3.4.21 \(\int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx\) [321]

3.4.21.1 Optimal result
3.4.21.2 Mathematica [A] (verified)
3.4.21.3 Rubi [A] (verified)
3.4.21.4 Maple [A] (verified)
3.4.21.5 Fricas [A] (verification not implemented)
3.4.21.6 Sympy [C] (verification not implemented)
3.4.21.7 Maxima [A] (verification not implemented)
3.4.21.8 Giac [A] (verification not implemented)
3.4.21.9 Mupad [B] (verification not implemented)

3.4.21.1 Optimal result

Integrand size = 19, antiderivative size = 106 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx=-\frac {1}{2} a c^4 (6-a x) \sqrt {1-a^2 x^2}+\frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {1}{2} a c^4 \arcsin (a x)+3 a c^4 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \]

output
1/3*a*c^4*(-a^2*x^2+1)^(3/2)-c^4*(-a^2*x^2+1)^(3/2)/x+1/2*a*c^4*arcsin(a*x 
)+3*a*c^4*arctanh((-a^2*x^2+1)^(1/2))-1/2*a*c^4*(-a*x+6)*(-a^2*x^2+1)^(1/2 
)
 
3.4.21.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx=-\frac {c^4 \left (6+16 a x-15 a^2 x^2-14 a^3 x^3+9 a^4 x^4-2 a^5 x^5+9 a x \sqrt {1-a^2 x^2} \arcsin (a x)+24 a x \sqrt {1-a^2 x^2} \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )-18 a x \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{6 x \sqrt {1-a^2 x^2}} \]

input
Integrate[(E^ArcTanh[a*x]*(c - a*c*x)^4)/x^2,x]
 
output
-1/6*(c^4*(6 + 16*a*x - 15*a^2*x^2 - 14*a^3*x^3 + 9*a^4*x^4 - 2*a^5*x^5 + 
9*a*x*Sqrt[1 - a^2*x^2]*ArcSin[a*x] + 24*a*x*Sqrt[1 - a^2*x^2]*ArcSin[Sqrt 
[1 - a*x]/Sqrt[2]] - 18*a*x*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])) 
/(x*Sqrt[1 - a^2*x^2])
 
3.4.21.3 Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {6678, 27, 540, 2340, 27, 535, 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.

\(\displaystyle \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx\)

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 535

\(\displaystyle c^4 \left (-a \left (\frac {1}{2} \int \frac {6-a x}{x \sqrt {1-a^2 x^2}}dx+\frac {1}{2} \sqrt {1-a^2 x^2} (6-a x)\right )+\frac {1}{3} a \left (1-a^2 x^2\right )^{3/2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{x}\right )\)

\(\Big \downarrow \) 538

\(\displaystyle c^4 \left (-a \left (\frac {1}{2} \left (6 \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-a \int \frac {1}{\sqrt {1-a^2 x^2}}dx\right )+\frac {1}{2} \sqrt {1-a^2 x^2} (6-a x)\right )+\frac {1}{3} a \left (1-a^2 x^2\right )^{3/2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{x}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle c^4 \left (-a \left (\frac {1}{2} \left (6 \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\arcsin (a x)\right )+\frac {1}{2} \sqrt {1-a^2 x^2} (6-a x)\right )+\frac {1}{3} a \left (1-a^2 x^2\right )^{3/2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{x}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle c^4 \left (-a \left (\frac {1}{2} \left (3 \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\arcsin (a x)\right )+\frac {1}{2} \sqrt {1-a^2 x^2} (6-a x)\right )+\frac {1}{3} a \left (1-a^2 x^2\right )^{3/2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{x}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c^4 \left (-a \left (\frac {1}{2} \left (-\frac {6 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}-\arcsin (a x)\right )+\frac {1}{2} \sqrt {1-a^2 x^2} (6-a x)\right )+\frac {1}{3} a \left (1-a^2 x^2\right )^{3/2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{x}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c^4 \left (-a \left (\frac {1}{2} \left (-6 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\arcsin (a x)\right )+\frac {1}{2} \sqrt {1-a^2 x^2} (6-a x)\right )+\frac {1}{3} a \left (1-a^2 x^2\right )^{3/2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{x}\right )\)

input
Int[(E^ArcTanh[a*x]*(c - a*c*x)^4)/x^2,x]
 
output
c^4*((a*(1 - a^2*x^2)^(3/2))/3 - (1 - a^2*x^2)^(3/2)/x - a*(((6 - a*x)*Sqr 
t[1 - a^2*x^2])/2 + (-ArcSin[a*x] - 6*ArcTanh[Sqrt[1 - a^2*x^2]])/2))
 

