\(\int \frac {e^{\text {arctanh}(a x)}}{x^2 (c-a c x)^3} \, dx\) [387]

Optimal result

Integrand size = 19, antiderivative size = 148 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 (c-a c x)^3} \, dx=\frac {2 \sqrt {1-a^2 x^2}}{5 c^3 x (1-a x)^3}+\frac {13 \sqrt {1-a^2 x^2}}{15 c^3 x (1-a x)^2}+\frac {94 a \sqrt {1-a^2 x^2}}{15 c^3 (1-a x)}-\frac {34 \sqrt {1-a^2 x^2}}{15 c^3 x (1-a x)}-\frac {4 a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c^3} \] Output:

2/5*(-a^2*x^2+1)^(1/2)/c^3/x/(-a*x+1)^3+13/15*(-a^2*x^2+1)^(1/2)/c^3/x/(-a 
*x+1)^2+94/15*a*(-a^2*x^2+1)^(1/2)/c^3/(-a*x+1)-34/15*(-a^2*x^2+1)^(1/2)/c 
^3/x/(-a*x+1)-4*a*arctanh((-a^2*x^2+1)^(1/2))/c^3
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 (c-a c x)^3} \, dx=\frac {-15+134 a x-73 a^2 x^2-128 a^3 x^3+94 a^4 x^4-60 a x (-1+a x)^2 \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{15 c^3 x (-1+a x)^2 \sqrt {1-a^2 x^2}} \] Input:

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

Output:

(-15 + 134*a*x - 73*a^2*x^2 - 128*a^3*x^3 + 94*a^4*x^4 - 60*a*x*(-1 + a*x) 
^2*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(15*c^3*x*(-1 + a*x)^2*Sq 
rt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.85, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {6678, 27, 570, 532, 25, 2336, 25, 2336, 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.

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 570

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

\(\Big \downarrow \) 532

\(\displaystyle \frac {\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{5} \int -\frac {27 a^2 x^2+20 a x+5}{x^2 \left (1-a^2 x^2\right )^{5/2}}dx}{c^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{5} \int \frac {27 a^2 x^2+20 a x+5}{x^2 \left (1-a^2 x^2\right )^{5/2}}dx+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {\frac {1}{5} \left (\frac {4 a (8 a x+5)}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{3} \int -\frac {64 a^2 x^2+60 a x+15}{x^2 \left (1-a^2 x^2\right )^{3/2}}dx\right )+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {64 a^2 x^2+60 a x+15}{x^2 \left (1-a^2 x^2\right )^{3/2}}dx+\frac {4 a (8 a x+5)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {a (79 a x+60)}{\sqrt {1-a^2 x^2}}-\int -\frac {15 (4 a x+1)}{x^2 \sqrt {1-a^2 x^2}}dx\right )+\frac {4 a (8 a x+5)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 \int \frac {4 a x+1}{x^2 \sqrt {1-a^2 x^2}}dx+\frac {a (79 a x+60)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a (8 a x+5)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 \left (4 a \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{x}\right )+\frac {a (79 a x+60)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a (8 a x+5)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 \left (2 a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\frac {\sqrt {1-a^2 x^2}}{x}\right )+\frac {a (79 a x+60)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a (8 a x+5)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 \left (-\frac {4 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a}-\frac {\sqrt {1-a^2 x^2}}{x}\right )+\frac {a (79 a x+60)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a (8 a x+5)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 \left (-4 a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{x}\right )+\frac {a (79 a x+60)}{\sqrt {1-a^2 x^2}}\right )+\frac {4 a (8 a x+5)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {8 a (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

Input:

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

Output:

((8*a*(1 + a*x))/(5*(1 - a^2*x^2)^(5/2)) + ((4*a*(5 + 8*a*x))/(3*(1 - a^2* 
x^2)^(3/2)) + ((a*(60 + 79*a*x))/Sqrt[1 - a^2*x^2] + 15*(-(Sqrt[1 - a^2*x^ 
2]/x) - 4*a*ArcTanh[Sqrt[1 - a^2*x^2]]))/3)/5)/c^3
 

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 570

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

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 6678

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.68

method	result	size
default	\(-\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{x}+4 a \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {4 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )^{2}}+\frac {5 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}}{c^{3}}\)	\(248\)
risch	\(\frac {a^{2} x^{2}-1}{x \sqrt {-a^{2} x^{2}+1}\, c^{3}}-\frac {a \left (4 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {4 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a^{2}}-\frac {3 \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{a}+\frac {4 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}\right )}{c^{3}}\)	\(308\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 (c-a c x)^3} \, dx=\frac {104 \, a^{4} x^{4} - 312 \, a^{3} x^{3} + 312 \, a^{2} x^{2} - 104 \, a x + 60 \, {\left (a^{4} x^{4} - 3 \, a^{3} x^{3} + 3 \, a^{2} x^{2} - a x\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (94 \, a^{3} x^{3} - 222 \, a^{2} x^{2} + 149 \, a x - 15\right )} \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{3} c^{3} x^{4} - 3 \, a^{2} c^{3} x^{3} + 3 \, a c^{3} x^{2} - c^{3} x\right )}} \] Input:

