\(\int e^{2 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx\) [1100]

Optimal result

Integrand size = 27, antiderivative size = 129 \[ \int e^{2 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=-\frac {4 \sqrt {c-a^2 c x^2}}{3 a^3}-\frac {7 x \sqrt {c-a^2 c x^2}}{8 a^2}-\frac {2 x^2 \sqrt {c-a^2 c x^2}}{3 a}-\frac {1}{4} x^3 \sqrt {c-a^2 c x^2}+\frac {7 \sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{8 a^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.68 \[ \int e^{2 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=-\frac {\sqrt {c-a^2 c x^2} \left (32+21 a x+16 a^2 x^2+6 a^3 x^3\right )+21 \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )}{24 a^3} \] Input:

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

Output:

-1/24*(Sqrt[c - a^2*c*x^2]*(32 + 21*a*x + 16*a^2*x^2 + 6*a^3*x^3) + 21*Sqr 
t[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/a^3
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.24, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {6701, 541, 25, 27, 533, 27, 533, 27, 455, 224, 216}

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.

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

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 541

\(\displaystyle c \left (-\frac {\int -\frac {a^2 c x^2 (8 a x+7)}{\sqrt {c-a^2 c x^2}}dx}{4 a^2 c}-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {\int \frac {a^2 c x^2 (8 a x+7)}{\sqrt {c-a^2 c x^2}}dx}{4 a^2 c}-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{4} \int \frac {x^2 (8 a x+7)}{\sqrt {c-a^2 c x^2}}dx-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 533

\(\displaystyle c \left (\frac {1}{4} \left (\frac {\int \frac {a c x (21 a x+16)}{\sqrt {c-a^2 c x^2}}dx}{3 a^2 c}-\frac {8 x^2 \sqrt {c-a^2 c x^2}}{3 a c}\right )-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{4} \left (\frac {\int \frac {x (21 a x+16)}{\sqrt {c-a^2 c x^2}}dx}{3 a}-\frac {8 x^2 \sqrt {c-a^2 c x^2}}{3 a c}\right )-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 533

\(\displaystyle c \left (\frac {1}{4} \left (\frac {\frac {\int \frac {a c (32 a x+21)}{\sqrt {c-a^2 c x^2}}dx}{2 a^2 c}-\frac {21 x \sqrt {c-a^2 c x^2}}{2 a c}}{3 a}-\frac {8 x^2 \sqrt {c-a^2 c x^2}}{3 a c}\right )-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{4} \left (\frac {\frac {\int \frac {32 a x+21}{\sqrt {c-a^2 c x^2}}dx}{2 a}-\frac {21 x \sqrt {c-a^2 c x^2}}{2 a c}}{3 a}-\frac {8 x^2 \sqrt {c-a^2 c x^2}}{3 a c}\right )-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 455

\(\displaystyle c \left (\frac {1}{4} \left (\frac {\frac {21 \int \frac {1}{\sqrt {c-a^2 c x^2}}dx-\frac {32 \sqrt {c-a^2 c x^2}}{a c}}{2 a}-\frac {21 x \sqrt {c-a^2 c x^2}}{2 a c}}{3 a}-\frac {8 x^2 \sqrt {c-a^2 c x^2}}{3 a c}\right )-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle c \left (\frac {1}{4} \left (\frac {\frac {21 \int \frac {1}{\frac {a^2 c x^2}{c-a^2 c x^2}+1}d\frac {x}{\sqrt {c-a^2 c x^2}}-\frac {32 \sqrt {c-a^2 c x^2}}{a c}}{2 a}-\frac {21 x \sqrt {c-a^2 c x^2}}{2 a c}}{3 a}-\frac {8 x^2 \sqrt {c-a^2 c x^2}}{3 a c}\right )-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle c \left (\frac {1}{4} \left (\frac {\frac {\frac {21 \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a \sqrt {c}}-\frac {32 \sqrt {c-a^2 c x^2}}{a c}}{2 a}-\frac {21 x \sqrt {c-a^2 c x^2}}{2 a c}}{3 a}-\frac {8 x^2 \sqrt {c-a^2 c x^2}}{3 a c}\right )-\frac {x^3 \sqrt {c-a^2 c x^2}}{4 c}\right )\)

Input:

