\(\int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^2 \, dx\) [788]

Optimal result

Integrand size = 27, antiderivative size = 101 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^2 \, dx=-\frac {5 \sqrt {c-\frac {c}{a^2 x^2}} x}{3 a^2}-\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^2}{a}-\frac {1}{3} \sqrt {c-\frac {c}{a^2 x^2}} x^3-\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {c}}\right )}{a^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.83 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^2 \, dx=-\frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left (\sqrt {-1+a^2 x^2} \left (5+3 a x+a^2 x^2\right )+3 \log \left (a x+\sqrt {-1+a^2 x^2}\right )\right )}{3 a^2 \sqrt {-1+a^2 x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6709, 541, 25, 27, 533, 27, 455, 223}

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.

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

\(\Big \downarrow \) 6709

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

\(\Big \downarrow \) 541

\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {\int -\frac {a^2 x (6 a x+5)}{\sqrt {1-a^2 x^2}}dx}{3 a^2}-\frac {1}{3} x^2 \sqrt {1-a^2 x^2}\right )}{\sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {\int \frac {a^2 x (6 a x+5)}{\sqrt {1-a^2 x^2}}dx}{3 a^2}-\frac {1}{3} x^2 \sqrt {1-a^2 x^2}\right )}{\sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {1}{3} \int \frac {x (6 a x+5)}{\sqrt {1-a^2 x^2}}dx-\frac {1}{3} x^2 \sqrt {1-a^2 x^2}\right )}{\sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 223

\(\displaystyle \frac {x \left (\frac {1}{3} \left (\frac {\frac {3 \arcsin (a x)}{a}-\frac {5 \sqrt {1-a^2 x^2}}{a}}{a}-\frac {3 x \sqrt {1-a^2 x^2}}{a}\right )-\frac {1}{3} x^2 \sqrt {1-a^2 x^2}\right ) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}\)

Input:

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

Output:

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

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 223

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

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 6709

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.24

method	result	size
risch	\(-\frac {\left (a^{2} x^{2}+3 a x +5\right ) x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{3 a^{2}}-\frac {\ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c}\right ) x \sqrt {c \left (a^{2} x^{2}-1\right )}\, \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{a \sqrt {a^{2} c}\, \left (a^{2} x^{2}-1\right )}\)	\(125\)
default	\(\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \left (-{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a^{3}-3 \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{2} c x +3 c^{\frac {3}{2}} \ln \left (x \sqrt {c}+\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\right )-6 c^{\frac {3}{2}} \ln \left (\frac {\sqrt {c}\, \sqrt {\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}}+x c}{\sqrt {c}}\right )-6 \sqrt {\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}}\, a c \right )}{3 \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{3} c}\)	\(174\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.02 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^2 \, dx=\left [-\frac {2 \, {\left (a^{3} x^{3} + 3 \, a^{2} x^{2} + 5 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 3 \, \sqrt {c} \log \left (2 \, a^{2} c x^{2} - 2 \, a^{2} \sqrt {c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right )}{6 \, a^{3}}, -\frac {{\left (a^{3} x^{3} + 3 \, a^{2} x^{2} + 5 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 3 \, \sqrt {-c} \arctan \left (\frac {a^{2} \sqrt {-c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right )}{3 \, a^{3}}\right ] \] Input:

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

Output:

[-1/6*(2*(a^3*x^3 + 3*a^2*x^2 + 5*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - 3 
*sqrt(c)*log(2*a^2*c*x^2 - 2*a^2*sqrt(c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x^2 
)) - c))/a^3, -1/3*((a^3*x^3 + 3*a^2*x^2 + 5*a*x)*sqrt((a^2*c*x^2 - c)/(a^ 
2*x^2)) - 3*sqrt(-c)*arctan(a^2*sqrt(-c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x^2 
))/(a^2*c*x^2 - c)))/a^3]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.15 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^2 \, dx=-\frac {1}{6} \, {\left (2 \, \sqrt {a^{2} c x^{2} - c} {\left (x {\left (\frac {x \mathrm {sgn}\left (x\right )}{a^{2}} + \frac {3 \, \mathrm {sgn}\left (x\right )}{a^{3}}\right )} + \frac {5 \, \mathrm {sgn}\left (x\right )}{a^{4}}\right )} - \frac {6 \, \sqrt {c} \log \left ({\left | -\sqrt {a^{2} c} x + \sqrt {a^{2} c x^{2} - c} \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{3} {\left | a \right |}} + \frac {{\left (3 \, a \sqrt {c} \log \left ({\left | c \right |}\right ) - 10 \, \sqrt {-c} {\left | a \right |}\right )} \mathrm {sgn}\left (x\right )}{a^{4} {\left | a \right |}}\right )} {\left | a \right |} \] Input:

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

Output:

-1/6*(2*sqrt(a^2*c*x^2 - c)*(x*(x*sgn(x)/a^2 + 3*sgn(x)/a^3) + 5*sgn(x)/a^ 
4) - 6*sqrt(c)*log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 - c)))*sgn(x)/(a^3* 
abs(a)) + (3*a*sqrt(c)*log(abs(c)) - 10*sqrt(-c)*abs(a))*sgn(x)/(a^4*abs(a 
)))*abs(a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.68 \[ \int e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^2 \, dx=\frac {\sqrt {c}\, \left (-\sqrt {a^{2} x^{2}-1}\, a^{2} x^{2}-3 \sqrt {a^{2} x^{2}-1}\, a x -5 \sqrt {a^{2} x^{2}-1}-3 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}-1}+a x \right )\right )}{3 a^{3}} \] Input:

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

Output:

(sqrt(c)*( - sqrt(a**2*x**2 - 1)*a**2*x**2 - 3*sqrt(a**2*x**2 - 1)*a*x - 5 
*sqrt(a**2*x**2 - 1) - 3*log(sqrt(a**2*x**2 - 1) + a*x)))/(3*a**3)