\(\int \frac {e^{\text {arctanh}(a x)} x^2}{(c-a^2 c x^2)^{5/2}} \, dx\) [1017]

Optimal result

Integrand size = 25, antiderivative size = 184 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{8 a^3 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{4 a^3 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a^3 c^2 (1+a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{8 a^3 c^2 \sqrt {c-a^2 c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.46 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-2+3 a x+a^2 x^2-(-1+a x)^2 (1+a x) \text {arctanh}(a x)\right )}{8 a^3 c^2 (-1+a x)^2 (1+a x) \sqrt {c-a^2 c x^2}} \] Input:

Integrate[(E^ArcTanh[a*x]*x^2)/(c - a^2*c*x^2)^(5/2),x]
 

Output:

(Sqrt[1 - a^2*x^2]*(-2 + 3*a*x + a^2*x^2 - (-1 + a*x)^2*(1 + a*x)*ArcTanh[ 
a*x]))/(8*a^3*c^2*(-1 + a*x)^2*(1 + a*x)*Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 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.

\(\displaystyle \int \frac {x^2 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6703

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

Int[(E^ArcTanh[a*x]*x^2)/(c - a^2*c*x^2)^(5/2),x]
 

Output:

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

Defintions of rubi rules used

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 6700

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

rule 6703

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {\left (\ln \left (a x +1\right ) x^{3} a^{3}-a^{3} \ln \left (a x -1\right ) x^{3}-\ln \left (a x +1\right ) x^{2} a^{2}+a^{2} \ln \left (a x -1\right ) x^{2}-2 a^{2} x^{2}-\ln \left (a x +1\right ) x a +a \ln \left (a x -1\right ) x -6 a x +\ln \left (a x +1\right )-\ln \left (a x -1\right )+4\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{16 \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right )^{2} \left (a x +1\right ) c^{3} a^{3}}\)	\(150\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVE 
RBOSE)
 

Output:

-1/16*(ln(a*x+1)*x^3*a^3-a^3*ln(a*x-1)*x^3-ln(a*x+1)*x^2*a^2+a^2*ln(a*x-1) 
*x^2-2*a^2*x^2-ln(a*x+1)*x*a+a*ln(a*x-1)*x-6*a*x+ln(a*x+1)-ln(a*x-1)+4)/(- 
a^2*x^2+1)^(1/2)/(a*x-1)^2/(a*x+1)*(-c*(a^2*x^2-1))^(1/2)/c^3/a^3
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.48 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\left [\frac {{\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} + 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) - 4 \, {\left (2 \, a^{3} x^{3} - a^{2} x^{2} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{32 \, {\left (a^{8} c^{3} x^{5} - a^{7} c^{3} x^{4} - 2 \, a^{6} c^{3} x^{3} + 2 \, a^{5} c^{3} x^{2} + a^{4} c^{3} x - a^{3} c^{3}\right )}}, -\frac {{\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) + 2 \, {\left (2 \, a^{3} x^{3} - a^{2} x^{2} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{16 \, {\left (a^{8} c^{3} x^{5} - a^{7} c^{3} x^{4} - 2 \, a^{6} c^{3} x^{3} + 2 \, a^{5} c^{3} x^{2} + a^{4} c^{3} x - a^{3} c^{3}\right )}}\right ] \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm 
="fricas")
 

Output:

[1/32*((a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*sqrt(c)*log(- 
(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a^2*c*x^2 + 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x^2 
 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1 
)) - 4*(2*a^3*x^3 - a^2*x^2 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1) 
)/(a^8*c^3*x^5 - a^7*c^3*x^4 - 2*a^6*c^3*x^3 + 2*a^5*c^3*x^2 + a^4*c^3*x - 
 a^3*c^3), -1/16*((a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*sq 
rt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*sqrt(-c)*x/(a^4* 
c*x^4 - c)) + 2*(2*a^3*x^3 - a^2*x^2 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2 
*x^2 + 1))/(a^8*c^3*x^5 - a^7*c^3*x^4 - 2*a^6*c^3*x^3 + 2*a^5*c^3*x^2 + a^ 
4*c^3*x - a^3*c^3)]
 

Sympy [F]

\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^{2} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**2/(-a**2*c*x**2+c)**(5/2),x)
 

Output:

Integral(x**2*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1 
))**(5/2)), x)
 

Maxima [F]

\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{2}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm 
="maxima")
 

Output:

integrate((a*x + 1)*x^2/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)), x)
 

Giac [F]

\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{2}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm 
="giac")
 

Output:

integrate((a*x + 1)*x^2/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^2\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:

int((x^2*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)),x)
 

Output:

int((x^2*(a*x + 1))/((c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-\mathrm {log}\left (a x -1\right ) a x +\mathrm {log}\left (a x -1\right )-\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+\mathrm {log}\left (a x +1\right ) a x -\mathrm {log}\left (a x +1\right )+2 a^{3} x^{3}+4 a x -2\right )}{16 a^{3} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(5/2),x)
 

Output:

(sqrt(c)*(log(a*x - 1)*a**3*x**3 - log(a*x - 1)*a**2*x**2 - log(a*x - 1)*a 
*x + log(a*x - 1) - log(a*x + 1)*a**3*x**3 + log(a*x + 1)*a**2*x**2 + log( 
a*x + 1)*a*x - log(a*x + 1) + 2*a**3*x**3 + 4*a*x - 2))/(16*a**3*c**3*(a** 
3*x**3 - a**2*x**2 - a*x + 1))