\(\int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{7/2}} \, dx\) [623]

Optimal result

Integrand size = 24, antiderivative size = 278 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (1-a x)^4 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{12 (1-a x)^3 \left (c-a^2 c x^2\right )^{7/2}}-\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1-a x)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{8 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}+\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1+a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7 \text {arctanh}(a x)}{32 \left (c-a^2 c x^2\right )^{7/2}} \] Output:

-1/16*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)^4/(-a^2*c*x^2+c)^(7/2)-1/12*a^6 
*(1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)^3/(-a^2*c*x^2+c)^(7/2)-3/32*a^6*(1-1/a^2 
/x^2)^(7/2)*x^7/(-a*x+1)^2/(-a^2*c*x^2+c)^(7/2)-1/8*a^6*(1-1/a^2/x^2)^(7/2 
)*x^7/(-a*x+1)/(-a^2*c*x^2+c)^(7/2)+1/32*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(a*x+ 
1)/(-a^2*c*x^2+c)^(7/2)-5/32*a^6*(1-1/a^2/x^2)^(7/2)*x^7*arctanh(a*x)/(-a^ 
2*c*x^2+c)^(7/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.36 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (32-15 a x-35 a^2 x^2+45 a^3 x^3-15 a^4 x^4+15 (-1+a x)^4 (1+a x) \text {arctanh}(a x)\right )}{96 c^3 (-1+a x)^4 (1+a x) \sqrt {c-a^2 c x^2}} \] Input:

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

Output:

(Sqrt[1 - 1/(a^2*x^2)]*x*(32 - 15*a*x - 35*a^2*x^2 + 45*a^3*x^3 - 15*a^4*x 
^4 + 15*(-1 + a*x)^4*(1 + a*x)*ArcTanh[a*x]))/(96*c^3*(-1 + a*x)^4*(1 + a* 
x)*Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.44, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6746, 6747, 25, 54, 2009}

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.

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

\(\Big \downarrow \) 6746

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

\(\Big \downarrow \) 6747

\(\displaystyle \frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int -\frac {1}{(1-a x)^5 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{7/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \frac {1}{(1-a x)^5 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{7/2}}\)

\(\Big \downarrow \) 54

\(\displaystyle -\frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \left (\frac {1}{8 (a x-1)^2}+\frac {1}{32 (a x+1)^2}-\frac {3}{16 (a x-1)^3}+\frac {1}{4 (a x-1)^4}-\frac {1}{4 (a x-1)^5}-\frac {5}{32 \left (a^2 x^2-1\right )}\right )dx}{\left (c-a^2 c x^2\right )^{7/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \left (\frac {5 \text {arctanh}(a x)}{32 a}+\frac {1}{8 a (1-a x)}-\frac {1}{32 a (a x+1)}+\frac {3}{32 a (1-a x)^2}+\frac {1}{12 a (1-a x)^3}+\frac {1}{16 a (1-a x)^4}\right )}{\left (c-a^2 c x^2\right )^{7/2}}\)

Input:

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

Output:

-((a^7*(1 - 1/(a^2*x^2))^(7/2)*x^7*(1/(16*a*(1 - a*x)^4) + 1/(12*a*(1 - a* 
x)^3) + 3/(32*a*(1 - a*x)^2) + 1/(8*a*(1 - a*x)) - 1/(32*a*(1 + a*x)) + (5 
*ArcTanh[a*x])/(32*a)))/(c - a^2*c*x^2)^(7/2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6746

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p)   Int[u*x^(2*p)*(1 - 
 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && 
EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]
 

rule 6747

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb 
ol] :> Simp[c^p/a^(2*p)   Int[(u/x^(2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p 
 + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !Inte 
gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (15 \ln \left (a x +1\right ) x^{5} a^{5}-15 \ln \left (a x -1\right ) x^{5} a^{5}-45 \ln \left (a x +1\right ) x^{4} a^{4}+45 \ln \left (a x -1\right ) x^{4} a^{4}-30 a^{4} x^{4}+30 \ln \left (a x +1\right ) x^{3} a^{3}-30 a^{3} \ln \left (a x -1\right ) x^{3}+90 a^{3} x^{3}+30 \ln \left (a x +1\right ) x^{2} a^{2}-30 a^{2} \ln \left (a x -1\right ) x^{2}-70 a^{2} x^{2}-45 \ln \left (a x +1\right ) x a +45 a \ln \left (a x -1\right ) x -30 a x +15 \ln \left (a x +1\right )-15 \ln \left (a x -1\right )+64\right )}{192 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x -1\right )^{2} \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{4} a}\)	\(241\)

Input:

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

Output:

