\(\int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx\) [335]

Optimal result

Integrand size = 19, antiderivative size = 102 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx=-\frac {5 a^2 c^3 \sqrt {1-a^2 x^2}}{8 x^2}-\frac {c^3 \left (1-a^2 x^2\right )^{3/2}}{4 x^4}+\frac {2 a c^3 \left (1-a^2 x^2\right )^{3/2}}{3 x^3}+\frac {5}{8} a^4 c^3 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx=\frac {c^3 \left (-6+16 a x-3 a^2 x^2-32 a^3 x^3+9 a^4 x^4+16 a^5 x^5+15 a^4 x^4 \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{24 x^4 \sqrt {1-a^2 x^2}} \] Input:

Integrate[(E^ArcTanh[a*x]*(c - a*c*x)^3)/x^5,x]
 

Output:

(c^3*(-6 + 16*a*x - 3*a^2*x^2 - 32*a^3*x^3 + 9*a^4*x^4 + 16*a^5*x^5 + 15*a 
^4*x^4*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]]))/(24*x^4*Sqrt[1 - a^2 
*x^2])
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {6678, 27, 540, 27, 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.

\(\displaystyle \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx\)

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(E^ArcTanh[a*x]*(c - a*c*x)^3)/x^5,x]
 

Output:

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

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {\left (16 a^{5} x^{5}+9 a^{4} x^{4}-32 a^{3} x^{3}-3 a^{2} x^{2}+16 a x -6\right ) c^{3}}{24 x^{4} \sqrt {-a^{2} x^{2}+1}}+\frac {5 a^{4} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right ) c^{3}}{8}\)	\(81\)
default	\(-c^{3} \left (-a^{4} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\sqrt {-a^{2} x^{2}+1}}{4 x^{4}}-\frac {3 a^{2} \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\right )}{4}+2 a \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{3 x^{3}}-\frac {2 a^{2} \sqrt {-a^{2} x^{2}+1}}{3 x}\right )+\frac {2 a^{3} \sqrt {-a^{2} x^{2}+1}}{x}\right )\)	\(144\)
meijerg	\(-\frac {a^{4} c^{3} \left (\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{2 \sqrt {\pi }}-\frac {2 a^{3} c^{3} \sqrt {-a^{2} x^{2}+1}}{x}+\frac {2 a \,c^{3} \left (2 a^{2} x^{2}+1\right ) \sqrt {-a^{2} x^{2}+1}}{3 x^{3}}+\frac {a^{4} c^{3} \left (-\frac {\sqrt {\pi }}{2 x^{4} a^{4}}-\frac {\sqrt {\pi }}{2 x^{2} a^{2}}+\frac {3 \left (\frac {7}{6}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{8}+\frac {\sqrt {\pi }\, \left (-7 a^{4} x^{4}+8 a^{2} x^{2}+8\right )}{16 a^{4} x^{4}}-\frac {\sqrt {\pi }\, \left (12 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{16 a^{4} x^{4}}-\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )}{4}\right )}{2 \sqrt {\pi }}\)	\(248\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^5,x,method=_RETURNVERBOSE)
 

Output:

1/24*(16*a^5*x^5+9*a^4*x^4-32*a^3*x^3-3*a^2*x^2+16*a*x-6)/x^4/(-a^2*x^2+1) 
^(1/2)*c^3+5/8*a^4*arctanh(1/(-a^2*x^2+1)^(1/2))*c^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx=-\frac {15 \, a^{4} c^{3} x^{4} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + {\left (16 \, a^{3} c^{3} x^{3} + 9 \, a^{2} c^{3} x^{2} - 16 \, a c^{3} x + 6 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{24 \, x^{4}} \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^5,x, algorithm="fricas 
")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.08 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.40 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx=- a^{4} c^{3} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) + 2 a^{3} c^{3} \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) - 2 a c^{3} \left (\begin {cases} - \frac {2 i a^{2} \sqrt {a^{2} x^{2} - 1}}{3 x} - \frac {i \sqrt {a^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {2 a^{2} \sqrt {- a^{2} x^{2} + 1}}{3 x} - \frac {\sqrt {- a^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) + c^{3} \left (\begin {cases} - \frac {3 a^{4} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{8} + \frac {3 a^{3}}{8 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {a}{8 x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {3 i a^{4} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{8} - \frac {3 i a^{3}}{8 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i a}{8 x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{4 a x^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)**3/x**5,x)
 

Output:

