\(\int \frac {(a+c x^2)^2}{\sqrt {-1+e x} \sqrt {1+e x}} \, dx\) [15]

Optimal result

Integrand size = 28, antiderivative size = 101 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {-1+e x} \sqrt {1+e x}} \, dx=\frac {c \left (3 c+8 a e^2\right ) x \sqrt {-1+e x} \sqrt {1+e x}}{8 e^4}+\frac {c^2 x^3 \sqrt {-1+e x} \sqrt {1+e x}}{4 e^2}+\frac {\left (3 c^2+8 a c e^2+8 a^2 e^4\right ) \text {arccosh}(e x)}{8 e^5} \] Output:

1/8*c*(8*a*e^2+3*c)*x*(e*x-1)^(1/2)*(e*x+1)^(1/2)/e^4+1/4*c^2*x^3*(e*x-1)^ 
(1/2)*(e*x+1)^(1/2)/e^2+1/8*(8*a^2*e^4+8*a*c*e^2+3*c^2)*arccosh(e*x)/e^5
 

Mathematica [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {-1+e x} \sqrt {1+e x}} \, dx=\frac {c e x \sqrt {-1+e x} \sqrt {1+e x} \left (8 a e^2+c \left (3+2 e^2 x^2\right )\right )+2 \left (3 c^2+8 a c e^2+8 a^2 e^4\right ) \text {arctanh}\left (\sqrt {\frac {-1+e x}{1+e x}}\right )}{8 e^5} \] Input:

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

Output:

(c*e*x*Sqrt[-1 + e*x]*Sqrt[1 + e*x]*(8*a*e^2 + c*(3 + 2*e^2*x^2)) + 2*(3*c 
^2 + 8*a*c*e^2 + 8*a^2*e^4)*ArcTanh[Sqrt[(-1 + e*x)/(1 + e*x)]])/(8*e^5)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.44, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {648, 318, 299, 224, 219}

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.

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

\(\Big \downarrow \) 648

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 299

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (\frac {\frac {\left (8 a^2 e^4+8 a c e^2+3 c^2\right ) \int \frac {1}{\sqrt {e^2 x^2-1}}dx}{2 e^2}+\frac {3 c x \sqrt {e^2 x^2-1} \left (2 a e^2+c\right )}{2 e^2}}{4 e^2}+\frac {c x \sqrt {e^2 x^2-1} \left (a+c x^2\right )}{4 e^2}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (\frac {\frac {\left (8 a^2 e^4+8 a c e^2+3 c^2\right ) \int \frac {1}{1-\frac {e^2 x^2}{e^2 x^2-1}}d\frac {x}{\sqrt {e^2 x^2-1}}}{2 e^2}+\frac {3 c x \sqrt {e^2 x^2-1} \left (2 a e^2+c\right )}{2 e^2}}{4 e^2}+\frac {c x \sqrt {e^2 x^2-1} \left (a+c x^2\right )}{4 e^2}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {e^2 x^2-1} \left (\frac {\frac {\left (8 a^2 e^4+8 a c e^2+3 c^2\right ) \text {arctanh}\left (\frac {e x}{\sqrt {e^2 x^2-1}}\right )}{2 e^3}+\frac {3 c x \sqrt {e^2 x^2-1} \left (2 a e^2+c\right )}{2 e^2}}{4 e^2}+\frac {c x \sqrt {e^2 x^2-1} \left (a+c x^2\right )}{4 e^2}\right )}{\sqrt {e x-1} \sqrt {e x+1}}\)

Input:

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

Output:

(Sqrt[-1 + e^2*x^2]*((c*x*(a + c*x^2)*Sqrt[-1 + e^2*x^2])/(4*e^2) + ((3*c* 
(c + 2*a*e^2)*x*Sqrt[-1 + e^2*x^2])/(2*e^2) + ((3*c^2 + 8*a*c*e^2 + 8*a^2* 
e^4)*ArcTanh[(e*x)/Sqrt[-1 + e^2*x^2]])/(2*e^3))/(4*e^2)))/(Sqrt[-1 + e*x] 
*Sqrt[1 + e*x])
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

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 [A] (verified)

