\(\int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx\) [166]

Optimal result

Integrand size = 16, antiderivative size = 105 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {5}{8} a c^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {2}{3} a^2 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3-\frac {1}{4} a^3 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4+\frac {5 c^3 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a} \] Output:

-5/8*a*c^3*(1-1/a^2/x^2)^(1/2)*x^2+2/3*a^2*c^3*(1-1/a^2/x^2)^(3/2)*x^3-1/4 
*a^3*c^3*(1-1/a^2/x^2)^(3/2)*x^4+5/8*c^3*arctanh((1-1/a^2/x^2)^(1/2))/a
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {c^3 \left (-a \sqrt {1-\frac {1}{a^2 x^2}} x \left (16+9 a x-16 a^2 x^2+6 a^3 x^3\right )+15 \log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )\right )}{24 a} \] Input:

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

Output:

(c^3*(-(a*Sqrt[1 - 1/(a^2*x^2)]*x*(16 + 9*a*x - 16*a^2*x^2 + 6*a^3*x^3)) + 
 15*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x]))/(24*a)
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6724, 27, 540, 534, 243, 51, 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

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

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 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 = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.14

method	result	size
risch	\(-\frac {\left (6 a^{3} x^{3}-16 a^{2} x^{2}+9 a x +16\right ) \left (a x -1\right ) c^{3}}{24 a \sqrt {\frac {a x -1}{a x +1}}}+\frac {5 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) c^{3} \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{8 \sqrt {a^{2}}\, \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\)	\(120\)
default	\(\frac {\left (a x -1\right ) c^{3} \left (-6 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a x +16 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-15 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a x +15 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a \right )}{24 a \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}\)	\(141\)

Input:

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

Output:

-1/24*(6*a^3*x^3-16*a^2*x^2+9*a*x+16)*(a*x-1)/a*c^3/((a*x-1)/(a*x+1))^(1/2 
)+5/8*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))^(1/2)/(a*x+1)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.10 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {15 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 15 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (6 \, a^{4} c^{3} x^{4} - 10 \, a^{3} c^{3} x^{3} - 7 \, a^{2} c^{3} x^{2} + 25 \, a c^{3} x + 16 \, c^{3}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{24 \, a} \] Input:

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

Output:

1/24*(15*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 15*c^3*log(sqrt((a*x - 1 
)/(a*x + 1)) - 1) - (6*a^4*c^3*x^4 - 10*a^3*c^3*x^3 - 7*a^2*c^3*x^2 + 25*a 
*c^3*x + 16*c^3)*sqrt((a*x - 1)/(a*x + 1)))/a
 

Sympy [F]

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

integrate(1/((a*x-1)/(a*x+1))**(1/2)*(-a*c*x+c)**3,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(-1/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. \(2 (89) = 178\).

Time = 0.04 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.10 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {1}{24} \, {\left (\frac {15 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {15 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac {2 \, {\left (15 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} + 73 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 55 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 15 \, c^{3} \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {4 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {6 \, {\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, {\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} - a^{2}}\right )} a \] Input:

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

Output:

1/24*(15*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 15*c^3*log(sqrt((a*x 
 - 1)/(a*x + 1)) - 1)/a^2 + 2*(15*c^3*((a*x - 1)/(a*x + 1))^(7/2) + 73*c^3 
*((a*x - 1)/(a*x + 1))^(5/2) - 55*c^3*((a*x - 1)/(a*x + 1))^(3/2) + 15*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 = 0.13 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.12 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {5 \, c^{3} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right )}{8 \, {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} - \frac {1}{24} \, \sqrt {a^{2} x^{2} - 1} {\left ({\left (2 \, {\left (\frac {3 \, a^{2} c^{3} x}{\mathrm {sgn}\left (a x + 1\right )} - \frac {8 \, a c^{3}}{\mathrm {sgn}\left (a x + 1\right )}\right )} x + \frac {9 \, c^{3}}{\mathrm {sgn}\left (a x + 1\right )}\right )} x + \frac {16 \, c^{3}}{a \mathrm {sgn}\left (a x + 1\right )}\right )} \] Input:

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

Output:

-5/8*c^3*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))/(abs(a)*sgn(a*x + 1)) - 1 
/24*sqrt(a^2*x^2 - 1)*((2*(3*a^2*c^3*x/sgn(a*x + 1) - 8*a*c^3/sgn(a*x + 1) 
)*x + 9*c^3/sgn(a*x + 1))*x + 16*c^3/(a*sgn(a*x + 1)))
 

Mupad [B] (verification not implemented)

Time = 13.76 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.69 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {5\,c^3\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4\,a}-\frac {\frac {5\,c^3\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{4}-\frac {55\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{12}+\frac {73\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{12}+\frac {5\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{4}}{a-\frac {4\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {6\,a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {4\,a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {a\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}} \] Input:

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

Output:

(5*c^3*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/(4*a) - ((5*c^3*((a*x - 1)/(a*x 
 + 1))^(1/2))/4 - (55*c^3*((a*x - 1)/(a*x + 1))^(3/2))/12 + (73*c^3*((a*x 
- 1)/(a*x + 1))^(5/2))/12 + (5*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)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.95 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {c^{3} \left (-6 \sqrt {a x +1}\, \sqrt {a x -1}\, a^{3} x^{3}+16 \sqrt {a x +1}\, \sqrt {a x -1}\, a^{2} x^{2}-9 \sqrt {a x +1}\, \sqrt {a x -1}\, a x -16 \sqrt {a x +1}\, \sqrt {a x -1}+30 \,\mathrm {log}\left (\frac {\sqrt {a x -1}+\sqrt {a x +1}}{\sqrt {2}}\right )\right )}{24 a} \] Input:

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

Output:

(c**3*( - 6*sqrt(a*x + 1)*sqrt(a*x - 1)*a**3*x**3 + 16*sqrt(a*x + 1)*sqrt( 
a*x - 1)*a**2*x**2 - 9*sqrt(a*x + 1)*sqrt(a*x - 1)*a*x - 16*sqrt(a*x + 1)* 
sqrt(a*x - 1) + 30*log((sqrt(a*x - 1) + sqrt(a*x + 1))/sqrt(2))))/(24*a)