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

Output:

1/15*(104*a^4*x^4 - 312*a^3*x^3 + 312*a^2*x^2 - 104*a*x + 60*(a^4*x^4 - 3* 
a^3*x^3 + 3*a^2*x^2 - a*x)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (94*a^3*x^3 - 
 222*a^2*x^2 + 149*a*x - 15)*sqrt(-a^2*x^2 + 1))/(a^3*c^3*x^4 - 3*a^2*c^3* 
x^3 + 3*a*c^3*x^2 - c^3*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (126) = 252\).

Time = 0.14 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.82 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 (c-a c x)^3} \, dx=-\frac {4 \, a^{2} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c^{3} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{2 \, c^{3} x {\left | a \right |}} - \frac {{\left (15 \, a^{2} - \frac {491 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{x} + \frac {1690 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{2} x^{2}} - \frac {2570 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{4} x^{3}} + \frac {1815 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{6} x^{4}} - \frac {555 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{5}}{a^{8} x^{5}}\right )} a^{2} x}{30 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \] Input:

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

Output:

-4*a^2*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c^3* 
abs(a)) - 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(c^3*x*abs(a)) - 1/30*(15*a^ 
2 - 491*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/x + 1690*(sqrt(-a^2*x^2 + 1)*abs(a 
) + a)^2/(a^2*x^2) - 2570*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^4*x^3) + 18 
15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^6*x^4) - 555*(sqrt(-a^2*x^2 + 1)*a 
bs(a) + a)^5/(a^8*x^5))*a^2*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^3*((sqrt( 
-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))
 

Mupad [B] (verification not implemented)

Time = 23.23 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 (c-a c x)^3} \, dx=\frac {19\,a^3\,\sqrt {1-a^2\,x^2}}{15\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}-\frac {\sqrt {1-a^2\,x^2}}{c^3\,x}+\frac {79\,a^2\,\sqrt {1-a^2\,x^2}}{15\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}+\frac {2\,a^2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )}+\frac {a\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c^3} \] Input:

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

Output:

(19*a^3*(1 - a^2*x^2)^(1/2))/(15*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) - 
(1 - a^2*x^2)^(1/2)/(c^3*x) + (a*atan((1 - a^2*x^2)^(1/2)*1i)*4i)/c^3 + (7 
9*a^2*(1 - a^2*x^2)^(1/2))/(15*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) - (c^3*(-a 
^2)^(1/2))/a)) + (2*a^2*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(3*c^3*x*(-a^ 
2)^(1/2) - (c^3*(-a^2)^(1/2))/a + a^2*c^3*x^3*(-a^2)^(1/2) - 3*a*c^3*x^2*( 
-a^2)^(1/2)))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 (c-a c x)^3} \, dx=\frac {120 \sqrt {-a^{2} x^{2}+1}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{3} x^{3}-240 \sqrt {-a^{2} x^{2}+1}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{2} x^{2}+120 \sqrt {-a^{2} x^{2}+1}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a x -61 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+190 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-171 \sqrt {-a^{2} x^{2}+1}\, a x +30 \sqrt {-a^{2} x^{2}+1}+120 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{4} x^{4}-360 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{3} x^{3}+360 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{2} x^{2}-120 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a x +315 a^{4} x^{4}-637 a^{3} x^{3}+235 a^{2} x^{2}+141 a x -30}{30 c^{3} x \left (\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\, a x +\sqrt {-a^{2} x^{2}+1}+a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:

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

Output:

(120*sqrt( - a**2*x**2 + 1)*log(tan(asin(a*x)/2))*a**3*x**3 - 240*sqrt( - 
a**2*x**2 + 1)*log(tan(asin(a*x)/2))*a**2*x**2 + 120*sqrt( - a**2*x**2 + 1 
)*log(tan(asin(a*x)/2))*a*x - 61*sqrt( - a**2*x**2 + 1)*a**3*x**3 + 190*sq 
rt( - a**2*x**2 + 1)*a**2*x**2 - 171*sqrt( - a**2*x**2 + 1)*a*x + 30*sqrt( 
 - a**2*x**2 + 1) + 120*log(tan(asin(a*x)/2))*a**4*x**4 - 360*log(tan(asin 
(a*x)/2))*a**3*x**3 + 360*log(tan(asin(a*x)/2))*a**2*x**2 - 120*log(tan(as 
in(a*x)/2))*a*x + 315*a**4*x**4 - 637*a**3*x**3 + 235*a**2*x**2 + 141*a*x 
- 30)/(30*c**3*x*(sqrt( - a**2*x**2 + 1)*a**2*x**2 - 2*sqrt( - a**2*x**2 + 
 1)*a*x + sqrt( - a**2*x**2 + 1) + a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))