Time = 0.82 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.22

method	result	size
risch	\(\frac {c x \left (2 x^{2} c \,e^{2}+8 a \,e^{2}+3 c \right ) \sqrt {e x -1}\, \sqrt {e x +1}}{8 e^{4}}+\frac {\left (8 a^{2} e^{4}+8 a c \,e^{2}+3 c^{2}\right ) \ln \left (\frac {e^{2} x}{\sqrt {e^{2}}}+\sqrt {e^{2} x^{2}-1}\right ) \sqrt {\left (e x -1\right ) \left (e x +1\right )}}{8 e^{4} \sqrt {e^{2}}\, \sqrt {e x -1}\, \sqrt {e x +1}}\)	\(123\)
default	\(\frac {\sqrt {e x -1}\, \sqrt {e x +1}\, \left (2 \,\operatorname {csgn}\left (e \right ) c^{2} e^{3} x^{3} \sqrt {e^{2} x^{2}-1}+8 \,\operatorname {csgn}\left (e \right ) e^{3} \sqrt {e^{2} x^{2}-1}\, a c x +8 \ln \left (\left (\sqrt {e^{2} x^{2}-1}\, \operatorname {csgn}\left (e \right )+e x \right ) \operatorname {csgn}\left (e \right )\right ) a^{2} e^{4}+3 \,\operatorname {csgn}\left (e \right ) e \sqrt {e^{2} x^{2}-1}\, c^{2} x +8 \ln \left (\left (\sqrt {e^{2} x^{2}-1}\, \operatorname {csgn}\left (e \right )+e x \right ) \operatorname {csgn}\left (e \right )\right ) a c \,e^{2}+3 \ln \left (\left (\sqrt {e^{2} x^{2}-1}\, \operatorname {csgn}\left (e \right )+e x \right ) \operatorname {csgn}\left (e \right )\right ) c^{2}\right ) \operatorname {csgn}\left (e \right )}{8 e^{5} \sqrt {e^{2} x^{2}-1}}\)	\(185\)

Input:

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

Output:

1/8*c*x*(2*c*e^2*x^2+8*a*e^2+3*c)*(e*x-1)^(1/2)*(e*x+1)^(1/2)/e^4+1/8*(8*a 
^2*e^4+8*a*c*e^2+3*c^2)/e^4*ln(e^2*x/(e^2)^(1/2)+(e^2*x^2-1)^(1/2))/(e^2)^ 
(1/2)*((e*x-1)*(e*x+1))^(1/2)/(e*x-1)^(1/2)/(e*x+1)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {-1+e x} \sqrt {1+e x}} \, dx=\frac {{\left (2 \, c^{2} e^{3} x^{3} + {\left (8 \, a c e^{3} + 3 \, c^{2} e\right )} x\right )} \sqrt {e x + 1} \sqrt {e x - 1} - {\left (8 \, a^{2} e^{4} + 8 \, a c e^{2} + 3 \, c^{2}\right )} \log \left (-e x + \sqrt {e x + 1} \sqrt {e x - 1}\right )}{8 \, e^{5}} \] Input:

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

Output:

1/8*((2*c^2*e^3*x^3 + (8*a*c*e^3 + 3*c^2*e)*x)*sqrt(e*x + 1)*sqrt(e*x - 1) 
 - (8*a^2*e^4 + 8*a*c*e^2 + 3*c^2)*log(-e*x + sqrt(e*x + 1)*sqrt(e*x - 1)) 
)/e^5
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.70 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {-1+e x} \sqrt {1+e x}} \, dx=\frac {\sqrt {e^{2} x^{2} - 1} c^{2} x^{3}}{4 \, e^{2}} + \frac {a^{2} \log \left (2 \, e^{2} x + 2 \, \sqrt {e^{2} x^{2} - 1} \sqrt {e^{2}}\right )}{\sqrt {e^{2}}} + \frac {\sqrt {e^{2} x^{2} - 1} a c x}{e^{2}} + \frac {a c \log \left (2 \, e^{2} x + 2 \, \sqrt {e^{2} x^{2} - 1} \sqrt {e^{2}}\right )}{\sqrt {e^{2}} e^{2}} + \frac {3 \, \sqrt {e^{2} x^{2} - 1} c^{2} x}{8 \, e^{4}} + \frac {3 \, c^{2} \log \left (2 \, e^{2} x + 2 \, \sqrt {e^{2} x^{2} - 1} \sqrt {e^{2}}\right )}{8 \, \sqrt {e^{2}} e^{4}} \] Input:

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

Output:

1/4*sqrt(e^2*x^2 - 1)*c^2*x^3/e^2 + a^2*log(2*e^2*x + 2*sqrt(e^2*x^2 - 1)* 
sqrt(e^2))/sqrt(e^2) + sqrt(e^2*x^2 - 1)*a*c*x/e^2 + a*c*log(2*e^2*x + 2*s 
qrt(e^2*x^2 - 1)*sqrt(e^2))/(sqrt(e^2)*e^2) + 3/8*sqrt(e^2*x^2 - 1)*c^2*x/ 
e^4 + 3/8*c^2*log(2*e^2*x + 2*sqrt(e^2*x^2 - 1)*sqrt(e^2))/(sqrt(e^2)*e^4)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.41 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {-1+e x} \sqrt {1+e x}} \, dx=\frac {{\left ({\left (e x + 1\right )} {\left (2 \, {\left (e x + 1\right )} {\left (\frac {{\left (e x + 1\right )} c^{2}}{e^{4}} - \frac {3 \, c^{2}}{e^{4}}\right )} + \frac {8 \, a c e^{18} + 9 \, c^{2} e^{16}}{e^{20}}\right )} - \frac {8 \, a c e^{18} + 5 \, c^{2} e^{16}}{e^{20}}\right )} \sqrt {e x + 1} \sqrt {e x - 1} - \frac {2 \, {\left (8 \, a^{2} e^{4} + 8 \, a c e^{2} + 3 \, c^{2}\right )} \log \left (\sqrt {e x + 1} - \sqrt {e x - 1}\right )}{e^{4}}}{8 \, e} \] Input:

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

Output:

1/8*(((e*x + 1)*(2*(e*x + 1)*((e*x + 1)*c^2/e^4 - 3*c^2/e^4) + (8*a*c*e^18 
 + 9*c^2*e^16)/e^20) - (8*a*c*e^18 + 5*c^2*e^16)/e^20)*sqrt(e*x + 1)*sqrt( 
e*x - 1) - 2*(8*a^2*e^4 + 8*a*c*e^2 + 3*c^2)*log(sqrt(e*x + 1) - sqrt(e*x 
- 1))/e^4)/e
 

Mupad [B] (verification not implemented)

Time = 18.76 (sec) , antiderivative size = 577, normalized size of antiderivative = 5.71 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {-1+e x} \sqrt {1+e x}} \, dx=\frac {-\frac {{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^{15}\,\left (\frac {3\,c^2}{2}+4\,a\,c\,e^2\right )}{{\left (\sqrt {e\,x+1}-1\right )}^{15}}+\frac {{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^3\,\left (\frac {23\,c^2}{2}-12\,a\,c\,e^2\right )}{{\left (\sqrt {e\,x+1}-1\right )}^3}+\frac {{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^{13}\,\left (\frac {23\,c^2}{2}-12\,a\,c\,e^2\right )}{{\left (\sqrt {e\,x+1}-1\right )}^{13}}+\frac {{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^5\,\left (\frac {333\,c^2}{2}+60\,a\,c\,e^2\right )}{{\left (\sqrt {e\,x+1}-1\right )}^5}+\frac {{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^{11}\,\left (\frac {333\,c^2}{2}+60\,a\,c\,e^2\right )}{{\left (\sqrt {e\,x+1}-1\right )}^{11}}+\frac {{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^7\,\left (\frac {671\,c^2}{2}-44\,a\,c\,e^2\right )}{{\left (\sqrt {e\,x+1}-1\right )}^7}+\frac {{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^9\,\left (\frac {671\,c^2}{2}-44\,a\,c\,e^2\right )}{{\left (\sqrt {e\,x+1}-1\right )}^9}-\frac {\left (\sqrt {e\,x-1}-\mathrm {i}\right )\,\left (\frac {3\,c^2}{2}+4\,a\,c\,e^2\right )}{\sqrt {e\,x+1}-1}}{e^5-\frac {8\,e^5\,{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {e\,x+1}-1\right )}^2}+\frac {28\,e^5\,{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {e\,x+1}-1\right )}^4}-\frac {56\,e^5\,{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {e\,x+1}-1\right )}^6}+\frac {70\,e^5\,{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {e\,x+1}-1\right )}^8}-\frac {56\,e^5\,{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^{10}}{{\left (\sqrt {e\,x+1}-1\right )}^{10}}+\frac {28\,e^5\,{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^{12}}{{\left (\sqrt {e\,x+1}-1\right )}^{12}}-\frac {8\,e^5\,{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^{14}}{{\left (\sqrt {e\,x+1}-1\right )}^{14}}+\frac {e^5\,{\left (\sqrt {e\,x-1}-\mathrm {i}\right )}^{16}}{{\left (\sqrt {e\,x+1}-1\right )}^{16}}}+\frac {\mathrm {atanh}\left (\frac {\sqrt {e\,x-1}-\mathrm {i}}{\sqrt {e\,x+1}-1}\right )\,\left (8\,a^2\,e^4+8\,a\,c\,e^2+3\,c^2\right )}{2\,e^5} \] Input:

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

Output:

((((e*x - 1)^(1/2) - 1i)^3*((23*c^2)/2 - 12*a*c*e^2))/((e*x + 1)^(1/2) - 1 
)^3 - (((e*x - 1)^(1/2) - 1i)^15*((3*c^2)/2 + 4*a*c*e^2))/((e*x + 1)^(1/2) 
 - 1)^15 + (((e*x - 1)^(1/2) - 1i)^13*((23*c^2)/2 - 12*a*c*e^2))/((e*x + 1 
)^(1/2) - 1)^13 + (((e*x - 1)^(1/2) - 1i)^5*((333*c^2)/2 + 60*a*c*e^2))/(( 
e*x + 1)^(1/2) - 1)^5 + (((e*x - 1)^(1/2) - 1i)^11*((333*c^2)/2 + 60*a*c*e 
^2))/((e*x + 1)^(1/2) - 1)^11 + (((e*x - 1)^(1/2) - 1i)^7*((671*c^2)/2 - 4 
4*a*c*e^2))/((e*x + 1)^(1/2) - 1)^7 + (((e*x - 1)^(1/2) - 1i)^9*((671*c^2) 
/2 - 44*a*c*e^2))/((e*x + 1)^(1/2) - 1)^9 - (((e*x - 1)^(1/2) - 1i)*((3*c^ 
2)/2 + 4*a*c*e^2))/((e*x + 1)^(1/2) - 1))/(e^5 - (8*e^5*((e*x - 1)^(1/2) - 
 1i)^2)/((e*x + 1)^(1/2) - 1)^2 + (28*e^5*((e*x - 1)^(1/2) - 1i)^4)/((e*x 
+ 1)^(1/2) - 1)^4 - (56*e^5*((e*x - 1)^(1/2) - 1i)^6)/((e*x + 1)^(1/2) - 1 
)^6 + (70*e^5*((e*x - 1)^(1/2) - 1i)^8)/((e*x + 1)^(1/2) - 1)^8 - (56*e^5* 
((e*x - 1)^(1/2) - 1i)^10)/((e*x + 1)^(1/2) - 1)^10 + (28*e^5*((e*x - 1)^( 
1/2) - 1i)^12)/((e*x + 1)^(1/2) - 1)^12 - (8*e^5*((e*x - 1)^(1/2) - 1i)^14 
)/((e*x + 1)^(1/2) - 1)^14 + (e^5*((e*x - 1)^(1/2) - 1i)^16)/((e*x + 1)^(1 
/2) - 1)^16) + (atanh(((e*x - 1)^(1/2) - 1i)/((e*x + 1)^(1/2) - 1))*(3*c^2 
 + 8*a^2*e^4 + 8*a*c*e^2))/(2*e^5)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+c x^2\right )^2}{\sqrt {-1+e x} \sqrt {1+e x}} \, dx=\frac {8 \sqrt {e x +1}\, \sqrt {e x -1}\, a c \,e^{3} x +2 \sqrt {e x +1}\, \sqrt {e x -1}\, c^{2} e^{3} x^{3}+3 \sqrt {e x +1}\, \sqrt {e x -1}\, c^{2} e x +16 \,\mathrm {log}\left (\frac {\sqrt {e x -1}+\sqrt {e x +1}}{\sqrt {2}}\right ) a^{2} e^{4}+16 \,\mathrm {log}\left (\frac {\sqrt {e x -1}+\sqrt {e x +1}}{\sqrt {2}}\right ) a c \,e^{2}+6 \,\mathrm {log}\left (\frac {\sqrt {e x -1}+\sqrt {e x +1}}{\sqrt {2}}\right ) c^{2}}{8 e^{5}} \] Input:

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

Output:

(8*sqrt(e*x + 1)*sqrt(e*x - 1)*a*c*e**3*x + 2*sqrt(e*x + 1)*sqrt(e*x - 1)* 
c**2*e**3*x**3 + 3*sqrt(e*x + 1)*sqrt(e*x - 1)*c**2*e*x + 16*log((sqrt(e*x 
 - 1) + sqrt(e*x + 1))/sqrt(2))*a**2*e**4 + 16*log((sqrt(e*x - 1) + sqrt(e 
*x + 1))/sqrt(2))*a*c*e**2 + 6*log((sqrt(e*x - 1) + sqrt(e*x + 1))/sqrt(2) 
)*c**2)/(8*e**5)