\(\int e^{-\coth ^{-1}(a x)} (c-a c x)^3 \, dx\) [201]

Optimal result

Integrand size = 18, antiderivative size = 127 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {20}{3} c^3 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {27}{8} a c^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {4}{3} a^2 c^3 \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {1}{4} a^3 c^3 \sqrt {1-\frac {1}{a^2 x^2}} x^4-\frac {35 c^3 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a} \] Output:

20/3*c^3*(1-1/a^2/x^2)^(1/2)*x-27/8*a*c^3*(1-1/a^2/x^2)^(1/2)*x^2+4/3*a^2* 
c^3*(1-1/a^2/x^2)^(1/2)*x^3-1/4*a^3*c^3*(1-1/a^2/x^2)^(1/2)*x^4-35/8*c^3*a 
rctanh((1-1/a^2/x^2)^(1/2))/a
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.57 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {c^3 \left (a \sqrt {1-\frac {1}{a^2 x^2}} x \left (160-81 a x+32 a^2 x^2-6 a^3 x^3\right )-105 \log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )\right )}{24 a} \] Input:

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

Output:

(c^3*(a*Sqrt[1 - 1/(a^2*x^2)]*x*(160 - 81*a*x + 32*a^2*x^2 - 6*a^3*x^3) - 
105*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x]))/(24*a)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6724, 27, 540, 2338, 2338, 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.

\(\displaystyle \int (c-a c x)^3 e^{-\coth ^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 6724

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 540

\(\displaystyle \frac {c^3 \left (-\frac {1}{4} \int \frac {\left (16 a^3-\frac {27 a^2}{x}+\frac {16 a}{x^2}-\frac {4}{x^3}\right ) x^4}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {1}{4} a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {c^3 \left (\frac {1}{4} \left (\frac {1}{3} \int \frac {\left (81 a^2-\frac {80 a}{x}+\frac {12}{x^2}\right ) x^3}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\frac {16}{3} a^3 x^3 \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{4} a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {c^3 \left (\frac {1}{4} \left (\frac {1}{3} \left (-\frac {1}{2} \int \frac {5 \left (32 a-\frac {21}{x}\right ) x^2}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {81}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {16}{3} a^3 x^3 \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{4} a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {c^3 \left (\frac {1}{4} \left (\frac {1}{3} \left (-\frac {5}{2} \int \frac {\left (32 a-\frac {21}{x}\right ) x^2}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {81}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {16}{3} a^3 x^3 \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{4} a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {c^3 \left (\frac {1}{4} \left (\frac {1}{3} \left (-\frac {5}{2} \left (-21 \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-32 a x \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {81}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {16}{3} a^3 x^3 \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{4} a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {c^3 \left (\frac {1}{4} \left (\frac {1}{3} \left (-\frac {5}{2} \left (-\frac {21}{2} \int \frac {x}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x^2}-32 a x \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {81}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {16}{3} a^3 x^3 \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{4} a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {c^3 \left (\frac {1}{4} \left (\frac {1}{3} \left (-\frac {5}{2} \left (21 a^2 \int \frac {1}{a^2-a^2 \sqrt {1-\frac {1}{a^2 x^2}}}d\sqrt {1-\frac {1}{a^2 x^2}}-32 a x \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {81}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {16}{3} a^3 x^3 \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{4} a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {c^3 \left (\frac {1}{4} \left (\frac {1}{3} \left (-\frac {5}{2} \left (21 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )-32 a x \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {81}{2} a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )+\frac {16}{3} a^3 x^3 \sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {1}{4} a^4 x^4 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\)

Input:

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

Output:

(c^3*(-1/4*(a^4*Sqrt[1 - 1/(a^2*x^2)]*x^4) + ((16*a^3*Sqrt[1 - 1/(a^2*x^2) 
]*x^3)/3 + ((-81*a^2*Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 - (5*(-32*a*Sqrt[1 - 1/( 
a^2*x^2)]*x + 21*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]))/2)/3)/4))/a
 

Defintions of rubi rules used

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 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 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 6724

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S 
imp[-d^n   Subst[Int[(d + c*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(p + 2)), x], 
 x, 1/x], x] /; FreeQ[{a, c, d}, x] && EqQ[a*c + d, 0] && IntegerQ[p] && In 
tegerQ[n]
 

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {\left (6 a^{3} x^{3}-32 a^{2} x^{2}+81 a x -160\right ) \left (a x +1\right ) c^{3} \sqrt {\frac {a x -1}{a x +1}}}{24 a}-\frac {35 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) c^{3} \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{8 \sqrt {a^{2}}\, \left (a x -1\right )}\)	\(120\)
default	\(\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) c^{3} \left (-6 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a x +32 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-87 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a x +192 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}+87 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -192 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right )}{24 a \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}\)	\(196\)

Input:

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

Output:

