\(\int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx\) [284]

Optimal result

Integrand size = 20, antiderivative size = 96 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=-\frac {1-3 a x}{4 a c^3 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {2} \sqrt {c-a c x}}\right )}{4 \sqrt {2} a c^{7/2}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=-\frac {(1-a x)^{3/2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {1}{2} (1+a x)\right )}{2 a c^2 \sqrt {1+a x} (c-a c x)^{3/2}} \] Input:

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^(7/2)),x]
 

Output:

-1/2*((1 - a*x)^(3/2)*Hypergeometric2F1[-1/2, 2, 1/2, (1 + a*x)/2])/(a*c^2 
*Sqrt[1 + a*x]*(c - a*c*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.31, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6677, 470, 467, 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.

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

\(\Big \downarrow \) 6677

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

\(\Big \downarrow \) 470

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

\(\Big \downarrow \) 467

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

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^(7/2)),x]
 

Output:

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

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 467

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

rule 470

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 
2*p + 2)/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 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 = 124, normalized size of antiderivative = 1.29

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, a x \sqrt {c \left (a x +1\right )}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {c \left (a x +1\right )}-6 \sqrt {c}\, a x +2 \sqrt {c}\right )}{8 c^{\frac {9}{2}} \left (a x -1\right )^{2} \left (a x +1\right ) a}\)	\(124\)

Input:

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

Output:

-1/8*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(9/2)*(3*arctanh(1/2*(c*(a*x+ 
1))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*a*x*(c*(a*x+1))^(1/2)-3*2^(1/2)*arctanh 
(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*(c*(a*x+1))^(1/2)-6*c^(1/2)*a*x+2* 
c^(1/2))/(a*x-1)^2/(a*x+1)/a
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 307, normalized size of antiderivative = 3.20 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (3 \, a x - 1\right )}}{16 \, {\left (a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - a^{2} c^{4} x + a c^{4}\right )}}, \frac {3 \, \sqrt {2} {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{2 \, {\left (a c x - c\right )}}\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (3 \, a x - 1\right )}}{8 \, {\left (a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - a^{2} c^{4} x + a c^{4}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=-\frac {{\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{a \sqrt {-c} c^{2}} + \frac {2 \, {\left (3 \, a c x - c\right )}}{{\left ({\left (a c x + c\right )}^{\frac {3}{2}} - 2 \, \sqrt {a c x + c} c\right )} a c^{2}}\right )} {\left | c \right |}}{8 \, c^{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^{7/2}} \, dx=\frac {\sqrt {c}\, \left (-12 \sqrt {a x +1}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right ) a x +12 \sqrt {a x +1}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )+9 \sqrt {a x +1}\, \sqrt {2}\, a x -9 \sqrt {a x +1}\, \sqrt {2}-24 a x +8\right )}{32 \sqrt {a x +1}\, a \,c^{4} \left (a x -1\right )} \] Input:

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

Output:

(sqrt(c)*( - 12*sqrt(a*x + 1)*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2 
))/2))*a*x + 12*sqrt(a*x + 1)*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2 
))/2)) + 9*sqrt(a*x + 1)*sqrt(2)*a*x - 9*sqrt(a*x + 1)*sqrt(2) - 24*a*x + 
8))/(32*sqrt(a*x + 1)*a*c**4*(a*x - 1))