3.4.21.3.1 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 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 535
Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_), x_Symbol] :> Sim 
p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p 
 + 1)   Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free 
Q[{a, b, c, d}, x] && GtQ[p, 0] && IntegerQ[2*p]
 

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 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]
 
3.4.21.4 Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.26

method result size
risch \(\frac {\left (a^{2} x^{2}-1\right ) c^{4}}{x \sqrt {-a^{2} x^{2}+1}}+\left (\frac {a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}-\frac {\sqrt {-a^{2} x^{2}+1}\, a^{3} x^{2}}{3}-\frac {8 \sqrt {-a^{2} x^{2}+1}\, a}{3}+3 a \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {3 \sqrt {-a^{2} x^{2}+1}\, a^{2} x}{2}\right ) c^{4}\) \(134\)
default \(c^{4} \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{x}+a^{5} \left (-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}\right )+\frac {2 a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-3 a^{4} \left (-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\right )-2 \sqrt {-a^{2} x^{2}+1}\, a +3 a \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )\) \(180\)
meijerg \(\frac {a \,c^{4} \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}\right )}{2 \sqrt {\pi }}+\frac {3 a^{2} c^{4} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}} \sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{2 \sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {a \,c^{4} \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{\sqrt {\pi }}+2 a \,c^{4} \arcsin \left (a x \right )-\frac {3 a \,c^{4} \left (-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )+\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }\right )}{2 \sqrt {\pi }}-\frac {c^{4} \sqrt {-a^{2} x^{2}+1}}{x}\) \(222\)

input
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^2,x,method=_RETURNVERBOSE)
 
output
(a^2*x^2-1)/x/(-a^2*x^2+1)^(1/2)*c^4+(1/2*a^2/(a^2)^(1/2)*arctan((a^2)^(1/ 
2)*x/(-a^2*x^2+1)^(1/2))-1/3*(-a^2*x^2+1)^(1/2)*a^3*x^2-8/3*(-a^2*x^2+1)^( 
1/2)*a+3*a*arctanh(1/(-a^2*x^2+1)^(1/2))+3/2*(-a^2*x^2+1)^(1/2)*a^2*x)*c^4
 
3.4.21.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx=-\frac {6 \, a c^{4} x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + 18 \, a c^{4} x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + 16 \, a c^{4} x + {\left (2 \, a^{3} c^{4} x^{3} - 9 \, a^{2} c^{4} x^{2} + 16 \, a c^{4} x + 6 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, x} \]

input
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^2,x, algorithm="fricas 
")
 
output
-1/6*(6*a*c^4*x*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + 18*a*c^4*x*log((s 
qrt(-a^2*x^2 + 1) - 1)/x) + 16*a*c^4*x + (2*a^3*c^4*x^3 - 9*a^2*c^4*x^2 + 
16*a*c^4*x + 6*c^4)*sqrt(-a^2*x^2 + 1))/x
 
3.4.21.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.02 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.73 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx=a^{5} c^{4} \left (\begin {cases} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{3 a^{2}} - \frac {2 \sqrt {- a^{2} x^{2} + 1}}{3 a^{4}} & \text {for}\: a^{2} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right ) - 3 a^{4} c^{4} \left (\begin {cases} - \frac {x \sqrt {- a^{2} x^{2} + 1}}{2 a^{2}} + \frac {\log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{2 a^{2} \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {x^{3}}{3} & \text {otherwise} \end {cases}\right ) + 2 a^{3} c^{4} \left (\begin {cases} - \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {for}\: a^{2} \neq 0 \\\frac {x^{2}}{2} & \text {otherwise} \end {cases}\right ) + 2 a^{2} c^{4} \left (\begin {cases} \frac {\log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{\sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\x & \text {otherwise} \end {cases}\right ) - 3 a c^{4} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) + c^{4} \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) \]