-1/24*(6*a^3*x^3-32*a^2*x^2+81*a*x-160)*(a*x+1)/a*c^3*((a*x-1)/(a*x+1))^(1 
/2)-35/8*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2-1)^(1/2))/(a^2)^(1/2)*c^3*((a*x-1)/ 
(a*x+1))^(1/2)/(a*x-1)*((a*x-1)*(a*x+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {105 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 105 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (6 \, a^{4} c^{3} x^{4} - 26 \, a^{3} c^{3} x^{3} + 49 \, a^{2} c^{3} x^{2} - 79 \, a c^{3} x - 160 \, c^{3}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{24 \, a} \] Input:

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

Output:

-1/24*(105*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 105*c^3*log(sqrt((a*x 
- 1)/(a*x + 1)) - 1) + (6*a^4*c^3*x^4 - 26*a^3*c^3*x^3 + 49*a^2*c^3*x^2 - 
79*a*c^3*x - 160*c^3)*sqrt((a*x - 1)/(a*x + 1)))/a
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (107) = 214\).

Time = 0.04 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.74 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {1}{24} \, {\left (\frac {105 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {105 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {2 \, {\left (279 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 511 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 385 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 105 \, c^{3} \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {4 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {6 \, {\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, {\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} - a^{2}}\right )} a \] Input:

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

Output:

-1/24*(105*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 105*c^3*log(sqrt(( 
a*x - 1)/(a*x + 1)) - 1)/a^2 - 2*(279*c^3*((a*x - 1)/(a*x + 1))^(7/2) - 51 
1*c^3*((a*x - 1)/(a*x + 1))^(5/2) + 385*c^3*((a*x - 1)/(a*x + 1))^(3/2) - 
105*c^3*sqrt((a*x - 1)/(a*x + 1)))/(4*(a*x - 1)*a^2/(a*x + 1) - 6*(a*x - 1 
)^2*a^2/(a*x + 1)^2 + 4*(a*x - 1)^3*a^2/(a*x + 1)^3 - (a*x - 1)^4*a^2/(a*x 
 + 1)^4 - a^2))*a
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.86 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {35 \, c^{3} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{8 \, {\left | a \right |}} + \frac {1}{24} \, \sqrt {a^{2} x^{2} - 1} {\left (\frac {160 \, c^{3} \mathrm {sgn}\left (a x + 1\right )}{a} - {\left (81 \, c^{3} \mathrm {sgn}\left (a x + 1\right ) + 2 \, {\left (3 \, a^{2} c^{3} x \mathrm {sgn}\left (a x + 1\right ) - 16 \, a c^{3} \mathrm {sgn}\left (a x + 1\right )\right )} x\right )} x\right )} \] Input:

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

Output:

35/8*c^3*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sgn(a*x + 1)/abs(a) + 1/2 
4*sqrt(a^2*x^2 - 1)*(160*c^3*sgn(a*x + 1)/a - (81*c^3*sgn(a*x + 1) + 2*(3* 
a^2*c^3*x*sgn(a*x + 1) - 16*a*c^3*sgn(a*x + 1))*x)*x)
 

Mupad [B] (verification not implemented)

Time = 13.56 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.39 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {\frac {35\,c^3\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{4}-\frac {385\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{12}+\frac {511\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{12}-\frac {93\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{4}}{a-\frac {4\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {6\,a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {4\,a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {a\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}}-\frac {35\,c^3\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4\,a} \] Input:

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

Output:

((35*c^3*((a*x - 1)/(a*x + 1))^(1/2))/4 - (385*c^3*((a*x - 1)/(a*x + 1))^( 
3/2))/12 + (511*c^3*((a*x - 1)/(a*x + 1))^(5/2))/12 - (93*c^3*((a*x - 1)/( 
a*x + 1))^(7/2))/4)/(a - (4*a*(a*x - 1))/(a*x + 1) + (6*a*(a*x - 1)^2)/(a* 
x + 1)^2 - (4*a*(a*x - 1)^3)/(a*x + 1)^3 + (a*(a*x - 1)^4)/(a*x + 1)^4) - 
(35*c^3*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/(4*a)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.79 \[ \int e^{-\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {c^{3} \left (-6 \sqrt {a x +1}\, \sqrt {a x -1}\, a^{3} x^{3}+32 \sqrt {a x +1}\, \sqrt {a x -1}\, a^{2} x^{2}-81 \sqrt {a x +1}\, \sqrt {a x -1}\, a x +160 \sqrt {a x +1}\, \sqrt {a x -1}-210 \,\mathrm {log}\left (\frac {\sqrt {a x -1}+\sqrt {a x +1}}{\sqrt {2}}\right )\right )}{24 a} \] Input:

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

Output:

(c**3*( - 6*sqrt(a*x + 1)*sqrt(a*x - 1)*a**3*x**3 + 32*sqrt(a*x + 1)*sqrt( 
a*x - 1)*a**2*x**2 - 81*sqrt(a*x + 1)*sqrt(a*x - 1)*a*x + 160*sqrt(a*x + 1 
)*sqrt(a*x - 1) - 210*log((sqrt(a*x - 1) + sqrt(a*x + 1))/sqrt(2))))/(24*a 
)