\(\int e^{3 \text {arctanh}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\) [674]

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

Optimal result

Integrand size = 20, antiderivative size = 73 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c \sqrt {1-a^2 x^2}}{a}+\frac {c \sqrt {1-a^2 x^2}}{a^2 x}-\frac {3 c \arcsin (a x)}{a}+\frac {3 c \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{a} \] Output:

c*(-a^2*x^2+1)^(1/2)/a+c*(-a^2*x^2+1)^(1/2)/a^2/x-3*c*arcsin(a*x)/a+3*c*ar 
ctanh((-a^2*x^2+1)^(1/2))/a
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c \left ((1+a x) \sqrt {1-a^2 x^2}-3 a x \arcsin (a x)+3 a x \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{a^2 x} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {6707, 6698, 540, 25, 2340, 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.

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

\(\Big \downarrow \) 6707

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

\(\Big \downarrow \) 6698

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

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 538

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((c*(-(a*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/x + 3*a*(ArcSin[a*x] - Ar 
cTanh[Sqrt[1 - a^2*x^2]])))/a^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 2340
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -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]
 

rule 6707
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb 
ol] :> Simp[d^p   Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x 
] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
 
Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.32

method result size
risch \(-\frac {\left (a^{2} x^{2}-1\right ) c}{x \sqrt {-a^{2} x^{2}+1}\, a^{2}}-\frac {\left (-\sqrt {-a^{2} x^{2}+1}-3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {3 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}\right ) c}{a}\) \(96\)
default \(\frac {c \left (a^{5} \left (-\frac {x^{2}}{a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {-a^{2} x^{2}+1}}\right )-\frac {4 a^{2} x}{\sqrt {-a^{2} x^{2}+1}}+\frac {1}{x \sqrt {-a^{2} x^{2}+1}}-3 a \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )+\frac {2 a}{\sqrt {-a^{2} x^{2}+1}}+3 a^{4} \left (\frac {x}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}\right )\right )}{a^{2}}\) \(182\)
meijerg \(-\frac {2 c x}{\sqrt {-a^{2} x^{2}+1}}+\frac {c \left (-2 a^{2} x^{2}+1\right )}{a^{2} x \sqrt {-a^{2} x^{2}+1}}+\frac {c \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right )}{4 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {2 c \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {3 c \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {3 c \left (\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{2}-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{a \sqrt {\pi }}\) \(253\)

Input:

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

Output:

-(a^2*x^2-1)/x/(-a^2*x^2+1)^(1/2)*c/a^2-1/a*(-(-a^2*x^2+1)^(1/2)-3*arctanh 
(1/(-a^2*x^2+1)^(1/2))+3*a/(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 = 82, normalized size of antiderivative = 1.12 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {6 \, a c x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - 3 \, a c x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + a c x + \sqrt {-a^{2} x^{2} + 1} {\left (a c x + c\right )}}{a^{2} x} \] Input:

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

Output:

(6*a*c*x*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - 3*a*c*x*log((sqrt(-a^2*x 
^2 + 1) - 1)/x) + a*c*x + sqrt(-a^2*x^2 + 1)*(a*c*x + c))/(a^2*x)
 

Sympy [A] (verification not implemented)

Time = 5.06 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.03 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=- a c \left (\begin {cases} - \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {for}\: a^{2} \neq 0 \\\frac {x^{2}}{2} & \text {otherwise} \end {cases}\right ) - 3 c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{\sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\x & \text {otherwise} \end {cases}\right ) - \frac {3 c \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right )}{a} - \frac {c \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right )}{a^{2}} \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (67) = 134\).

Time = 0.11 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.74 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-a^{3} c {\left (\frac {x^{2}}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {2}{\sqrt {-a^{2} x^{2} + 1} a^{4}}\right )} + 3 \, a^{2} c {\left (\frac {x}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {\arcsin \left (a x\right )}{a^{3}}\right )} - \frac {2 \, c x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {3 \, c {\left (\frac {1}{\sqrt {-a^{2} x^{2} + 1}} - \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )\right )}}{a} - \frac {{\left (\frac {2 \, a^{2} x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} x}\right )} c}{a^{2}} + \frac {2 \, c}{\sqrt {-a^{2} x^{2} + 1} a} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.77 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {a^{2} c x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} - \frac {3 \, c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} + \frac {3 \, c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c}{2 \, a^{2} x {\left | a \right |}} \] Input:

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

Output:

-1/2*a^2*c*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*abs(a)) - 3*c*arcsin(a*x)*sg 
n(a)/abs(a) + 3*c*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs 
(x)))/abs(a) + sqrt(-a^2*x^2 + 1)*c/a + 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a 
)*c/(a^2*x*abs(a))
 

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.11 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c\,\sqrt {1-a^2\,x^2}}{a}-\frac {3\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {c\,\sqrt {1-a^2\,x^2}}{a^2\,x}-\frac {c\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{a} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.12 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c \left (-6 \mathit {asin} \left (a x \right ) a x +2 \sqrt {-a^{2} x^{2}+1}\, a x +2 \sqrt {-a^{2} x^{2}+1}-3 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}-1\right ) a x +3 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}+1\right ) a x \right )}{2 a^{2} x} \] Input:

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

Output:

(c*( - 6*asin(a*x)*a*x + 2*sqrt( - a**2*x**2 + 1)*a*x + 2*sqrt( - a**2*x** 
2 + 1) - 3*log(sqrt( - a**2*x**2 + 1) - 1)*a*x + 3*log(sqrt( - a**2*x**2 + 
 1) + 1)*a*x))/(2*a**2*x)