\(\int e^{\text {arctanh}(a x)} (c-\frac {c}{a^2 x^2})^{7/2} \, dx\) [712]

Optimal result

Integrand size = 22, antiderivative size = 299 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=-\frac {\left (c-\frac {c}{a^2 x^2}\right )^{7/2} x}{6 \left (1-a^2 x^2\right )^{7/2}}-\frac {a \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^2}{5 \left (1-a^2 x^2\right )^{7/2}}+\frac {3 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^3}{4 \left (1-a^2 x^2\right )^{7/2}}+\frac {a^3 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^4}{\left (1-a^2 x^2\right )^{7/2}}-\frac {3 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^5}{2 \left (1-a^2 x^2\right )^{7/2}}-\frac {3 a^5 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^6}{\left (1-a^2 x^2\right )^{7/2}}-\frac {a^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^8}{\left (1-a^2 x^2\right )^{7/2}}-\frac {a^6 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 \log (x)}{\left (1-a^2 x^2\right )^{7/2}} \] Output:

-1/6*(c-c/a^2/x^2)^(7/2)*x/(-a^2*x^2+1)^(7/2)-1/5*a*(c-c/a^2/x^2)^(7/2)*x^ 
2/(-a^2*x^2+1)^(7/2)+3/4*a^2*(c-c/a^2/x^2)^(7/2)*x^3/(-a^2*x^2+1)^(7/2)+a^ 
3*(c-c/a^2/x^2)^(7/2)*x^4/(-a^2*x^2+1)^(7/2)-3/2*a^4*(c-c/a^2/x^2)^(7/2)*x 
^5/(-a^2*x^2+1)^(7/2)-3*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)-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 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \left (10+12 a x-45 a^2 x^2-60 a^3 x^3+90 a^4 x^4+180 a^5 x^5+60 a^7 x^7+60 a^6 x^6 \log (x)\right )}{60 a^6 x^5 \sqrt {1-a^2 x^2}} \] Input:

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

Output:

(c^3*Sqrt[c - c/(a^2*x^2)]*(10 + 12*a*x - 45*a^2*x^2 - 60*a^3*x^3 + 90*a^4 
*x^4 + 180*a^5*x^5 + 60*a^7*x^7 + 60*a^6*x^6*Log[x]))/(60*a^6*x^5*Sqrt[1 - 
 a^2*x^2])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.32, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, 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.

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

\(\Big \downarrow \) 6710

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

Int[E^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 - a/(5*x^5) + (3*a^2)/(4*x^4) + a 
^3/x^3 - (3*a^4)/(2*x^2) - (3*a^5)/x - a^7*x - 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 = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.34

method	result	size
default	\(\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {7}{2}} x \left (60 a^{7} x^{7}+60 a^{6} \ln \left (x \right ) x^{6}+180 a^{5} x^{5}+90 a^{4} x^{4}-60 a^{3} x^{3}-45 a^{2} x^{2}+12 a x +10\right )}{60 \left (a^{2} x^{2}-1\right )^{3} \sqrt {-a^{2} x^{2}+1}}\)	\(102\)

Input:

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

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+60*a^6*ln(x)*x^6+180*a^5*x^5+90*a^4*x^4-60*a^3*x^3-45*a^2*x^2+12*a 
*x+10)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.82 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\left [\frac {30 \, {\left (a^{7} c^{3} x^{7} - a^{5} c^{3} x^{5}\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} + {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) - {\left (60 \, a^{7} c^{3} x^{7} + 180 \, a^{5} c^{3} x^{5} + 90 \, a^{4} c^{3} x^{4} - {\left (60 \, a^{7} + 180 \, a^{5} + 90 \, a^{4} - 60 \, a^{3} - 45 \, a^{2} + 12 \, a + 10\right )} c^{3} x^{6} - 60 \, a^{3} c^{3} x^{3} - 45 \, a^{2} c^{3} x^{2} + 12 \, a c^{3} x + 10 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{60 \, {\left (a^{8} x^{7} - a^{6} x^{5}\right )}}, -\frac {60 \, {\left (a^{7} c^{3} x^{7} - a^{5} c^{3} x^{5}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} - a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} + {\left (a^{2} - 1\right )} c x^{2} - c}\right ) + {\left (60 \, a^{7} c^{3} x^{7} + 180 \, a^{5} c^{3} x^{5} + 90 \, a^{4} c^{3} x^{4} - {\left (60 \, a^{7} + 180 \, a^{5} + 90 \, a^{4} - 60 \, a^{3} - 45 \, a^{2} + 12 \, a + 10\right )} c^{3} x^{6} - 60 \, a^{3} c^{3} x^{3} - 45 \, a^{2} c^{3} x^{2} + 12 \, a c^{3} x + 10 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{60 \, {\left (a^{8} x^{7} - a^{6} x^{5}\right )}}\right ] \] Input:

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

Output:

[1/60*(30*(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 + 180*a^5*c^3*x^5 + 90*a 
^4*c^3*x^4 - (60*a^7 + 180*a^5 + 90*a^4 - 60*a^3 - 45*a^2 + 12*a + 10)*c^3 
*x^6 - 60*a^3*c^3*x^3 - 45*a^2*c^3*x^2 + 12*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*(60*(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)*s 
qrt(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 + 180*a^5*c^3*x^5 + 90*a^4*c^3*x^4 - (60*a^7 + 180*a^5 + 
 90*a^4 - 60*a^3 - 45*a^2 + 12*a + 10)*c^3*x^6 - 60*a^3*c^3*x^3 - 45*a^2*c 
^3*x^2 + 12*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]

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

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.37 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {1}{60} \, {\left (\frac {60 \, c^{3} x \mathrm {sgn}\left (x\right )}{a} + \frac {60 \, c^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{2}} + \frac {180 \, a^{5} c^{3} x^{5} \mathrm {sgn}\left (x\right ) + 90 \, a^{4} c^{3} x^{4} \mathrm {sgn}\left (x\right ) - 60 \, a^{3} c^{3} x^{3} \mathrm {sgn}\left (x\right ) - 45 \, a^{2} c^{3} x^{2} \mathrm {sgn}\left (x\right ) + 12 \, a c^{3} x \mathrm {sgn}\left (x\right ) + 10 \, c^{3} \mathrm {sgn}\left (x\right )}{a^{8} x^{6}}\right )} \sqrt {-c} {\left | a \right |} \] Input:

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

Output:

1/60*(60*c^3*x*sgn(x)/a + 60*c^3*log(abs(x))*sgn(x)/a^2 + (180*a^5*c^3*x^5 
*sgn(x) + 90*a^4*c^3*x^4*sgn(x) - 60*a^3*c^3*x^3*sgn(x) - 45*a^2*c^3*x^2*s 
gn(x) + 12*a*c^3*x*sgn(x) + 10*c^3*sgn(x))/(a^8*x^6))*sqrt(-c)*abs(a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.23 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {\sqrt {c}\, c^{3} i \left (60 \,\mathrm {log}\left (x \right ) a^{6} x^{6}+60 a^{7} x^{7}+180 a^{5} x^{5}+90 a^{4} x^{4}-60 a^{3} x^{3}-45 a^{2} x^{2}+12 a x +10\right )}{60 a^{7} x^{6}} \] Input:

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

Output:

(sqrt(c)*c**3*i*(60*log(x)*a**6*x**6 + 60*a**7*x**7 + 180*a**5*x**5 + 90*a 
**4*x**4 - 60*a**3*x**3 - 45*a**2*x**2 + 12*a*x + 10))/(60*a**7*x**6)