\(\int \frac {e^{3 \text {arctanh}(a x)} (c-a^2 c x^2)}{x^3} \, dx\) [1180]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^3} \, dx=-\frac {c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {3 a c \sqrt {1-a^2 x^2}}{x}+a^2 c \arcsin (a x)-\frac {7}{2} a^2 c \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.79 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^3} \, dx=\frac {1}{2} c \left (-\frac {(1+6 a x) \sqrt {1-a^2 x^2}}{x^2}+2 a^2 \arcsin (a x)-7 a^2 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right ) \] Input:

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

Output:

(c*(-(((1 + 6*a*x)*Sqrt[1 - a^2*x^2])/x^2) + 2*a^2*ArcSin[a*x] - 7*a^2*Arc 
Tanh[Sqrt[1 - a^2*x^2]]))/2
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {6698, 540, 25, 2338, 25, 27, 538, 223, 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.

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

\(\Big \downarrow \) 6698

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

\(\Big \downarrow \) 540

\(\displaystyle c \left (-\frac {1}{2} \int -\frac {2 x^2 a^3+7 x a^2+6 a}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {1}{2} \int \frac {2 x^2 a^3+7 x a^2+6 a}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 2338

\(\displaystyle c \left (\frac {1}{2} \left (-\int -\frac {a^2 (2 a x+7)}{x \sqrt {1-a^2 x^2}}dx-\frac {6 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {1}{2} \left (\int \frac {a^2 (2 a x+7)}{x \sqrt {1-a^2 x^2}}dx-\frac {6 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{2} \left (a^2 \int \frac {2 a x+7}{x \sqrt {1-a^2 x^2}}dx-\frac {6 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 538

\(\displaystyle c \left (\frac {1}{2} \left (a^2 \left (2 a \int \frac {1}{\sqrt {1-a^2 x^2}}dx+7 \int \frac {1}{x \sqrt {1-a^2 x^2}}dx\right )-\frac {6 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle c \left (\frac {1}{2} \left (a^2 \left (7 \int \frac {1}{x \sqrt {1-a^2 x^2}}dx+2 \arcsin (a x)\right )-\frac {6 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle c \left (\frac {1}{2} \left (a^2 \left (\frac {7}{2} \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2+2 \arcsin (a x)\right )-\frac {6 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c \left (\frac {1}{2} \left (a^2 \left (2 \arcsin (a x)-\frac {7 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}\right )-\frac {6 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c \left (\frac {1}{2} \left (a^2 \left (2 \arcsin (a x)-7 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )-\frac {6 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )\)

Input:

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

Output:

c*(-1/2*Sqrt[1 - a^2*x^2]/x^2 + ((-6*a*Sqrt[1 - a^2*x^2])/x + a^2*(2*ArcSi 
n[a*x] - 7*ArcTanh[Sqrt[1 - a^2*x^2]]))/2)
 

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[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 538

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

rule 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

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 6698

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[c^p   Int[x^m*(1 - a^2*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 + 1)/2, 0] &&  !IntegerQ[p - n/2]
 

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.20

method	result	size
risch	\(\frac {\left (6 a^{3} x^{3}+a^{2} x^{2}-6 a x -1\right ) c}{2 x^{2} \sqrt {-a^{2} x^{2}+1}}+\left (-\frac {7 a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {a^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}\right ) c\)	\(91\)
default	\(-c \left (a^{5} \left (\frac {x}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}\right )+\frac {2 a^{3} x}{\sqrt {-a^{2} x^{2}+1}}+\frac {1}{2 x^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {7 a^{2} \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{2}-3 a \left (-\frac {1}{x \sqrt {-a^{2} x^{2}+1}}+\frac {2 a^{2} x}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {3 a^{2}}{\sqrt {-a^{2} x^{2}+1}}\right )\)	\(181\)
meijerg	\(\frac {2 a^{2} c \left (\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{2}-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{\sqrt {\pi }}-\frac {a^{2} c \left (\frac {\sqrt {\pi }}{2 x^{2} a^{2}}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }\, \left (-20 a^{2} x^{2}+8\right )}{16 a^{2} x^{2}}+\frac {\sqrt {\pi }\, \left (-24 a^{2} x^{2}+8\right )}{16 a^{2} x^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )}{2}\right )}{\sqrt {\pi }}+\frac {a^{3} c \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {2 a^{3} c x}{\sqrt {-a^{2} x^{2}+1}}+\frac {3 a^{2} c \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {\pi }}-\frac {3 a c \left (-2 a^{2} x^{2}+1\right )}{x \sqrt {-a^{2} x^{2}+1}}\)	\(334\)

Input:

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

Output:

1/2*(6*a^3*x^3+a^2*x^2-6*a*x-1)/x^2/(-a^2*x^2+1)^(1/2)*c+(-7/2*a^2*arctanh 
(1/(-a^2*x^2+1)^(1/2))+a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^( 
1/2)))*c
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.12 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^3} \, dx=-\frac {4 \, a^{2} c x^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - 7 \, a^{2} c x^{2} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (6 \, a c x + c\right )}}{2 \, x^{2}} \] Input:

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

Output:

-1/2*(4*a^2*c*x^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - 7*a^2*c*x^2*log 
((sqrt(-a^2*x^2 + 1) - 1)/x) + sqrt(-a^2*x^2 + 1)*(6*a*c*x + c))/x^2
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.02 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.88 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^3} \, dx=a^{3} c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{\sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\x & \text {otherwise} \end {cases}\right ) + 3 a^{2} c \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) + 3 a c \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} + \frac {a}{2 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{2 a x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

a**3*c*Piecewise((log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/sqrt 
(-a**2), Ne(a**2, 0)), (x, True)) + 3*a**2*c*Piecewise((-acosh(1/(a*x)), 1 
/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) + 3*a*c*Piecewise((-I*sqrt( 
a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) + 
c*Piecewise((-a**2*acosh(1/(a*x))/2 + a/(2*x*sqrt(-1 + 1/(a**2*x**2))) - 1 
/(2*a*x**3*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (I*a**2*asin( 
1/(a*x))/2 - I*a*sqrt(1 - 1/(a**2*x**2))/(2*x), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.53 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^3} \, dx=\frac {3 \, a^{3} c x}{\sqrt {-a^{2} x^{2} + 1}} + a^{2} c \arcsin \left (a x\right ) - \frac {7}{2} \, a^{2} c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {a^{2} c}{2 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {3 \, a c}{\sqrt {-a^{2} x^{2} + 1} x} - \frac {c}{2 \, \sqrt {-a^{2} x^{2} + 1} x^{2}} \] Input:

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

Output:

3*a^3*c*x/sqrt(-a^2*x^2 + 1) + a^2*c*arcsin(a*x) - 7/2*a^2*c*log(2*sqrt(-a 
^2*x^2 + 1)/abs(x) + 2/abs(x)) + 1/2*a^2*c/sqrt(-a^2*x^2 + 1) - 3*a*c/(sqr 
t(-a^2*x^2 + 1)*x) - 1/2*c/(sqrt(-a^2*x^2 + 1)*x^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (66) = 132\).

Time = 0.17 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.36 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^3} \, dx=\frac {a^{3} c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} + \frac {{\left (a^{3} c + \frac {12 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a c}{x}\right )} a^{4} x^{2}}{8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} {\left | a \right |}} - \frac {7 \, a^{3} c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} - \frac {\frac {12 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a c {\left | a \right |}}{x} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c {\left | a \right |}}{a x^{2}}}{8 \, a^{2}} \] Input:

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

Output:

a^3*c*arcsin(a*x)*sgn(a)/abs(a) + 1/8*(a^3*c + 12*(sqrt(-a^2*x^2 + 1)*abs( 
a) + a)*a*c/x)*a^4*x^2/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*abs(a)) - 7/2*a^ 
3*c*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 
 1/8*(12*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a*c*abs(a)/x + (sqrt(-a^2*x^2 + 1 
)*abs(a) + a)^2*c*abs(a)/(a*x^2))/a^2
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^3} \, dx=\frac {a^3\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}-\frac {3\,a\,c\,\sqrt {1-a^2\,x^2}}{x}-\frac {c\,\sqrt {1-a^2\,x^2}}{2\,x^2}+\frac {a^2\,c\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{2} \] Input:

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

Output:

(a^2*c*atan((1 - a^2*x^2)^(1/2)*1i)*7i)/2 - (c*(1 - a^2*x^2)^(1/2))/(2*x^2 
) - (3*a*c*(1 - a^2*x^2)^(1/2))/x + (a^3*c*asinh(x*(-a^2)^(1/2)))/(-a^2)^( 
1/2)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.83 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^3} \, dx=\frac {c \left (2 \mathit {asin} \left (a x \right ) a^{2} x^{2}-6 \sqrt {-a^{2} x^{2}+1}\, a x -\sqrt {-a^{2} x^{2}+1}+7 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{2} x^{2}\right )}{2 x^{2}} \] Input:

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

Output:

(c*(2*asin(a*x)*a**2*x**2 - 6*sqrt( - a**2*x**2 + 1)*a*x - sqrt( - a**2*x* 
*2 + 1) + 7*log(tan(asin(a*x)/2))*a**2*x**2))/(2*x**2)