3.3.33 \(\int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\) [233]

3.3.33.1 Optimal result
3.3.33.2 Mathematica [A] (verified)
3.3.33.3 Rubi [A] (verified)
3.3.33.4 Maple [A] (verified)
3.3.33.5 Fricas [A] (verification not implemented)
3.3.33.6 Sympy [F(-1)]
3.3.33.7 Maxima [F]
3.3.33.8 Giac [A] (verification not implemented)
3.3.33.9 Mupad [F(-1)]

3.3.33.1 Optimal result

Integrand size = 18, antiderivative size = 193 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {a^2 \left (1-\frac {1}{a x}\right )^{5/2} \sqrt {1+\frac {1}{a x}} x^2}{8 \left (a-\frac {1}{x}\right ) (c-a c x)^{5/2}}-\frac {a^3 \left (1-\frac {1}{a x}\right )^{5/2} \left (1+\frac {1}{a x}\right )^{3/2} x^2}{4 \left (a-\frac {1}{x}\right )^2 (c-a c x)^{5/2}}+\frac {a^{3/2} \left (1-\frac {1}{a x}\right )^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {1+\frac {1}{a x}}}\right )}{8 \sqrt {2} \left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}} \]

output
-1/4*a^3*(1-1/a/x)^(5/2)*(1+1/a/x)^(3/2)*x^2/(a-1/x)^2/(-a*c*x+c)^(5/2)+1/ 
16*a^(3/2)*(1-1/a/x)^(5/2)*arctanh(2^(1/2)*(1/x)^(1/2)/a^(1/2)/(1+1/a/x)^( 
1/2))/(1/x)^(5/2)/(-a*c*x+c)^(5/2)*2^(1/2)+1/8*a^2*(1-1/a/x)^(5/2)*x^2*(1+ 
1/a/x)^(1/2)/(a-1/x)/(-a*c*x+c)^(5/2)
 
3.3.33.2 Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.64 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\sqrt {1-\frac {1}{a x}} x \left (-2 \sqrt {a} \sqrt {1+\frac {1}{a x}} (3+a x)+\sqrt {2} \sqrt {\frac {1}{x}} (-1+a x)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {1+\frac {1}{a x}}}\right )\right )}{16 \sqrt {a} c^2 (-1+a x)^2 \sqrt {c-a c x}} \]

input
Integrate[E^ArcCoth[a*x]/(c - a*c*x)^(5/2),x]
 
output
(Sqrt[1 - 1/(a*x)]*x*(-2*Sqrt[a]*Sqrt[1 + 1/(a*x)]*(3 + a*x) + Sqrt[2]*Sqr 
t[x^(-1)]*(-1 + a*x)^2*ArcTanh[(Sqrt[2]*Sqrt[x^(-1)])/(Sqrt[a]*Sqrt[1 + 1/ 
(a*x)])]))/(16*Sqrt[a]*c^2*(-1 + a*x)^2*Sqrt[c - a*c*x])
 
3.3.33.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6727, 27, 105, 105, 104, 219}

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.

\(\displaystyle \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\)

\(\Big \downarrow \) 6727

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 219

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

input
Int[E^ArcCoth[a*x]/(c - a*c*x)^(5/2),x]
 
output
-((a^3*(1 - 1/(a*x))^(5/2)*(((1 + 1/(a*x))^(3/2)*Sqrt[x^(-1)])/(4*(a - x^( 
-1))^2) + (-((Sqrt[1 + 1/(a*x)]*Sqrt[x^(-1)])/(a*(a - x^(-1)))) - ArcTanh[ 
(Sqrt[2]*Sqrt[x^(-1)])/(Sqrt[a]*Sqrt[1 + 1/(a*x)])]/(Sqrt[2]*a^(3/2)))/8)) 
/((x^(-1))^(5/2)*(c - a*c*x)^(5/2)))
 

3.3.33.3.1 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 104
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 6727
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Si 
mp[(-(1/x)^p)*((c + d*x)^p/(1 + c/(d*x))^p)   Subst[Int[((1 + c*(x/d))^p*(( 
1 + x/a)^(n/2)/x^(p + 2)))/(1 - x/a)^(n/2), x], x, 1/x], x] /; FreeQ[{a, c, 
 d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[p]
 
