\(\int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx\) [274]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 185 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx=-\frac {a^2 \left (1-\frac {1}{a x}\right )^{7/2} x^2}{2 \left (a-\frac {1}{x}\right ) \sqrt {1+\frac {1}{a x}} (c-a c x)^{7/2}}-\frac {3 a^2 \left (1-\frac {1}{a x}\right )^{7/2} x^3}{4 \sqrt {1+\frac {1}{a x}} (c-a c x)^{7/2}}+\frac {3 \left (1-\frac {1}{a x}\right )^{7/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}\right )}{4 \sqrt {2} a \left (\frac {1}{a x}\right )^{7/2} (c-a c x)^{7/2}} \] Output:

-1/2*a^2*(1-1/a/x)^(7/2)*x^2/(a-1/x)/(1+1/a/x)^(1/2)/(-a*c*x+c)^(7/2)-3/4* 
a^2*(1-1/a/x)^(7/2)*x^3/(1+1/a/x)^(1/2)/(-a*c*x+c)^(7/2)+3/8*(1-1/a/x)^(7/ 
2)*arctanh(2^(1/2)*(1/a/x)^(1/2)/(1+1/a/x)^(1/2))*2^(1/2)/a/(1/a/x)^(7/2)/ 
(-a*c*x+c)^(7/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx=\frac {\sqrt {1-\frac {1}{a x}} \left (-2+6 a x-\frac {3 \sqrt {2} \sqrt {a} \sqrt {1+\frac {1}{a x}} (-1+a x) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {1+\frac {1}{a x}}}\right )}{\sqrt {\frac {1}{x}}}\right )}{8 a c^3 \sqrt {1+\frac {1}{a x}} (-1+a x) \sqrt {c-a c x}} \] Input:

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^(7/2)),x]
 

Output:

(Sqrt[1 - 1/(a*x)]*(-2 + 6*a*x - (3*Sqrt[2]*Sqrt[a]*Sqrt[1 + 1/(a*x)]*(-1 
+ a*x)*ArcTanh[(Sqrt[2]*Sqrt[x^(-1)])/(Sqrt[a]*Sqrt[1 + 1/(a*x)])])/Sqrt[x 
^(-1)]))/(8*a*c^3*Sqrt[1 + 1/(a*x)]*(-1 + a*x)*Sqrt[c - a*c*x])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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.

\(\displaystyle \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx\)

\(\Big \downarrow \) 6727

\(\displaystyle -\frac {\left (1-\frac {1}{a x}\right )^{7/2} \int \frac {a^2 \left (\frac {1}{x}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^2 \left (1+\frac {1}{a x}\right )^{3/2}}d\frac {1}{x}}{\left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a^2 \left (1-\frac {1}{a x}\right )^{7/2} \int \frac {\left (\frac {1}{x}\right )^{3/2}}{\left (a-\frac {1}{x}\right )^2 \left (1+\frac {1}{a x}\right )^{3/2}}d\frac {1}{x}}{\left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {a^2 \left (1-\frac {1}{a x}\right )^{7/2} \left (\frac {\left (\frac {1}{x}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right ) \sqrt {\frac {1}{a x}+1}}-\frac {3}{4} \int \frac {\sqrt {\frac {1}{x}}}{\left (a-\frac {1}{x}\right ) \left (1+\frac {1}{a x}\right )^{3/2}}d\frac {1}{x}\right )}{\left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {a^2 \left (1-\frac {1}{a x}\right )^{7/2} \left (\frac {\left (\frac {1}{x}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right ) \sqrt {\frac {1}{a x}+1}}-\frac {3}{4} \left (\frac {1}{2} a \int \frac {1}{\left (a-\frac {1}{x}\right ) \sqrt {1+\frac {1}{a x}} \sqrt {\frac {1}{x}}}d\frac {1}{x}-\frac {\sqrt {\frac {1}{x}}}{\sqrt {\frac {1}{a x}+1}}\right )\right )}{\left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {a^2 \left (1-\frac {1}{a x}\right )^{7/2} \left (\frac {\left (\frac {1}{x}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right ) \sqrt {\frac {1}{a x}+1}}-\frac {3}{4} \left (a \int \frac {1}{a-\frac {2}{x^2}}d\frac {\sqrt {\frac {1}{x}}}{\sqrt {1+\frac {1}{a x}}}-\frac {\sqrt {\frac {1}{x}}}{\sqrt {\frac {1}{a x}+1}}\right )\right )}{\left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {a^2 \left (1-\frac {1}{a x}\right )^{7/2} \left (\frac {\left (\frac {1}{x}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right ) \sqrt {\frac {1}{a x}+1}}-\frac {3}{4} \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {\frac {1}{a x}+1}}\right )}{\sqrt {2}}-\frac {\sqrt {\frac {1}{x}}}{\sqrt {\frac {1}{a x}+1}}\right )\right )}{\left (\frac {1}{x}\right )^{7/2} (c-a c x)^{7/2}}\)

