\(\int e^{3 \text {arctanh}(a x)} (c-\frac {c}{a^2 x^2})^{7/2} \, dx\) [728]

Optimal result

Integrand size = 24, antiderivative size = 301 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=-\frac {\left (c-\frac {c}{a^2 x^2}\right )^{7/2} x}{6 \left (1-a^2 x^2\right )^{7/2}}-\frac {3 a \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^2}{5 \left (1-a^2 x^2\right )^{7/2}}-\frac {a^2 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^3}{4 \left (1-a^2 x^2\right )^{7/2}}+\frac {5 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^4}{3 \left (1-a^2 x^2\right )^{7/2}}+\frac {5 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^5}{2 \left (1-a^2 x^2\right )^{7/2}}-\frac {a^5 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^6}{\left (1-a^2 x^2\right )^{7/2}}+\frac {a^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^8}{\left (1-a^2 x^2\right )^{7/2}}+\frac {3 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 \log (x)}{\left (1-a^2 x^2\right )^{7/2}} \] Output:

-1/6*(c-c/a^2/x^2)^(7/2)*x/(-a^2*x^2+1)^(7/2)-3/5*a*(c-c/a^2/x^2)^(7/2)*x^ 
2/(-a^2*x^2+1)^(7/2)-1/4*a^2*(c-c/a^2/x^2)^(7/2)*x^3/(-a^2*x^2+1)^(7/2)+5/ 
3*a^3*(c-c/a^2/x^2)^(7/2)*x^4/(-a^2*x^2+1)^(7/2)+5/2*a^4*(c-c/a^2/x^2)^(7/ 
2)*x^5/(-a^2*x^2+1)^(7/2)-a^5*(c-c/a^2/x^2)^(7/2)*x^6/(-a^2*x^2+1)^(7/2)+a 
^7*(c-c/a^2/x^2)^(7/2)*x^8/(-a^2*x^2+1)^(7/2)+3*a^6*(c-c/a^2/x^2)^(7/2)*x^ 
7*ln(x)/(-a^2*x^2+1)^(7/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.33 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=-\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \left (-10-36 a x-15 a^2 x^2+100 a^3 x^3+150 a^4 x^4-60 a^5 x^5+60 a^7 x^7+180 a^6 x^6 \log (x)\right )}{60 a^6 x^5 \sqrt {1-a^2 x^2}} \] Input:

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

Output:

-1/60*(c^3*Sqrt[c - c/(a^2*x^2)]*(-10 - 36*a*x - 15*a^2*x^2 + 100*a^3*x^3 
+ 150*a^4*x^4 - 60*a^5*x^5 + 60*a^7*x^7 + 180*a^6*x^6*Log[x]))/(a^6*x^5*Sq 
rt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6710, 6700, 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.

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

\(\Big \downarrow \) 6710

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \int \left (a^7+\frac {3 a^6}{x}+\frac {a^5}{x^2}-\frac {5 a^4}{x^3}-\frac {5 a^3}{x^4}+\frac {a^2}{x^5}+\frac {3 a}{x^6}+\frac {1}{x^7}\right )dx}{\left (1-a^2 x^2\right )^{7/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \left (a^7 x+3 a^6 \log (x)-\frac {a^5}{x}+\frac {5 a^4}{2 x^2}+\frac {5 a^3}{3 x^3}-\frac {a^2}{4 x^4}-\frac {3 a}{5 x^5}-\frac {1}{6 x^6}\right )}{\left (1-a^2 x^2\right )^{7/2}}\)

Input:

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

Output:

((c - c/(a^2*x^2))^(7/2)*x^7*(-1/6*1/x^6 - (3*a)/(5*x^5) - a^2/(4*x^4) + ( 
5*a^3)/(3*x^3) + (5*a^4)/(2*x^2) - a^5/x + a^7*x + 3*a^6*Log[x]))/(1 - a^2 
*x^2)^(7/2)
 

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 6700

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[c^p   Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], 
 x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || 
 GtQ[c, 0])
 

rule 6710

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.34

method	result	size
default	\(-\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {7}{2}} x \left (60 a^{7} x^{7}+180 a^{6} \ln \left (x \right ) x^{6}-60 a^{5} x^{5}+150 a^{4} x^{4}+100 a^{3} x^{3}-15 a^{2} x^{2}-36 a x -10\right )}{60 \left (a^{2} x^{2}-1\right )^{3} \sqrt {-a^{2} x^{2}+1}}\)	\(102\)

Input:

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

Output:

-1/60*(c*(a^2*x^2-1)/a^2/x^2)^(7/2)*x/(a^2*x^2-1)^3/(-a^2*x^2+1)^(1/2)*(60 
*a^7*x^7+180*a^6*ln(x)*x^6-60*a^5*x^5+150*a^4*x^4+100*a^3*x^3-15*a^2*x^2-3 
6*a*x-10)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.81 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\left [\frac {90 \, {\left (a^{7} c^{3} x^{7} - a^{5} c^{3} x^{5}\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} - {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) + {\left (60 \, a^{7} c^{3} x^{7} - 60 \, a^{5} c^{3} x^{5} + 150 \, a^{4} c^{3} x^{4} - {\left (60 \, a^{7} - 60 \, a^{5} + 150 \, a^{4} + 100 \, a^{3} - 15 \, a^{2} - 36 \, a - 10\right )} c^{3} x^{6} + 100 \, a^{3} c^{3} x^{3} - 15 \, a^{2} c^{3} x^{2} - 36 \, a c^{3} x - 10 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{60 \, {\left (a^{8} x^{7} - a^{6} x^{5}\right )}}, \frac {180 \, {\left (a^{7} c^{3} x^{7} - a^{5} c^{3} x^{5}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} - a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} + {\left (a^{2} - 1\right )} c x^{2} - c}\right ) + {\left (60 \, a^{7} c^{3} x^{7} - 60 \, a^{5} c^{3} x^{5} + 150 \, a^{4} c^{3} x^{4} - {\left (60 \, a^{7} - 60 \, a^{5} + 150 \, a^{4} + 100 \, a^{3} - 15 \, a^{2} - 36 \, a - 10\right )} c^{3} x^{6} + 100 \, a^{3} c^{3} x^{3} - 15 \, a^{2} c^{3} x^{2} - 36 \, a c^{3} x - 10 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{60 \, {\left (a^{8} x^{7} - a^{6} x^{5}\right )}}\right ] \] Input:

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

Output:

[1/60*(90*(a^7*c^3*x^7 - a^5*c^3*x^5)*sqrt(-c)*log((a^2*c*x^6 + a^2*c*x^2 
- c*x^4 - (a*x^5 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/( 
a^2*x^2)) - c)/(a^2*x^4 - x^2)) + (60*a^7*c^3*x^7 - 60*a^5*c^3*x^5 + 150*a 
^4*c^3*x^4 - (60*a^7 - 60*a^5 + 150*a^4 + 100*a^3 - 15*a^2 - 36*a - 10)*c^ 
3*x^6 + 100*a^3*c^3*x^3 - 15*a^2*c^3*x^2 - 36*a*c^3*x - 10*c^3)*sqrt(-a^2* 
x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^8*x^7 - a^6*x^5), 1/60*(180*( 
a^7*c^3*x^7 - a^5*c^3*x^5)*sqrt(c)*arctan(sqrt(-a^2*x^2 + 1)*(a*x^3 - a*x) 
*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^4 + (a^2 - 1)*c*x^2 - c) 
) + (60*a^7*c^3*x^7 - 60*a^5*c^3*x^5 + 150*a^4*c^3*x^4 - (60*a^7 - 60*a^5 
+ 150*a^4 + 100*a^3 - 15*a^2 - 36*a - 10)*c^3*x^6 + 100*a^3*c^3*x^3 - 15*a 
^2*c^3*x^2 - 36*a*c^3*x - 10*c^3)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/ 
(a^2*x^2)))/(a^8*x^7 - a^6*x^5)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.37 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=-\frac {1}{60} \, {\left (\frac {60 \, c^{3} x \mathrm {sgn}\left (x\right )}{a} + \frac {180 \, c^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{2}} - \frac {60 \, a^{5} c^{3} x^{5} \mathrm {sgn}\left (x\right ) - 150 \, a^{4} c^{3} x^{4} \mathrm {sgn}\left (x\right ) - 100 \, a^{3} c^{3} x^{3} \mathrm {sgn}\left (x\right ) + 15 \, a^{2} c^{3} x^{2} \mathrm {sgn}\left (x\right ) + 36 \, a c^{3} x \mathrm {sgn}\left (x\right ) + 10 \, c^{3} \mathrm {sgn}\left (x\right )}{a^{8} x^{6}}\right )} \sqrt {-c} {\left | a \right |} \] Input:

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

Output:

-1/60*(60*c^3*x*sgn(x)/a + 180*c^3*log(abs(x))*sgn(x)/a^2 - (60*a^5*c^3*x^ 
5*sgn(x) - 150*a^4*c^3*x^4*sgn(x) - 100*a^3*c^3*x^3*sgn(x) + 15*a^2*c^3*x^ 
2*sgn(x) + 36*a*c^3*x*sgn(x) + 10*c^3*sgn(x))/(a^8*x^6))*sqrt(-c)*abs(a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.23 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {\sqrt {c}\, c^{3} i \left (-180 \,\mathrm {log}\left (x \right ) a^{6} x^{6}-60 a^{7} x^{7}+60 a^{5} x^{5}-150 a^{4} x^{4}-100 a^{3} x^{3}+15 a^{2} x^{2}+36 a x +10\right )}{60 a^{7} x^{6}} \] Input:

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

Output:

(sqrt(c)*c**3*i*( - 180*log(x)*a**6*x**6 - 60*a**7*x**7 + 60*a**5*x**5 - 1 
50*a**4*x**4 - 100*a**3*x**3 + 15*a**2*x**2 + 36*a*x + 10))/(60*a**7*x**6)