\(\int \frac {e^{2 \text {arctanh}(a x)}}{x^3 (c-a^2 c x^2)^{3/2}} \, dx\) [1144]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 162 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {2 a^2 (1+a x)}{3 \left (c-a^2 c x^2\right )^{3/2}}+\frac {7 a^2}{2 c \sqrt {c-a^2 c x^2}}-\frac {1}{2 c x^2 \sqrt {c-a^2 c x^2}}+\frac {10 a^3 x}{3 c \sqrt {c-a^2 c x^2}}-\frac {2 a \sqrt {c-a^2 c x^2}}{c^2 x}-\frac {7 a^2 \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{2 c^{3/2}} \] Output:

2/3*a^2*(a*x+1)/(-a^2*c*x^2+c)^(3/2)+7/2*a^2/c/(-a^2*c*x^2+c)^(1/2)-1/2/c/ 
x^2/(-a^2*c*x^2+c)^(1/2)+10/3*a^3*x/c/(-a^2*c*x^2+c)^(1/2)-2*a*(-a^2*c*x^2 
+c)^(1/2)/c^2/x-7/2*a^2*arctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.65 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c-a^2 c x^2} \left (3+6 a x-43 a^2 x^2+32 a^3 x^3\right )}{6 c^2 x^2 (-1+a x)^2}+\frac {7 a^2 \log (x)}{2 c^{3/2}}-\frac {7 a^2 \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right )}{2 c^{3/2}} \] Input:

Integrate[E^(2*ArcTanh[a*x])/(x^3*(c - a^2*c*x^2)^(3/2)),x]
 

Output:

-1/6*(Sqrt[c - a^2*c*x^2]*(3 + 6*a*x - 43*a^2*x^2 + 32*a^3*x^3))/(c^2*x^2* 
(-1 + a*x)^2) + (7*a^2*Log[x])/(2*c^(3/2)) - (7*a^2*Log[c + Sqrt[c]*Sqrt[c 
 - a^2*c*x^2]])/(2*c^(3/2))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6701, 532, 25, 2336, 27, 2338, 25, 27, 534, 243, 73, 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.

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

\(\Big \downarrow \) 6701

\(\displaystyle c \int \frac {(a x+1)^2}{x^3 \left (c-a^2 c x^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 532

\(\displaystyle c \left (\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\int -\frac {4 a^3 x^3+6 a^2 x^2+6 a x+3}{x^3 \left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {\int \frac {4 a^3 x^3+6 a^2 x^2+6 a x+3}{x^3 \left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 2336

\(\displaystyle c \left (\frac {\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}-\frac {\int -\frac {3 \left (3 a^2 x^2+2 a x+1\right )}{x^3 \sqrt {c-a^2 c x^2}}dx}{c}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {\frac {3 \int \frac {3 a^2 x^2+2 a x+1}{x^3 \sqrt {c-a^2 c x^2}}dx}{c}+\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 2338

\(\displaystyle c \left (\frac {\frac {3 \left (-\frac {\int -\frac {a c (7 a x+4)}{x^2 \sqrt {c-a^2 c x^2}}dx}{2 c}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )}{c}+\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {\frac {3 \left (\frac {\int \frac {a c (7 a x+4)}{x^2 \sqrt {c-a^2 c x^2}}dx}{2 c}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )}{c}+\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {\frac {3 \left (\frac {1}{2} a \int \frac {7 a x+4}{x^2 \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )}{c}+\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 534

\(\displaystyle c \left (\frac {\frac {3 \left (\frac {1}{2} a \left (7 a \int \frac {1}{x \sqrt {c-a^2 c x^2}}dx-\frac {4 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )}{c}+\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle c \left (\frac {\frac {3 \left (\frac {1}{2} a \left (\frac {7}{2} a \int \frac {1}{x^2 \sqrt {c-a^2 c x^2}}dx^2-\frac {4 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )}{c}+\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c \left (\frac {\frac {3 \left (\frac {1}{2} a \left (-\frac {7 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2 c}}d\sqrt {c-a^2 c x^2}}{a c}-\frac {4 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )}{c}+\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c \left (\frac {\frac {3 \left (\frac {1}{2} a \left (-\frac {7 a \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {4 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )}{c}+\frac {a^2 (10 a x+9)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a^2 (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

Input:

Int[E^(2*ArcTanh[a*x])/(x^3*(c - a^2*c*x^2)^(3/2)),x]
 

Output:

c*((2*a^2*(1 + a*x))/(3*c*(c - a^2*c*x^2)^(3/2)) + ((a^2*(9 + 10*a*x))/(c* 
Sqrt[c - a^2*c*x^2]) + (3*(-1/2*Sqrt[c - a^2*c*x^2]/(c*x^2) + (a*((-4*Sqrt 
[c - a^2*c*x^2])/(c*x) - (7*a*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]])/Sqrt[c 
]))/2))/c)/(3*c))
 

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 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 243
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 532
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 534
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 2336
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 2338
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

rule 6701
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[c^(n/2)   Int[x^m*(c + d*x^2)^(p - n/2)*(1 + a*x)^n, x], x] 
 /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ 
[c, 0]) && IGtQ[n/2, 0]
 
Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06

method result size
risch \(\frac {4 a^{3} x^{3}+a^{2} x^{2}-4 a x -1}{2 x^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, c}+\frac {-\frac {7 a^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{2 \sqrt {c}}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{3 c \left (x -\frac {1}{a}\right )^{2}}-\frac {10 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{3 c \left (x -\frac {1}{a}\right )}}{c}\) \(172\)
default \(-\frac {1}{2 c \,x^{2} \sqrt {-a^{2} c \,x^{2}+c}}+\frac {7 a^{2} \left (\frac {1}{c \sqrt {-a^{2} c \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}\right )}{2}+2 a \left (-\frac {1}{c x \sqrt {-a^{2} c \,x^{2}+c}}+\frac {2 a^{2} x}{c \sqrt {-a^{2} c \,x^{2}+c}}\right )-2 a^{2} \left (\frac {1}{3 a c \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}+\frac {-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c}{3 a \,c^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )\) \(232\)

Input:

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

Output:

1/2*(4*a^3*x^3+a^2*x^2-4*a*x-1)/x^2/(-c*(a^2*x^2-1))^(1/2)/c+(-7/2*a^2/c^( 
1/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)+1/3/c/(x-1/a)^2*(-(x-1/a)^ 
2*a^2*c-2*(x-1/a)*a*c)^(1/2)-10/3*a/c/(x-1/a)*(-(x-1/a)^2*a^2*c-2*(x-1/a)* 
a*c)^(1/2))/c
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.58 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\left [\frac {21 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - 2 \, {\left (32 \, a^{3} x^{3} - 43 \, a^{2} x^{2} + 6 \, a x + 3\right )} \sqrt {-a^{2} c x^{2} + c}}{12 \, {\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{2}\right )}}, \frac {21 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) - {\left (32 \, a^{3} x^{3} - 43 \, a^{2} x^{2} + 6 \, a x + 3\right )} \sqrt {-a^{2} c x^{2} + c}}{6 \, {\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/12*(21*(a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*sqrt(c)*log(-(a^2*c*x^2 + 2*sqrt 
(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) - 2*(32*a^3*x^3 - 43*a^2*x^2 + 6*a*x 
+ 3)*sqrt(-a^2*c*x^2 + c))/(a^2*c^2*x^4 - 2*a*c^2*x^3 + c^2*x^2), 1/6*(21* 
(a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*sqrt(-c)*arctan(sqrt(-a^2*c*x^2 + c)*sqrt( 
-c)/c) - (32*a^3*x^3 - 43*a^2*x^2 + 6*a*x + 3)*sqrt(-a^2*c*x^2 + c))/(a^2* 
c^2*x^4 - 2*a*c^2*x^3 + c^2*x^2)]
 

Sympy [F]

\[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=- \int \frac {a x}{- a^{3} c x^{6} \sqrt {- a^{2} c x^{2} + c} + a^{2} c x^{5} \sqrt {- a^{2} c x^{2} + c} + a c x^{4} \sqrt {- a^{2} c x^{2} + c} - c x^{3} \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {1}{- a^{3} c x^{6} \sqrt {- a^{2} c x^{2} + c} + a^{2} c x^{5} \sqrt {- a^{2} c x^{2} + c} + a c x^{4} \sqrt {- a^{2} c x^{2} + c} - c x^{3} \sqrt {- a^{2} c x^{2} + c}}\, dx \] Input:

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

Output:

-Integral(a*x/(-a**3*c*x**6*sqrt(-a**2*c*x**2 + c) + a**2*c*x**5*sqrt(-a** 
2*c*x**2 + c) + a*c*x**4*sqrt(-a**2*c*x**2 + c) - c*x**3*sqrt(-a**2*c*x**2 
 + c)), x) - Integral(1/(-a**3*c*x**6*sqrt(-a**2*c*x**2 + c) + a**2*c*x**5 
*sqrt(-a**2*c*x**2 + c) + a*c*x**4*sqrt(-a**2*c*x**2 + c) - c*x**3*sqrt(-a 
**2*c*x**2 + c)), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.26 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=a^{4} c^{2} {\left (\frac {7 \, \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c} c^{3}} - \frac {{\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{3} a - 4 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} \sqrt {-c} {\left | a \right |} + {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )} a c + 4 \, \sqrt {-c} c {\left | a \right |}}{{\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{2} a^{3} c^{3}}\right )} \] Input:

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

Output:

a^4*c^2*(7*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/(a^2* 
sqrt(-c)*c^3) - ((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^3*a - 4*(sqrt(-a^ 
2*c)*x - sqrt(-a^2*c*x^2 + c))^2*sqrt(-c)*abs(a) + (sqrt(-a^2*c)*x - sqrt( 
-a^2*c*x^2 + c))*a*c + 4*sqrt(-c)*c*abs(a))/(((sqrt(-a^2*c)*x - sqrt(-a^2* 
c*x^2 + c))^2 - c)^2*a^3*c^3))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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