input
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)**4/x**2,x)
 
output
a**5*c**4*Piecewise((-x**2*sqrt(-a**2*x**2 + 1)/(3*a**2) - 2*sqrt(-a**2*x* 
*2 + 1)/(3*a**4), Ne(a**2, 0)), (x**4/4, True)) - 3*a**4*c**4*Piecewise((- 
x*sqrt(-a**2*x**2 + 1)/(2*a**2) + log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2 
*x**2 + 1))/(2*a**2*sqrt(-a**2)), Ne(a**2, 0)), (x**3/3, True)) + 2*a**3*c 
**4*Piecewise((-sqrt(-a**2*x**2 + 1)/a**2, Ne(a**2, 0)), (x**2/2, True)) + 
 2*a**2*c**4*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*c**4*Piecewise((-acosh(1/(a* 
x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) + c**4*Piecewise((-I* 
sqrt(a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True 
))
 
3.4.21.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx=-\frac {1}{3} \, \sqrt {-a^{2} x^{2} + 1} a^{3} c^{4} x^{2} + \frac {3}{2} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{4} x + \frac {1}{2} \, a c^{4} \arcsin \left (a x\right ) + 3 \, a c^{4} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {8}{3} \, \sqrt {-a^{2} x^{2} + 1} a c^{4} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{4}}{x} \]

input
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^2,x, algorithm="maxima 
")
 
output
-1/3*sqrt(-a^2*x^2 + 1)*a^3*c^4*x^2 + 3/2*sqrt(-a^2*x^2 + 1)*a^2*c^4*x + 1 
/2*a*c^4*arcsin(a*x) + 3*a*c^4*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) 
 - 8/3*sqrt(-a^2*x^2 + 1)*a*c^4 - sqrt(-a^2*x^2 + 1)*c^4/x
 
3.4.21.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.55 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx=\frac {a^{4} c^{4} x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} + \frac {a^{2} c^{4} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{2 \, {\left | a \right |}} + \frac {3 \, a^{2} c^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} - \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{4}}{2 \, x {\left | a \right |}} - \frac {1}{6} \, {\left (16 \, a c^{4} + {\left (2 \, a^{3} c^{4} x - 9 \, a^{2} c^{4}\right )} x\right )} \sqrt {-a^{2} x^{2} + 1} \]

input
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^2,x, algorithm="giac")
 
output
1/2*a^4*c^4*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*abs(a)) + 1/2*a^2*c^4*arcsi 
n(a*x)*sgn(a)/abs(a) + 3*a^2*c^4*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) 
- 2*a)/(a^2*abs(x)))/abs(a) - 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^4/(x*a 
bs(a)) - 1/6*(16*a*c^4 + (2*a^3*c^4*x - 9*a^2*c^4)*x)*sqrt(-a^2*x^2 + 1)
 
3.4.21.9 Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^2} \, dx=\frac {3\,a^2\,c^4\,x\,\sqrt {1-a^2\,x^2}}{2}-\frac {c^4\,\sqrt {1-a^2\,x^2}}{x}-\frac {8\,a\,c^4\,\sqrt {1-a^2\,x^2}}{3}+\frac {a^2\,c^4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}}-\frac {a^3\,c^4\,x^2\,\sqrt {1-a^2\,x^2}}{3}-a\,c^4\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,3{}\mathrm {i} \]

input
int(((c - a*c*x)^4*(a*x + 1))/(x^2*(1 - a^2*x^2)^(1/2)),x)
 
output
(3*a^2*c^4*x*(1 - a^2*x^2)^(1/2))/2 - (c^4*(1 - a^2*x^2)^(1/2))/x - a*c^4* 
atan((1 - a^2*x^2)^(1/2)*1i)*3i - (8*a*c^4*(1 - a^2*x^2)^(1/2))/3 + (a^2*c 
^4*asinh(x*(-a^2)^(1/2)))/(2*(-a^2)^(1/2)) - (a^3*c^4*x^2*(1 - a^2*x^2)^(1 
/2))/3