[a]*Sqrt[-1 + e*x]*Sqrt[1 + e*x])])/(Sqrt[a]*(c + a*e^2) 
^(3/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {648, 296, 291, 221}

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.

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

\(\Big \downarrow \) 648

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

\(\Big \downarrow \) 296

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (-\frac {c \int \frac {1}{\left (c x^2+a\right ) \sqrt {e^2 x^2-1}}dx}{a e^2+c}-\frac {e^2 x}{\sqrt {e^2 x^2-1} \left (a e^2+c\right )}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (-\frac {c \int \frac {1}{a-\frac {\left (a e^2+c\right ) x^2}{e^2 x^2-1}}d\frac {x}{\sqrt {e^2 x^2-1}}}{a e^2+c}-\frac {e^2 x}{\sqrt {e^2 x^2-1} \left (a e^2+c\right )}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (-\frac {c \text {arctanh}\left (\frac {x \sqrt {a e^2+c}}{\sqrt {a} \sqrt {e^2 x^2-1}}\right )}{\sqrt {a} \left (a e^2+c\right )^{3/2}}-\frac {e^2 x}{\sqrt {e^2 x^2-1} \left (a e^2+c\right )}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

Input:

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

Output:

(Sqrt[-1 + e^2*x^2]*(-((e^2*x)/((c + a*e^2)*Sqrt[-1 + e^2*x^2])) - (c*ArcT 
anh[(Sqrt[c + a*e^2]*x)/(Sqrt[a]*Sqrt[-1 + e^2*x^2])])/(Sqrt[a]*(c + a*e^2 
)^(3/2))))/(Sqrt[-1 + e*x]*Sqrt[1 + e*x])
 

Defintions of rubi rules used

rule 221

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

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 296

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d))   Int[ 
(a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N 
eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1 
]) && NeQ[p, -1]
 

rule 648

Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) 
^2)^(p_), x_Symbol] :> Simp[(c + d*x)^FracPart[m]*((e + f*x)^FracPart[m]/(c 
*e + d*f*x^2)^FracPart[m])   Int[(c*e + d*f*x^2)^m*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m, n] && EqQ[d*e + c*f, 0] && 
  !(EqQ[p, 2] && LtQ[m, -1])
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.60 (sec) , antiderivative size = 687, normalized size of antiderivative = 5.92

method	result	size
default	\(\frac {c^{2} \operatorname {csgn}\left (e \right )^{2} \left (\ln \left (\frac {2 \sqrt {-a c}\, e^{2} x +2 \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c -2 c}{c x -\sqrt {-a c}}\right ) a c \,e^{4} x^{2}-\ln \left (\frac {-2 \sqrt {-a c}\, e^{2} x +2 \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c -2 c}{c x +\sqrt {-a c}}\right ) a c \,e^{4} x^{2}-2 a \sqrt {-a c}\, \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, e^{4} x +\ln \left (\frac {2 \sqrt {-a c}\, e^{2} x +2 \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c -2 c}{c x -\sqrt {-a c}}\right ) c^{2} e^{2} x^{2}-\ln \left (\frac {-2 \sqrt {-a c}\, e^{2} x +2 \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c -2 c}{c x +\sqrt {-a c}}\right ) c^{2} e^{2} x^{2}-2 \sqrt {-a c}\, \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c \,e^{2} x -\ln \left (\frac {2 \sqrt {-a c}\, e^{2} x +2 \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c -2 c}{c x -\sqrt {-a c}}\right ) a c \,e^{2}+\ln \left (\frac {-2 \sqrt {-a c}\, e^{2} x +2 \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c -2 c}{c x +\sqrt {-a c}}\right ) a c \,e^{2}-\ln \left (\frac {2 \sqrt {-a c}\, e^{2} x +2 \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c -2 c}{c x -\sqrt {-a c}}\right ) c^{2}+\ln \left (\frac {-2 \sqrt {-a c}\, e^{2} x +2 \sqrt {e^{2} x^{2}-1}\, \sqrt {-\frac {a \,e^{2}+c}{c}}\, c -2 c}{c x +\sqrt {-a c}}\right ) c^{2}\right )}{2 \sqrt {-\frac {a \,e^{2}+c}{c}}\, \sqrt {-a c}\, \left (-e \sqrt {-a c}+c \right )^{2} \left (e \sqrt {-a c}+c \right )^{2} \sqrt {e^{2} x^{2}-1}\, \sqrt {e x +1}\, \sqrt {e x -1}}\)	\(687\)

