\(\int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx\) [343]

Optimal result

Integrand size = 19, antiderivative size = 136 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx=\frac {5}{2} a^2 c^4 \sqrt {1-a^2 x^2}+\frac {5}{2} a^3 c^4 x \sqrt {1-a^2 x^2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{2 x^2}+\frac {3 a c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {5}{2} a^2 c^4 \arcsin (a x)-\frac {5}{2} a^2 c^4 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx=\frac {1}{4} c^4 \left (\frac {2 (1+a x)^2 \left (-1+8 a x-8 a^2 x^2+a^3 x^3\right )}{x^2 \sqrt {1-a^2 x^2}}+5 a^2 \arcsin (a x)-10 a^2 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )-10 a^2 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right ) \] Input:

Integrate[(E^ArcTanh[a*x]*(c - a*c*x)^4)/x^3,x]
 

Output:

(c^4*((2*(1 + a*x)^2*(-1 + 8*a*x - 8*a^2*x^2 + a^3*x^3))/(x^2*Sqrt[1 - a^2 
*x^2]) + 5*a^2*ArcSin[a*x] - 10*a^2*ArcSin[Sqrt[1 - a*x]/Sqrt[2]] - 10*a^2 
*ArcTanh[Sqrt[1 - a^2*x^2]]))/4
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.72, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {6678, 27, 540, 2338, 27, 535, 27, 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.

\(\displaystyle \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx\)

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 540

\(\displaystyle c^4 \left (-\frac {1}{2} \int \frac {\sqrt {1-a^2 x^2} \left (2 x^2 a^3-5 x a^2+6 a\right )}{x^2}dx-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 2338

\(\displaystyle c^4 \left (\frac {1}{2} \left (\int \frac {5 a^2 (2 a x+1) \sqrt {1-a^2 x^2}}{x}dx+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c^4 \left (\frac {1}{2} \left (5 a^2 \int \frac {(2 a x+1) \sqrt {1-a^2 x^2}}{x}dx+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 535

\(\displaystyle c^4 \left (\frac {1}{2} \left (5 a^2 \left (\frac {1}{2} \int \frac {2 (a x+1)}{x \sqrt {1-a^2 x^2}}dx+\sqrt {1-a^2 x^2} (a x+1)\right )+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c^4 \left (\frac {1}{2} \left (5 a^2 \left (\int \frac {a x+1}{x \sqrt {1-a^2 x^2}}dx+\sqrt {1-a^2 x^2} (a x+1)\right )+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 538

\(\displaystyle c^4 \left (\frac {1}{2} \left (5 a^2 \left (a \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\int \frac {1}{x \sqrt {1-a^2 x^2}}dx+\sqrt {1-a^2 x^2} (a x+1)\right )+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle c^4 \left (\frac {1}{2} \left (5 a^2 \left (\int \frac {1}{x \sqrt {1-a^2 x^2}}dx+\sqrt {1-a^2 x^2} (a x+1)+\arcsin (a x)\right )+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle c^4 \left (\frac {1}{2} \left (5 a^2 \left (\frac {1}{2} \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2+\sqrt {1-a^2 x^2} (a x+1)+\arcsin (a x)\right )+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c^4 \left (\frac {1}{2} \left (5 a^2 \left (-\frac {\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}+\sqrt {1-a^2 x^2} (a x+1)+\arcsin (a x)\right )+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c^4 \left (\frac {1}{2} \left (5 a^2 \left (-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )+\sqrt {1-a^2 x^2} (a x+1)+\arcsin (a x)\right )+\frac {6 a \left (1-a^2 x^2\right )^{3/2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{2 x^2}\right )\)

Input:

Int[(E^ArcTanh[a*x]*(c - a*c*x)^4)/x^3,x]
 