-1/192/((a*x-1)/(a*x+1))^(3/2)/(a*x-1)^2/(a*x+1)^2*(-c*(a^2*x^2-1))^(1/2)* 
(15*ln(a*x+1)*x^5*a^5-15*ln(a*x-1)*x^5*a^5-45*ln(a*x+1)*x^4*a^4+45*ln(a*x- 
1)*x^4*a^4-30*a^4*x^4+30*ln(a*x+1)*x^3*a^3-30*a^3*ln(a*x-1)*x^3+90*a^3*x^3 
+30*ln(a*x+1)*x^2*a^2-30*a^2*ln(a*x-1)*x^2-70*a^2*x^2-45*ln(a*x+1)*x*a+45* 
a*ln(a*x-1)*x-30*a*x+15*ln(a*x+1)-15*ln(a*x-1)+64)/(a^2*x^2-1)/c^4/a
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.68 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {15 \, {\left (a^{6} x^{5} - 3 \, a^{5} x^{4} + 2 \, a^{4} x^{3} + 2 \, a^{3} x^{2} - 3 \, a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, {\left (15 \, a^{4} x^{4} - 45 \, a^{3} x^{3} + 35 \, a^{2} x^{2} + 15 \, a x - 32\right )} \sqrt {-a^{2} c}}{192 \, {\left (a^{7} c^{4} x^{5} - 3 \, a^{6} c^{4} x^{4} + 2 \, a^{5} c^{4} x^{3} + 2 \, a^{4} c^{4} x^{2} - 3 \, a^{3} c^{4} x + a^{2} c^{4}\right )}} \] Input:

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

Output:

-1/192*(15*(a^6*x^5 - 3*a^5*x^4 + 2*a^4*x^3 + 2*a^3*x^2 - 3*a^2*x + a)*sqr 
t(-c)*log((a^2*c*x^2 - 2*sqrt(-a^2*c)*sqrt(-c)*x + c)/(a^2*x^2 - 1)) - 2*( 
15*a^4*x^4 - 45*a^3*x^3 + 35*a^2*x^2 + 15*a*x - 32)*sqrt(-a^2*c))/(a^7*c^4 
*x^5 - 3*a^6*c^4*x^4 + 2*a^5*c^4*x^3 + 2*a^4*c^4*x^2 - 3*a^3*c^4*x + a^2*c 
^4)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.42 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {c}\, i \left (15 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{5} x^{5}-45 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{4} x^{4}+30 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}+30 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}-45 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +15 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )+15 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{5} x^{5}-45 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{4} x^{4}+30 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}+30 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}-45 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x +15 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )-30 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{5} x^{5}+90 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{4} x^{4}-60 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}-60 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}+90 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x -30 \,\mathrm {log}\left (\sqrt {-a x +1}\right )+15 a^{5} x^{5}-75 a^{4} x^{4}+120 a^{3} x^{3}-40 a^{2} x^{2}-75 a x +79\right )}{192 a \,c^{4} \left (a^{5} x^{5}-3 a^{4} x^{4}+2 a^{3} x^{3}+2 a^{2} x^{2}-3 a x +1\right )} \] Input:

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

Output:

(sqrt(c)*i*(15*log(sqrt( - a*x + 1) - sqrt(2))*a**5*x**5 - 45*log(sqrt( - 
a*x + 1) - sqrt(2))*a**4*x**4 + 30*log(sqrt( - a*x + 1) - sqrt(2))*a**3*x* 
*3 + 30*log(sqrt( - a*x + 1) - sqrt(2))*a**2*x**2 - 45*log(sqrt( - a*x + 1 
) - sqrt(2))*a*x + 15*log(sqrt( - a*x + 1) - sqrt(2)) + 15*log(sqrt( - a*x 
 + 1) + sqrt(2))*a**5*x**5 - 45*log(sqrt( - a*x + 1) + sqrt(2))*a**4*x**4 
+ 30*log(sqrt( - a*x + 1) + sqrt(2))*a**3*x**3 + 30*log(sqrt( - a*x + 1) + 
 sqrt(2))*a**2*x**2 - 45*log(sqrt( - a*x + 1) + sqrt(2))*a*x + 15*log(sqrt 
( - a*x + 1) + sqrt(2)) - 30*log(sqrt( - a*x + 1))*a**5*x**5 + 90*log(sqrt 
( - a*x + 1))*a**4*x**4 - 60*log(sqrt( - a*x + 1))*a**3*x**3 - 60*log(sqrt 
( - a*x + 1))*a**2*x**2 + 90*log(sqrt( - a*x + 1))*a*x - 30*log(sqrt( - a* 
x + 1)) + 15*a**5*x**5 - 75*a**4*x**4 + 120*a**3*x**3 - 40*a**2*x**2 - 75* 
a*x + 79))/(192*a*c**4*(a**5*x**5 - 3*a**4*x**4 + 2*a**3*x**3 + 2*a**2*x** 
2 - 3*a*x + 1))