Input:

int(1/(e*x-1)^(3/2)/(e*x+1)^(3/2)/(c*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

1/2*c^2*csgn(e)^2*(ln(2*((-a*c)^(1/2)*e^2*x+(e^2*x^2-1)^(1/2)*(-(a*e^2+c)/ 
c)^(1/2)*c-c)/(c*x-(-a*c)^(1/2)))*a*c*e^4*x^2-ln(2*(-(-a*c)^(1/2)*e^2*x+(e 
^2*x^2-1)^(1/2)*(-(a*e^2+c)/c)^(1/2)*c-c)/(c*x+(-a*c)^(1/2)))*a*c*e^4*x^2- 
2*a*(-a*c)^(1/2)*(e^2*x^2-1)^(1/2)*(-(a*e^2+c)/c)^(1/2)*e^4*x+ln(2*((-a*c) 
^(1/2)*e^2*x+(e^2*x^2-1)^(1/2)*(-(a*e^2+c)/c)^(1/2)*c-c)/(c*x-(-a*c)^(1/2) 
))*c^2*e^2*x^2-ln(2*(-(-a*c)^(1/2)*e^2*x+(e^2*x^2-1)^(1/2)*(-(a*e^2+c)/c)^ 
(1/2)*c-c)/(c*x+(-a*c)^(1/2)))*c^2*e^2*x^2-2*(-a*c)^(1/2)*(e^2*x^2-1)^(1/2 
)*(-(a*e^2+c)/c)^(1/2)*c*e^2*x-ln(2*((-a*c)^(1/2)*e^2*x+(e^2*x^2-1)^(1/2)* 
(-(a*e^2+c)/c)^(1/2)*c-c)/(c*x-(-a*c)^(1/2)))*a*c*e^2+ln(2*(-(-a*c)^(1/2)* 
e^2*x+(e^2*x^2-1)^(1/2)*(-(a*e^2+c)/c)^(1/2)*c-c)/(c*x+(-a*c)^(1/2)))*a*c* 
e^2-ln(2*((-a*c)^(1/2)*e^2*x+(e^2*x^2-1)^(1/2)*(-(a*e^2+c)/c)^(1/2)*c-c)/( 
c*x-(-a*c)^(1/2)))*c^2+ln(2*(-(-a*c)^(1/2)*e^2*x+(e^2*x^2-1)^(1/2)*(-(a*e^ 
2+c)/c)^(1/2)*c-c)/(c*x+(-a*c)^(1/2)))*c^2)/(-(a*e^2+c)/c)^(1/2)/(-a*c)^(1 
/2)/(-e*(-a*c)^(1/2)+c)^2/(e*(-a*c)^(1/2)+c)^2/(e^2*x^2-1)^(1/2)/(e*x+1)^( 
1/2)/(e*x-1)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (96) = 192\).

