\(\int \frac {1}{(-1+e x)^{3/2} (1+e x)^{3/2} (a+c x^2)^2} \, dx\) [24]

Optimal result

Integrand size = 28, antiderivative size = 184 \[ \int \frac {1}{(-1+e x)^{3/2} (1+e x)^{3/2} \left (a+c x^2\right )^2} \, dx=\frac {e^2 \left (c-2 a e^2\right ) x}{2 a \left (c+a e^2\right )^2 \sqrt {-1+e x} \sqrt {1+e x}}-\frac {c x}{2 a \left (c+a e^2\right ) \sqrt {-1+e x} \sqrt {1+e x} \left (a+c x^2\right )}-\frac {c \left (c+4 a e^2\right ) \sqrt {-1+e^2 x^2} \text {arctanh}\left (\frac {\sqrt {c+a e^2} x}{\sqrt {a} \sqrt {-1+e^2 x^2}}\right )}{2 a^{3/2} \left (c+a e^2\right )^{5/2} \sqrt {-1+e x} \sqrt {1+e x}} \] Output:

1/2*e^2*(-2*a*e^2+c)*x/a/(a*e^2+c)^2/(e*x-1)^(1/2)/(e*x+1)^(1/2)-1/2*c*x/a 
/(a*e^2+c)/(e*x-1)^(1/2)/(e*x+1)^(1/2)/(c*x^2+a)-1/2*c*(4*a*e^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^(3/2)/(a 
*e^2+c)^(5/2)/(e*x-1)^(1/2)/(e*x+1)^(1/2)
 

Mathematica [A] (verified)

Time = 1.67 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(-1+e x)^{3/2} (1+e x)^{3/2} \left (a+c x^2\right )^2} \, dx=-\frac {x \left (2 a^2 e^4+2 a c e^4 x^2+c^2 \left (1-e^2 x^2\right )\right )}{2 a \left (c+a e^2\right )^2 \sqrt {-1+e x} \sqrt {1+e x} \left (a+c x^2\right )}-\frac {c \left (c+4 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c+a e^2} x}{\sqrt {a} \sqrt {-1+e x} \sqrt {1+e x}}\right )}{2 a^{3/2} \left (c+a e^2\right )^{5/2}} \] Input:

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

Output:

-1/2*(x*(2*a^2*e^4 + 2*a*c*e^4*x^2 + c^2*(1 - e^2*x^2)))/(a*(c + a*e^2)^2* 
Sqrt[-1 + e*x]*Sqrt[1 + e*x]*(a + c*x^2)) - (c*(c + 4*a*e^2)*ArcTanh[(Sqrt 
[c + a*e^2]*x)/(Sqrt[a]*Sqrt[-1 + e*x]*Sqrt[1 + e*x])])/(2*a^(3/2)*(c + a* 
e^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {648, 316, 402, 25, 27, 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.

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

\(\Big \downarrow \) 648

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (\frac {\frac {\int -\frac {c \left (4 a e^2+c\right )}{\left (c x^2+a\right ) \sqrt {e^2 x^2-1}}dx}{a e^2+c}+\frac {e^2 x \left (c-2 a e^2\right )}{\sqrt {e^2 x^2-1} \left (a e^2+c\right )}}{2 a \left (a e^2+c\right )}-\frac {c x}{2 a \sqrt {e^2 x^2-1} \left (a e^2+c\right ) \left (a+c x^2\right )}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (\frac {\frac {e^2 x \left (c-2 a e^2\right )}{\sqrt {e^2 x^2-1} \left (a e^2+c\right )}-\frac {\int \frac {c \left (4 a e^2+c\right )}{\left (c x^2+a\right ) \sqrt {e^2 x^2-1}}dx}{a e^2+c}}{2 a \left (a e^2+c\right )}-\frac {c x}{2 a \sqrt {e^2 x^2-1} \left (a e^2+c\right ) \left (a+c x^2\right )}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (\frac {\frac {e^2 x \left (c-2 a e^2\right )}{\sqrt {e^2 x^2-1} \left (a e^2+c\right )}-\frac {c \left (4 a e^2+c\right ) \int \frac {1}{\left (c x^2+a\right ) \sqrt {e^2 x^2-1}}dx}{a e^2+c}}{2 a \left (a e^2+c\right )}-\frac {c x}{2 a \sqrt {e^2 x^2-1} \left (a e^2+c\right ) \left (a+c x^2\right )}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (\frac {\frac {e^2 x \left (c-2 a e^2\right )}{\sqrt {e^2 x^2-1} \left (a e^2+c\right )}-\frac {c \left (4 a e^2+c\right ) \int \frac {1}{a-\frac {\left (a e^2+c\right ) x^2}{e^2 x^2-1}}d\frac {x}{\sqrt {e^2 x^2-1}}}{a e^2+c}}{2 a \left (a e^2+c\right )}-\frac {c x}{2 a \sqrt {e^2 x^2-1} \left (a e^2+c\right ) \left (a+c x^2\right )}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (\frac {\frac {e^2 x \left (c-2 a e^2\right )}{\sqrt {e^2 x^2-1} \left (a e^2+c\right )}-\frac {c \left (4 a e^2+c\right ) \text {arctanh}\left (\frac {x \sqrt {a e^2+c}}{\sqrt {a} \sqrt {e^2 x^2-1}}\right )}{\sqrt {a} \left (a e^2+c\right )^{3/2}}}{2 a \left (a e^2+c\right )}-\frac {c x}{2 a \sqrt {e^2 x^2-1} \left (a e^2+c\right ) \left (a+c x^2\right )}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

Input:

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

Output:

