\(\int e^{3 \text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx\) [407]

Optimal result

Integrand size = 23, antiderivative size = 157 \[ \int e^{3 \text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=-\frac {4 \sqrt {1+a x} \sqrt {c-a c x}}{a^3 \sqrt {1-a x}}-\frac {2 (1+a x)^{3/2} \sqrt {c-a c x}}{3 a^3 \sqrt {1-a x}}-\frac {2 (1+a x)^{7/2} \sqrt {c-a c x}}{7 a^3 \sqrt {1-a x}}+\frac {4 \sqrt {2} \sqrt {c-a c x} \text {arctanh}\left (\frac {\sqrt {1+a x}}{\sqrt {2}}\right )}{a^3 \sqrt {1-a x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.54 \[ \int e^{3 \text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=-\frac {2 \sqrt {c-a c x} \left (\sqrt {1+a x} \left (52+16 a x+9 a^2 x^2+3 a^3 x^3\right )-42 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+a x}}{\sqrt {2}}\right )\right )}{21 a^3 \sqrt {1-a x}} \] Input:

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

Output:

(-2*Sqrt[c - a*c*x]*(Sqrt[1 + a*x]*(52 + 16*a*x + 9*a^2*x^2 + 3*a^3*x^3) - 
 42*Sqrt[2]*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]]))/(21*a^3*Sqrt[1 - a*x])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6680, 37, 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.

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

\(\Big \downarrow \) 6680

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

\(\Big \downarrow \) 37

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 37

Int[(u_.)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> S 
imp[(a + b*x)^m/(c + d*x)^m   Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, 
b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !SimplerQ[a + b*x, c + d*x]
 

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 6680

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (-3 a^{3} x^{3} \sqrt {c \left (a x +1\right )}-9 a^{2} x^{2} \sqrt {c \left (a x +1\right )}+42 \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-16 a x \sqrt {c \left (a x +1\right )}-52 \sqrt {c \left (a x +1\right )}\right )}{21 \left (a x -1\right ) \sqrt {c \left (a x +1\right )}\, a^{3}}\)	\(129\)
risch	\(\frac {2 \left (3 a^{3} x^{3}+9 a^{2} x^{2}+16 a x +52\right ) \left (a x +1\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{21 a^{3} \sqrt {c \left (a x +1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}-\frac {4 \sqrt {2}\, \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right )}{a^{3} \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}\)	\(170\)

Input:

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

Output:

-2/21*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(-3*a^3*x^3*(c*(a*x+1))^(1/2)- 
9*a^2*x^2*(c*(a*x+1))^(1/2)+42*c^(1/2)*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/ 
2)*2^(1/2)/c^(1/2))-16*a*x*(c*(a*x+1))^(1/2)-52*(c*(a*x+1))^(1/2))/(a*x-1) 
/(c*(a*x+1))^(1/2)/a^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.62 \[ \int e^{3 \text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\left [\frac {2 \, {\left (21 \, \sqrt {2} {\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + {\left (3 \, a^{3} x^{3} + 9 \, a^{2} x^{2} + 16 \, a x + 52\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}\right )}}{21 \, {\left (a^{4} x - a^{3}\right )}}, \frac {2 \, {\left (42 \, \sqrt {2} {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{2 \, {\left (a c x - c\right )}}\right ) + {\left (3 \, a^{3} x^{3} + 9 \, a^{2} x^{2} + 16 \, a x + 52\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}\right )}}{21 \, {\left (a^{4} x - a^{3}\right )}}\right ] \] Input:

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

Output:

[2/21*(21*sqrt(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)) 
+ (3*a^3*x^3 + 9*a^2*x^2 + 16*a*x + 52)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c 
))/(a^4*x - a^3), 2/21*(42*sqrt(2)*(a*x - 1)*sqrt(-c)*arctan(1/2*sqrt(2)*s 
qrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a*c*x - c)) + (3*a^3*x^3 + 9* 
a^2*x^2 + 16*a*x + 52)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^4*x - a^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.83 \[ \int e^{3 \text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {2 \, c^{2} {\left (\frac {2 \, \sqrt {2} {\left (21 \, c \arctan \left (\frac {\sqrt {c}}{\sqrt {-c}}\right ) + 40 \, \sqrt {-c} \sqrt {c}\right )}}{a^{2} \sqrt {-c} c} - \frac {\frac {42 \, \sqrt {2} c^{4} \arctan \left (\frac {\sqrt {2} \sqrt {a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} + 3 \, {\left (a c x + c\right )}^{\frac {7}{2}} + 7 \, {\left (a c x + c\right )}^{\frac {3}{2}} c^{2} + 42 \, \sqrt {a c x + c} c^{3}}{a^{2} c^{4}}\right )}}{21 \, a {\left | c \right |}} \] Input:

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

Output:

2/21*c^2*(2*sqrt(2)*(21*c*arctan(sqrt(c)/sqrt(-c)) + 40*sqrt(-c)*sqrt(c))/ 
(a^2*sqrt(-c)*c) - (42*sqrt(2)*c^4*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt 
(-c))/sqrt(-c) + 3*(a*c*x + c)^(7/2) + 7*(a*c*x + c)^(3/2)*c^2 + 42*sqrt(a 
*c*x + c)*c^3)/(a^2*c^4))/(a*abs(c))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.50 \[ \int e^{3 \text {arctanh}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c}\, \left (-3 \sqrt {a x +1}\, a^{3} x^{3}-9 \sqrt {a x +1}\, a^{2} x^{2}-16 \sqrt {a x +1}\, a x -52 \sqrt {a x +1}-42 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )+80 \sqrt {2}\right )}{21 a^{3}} \] Input:

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

Output:

(2*sqrt(c)*( - 3*sqrt(a*x + 1)*a**3*x**3 - 9*sqrt(a*x + 1)*a**2*x**2 - 16* 
sqrt(a*x + 1)*a*x - 52*sqrt(a*x + 1) - 42*sqrt(2)*log(tan(asin(sqrt( - a*x 
 + 1)/sqrt(2))/2)) + 80*sqrt(2)))/(21*a**3)