\(\int e^{\coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{5/2} \, dx\) [807]

Optimal result

Integrand size = 22, antiderivative size = 236 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{3 a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {2 c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.31 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} \left (-\frac {1}{4 a^5 x^4}-\frac {1}{3 a^4 x^3}+\frac {1}{a^3 x^2}+\frac {2}{a^2 x}+x+\frac {\log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}} \] Input:

Integrate[E^ArcCoth[a*x]*(c - c/(a^2*x^2))^(5/2),x]
 

Output:

((c - c/(a^2*x^2))^(5/2)*(-1/4*1/(a^5*x^4) - 1/(3*a^4*x^3) + 1/(a^3*x^2) + 
 2/(a^2*x) + x + Log[x]/a))/(1 - 1/(a^2*x^2))^(5/2)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6751, 6747, 99, 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.

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

\(\Big \downarrow \) 6751

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

\(\Big \downarrow \) 6747

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

Int[E^ArcCoth[a*x]*(c - c/(a^2*x^2))^(5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

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

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]
 

rule 6751

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.41

method	result	size
default	\(\frac {\left (12 a^{5} x^{5}+12 \ln \left (x \right ) x^{4} a^{4}+24 a^{3} x^{3}+12 a^{2} x^{2}-4 a x -3\right ) {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}} x}{12 \left (a x +1\right ) \left (a^{2} x^{2}-1\right )^{2} \sqrt {\frac {a x -1}{a x +1}}}\)	\(96\)

Input:

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

Output:

1/12*(12*a^5*x^5+12*ln(x)*x^4*a^4+24*a^3*x^3+12*a^2*x^2-4*a*x-3)*(c*(a^2*x 
^2-1)/a^2/x^2)^(5/2)*x/(a*x+1)/(a^2*x^2-1)^2/((a*x-1)/(a*x+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.31 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {{\left (12 \, a^{5} c^{2} x^{5} + 12 \, a^{4} c^{2} x^{4} \log \left (x\right ) + 24 \, a^{3} c^{2} x^{3} + 12 \, a^{2} c^{2} x^{2} - 4 \, a c^{2} x - 3 \, c^{2}\right )} \sqrt {a^{2} c}}{12 \, a^{6} x^{4}} \] Input:

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

Output:

1/12*(12*a^5*c^2*x^5 + 12*a^4*c^2*x^4*log(x) + 24*a^3*c^2*x^3 + 12*a^2*c^2 
*x^2 - 4*a*c^2*x - 3*c^2)*sqrt(a^2*c)/(a^6*x^4)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.45 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {1}{6} \, {\left (\frac {12 \, c^{2} x \mathrm {sgn}\left (x\right )}{a \mathrm {sgn}\left (a x + 1\right )} + \frac {12 \, c^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{2} \mathrm {sgn}\left (a x + 1\right )} + \frac {24 \, a^{3} c^{2} x^{3} \mathrm {sgn}\left (x\right ) + 12 \, a^{2} c^{2} x^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c^{2} x \mathrm {sgn}\left (x\right ) - 3 \, c^{2} \mathrm {sgn}\left (x\right )}{a^{6} x^{4} \mathrm {sgn}\left (a x + 1\right )}\right )} \sqrt {c} {\left | a \right |} \] Input:

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

Output:

1/6*(12*c^2*x*sgn(x)/(a*sgn(a*x + 1)) + 12*c^2*log(abs(x))*sgn(x)/(a^2*sgn 
(a*x + 1)) + (24*a^3*c^2*x^3*sgn(x) + 12*a^2*c^2*x^2*sgn(x) - 4*a*c^2*x*sg 
n(x) - 3*c^2*sgn(x))/(a^6*x^4*sgn(a*x + 1)))*sqrt(c)*abs(a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.22 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} \left (12 \,\mathrm {log}\left (x \right ) a^{4} x^{4}+12 a^{5} x^{5}+24 a^{3} x^{3}+12 a^{2} x^{2}-4 a x -3\right )}{12 a^{5} x^{4}} \] Input:

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

Output:

(sqrt(c)*c**2*(12*log(x)*a**4*x**4 + 12*a**5*x**5 + 24*a**3*x**3 + 12*a**2 
*x**2 - 4*a*x - 3))/(12*a**5*x**4)