Input:

Int[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^(7/2)),x]
 

Output:

-((a^2*(1 - 1/(a*x))^(7/2)*((x^(-1))^(3/2)/(2*(a - x^(-1))*Sqrt[1 + 1/(a*x 
)]) - (3*(-(Sqrt[x^(-1)]/Sqrt[1 + 1/(a*x)]) + (Sqrt[a]*ArcTanh[(Sqrt[2]*Sq 
rt[x^(-1)])/(Sqrt[a]*Sqrt[1 + 1/(a*x)])])/Sqrt[2]))/4))/((x^(-1))^(7/2)*(c 
 - a*c*x)^(7/2)))
 

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 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]
 
Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.70

method result size
default \(-\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \sqrt {-c \left (a x -1\right )}\, \left (3 \sqrt {2}\, \arctan \left (\frac {\sqrt {-c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a x \sqrt {-c \left (a x +1\right )}-3 \arctan \left (\frac {\sqrt {-c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sqrt {-c \left (a x +1\right )}+6 \sqrt {c}\, a x -2 \sqrt {c}\right )}{8 \left (a x -1\right )^{3} c^{\frac {9}{2}} a}\) \(129\)

Input:

int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.57 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {-c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, \sqrt {-a c x + c} {\left (3 \, a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{16 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}}, \frac {3 \, \sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, {\left (a c x - c\right )}}\right ) - 2 \, \sqrt {-a c x + c} {\left (3 \, a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{8 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}}\right ] \] Input:

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(7/2),x, algorithm="fricas")
 

Output:

[-1/16*(3*sqrt(2)*(a^2*x^2 - 2*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*sqrt(-a*c*x + c)*(3*a*x - 1)*sqrt((a*x 
- 1)/(a*x + 1)))/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4), 1/8*(3*sqrt(2)*(a^2* 
x^2 - 2*a*x + 1)*sqrt(c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*(a*x + 1)*sqr 
t(c)*sqrt((a*x - 1)/(a*x + 1))/(a*c*x - c)) - 2*sqrt(-a*c*x + c)*(3*a*x - 
1)*sqrt((a*x - 1)/(a*x + 1)))/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx=\int { \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (-a c x + c\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(7/2),x, algorithm="maxima")
 

Output:

integrate(((a*x - 1)/(a*x + 1))^(3/2)/(-a*c*x + c)^(7/2), x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.49 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx=-\frac {{\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x - c}}{2 \, \sqrt {c}}\right )}{a c^{\frac {5}{2}}} - \frac {2 \, {\left (3 \, a c x - c\right )}}{{\left ({\left (-a c x - c\right )}^{\frac {3}{2}} + 2 \, \sqrt {-a c x - c} c\right )} a c^{2}}\right )} {\left | c \right |} \mathrm {sgn}\left (a x + 1\right )}{8 \, c^{2}} \] Input:

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(7/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx=\int \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{{\left (c-a\,c\,x\right )}^{7/2}} \,d x \] Input:

int(((a*x - 1)/(a*x + 1))^(3/2)/(c - a*c*x)^(7/2),x)
 

Output:

int(((a*x - 1)/(a*x + 1))^(3/2)/(c - a*c*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.59 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{7/2}} \, dx=\frac {\sqrt {c}\, i \left (12 \sqrt {a x +1}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a x -12 \sqrt {a x +1}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )-9 \sqrt {a x +1}\, \sqrt {2}\, a x +9 \sqrt {a x +1}\, \sqrt {2}+24 a x -8\right )}{32 \sqrt {a x +1}\, a \,c^{4} \left (a x -1\right )} \] Input:

int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(7/2),x)
 

Output:

(sqrt(c)*i*(12*sqrt(a*x + 1)*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2) 
)/2))*a*x - 12*sqrt(a*x + 1)*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2) 
)/2)) - 9*sqrt(a*x + 1)*sqrt(2)*a*x + 9*sqrt(a*x + 1)*sqrt(2) + 24*a*x - 8 
))/(32*sqrt(a*x + 1)*a*c**4*(a*x - 1))