Time = 0.12 (sec) , antiderivative size = 506, normalized size of antiderivative = 4.36 \[ \int \frac {1}{(-1+e x)^{3/2} (1+e x)^{3/2} \left (a+c x^2\right )} \, dx=\left [-\frac {2 \, a^{2} e^{3} + 2 \, a c e - 2 \, {\left (a^{2} e^{4} + a c e^{2}\right )} \sqrt {e x + 1} \sqrt {e x - 1} x - 2 \, {\left (a^{2} e^{5} + a c e^{3}\right )} x^{2} + {\left (c e^{2} x^{2} - c\right )} \sqrt {a^{2} e^{2} + a c} \log \left (-\frac {2 \, a^{2} e^{2} - {\left (4 \, a^{2} e^{4} + 4 \, a c e^{2} + c^{2}\right )} x^{2} + 2 \, {\left (\sqrt {a^{2} e^{2} + a c} {\left (2 \, a e^{2} + c\right )} x - 2 \, {\left (a^{2} e^{3} + a c e\right )} x\right )} \sqrt {e x + 1} \sqrt {e x - 1} + a c + 2 \, \sqrt {a^{2} e^{2} + a c} {\left ({\left (2 \, a e^{3} + c e\right )} x^{2} - a e\right )}}{c x^{2} + a}\right )}{2 \, {\left (a^{3} e^{4} + 2 \, a^{2} c e^{2} + a c^{2} - {\left (a^{3} e^{6} + 2 \, a^{2} c e^{4} + a c^{2} e^{2}\right )} x^{2}\right )}}, -\frac {a^{2} e^{3} + a c e - {\left (a^{2} e^{4} + a c e^{2}\right )} \sqrt {e x + 1} \sqrt {e x - 1} x - {\left (a^{2} e^{5} + a c e^{3}\right )} x^{2} - {\left (c e^{2} x^{2} - c\right )} \sqrt {-a^{2} e^{2} - a c} \arctan \left (\frac {\sqrt {-a^{2} e^{2} - a c} \sqrt {e x + 1} \sqrt {e x - 1} c x - \sqrt {-a^{2} e^{2} - a c} {\left (c e x^{2} + a e\right )}}{a^{2} e^{2} + a c}\right )}{a^{3} e^{4} + 2 \, a^{2} c e^{2} + a c^{2} - {\left (a^{3} e^{6} + 2 \, a^{2} c e^{4} + a c^{2} e^{2}\right )} x^{2}}\right ] \] Input:

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

Output:

[-1/2*(2*a^2*e^3 + 2*a*c*e - 2*(a^2*e^4 + a*c*e^2)*sqrt(e*x + 1)*sqrt(e*x 
- 1)*x - 2*(a^2*e^5 + a*c*e^3)*x^2 + (c*e^2*x^2 - c)*sqrt(a^2*e^2 + a*c)*l 
og(-(2*a^2*e^2 - (4*a^2*e^4 + 4*a*c*e^2 + c^2)*x^2 + 2*(sqrt(a^2*e^2 + a*c 
)*(2*a*e^2 + c)*x - 2*(a^2*e^3 + a*c*e)*x)*sqrt(e*x + 1)*sqrt(e*x - 1) + a 
*c + 2*sqrt(a^2*e^2 + a*c)*((2*a*e^3 + c*e)*x^2 - a*e))/(c*x^2 + a)))/(a^3 
*e^4 + 2*a^2*c*e^2 + a*c^2 - (a^3*e^6 + 2*a^2*c*e^4 + a*c^2*e^2)*x^2), -(a 
^2*e^3 + a*c*e - (a^2*e^4 + a*c*e^2)*sqrt(e*x + 1)*sqrt(e*x - 1)*x - (a^2* 
e^5 + a*c*e^3)*x^2 - (c*e^2*x^2 - c)*sqrt(-a^2*e^2 - a*c)*arctan((sqrt(-a^ 
2*e^2 - a*c)*sqrt(e*x + 1)*sqrt(e*x - 1)*c*x - sqrt(-a^2*e^2 - a*c)*(c*e*x 
^2 + a*e))/(a^2*e^2 + a*c)))/(a^3*e^4 + 2*a^2*c*e^2 + a*c^2 - (a^3*e^6 + 2 
*a^2*c*e^4 + a*c^2*e^2)*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(1/(sqrt(e*x + 1)*sqrt(e*x - 1)*a*e**2*x**2 - sqrt(e*x + 1)*sqrt(e*x - 
1)*a + sqrt(e*x + 1)*sqrt(e*x - 1)*c*e**2*x**4 - sqrt(e*x + 1)*sqrt(e*x - 
1)*c*x**2),x)