-a**4*c**3*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a 
*x)), True)) + 2*a**3*c**3*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x 
**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) - 2*a*c**3*Piecewise((-2*I*a** 
2*sqrt(a**2*x**2 - 1)/(3*x) - I*sqrt(a**2*x**2 - 1)/(3*x**3), Abs(a**2*x** 
2) > 1), (-2*a**2*sqrt(-a**2*x**2 + 1)/(3*x) - sqrt(-a**2*x**2 + 1)/(3*x** 
3), True)) + c**3*Piecewise((-3*a**4*acosh(1/(a*x))/8 + 3*a**3/(8*x*sqrt(- 
1 + 1/(a**2*x**2))) - a/(8*x**3*sqrt(-1 + 1/(a**2*x**2))) - 1/(4*a*x**5*sq 
rt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (3*I*a**4*asin(1/(a*x))/8 
- 3*I*a**3/(8*x*sqrt(1 - 1/(a**2*x**2))) + I*a/(8*x**3*sqrt(1 - 1/(a**2*x* 
*2))) + I/(4*a*x**5*sqrt(1 - 1/(a**2*x**2))), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx=\frac {5}{8} \, a^{4} c^{3} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{3} c^{3}}{3 \, x} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{3}}{8 \, x^{2}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a c^{3}}{3 \, x^{3}} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{3}}{4 \, x^{4}} \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^5,x, algorithm="maxima 
")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (86) = 172\).

Time = 0.13 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.94 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx=\frac {{\left (3 \, a^{5} c^{3} - \frac {16 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{3} c^{3}}{x} + \frac {24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a c^{3}}{x^{2}} + \frac {48 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c^{3}}{a x^{3}}\right )} a^{8} x^{4}}{192 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} {\left | a \right |}} + \frac {5 \, a^{5} c^{3} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{8 \, {\left | a \right |}} - \frac {\frac {48 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{5} c^{3} {\left | a \right |}}{x} + \frac {24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{3} c^{3} {\left | a \right |}}{x^{2}} - \frac {16 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} a c^{3} {\left | a \right |}}{x^{3}} + \frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} c^{3} {\left | a \right |}}{a x^{4}}}{192 \, a^{4}} \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^5,x, algorithm="giac")
 

Output:

1/192*(3*a^5*c^3 - 16*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^3*c^3/x + 24*(sqrt 
(-a^2*x^2 + 1)*abs(a) + a)^2*a*c^3/x^2 + 48*(sqrt(-a^2*x^2 + 1)*abs(a) + a 
)^3*c^3/(a*x^3))*a^8*x^4/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*abs(a)) + 5/8* 
a^5*c^3*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs( 
a) - 1/192*(48*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^5*c^3*abs(a)/x + 24*(sqrt 
(-a^2*x^2 + 1)*abs(a) + a)^2*a^3*c^3*abs(a)/x^2 - 16*(sqrt(-a^2*x^2 + 1)*a 
bs(a) + a)^3*a*c^3*abs(a)/x^3 + 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c^3*ab 
s(a)/(a*x^4))/a^4
 

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx=\frac {2\,a\,c^3\,\sqrt {1-a^2\,x^2}}{3\,x^3}-\frac {c^3\,\sqrt {1-a^2\,x^2}}{4\,x^4}-\frac {3\,a^2\,c^3\,\sqrt {1-a^2\,x^2}}{8\,x^2}-\frac {2\,a^3\,c^3\,\sqrt {1-a^2\,x^2}}{3\,x}-\frac {a^4\,c^3\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{8} \] Input:

int(((c - a*c*x)^3*(a*x + 1))/(x^5*(1 - a^2*x^2)^(1/2)),x)
 

Output:

(2*a*c^3*(1 - a^2*x^2)^(1/2))/(3*x^3) - (c^3*(1 - a^2*x^2)^(1/2))/(4*x^4) 
- (a^4*c^3*atan((1 - a^2*x^2)^(1/2)*1i)*5i)/8 - (3*a^2*c^3*(1 - a^2*x^2)^( 
1/2))/(8*x^2) - (2*a^3*c^3*(1 - a^2*x^2)^(1/2))/(3*x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^3}{x^5} \, dx=\frac {c^{3} \left (-16 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-9 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+16 \sqrt {-a^{2} x^{2}+1}\, a x -6 \sqrt {-a^{2} x^{2}+1}-15 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{4} x^{4}\right )}{24 x^{4}} \] Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^3/x^5,x)
 

Output:

(c**3*( - 16*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 9*sqrt( - a**2*x**2 + 1)*a 
**2*x**2 + 16*sqrt( - a**2*x**2 + 1)*a*x - 6*sqrt( - a**2*x**2 + 1) - 15*l 
og(tan(asin(a*x)/2))*a**4*x**4))/(24*x**4)