\(\int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^3} \, dx\) [371]

Optimal result

Integrand size = 19, antiderivative size = 108 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^3} \, dx=\frac {2 \sqrt {1-a^2 x^2}}{5 c^3 (1-a x)^3}+\frac {3 \sqrt {1-a^2 x^2}}{5 c^3 (1-a x)^2}+\frac {8 \sqrt {1-a^2 x^2}}{5 c^3 (1-a x)}-\frac {\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c^3} \] Output:

2/5*(-a^2*x^2+1)^(1/2)/c^3/(-a*x+1)^3+3/5*(-a^2*x^2+1)^(1/2)/c^3/(-a*x+1)^ 
2+8/5*(-a^2*x^2+1)^(1/2)/c^3/(-a*x+1)-arctanh((-a^2*x^2+1)^(1/2))/c^3
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.66 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^3} \, dx=\frac {16+60 a x+5 a^2 x^2-60 a^3 x^3+24 a^5 x^5+3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},1-a^2 x^2\right )}{15 c^3 \left (1-a^2 x^2\right )^{5/2}} \] Input:

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

Output:

(16 + 60*a*x + 5*a^2*x^2 - 60*a^3*x^3 + 24*a^5*x^5 + 3*Hypergeometric2F1[- 
5/2, 1, -3/2, 1 - a^2*x^2])/(15*c^3*(1 - a^2*x^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.82, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {6678, 27, 570, 532, 25, 2336, 27, 532, 27, 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.

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 570

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

\(\Big \downarrow \) 532

\(\displaystyle \frac {\frac {8 (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{5} \int -\frac {-5 a^2 x^2+12 a x+5}{x \left (1-a^2 x^2\right )^{5/2}}dx}{c^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{5} \int \frac {-5 a^2 x^2+12 a x+5}{x \left (1-a^2 x^2\right )^{5/2}}dx+\frac {8 (a x+1)}{5 \left (1-a^2 x^2\right )^{5/2}}}{c^3}\)

\(\Big \downarrow \) 2336

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 532

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Leaf count of result is larger than twice the leaf count of optimal. \(274\) vs. \(2(94)=188\).

Time = 0.25 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.55

method	result	size
default	\(-\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {4 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a^{2}}-\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a}}{c^{3}}\)	\(275\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^3} \, dx=\frac {13 \, a^{3} x^{3} - 39 \, a^{2} x^{2} + 39 \, a x + 5 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (8 \, a^{2} x^{2} - 19 \, a x + 13\right )} \sqrt {-a^{2} x^{2} + 1} - 13}{5 \, {\left (a^{3} c^{3} x^{3} - 3 \, a^{2} c^{3} x^{2} + 3 \, a c^{3} x - c^{3}\right )}} \] Input:

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

Output:

1/5*(13*a^3*x^3 - 39*a^2*x^2 + 39*a*x + 5*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1 
)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (8*a^2*x^2 - 19*a*x + 13)*sqrt(-a^2*x^ 
2 + 1) - 13)/(a^3*c^3*x^3 - 3*a^2*c^3*x^2 + 3*a*c^3*x - c^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (91) = 182\).

Time = 0.13 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.75 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^3} \, dx=-\frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c^{3} {\left | a \right |}} + \frac {2 \, {\left (13 \, a - \frac {45 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a x} + \frac {75 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{3} x^{2}} - \frac {55 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{5} x^{3}} + \frac {20 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{7} x^{4}}\right )}}{5 \, c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \] Input:

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

Output:

-a*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c^3*abs( 
a)) + 2/5*(13*a - 45*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a*x) + 75*(sqrt(-a^2 
*x^2 + 1)*abs(a) + a)^2/(a^3*x^2) - 55*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/( 
a^5*x^3) + 20*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^7*x^4))/(c^3*((sqrt(-a^ 
2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))
 

Mupad [B] (verification not implemented)

Time = 14.08 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.94 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^3} \, dx=\frac {3\,a^2\,\sqrt {1-a^2\,x^2}}{5\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {8\,a\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}+\frac {2\,a\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )}+\frac {\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^3} \] Input:

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

Output:

(atan((1 - a^2*x^2)^(1/2)*1i)*1i)/c^3 + (3*a^2*(1 - a^2*x^2)^(1/2))/(5*(a^ 
2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) + (8*a*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^ 
(1/2)*(c^3*x*(-a^2)^(1/2) - (c^3*(-a^2)^(1/2))/a)) + (2*a*(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 = 0.15 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.15 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^3} \, dx=\frac {-16 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+38 \sqrt {-a^{2} x^{2}+1}\, a x -26 \sqrt {-a^{2} x^{2}+1}+5 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}-1\right ) a^{3} x^{3}-15 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}-1\right ) a^{2} x^{2}+15 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}-1\right ) a x -5 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}-1\right )-5 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}+1\right ) a^{3} x^{3}+15 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}+1\right ) a^{2} x^{2}-15 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}+1\right ) a x +5 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}+1\right )}{10 c^{3} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:

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

Output:

( - 16*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 38*sqrt( - a**2*x**2 + 1)*a*x - 
26*sqrt( - a**2*x**2 + 1) + 5*log(sqrt( - a**2*x**2 + 1) - 1)*a**3*x**3 - 
15*log(sqrt( - a**2*x**2 + 1) - 1)*a**2*x**2 + 15*log(sqrt( - a**2*x**2 + 
1) - 1)*a*x - 5*log(sqrt( - a**2*x**2 + 1) - 1) - 5*log(sqrt( - a**2*x**2 
+ 1) + 1)*a**3*x**3 + 15*log(sqrt( - a**2*x**2 + 1) + 1)*a**2*x**2 - 15*lo 
g(sqrt( - a**2*x**2 + 1) + 1)*a*x + 5*log(sqrt( - a**2*x**2 + 1) + 1))/(10 
*c**3*(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))