Output:

c^4*(-1/2*(1 - a^2*x^2)^(3/2)/x^2 + ((6*a*(1 - a^2*x^2)^(3/2))/x + 5*a^2*( 
(1 + a*x)*Sqrt[1 - a^2*x^2] + ArcSin[a*x] - ArcTanh[Sqrt[1 - a^2*x^2]]))/2 
)
 

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 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 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 = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {\left (6 a^{3} x^{3}-a^{2} x^{2}-6 a x +1\right ) c^{4}}{2 x^{2} \sqrt {-a^{2} x^{2}+1}}+\left (-\frac {5 a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {5 a^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}+3 a^{2} \sqrt {-a^{2} x^{2}+1}-\frac {a^{3} x \sqrt {-a^{2} x^{2}+1}}{2}\right ) c^{4}\)	\(132\)
default	\(c^{4} \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {5 a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+a^{5} \left (-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\right )+\frac {2 a^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+\frac {3 a \sqrt {-a^{2} x^{2}+1}}{x}+3 a^{2} \sqrt {-a^{2} x^{2}+1}\right )\)	\(159\)
meijerg	\(-\frac {a^{3} c^{4} \left (-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}} \sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{2 \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {3 a^{2} c^{4} \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{2 \sqrt {\pi }}+2 a^{2} c^{4} \arcsin \left (a x \right )+\frac {a^{2} c^{4} \left (\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{\sqrt {\pi }}+\frac {3 a \,c^{4} \sqrt {-a^{2} x^{2}+1}}{x}-\frac {a^{2} c^{4} \left (\frac {\sqrt {\pi }}{x^{2} a^{2}}-\frac {\left (1-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{2}-\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right )}{8 a^{2} x^{2}}+\frac {\sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}}{a^{2} x^{2}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{2 \sqrt {\pi }}\)	\(293\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^3,x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-1/2*(6*a^3*x^3-a^2*x^2-6*a*x+1)/x^2/(-a^2*x^2+1)^(1/2)*c^4+(-5/2*a^2*arct 
anh(1/(-a^2*x^2+1)^(1/2))+5/2*a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x 
^2+1)^(1/2))+3*a^2*(-a^2*x^2+1)^(1/2)-1/2*a^3*x*(-a^2*x^2+1)^(1/2))*c^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx=-\frac {10 \, a^{2} c^{4} x^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - 5 \, a^{2} c^{4} x^{2} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - 6 \, a^{2} c^{4} x^{2} + {\left (a^{3} c^{4} x^{3} - 6 \, a^{2} c^{4} x^{2} - 6 \, a c^{4} x + c^{4}\right )} \sqrt {-a^{2} x^{2} + 1}}{2 \, x^{2}} \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^3,x, algorithm="fricas 
")
 

Output:

-1/2*(10*a^2*c^4*x^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - 5*a^2*c^4*x^ 
2*log((sqrt(-a^2*x^2 + 1) - 1)/x) - 6*a^2*c^4*x^2 + (a^3*c^4*x^3 - 6*a^2*c 
^4*x^2 - 6*a*c^4*x + c^4)*sqrt(-a^2*x^2 + 1))/x^2
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.89 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.48 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx=a^{5} c^{4} \left (\begin {cases} - \frac {x \sqrt {- a^{2} x^{2} + 1}}{2 a^{2}} + \frac {\log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{2 a^{2} \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {x^{3}}{3} & \text {otherwise} \end {cases}\right ) - 3 a^{4} c^{4} \left (\begin {cases} - \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {for}\: a^{2} \neq 0 \\\frac {x^{2}}{2} & \text {otherwise} \end {cases}\right ) + 2 a^{3} c^{4} \left (\begin {cases} \frac {\log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{\sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\x & \text {otherwise} \end {cases}\right ) + 2 a^{2} c^{4} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) - 3 a c^{4} \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) + c^{4} \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} + \frac {a}{2 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{2 a x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)**4/x**3,x)
 

