\(\int \frac {e^{3 \text {arctanh}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\) [1208]

Optimal result

Integrand size = 24, antiderivative size = 185 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{6 a c^2 (1-a x)^3 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.39 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-10+9 a x-3 a^2 x^2+3 (-1+a x)^3 \text {arctanh}(a x)\right )}{24 a c^2 (-1+a x)^3 \sqrt {c-a^2 c x^2}} \] Input:

Integrate[E^(3*ArcTanh[a*x])/(c - a^2*c*x^2)^(5/2),x]
 

Output:

(Sqrt[1 - a^2*x^2]*(-10 + 9*a*x - 3*a^2*x^2 + 3*(-1 + a*x)^3*ArcTanh[a*x]) 
)/(24*a*c^2*(-1 + a*x)^3*Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6693, 6690, 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.

\(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6693

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{3 \text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 6690

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

\(\Big \downarrow \) 54

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{8 (a x-1)^2}-\frac {1}{4 (a x-1)^3}+\frac {1}{2 (a x-1)^4}-\frac {1}{8 \left (a^2 x^2-1\right )}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {\text {arctanh}(a x)}{8 a}+\frac {1}{8 a (1-a x)}+\frac {1}{8 a (1-a x)^2}+\frac {1}{6 a (1-a x)^3}\right )}{c^2 \sqrt {c-a^2 c x^2}}\)

Input:

Int[E^(3*ArcTanh[a*x])/(c - a^2*c*x^2)^(5/2),x]
 

Output:

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

Defintions of rubi rules used

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 6690

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

rule 6693

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {\left (3 \ln \left (a x +1\right ) x^{3} a^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}-9 \ln \left (a x +1\right ) x^{2} a^{2}+9 a^{2} \ln \left (a x -1\right ) x^{2}-6 a^{2} x^{2}+9 \ln \left (a x +1\right ) x a -9 a \ln \left (a x -1\right ) x +18 a x -3 \ln \left (a x +1\right )+3 \ln \left (a x -1\right )-20\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{48 \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right )^{3} c^{3} a}\)	\(148\)

Input:

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/48*(3*ln(a*x+1)*x^3*a^3-3*a^3*ln(a*x-1)*x^3-9*ln(a*x+1)*x^2*a^2+9*a^2*ln 
(a*x-1)*x^2-6*a^2*x^2+9*ln(a*x+1)*x*a-9*a*ln(a*x-1)*x+18*a*x-3*ln(a*x+1)+3 
*ln(a*x-1)-20)/(-a^2*x^2+1)^(1/2)/(a*x-1)^3*(-c*(a^2*x^2-1))^(1/2)/c^3/a
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.48 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{5} x^{5} - 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 3 \, a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) + 4 \, {\left (10 \, a^{3} x^{3} - 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{96 \, {\left (a^{6} c^{3} x^{5} - 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x + a c^{3}\right )}}, \frac {3 \, {\left (a^{5} x^{5} - 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 3 \, a x + 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) + 2 \, {\left (10 \, a^{3} x^{3} - 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{48 \, {\left (a^{6} c^{3} x^{5} - 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x + a c^{3}\right )}}\right ] \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm=" 
fricas")
 

Output:

[1/96*(3*(a^5*x^5 - 3*a^4*x^4 + 2*a^3*x^3 + 2*a^2*x^2 - 3*a*x + 1)*sqrt(c) 
*log(-(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a^2*c*x^2 - 4*(a^3*x^3 + a*x)*sqrt(-a^2 
*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x 
^2 - 1)) + 4*(10*a^3*x^3 - 27*a^2*x^2 + 21*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt( 
-a^2*x^2 + 1))/(a^6*c^3*x^5 - 3*a^5*c^3*x^4 + 2*a^4*c^3*x^3 + 2*a^3*c^3*x^ 
2 - 3*a^2*c^3*x + a*c^3), 1/48*(3*(a^5*x^5 - 3*a^4*x^4 + 2*a^3*x^3 + 2*a^2 
*x^2 - 3*a*x + 1)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1 
)*a*sqrt(-c)*x/(a^4*c*x^4 - c)) + 2*(10*a^3*x^3 - 27*a^2*x^2 + 21*a*x)*sqr 
t(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 - 3*a^5*c^3*x^4 + 2*a^4 
*c^3*x^3 + 2*a^3*c^3*x^2 - 3*a^2*c^3*x + a*c^3)]
 

Sympy [F]

\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)/(-a**2*c*x**2+c)**(5/2),x)
 

Output:

Integral((a*x + 1)**3/((-(a*x - 1)*(a*x + 1))**(3/2)*(-c*(a*x - 1)*(a*x + 
1))**(5/2)), x)
 

Maxima [F]

\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm=" 
maxima")
 

Output:

integrate((a*x + 1)^3/((-a^2*c*x^2 + c)^(5/2)*(-a^2*x^2 + 1)^(3/2)), x)
 

Giac [F]

\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm=" 
giac")
 

Output:

integrate((a*x + 1)^3/((-a^2*c*x^2 + c)^(5/2)*(-a^2*x^2 + 1)^(3/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {{\left (a\,x+1\right )}^3}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \] Input:

int((a*x + 1)^3/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(3/2)),x)
 

Output:

int((a*x + 1)^3/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.75 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-3 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+9 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-9 \,\mathrm {log}\left (a x -1\right ) a x +3 \,\mathrm {log}\left (a x -1\right )+3 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-9 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+9 \,\mathrm {log}\left (a x +1\right ) a x -3 \,\mathrm {log}\left (a x +1\right )-2 a^{3} x^{3}+12 a x -18\right )}{48 a \,c^{3} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x)
 

Output:

(sqrt(c)*( - 3*log(a*x - 1)*a**3*x**3 + 9*log(a*x - 1)*a**2*x**2 - 9*log(a 
*x - 1)*a*x + 3*log(a*x - 1) + 3*log(a*x + 1)*a**3*x**3 - 9*log(a*x + 1)*a 
**2*x**2 + 9*log(a*x + 1)*a*x - 3*log(a*x + 1) - 2*a**3*x**3 + 12*a*x - 18 
))/(48*a*c**3*(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))