(Sqrt[-1 + e^2*x^2]*(-1/2*(c*x)/(a*(c + a*e^2)*(a + c*x^2)*Sqrt[-1 + e^2*x 
^2]) + ((e^2*(c - 2*a*e^2)*x)/((c + a*e^2)*Sqrt[-1 + e^2*x^2]) - (c*(c + 4 
*a*e^2)*ArcTanh[(Sqrt[c + a*e^2]*x)/(Sqrt[a]*Sqrt[-1 + e^2*x^2])])/(Sqrt[a 
]*(c + a*e^2)^(3/2)))/(2*a*(c + a*e^2))))/(Sqrt[-1 + e*x]*Sqrt[1 + e*x])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 316

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[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[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 = 1.71 (sec) , antiderivative size = 1781, normalized size of antiderivative = 9.68

method	result	size
default	\(\text {Expression too large to display}\)	\(1781\)

Input:

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

Output:

1/4*(-4*a^2*c*e^4*x*(-a*c)^(1/2)*(-(a*e^2+c)/c)^(1/2)*(e^2*x^2-1)^(1/2)-2* 
a*c^2*e^2*x*(-a*c)^(1/2)*(-(a*e^2+c)/c)^(1/2)*(e^2*x^2-1)^(1/2)+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^4*e^2*x^4-4*a^2*c*e^6*x^3*(-a*c)^(1/2)*(-(a*e^2+c)/c)^(1/2)*(e^2*x^ 
2-1)^(1/2)-2*a*c^2*e^4*x^3*(-a*c)^(1/2)*(-(a*e^2+c)/c)^(1/2)*(e^2*x^2-1)^( 
1/2)+4*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^3*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^4*e^2*x^4-4*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^3*c*e^4+4*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^3*c*e^4-5*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^2*c^2*e^2 
+5*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^2*c^2*e^2-4*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^2*c^2*e^6*x^4+4*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^3*c*e^6*x^2+5*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^3*e^4*x^4-4*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^3 
*c*e^6*x^2-5*ln(2*(-(-a*c)^(1/2)*e^2*x+(e^2*x^2-1)^(1/2)*(-(a*e^2+c)/c)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 468 vs. \(2 (156) = 312\).

Time = 0.13 (sec) , antiderivative size = 999, normalized size of antiderivative = 5.43 \[ \int \frac {1}{(-1+e x)^{3/2} (1+e x)^{3/2} \left (a+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/4*(4*a^4*e^5 + 2*a^3*c*e^3 - 2*a^2*c^2*e - 2*(2*a^3*c*e^7 + a^2*c^2*e^ 
5 - a*c^3*e^3)*x^4 - 2*(2*a^4*e^7 - a^3*c*e^5 - 2*a^2*c^2*e^3 + a*c^3*e)*x 
^2 - (4*a^2*c*e^2 - (4*a*c^2*e^4 + c^3*e^2)*x^4 + a*c^2 - (4*a^2*c*e^4 - 3 
*a*c^2*e^2 - c^3)*x^2)*sqrt(a^2*e^2 + a*c)*log(-(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)) - 2*((2*a^3*c*e^6 + a^2*c^2*e^4 - a 
*c^3*e^2)*x^3 + (2*a^4*e^6 + 2*a^3*c*e^4 + a^2*c^2*e^2 + a*c^3)*x)*sqrt(e* 
x + 1)*sqrt(e*x - 1))/(a^6*e^6 + 3*a^5*c*e^4 + 3*a^4*c^2*e^2 + a^3*c^3 - ( 
a^5*c*e^8 + 3*a^4*c^2*e^6 + 3*a^3*c^3*e^4 + a^2*c^4*e^2)*x^4 - (a^6*e^8 + 
2*a^5*c*e^6 - 2*a^3*c^3*e^2 - a^2*c^4)*x^2), -1/2*(2*a^4*e^5 + a^3*c*e^3 - 
 a^2*c^2*e - (2*a^3*c*e^7 + a^2*c^2*e^5 - a*c^3*e^3)*x^4 - (2*a^4*e^7 - a^ 
3*c*e^5 - 2*a^2*c^2*e^3 + a*c^3*e)*x^2 + (4*a^2*c*e^2 - (4*a*c^2*e^4 + c^3 
*e^2)*x^4 + a*c^2 - (4*a^2*c*e^4 - 3*a*c^2*e^2 - c^3)*x^2)*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)) - ((2*a^3*c*e^6 + a^2*c^ 
2*e^4 - a*c^3*e^2)*x^3 + (2*a^4*e^6 + 2*a^3*c*e^4 + a^2*c^2*e^2 + a*c^3)*x 
)*sqrt(e*x + 1)*sqrt(e*x - 1))/(a^6*e^6 + 3*a^5*c*e^4 + 3*a^4*c^2*e^2 + a^ 
3*c^3 - (a^5*c*e^8 + 3*a^4*c^2*e^6 + 3*a^3*c^3*e^4 + a^2*c^4*e^2)*x^4 - (a 
^6*e^8 + 2*a^5*c*e^6 - 2*a^3*c^3*e^2 - a^2*c^4)*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(1/(sqrt(e*x + 1)*sqrt(e*x - 1)*a**2*e**2*x**2 - sqrt(e*x + 1)*sqrt(e*x 
 - 1)*a**2 + 2*sqrt(e*x + 1)*sqrt(e*x - 1)*a*c*e**2*x**4 - 2*sqrt(e*x + 1) 
*sqrt(e*x - 1)*a*c*x**2 + sqrt(e*x + 1)*sqrt(e*x - 1)*c**2*e**2*x**6 - sqr 
t(e*x + 1)*sqrt(e*x - 1)*c**2*x**4),x)