3.3.33.4 Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.87

method result size
default \(\frac {\sqrt {-c \left (a x -1\right )}\, \left (-\sqrt {2}\, \arctan \left (\frac {\sqrt {-c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a^{2} c \,x^{2}+2 \sqrt {2}\, \arctan \left (\frac {\sqrt {-c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a c x +2 a x \sqrt {c}\, \sqrt {-c \left (a x +1\right )}-\sqrt {2}\, \arctan \left (\frac {\sqrt {-c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +6 \sqrt {-c \left (a x +1\right )}\, \sqrt {c}\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right )^{2} c^{\frac {7}{2}} \sqrt {-c \left (a x +1\right )}\, a}\) \(167\)

input
int(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^(5/2),x,method=_RETURNVERBOSE)
 
output
1/16*(-c*(a*x-1))^(1/2)*(-2^(1/2)*arctan(1/2*(-c*(a*x+1))^(1/2)*2^(1/2)/c^ 
(1/2))*a^2*c*x^2+2*2^(1/2)*arctan(1/2*(-c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))* 
a*c*x+2*a*x*c^(1/2)*(-c*(a*x+1))^(1/2)-2^(1/2)*arctan(1/2*(-c*(a*x+1))^(1/ 
2)*2^(1/2)/c^(1/2))*c+6*(-c*(a*x+1))^(1/2)*c^(1/2))/((a*x-1)/(a*x+1))^(1/2 
)/(a*x-1)^2/c^(7/2)/(-c*(a*x+1))^(1/2)/a
 
3.3.33.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.75 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\left [-\frac {\sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {-c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 4 \, {\left (a^{2} x^{2} + 4 \, a x + 3\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{32 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}}, -\frac {\sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}}}{a c x - c}\right ) - 2 \, {\left (a^{2} x^{2} + 4 \, a x + 3\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{16 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}}\right ] \]

input
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^(5/2),x, algorithm="fricas" 
)
 
output
[-1/32*(sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(-c)*log(-(a^2*c*x^2 
 + 2*a*c*x - 2*sqrt(2)*sqrt(-a*c*x + c)*(a*x + 1)*sqrt(-c)*sqrt((a*x - 1)/ 
(a*x + 1)) - 3*c)/(a^2*x^2 - 2*a*x + 1)) - 4*(a^2*x^2 + 4*a*x + 3)*sqrt(-a 
*c*x + c)*sqrt((a*x - 1)/(a*x + 1)))/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2* 
c^3*x - a*c^3), -1/16*(sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(c)*a 
rctan(sqrt(2)*sqrt(-a*c*x + c)*sqrt(c)*sqrt((a*x - 1)/(a*x + 1))/(a*c*x - 
c)) - 2*(a^2*x^2 + 4*a*x + 3)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1)))/ 
(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)]
 
3.3.33.6 Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\text {Timed out} \]

input
integrate(1/((a*x-1)/(a*x+1))**(1/2)/(-a*c*x+c)**(5/2),x)
 
output
Timed out
 
3.3.33.7 Maxima [F]

\[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\int { \frac {1}{{\left (-a c x + c\right )}^{\frac {5}{2}} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]

input
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^(5/2),x, algorithm="maxima" 
)
 
output
integrate(1/((-a*c*x + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)
 
3.3.33.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.40 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x - c}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (-a c x - c\right )}^{\frac {3}{2}} - 2 \, \sqrt {-a c x - c} c\right )}}{{\left (a c x - c\right )}^{2} c}}{16 \, a {\left | c \right |}} \]

input
integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^(5/2),x, algorithm="giac")
 
output
1/16*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*c*x - c)/sqrt(c))/c^(3/2) + 2*((- 
a*c*x - c)^(3/2) - 2*sqrt(-a*c*x - c)*c)/((a*c*x - c)^2*c))/(a*abs(c))
 
3.3.33.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\int \frac {1}{{\left (c-a\,c\,x\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

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