\(\int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx\) [257]

Optimal result

Integrand size = 20, antiderivative size = 122 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {3 \sqrt {1-a^2 x^2}}{4 a (c-a c x)^{3/2}}+\frac {c^2 \left (1-a^2 x^2\right )^{3/2}}{2 a (c-a c x)^{7/2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{4 \sqrt {2} a c^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\frac {\sqrt {c-a c x} \left (-2+8 a x+10 a^2 x^2+3 (-1+a x)^2 \sqrt {2+2 a x} \text {arctanh}\left (\frac {\sqrt {1+a x}}{\sqrt {2}}\right )\right )}{8 a c^2 (-1+a x)^2 \sqrt {1-a^2 x^2}} \] Input:

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

Output:

(Sqrt[c - a*c*x]*(-2 + 8*a*x + 10*a^2*x^2 + 3*(-1 + a*x)^2*Sqrt[2 + 2*a*x] 
*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]]))/(8*a*c^2*(-1 + a*x)^2*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6677, 465, 465, 471, 221}

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.

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

\(\Big \downarrow \) 6677

\(\displaystyle c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(c-a c x)^{9/2}}dx\)

\(\Big \downarrow \) 465

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

\(\Big \downarrow \) 465

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

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

\(\displaystyle c^3 \left (\frac {\left (1-a^2 x^2\right )^{3/2}}{2 a c (c-a c x)^{7/2}}-\frac {3 \left (\frac {\sqrt {1-a^2 x^2}}{a c (c-a c x)^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{\sqrt {2} a c^{5/2}}\right )}{4 c^2}\right )\)

Input:

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

Output:

c^3*((1 - a^2*x^2)^(3/2)/(2*a*c*(c - a*c*x)^(7/2)) - (3*(Sqrt[1 - a^2*x^2] 
/(a*c*(c - a*c*x)^(3/2)) - ArcTanh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sq 
rt[c - a*c*x])]/(Sqrt[2]*a*c^(5/2))))/(4*c^2))
 

Defintions of rubi rules used

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 465

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

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

rule 6677

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a^{2} c \,x^{2}-6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) a c x +10 a x \sqrt {c \left (a x +1\right )}\, \sqrt {c}+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -2 \sqrt {c \left (a x +1\right )}\, \sqrt {c}\right )}{8 c^{\frac {5}{2}} \left (a x -1\right )^{3} \sqrt {c \left (a x +1\right )}\, a}\)	\(158\)

Input:

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

Output:

-1/8*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(5/2)*(3*2^(1/2)*arctanh(1/2* 
(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*a^2*c*x^2-6*2^(1/2)*arctanh(1/2*(c*(a*x 
+1))^(1/2)*2^(1/2)/c^(1/2))*a*c*x+10*a*x*(c*(a*x+1))^(1/2)*c^(1/2)+3*2^(1/ 
2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*c-2*(c*(a*x+1))^(1/2)*c^ 
(1/2))/(a*x-1)^3/(c*(a*x+1))^(1/2)/a
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.53 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 4 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (5 \, a x - 1\right )}}{16 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}}, \frac {3 \, \sqrt {2} {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{2 \, {\left (a c x - c\right )}}\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (5 \, a x - 1\right )}}{8 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}}\right ] \] Input:

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

Output:

[1/16*(3*sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(c)*log(-(a^2*c*x^2 
 + 2*a*c*x - 2*sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/ 
(a^2*x^2 - 2*a*x + 1)) - 4*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(5*a*x - 1) 
)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2), 1/8*(3*sqrt(2)*(a^3 
*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a^2*x^2 + 
1)*sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) - 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c 
*x + c)*(5*a*x - 1))/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.60 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} - \frac {2 \, {\left (5 \, {\left (a c x + c\right )}^{\frac {3}{2}} - 6 \, \sqrt {a c x + c} c\right )}}{{\left (a c x - c\right )}^{2}}}{8 \, a {\left | c \right |}} \] Input:

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

Output:

-1/8*(3*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/sqrt(-c) - 2* 
(5*(a*c*x + c)^(3/2) - 6*sqrt(a*c*x + c)*c)/(a*c*x - c)^2)/(a*abs(c))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{(c-a c x)^{3/2}} \, dx=\frac {\sqrt {c}\, \left (10 \sqrt {a x +1}\, a x -2 \sqrt {a x +1}-3 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a^{2} x^{2}+6 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a x -3 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )\right )}{8 a \,c^{2} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:

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

Output:

(sqrt(c)*(10*sqrt(a*x + 1)*a*x - 2*sqrt(a*x + 1) - 3*sqrt(2)*log(tan(asin( 
sqrt( - a*x + 1)/sqrt(2))/2))*a**2*x**2 + 6*sqrt(2)*log(tan(asin(sqrt( - a 
*x + 1)/sqrt(2))/2))*a*x - 3*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2) 
)/2))))/(8*a*c**2*(a**2*x**2 - 2*a*x + 1))