Output:

a**5*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)) - 3*a**4*c**4*Piecewise((-sqrt(-a**2*x**2 + 1)/a**2, Ne(a**2, 
0)), (x**2/2, True)) + 2*a**3*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)) + 2*a**2*c** 
4*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), Tru 
e)) - 3*a*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)) + c**4*Piecewise((-a**2*acosh(1/(a*x))/2 + 
 a/(2*x*sqrt(-1 + 1/(a**2*x**2))) - 1/(2*a*x**3*sqrt(-1 + 1/(a**2*x**2))), 
 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/(a*x))/2 - I*a*sqrt(1 - 1/(a**2*x** 
2))/(2*x), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx=-\frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} a^{3} c^{4} x + \frac {5}{2} \, a^{2} c^{4} \arcsin \left (a x\right ) - \frac {5}{2} \, a^{2} c^{4} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + 3 \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{4} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} a c^{4}}{x} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{4}}{2 \, x^{2}} \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^3,x, algorithm="maxima 
")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx=\frac {5 \, a^{3} c^{4} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{2 \, {\left | a \right |}} - \frac {5 \, a^{3} c^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} + \frac {{\left (a^{3} c^{4} - \frac {12 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a c^{4}}{x}\right )} a^{4} x^{2}}{8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} {\left | a \right |}} - \frac {1}{2} \, {\left (a^{3} c^{4} x - 6 \, a^{2} c^{4}\right )} \sqrt {-a^{2} x^{2} + 1} + \frac {\frac {12 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a c^{4} {\left | a \right |}}{x} - \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{4} {\left | a \right |}}{a x^{2}}}{8 \, a^{2}} \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^3,x, algorithm="giac")
 

Output:

5/2*a^3*c^4*arcsin(a*x)*sgn(a)/abs(a) - 5/2*a^3*c^4*log(1/2*abs(-2*sqrt(-a 
^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + 1/8*(a^3*c^4 - 12*(sqrt(- 
a^2*x^2 + 1)*abs(a) + a)*a*c^4/x)*a^4*x^2/((sqrt(-a^2*x^2 + 1)*abs(a) + a) 
^2*abs(a)) - 1/2*(a^3*c^4*x - 6*a^2*c^4)*sqrt(-a^2*x^2 + 1) + 1/8*(12*(sqr 
t(-a^2*x^2 + 1)*abs(a) + a)*a*c^4*abs(a)/x - (sqrt(-a^2*x^2 + 1)*abs(a) + 
a)^2*c^4*abs(a)/(a*x^2))/a^2
 

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx=3\,a^2\,c^4\,\sqrt {1-a^2\,x^2}-\frac {c^4\,\sqrt {1-a^2\,x^2}}{2\,x^2}+\frac {3\,a\,c^4\,\sqrt {1-a^2\,x^2}}{x}-\frac {a^3\,c^4\,x\,\sqrt {1-a^2\,x^2}}{2}+\frac {5\,a^3\,c^4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}}+\frac {a^2\,c^4\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{2} \] Input:

int(((c - a*c*x)^4*(a*x + 1))/(x^3*(1 - a^2*x^2)^(1/2)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)^4}{x^3} \, dx=\frac {c^{4} \left (10 \mathit {asin} \left (a x \right ) a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+12 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+12 \sqrt {-a^{2} x^{2}+1}\, a x -2 \sqrt {-a^{2} x^{2}+1}+10 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{2} x^{2}-11 a^{2} x^{2}\right )}{4 x^{2}} \] Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^4/x^3,x)
 

Output:

(c**4*(10*asin(a*x)*a**2*x**2 - 2*sqrt( - a**2*x**2 + 1)*a**3*x**3 + 12*sq 
rt( - a**2*x**2 + 1)*a**2*x**2 + 12*sqrt( - a**2*x**2 + 1)*a*x - 2*sqrt( - 
 a**2*x**2 + 1) + 10*log(tan(asin(a*x)/2))*a**2*x**2 - 11*a**2*x**2))/(4*x 
**2)