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

Output:

c*(-1/4*(x^3*Sqrt[c - a^2*c*x^2])/c + ((-8*x^2*Sqrt[c - a^2*c*x^2])/(3*a*c 
) + ((-21*x*Sqrt[c - a^2*c*x^2])/(2*a*c) + ((-32*Sqrt[c - a^2*c*x^2])/(a*c 
) + (21*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(a*Sqrt[c]))/(2*a))/(3* 
a))/4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

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

rule 533

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

rule 541

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 6701

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {\left (6 a^{3} x^{3}+16 a^{2} x^{2}+21 a x +32\right ) \left (a^{2} x^{2}-1\right ) c}{24 a^{3} \sqrt {-c \left (a^{2} x^{2}-1\right )}}+\frac {7 \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c}{8 a^{2} \sqrt {a^{2} c}}\)	\(89\)
default	\(\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4 a^{2} c}-\frac {9 \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4 a^{2}}+\frac {2 \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3 a^{3} c}-\frac {2 \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}-\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )}{a^{3}}\)	\(185\)

Input:

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

Output:

1/24*(6*a^3*x^3+16*a^2*x^2+21*a*x+32)*(a^2*x^2-1)/a^3/(-c*(a^2*x^2-1))^(1/ 
2)*c+7/8/a^2/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))*c
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.30 \[ \int e^{2 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\left [-\frac {2 \, {\left (6 \, a^{3} x^{3} + 16 \, a^{2} x^{2} + 21 \, a x + 32\right )} \sqrt {-a^{2} c x^{2} + c} - 21 \, \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right )}{48 \, a^{3}}, -\frac {{\left (6 \, a^{3} x^{3} + 16 \, a^{2} x^{2} + 21 \, a x + 32\right )} \sqrt {-a^{2} c x^{2} + c} + 21 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right )}{24 \, a^{3}}\right ] \] Input:

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

Output:

[-1/48*(2*(6*a^3*x^3 + 16*a^2*x^2 + 21*a*x + 32)*sqrt(-a^2*c*x^2 + c) - 21 
*sqrt(-c)*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c))/a^3, 
 -1/24*((6*a^3*x^3 + 16*a^2*x^2 + 21*a*x + 32)*sqrt(-a^2*c*x^2 + c) + 21*s 
qrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)))/a^3]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74 \[ \int e^{2 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=-\frac {1}{24} \, a {\left (\frac {27 \, \sqrt {-a^{2} c x^{2} + c} x}{a^{3}} - \frac {6 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x}{a^{3} c} - \frac {21 \, \sqrt {c} \arcsin \left (a x\right )}{a^{4}} + \frac {48 \, \sqrt {-a^{2} c x^{2} + c}}{a^{4}} - \frac {16 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{a^{4} c}\right )} \] Input:

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

Output:

-1/24*a*(27*sqrt(-a^2*c*x^2 + c)*x/a^3 - 6*(-a^2*c*x^2 + c)^(3/2)*x/(a^3*c 
) - 21*sqrt(c)*arcsin(a*x)/a^4 + 48*sqrt(-a^2*c*x^2 + c)/a^4 - 16*(-a^2*c* 
x^2 + c)^(3/2)/(a^4*c))
 

Giac [A] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.65 \[ \int e^{2 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=-\frac {1}{24} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (2 \, {\left (3 \, x + \frac {8}{a}\right )} x + \frac {21}{a^{2}}\right )} x + \frac {32}{a^{3}}\right )} - \frac {7 \, c \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{8 \, a^{2} \sqrt {-c} {\left | a \right |}} \] Input:

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

Output:

-1/24*sqrt(-a^2*c*x^2 + c)*((2*(3*x + 8/a)*x + 21/a^2)*x + 32/a^3) - 7/8*c 
*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(a^2*sqrt(-c)*abs(a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.63 \[ \int e^{2 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c}\, \left (21 \mathit {asin} \left (a x \right )-6 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-16 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-21 \sqrt {-a^{2} x^{2}+1}\, a x -32 \sqrt {-a^{2} x^{2}+1}+32\right )}{24 a^{3}} \] Input:

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

Output:

(sqrt(c)*(21*asin(a*x) - 6*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 16*sqrt( - a 
**2*x**2 + 1)*a**2*x**2 - 21*sqrt( - a**2*x**2 + 1)*a*x - 32*sqrt( - a**2* 
x**2 + 1) + 32))/(24*a**3)