\(\int e^{3 \text {arctanh}(a x)} x \sqrt {c-a c x} \, dx\) [408]

Optimal result

Integrand size = 21, antiderivative size = 157 \[ \int e^{3 \text {arctanh}(a x)} x \sqrt {c-a c x} \, dx=-\frac {4 \sqrt {1+a x} \sqrt {c-a c x}}{a^2 \sqrt {1-a x}}-\frac {2 (1+a x)^{3/2} \sqrt {c-a c x}}{3 a^2 \sqrt {1-a x}}-\frac {2 (1+a x)^{5/2} \sqrt {c-a c x}}{5 a^2 \sqrt {1-a x}}+\frac {4 \sqrt {2} \sqrt {c-a c x} \text {arctanh}\left (\frac {\sqrt {1+a x}}{\sqrt {2}}\right )}{a^2 \sqrt {1-a x}} \] Output:

-4*(a*x+1)^(1/2)*(-a*c*x+c)^(1/2)/a^2/(-a*x+1)^(1/2)-2/3*(a*x+1)^(3/2)*(-a 
*c*x+c)^(1/2)/a^2/(-a*x+1)^(1/2)-2/5*(a*x+1)^(5/2)*(-a*c*x+c)^(1/2)/a^2/(- 
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^2/(-a*x+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.48 \[ \int e^{3 \text {arctanh}(a x)} x \sqrt {c-a c x} \, dx=-\frac {2 \sqrt {c-a c x} \left (\sqrt {1+a x} \left (38+11 a x+3 a^2 x^2\right )-30 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+a x}}{\sqrt {2}}\right )\right )}{15 a^2 \sqrt {1-a x}} \] Input:

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

Output:

(-2*Sqrt[c - a*c*x]*(Sqrt[1 + a*x]*(38 + 11*a*x + 3*a^2*x^2) - 30*Sqrt[2]* 
ArcTanh[Sqrt[1 + a*x]/Sqrt[2]]))/(15*a^2*Sqrt[1 - a*x])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6680, 37, 90, 60, 60, 73, 219}

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.

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

\(\Big \downarrow \) 6680

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

\(\Big \downarrow \) 37

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(Sqrt[c - a*c*x]*((-2*(1 + a*x)^(5/2))/(5*a^2) + ((-2*(1 + a*x)^(3/2))/(3* 
a) + 2*((-2*Sqrt[1 + a*x])/a + (2*Sqrt[2]*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]])/ 
a))/a))/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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 = 112, normalized size of antiderivative = 0.71

method	result	size
default	\(-\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (-3 a^{2} x^{2} \sqrt {c \left (a x +1\right )}+30 \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-11 a x \sqrt {c \left (a x +1\right )}-38 \sqrt {c \left (a x +1\right )}\right )}{15 \left (a x -1\right ) \sqrt {c \left (a x +1\right )}\, a^{2}}\)	\(112\)
risch	\(\frac {2 \left (3 a^{2} x^{2}+11 a x +38\right ) \left (a x +1\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{15 a^{2} \sqrt {c \left (a x +1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}-\frac {4 \sqrt {2}\, \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right )}{a^{2} \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}\)	\(162\)

Input:

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

Output:

-2/15*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(-3*a^2*x^2*(c*(a*x+1))^(1/2)+ 
30*c^(1/2)*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))-11*a*x*( 
c*(a*x+1))^(1/2)-38*(c*(a*x+1))^(1/2))/(a*x-1)/(c*(a*x+1))^(1/2)/a^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.52 \[ \int e^{3 \text {arctanh}(a x)} x \sqrt {c-a c x} \, dx=\left [\frac {2 \, {\left (15 \, \sqrt {2} {\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + {\left (3 \, a^{2} x^{2} + 11 \, a x + 38\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}\right )}}{15 \, {\left (a^{3} x - a^{2}\right )}}, \frac {2 \, {\left (30 \, \sqrt {2} {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{2 \, {\left (a c x - c\right )}}\right ) + {\left (3 \, a^{2} x^{2} + 11 \, a x + 38\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}\right )}}{15 \, {\left (a^{3} x - a^{2}\right )}}\right ] \] Input:

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

Output:

[2/15*(15*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^2*x^2 + 11*a*x + 38)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^3*x - 
a^2), 2/15*(30*sqrt(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)) + (3*a^2*x^2 + 11*a*x + 38)*s 
qrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^3*x - a^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int e^{3 \text {arctanh}(a x)} x \sqrt {c-a c x} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.41 \[ \int e^{3 \text {arctanh}(a x)} x \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c}\, \left (-3 \sqrt {a x +1}\, a^{2} x^{2}-11 \sqrt {a x +1}\, a x -38 \sqrt {a x +1}-30 \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )+52 \sqrt {2}\right )}{15 a^{2}} \] Input:

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

Output:

(2*sqrt(c)*( - 3*sqrt(a*x + 1)*a**2*x**2 - 11*sqrt(a*x + 1)*a*x - 38*sqrt( 
a*x + 1) - 30*sqrt(2)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2)) + 52*sqrt 
(2)))/(15*a**2)