\(\int \frac {e^{n \text {arctanh}(a x)} x^2}{(c-a^2 c x^2)^{3/2}} \, dx\) [1380]

Optimal result

Integrand size = 27, antiderivative size = 153 \[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {e^{n \text {arctanh}(a x)} (n-a x)}{a^3 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}}+\frac {2^{\frac {1+n}{2}} (1-a x)^{\frac {1-n}{2}} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-a x)\right )}{a^3 c (1-n) \sqrt {c-a^2 c x^2}} \] Output:

-exp(n*arctanh(a*x))*(-a*x+n)/a^3/c/(-n^2+1)/(-a^2*c*x^2+c)^(1/2)+2^(1/2+1 
/2*n)*(-a*x+1)^(1/2-1/2*n)*(-a^2*x^2+1)^(1/2)*hypergeom([1/2-1/2*n, 1/2-1/ 
2*n],[3/2-1/2*n],-1/2*a*x+1/2)/a^3/c/(1-n)/(-a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.26 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.01 \[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {(1-a x)^{\frac {1}{2} (-1-n)} \sqrt {1-a^2 x^2} \left ((n-a x) (1+a x)^{n/2}+2^{\frac {1+n}{2}} (1+n) (-1+a x) \sqrt {1+a x} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-a x)\right )\right )}{a^3 c (-1+n) (1+n) \sqrt {1+a x} \sqrt {c-a^2 c x^2}} \] Input:

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

Output:

((1 - a*x)^((-1 - n)/2)*Sqrt[1 - a^2*x^2]*((n - a*x)*(1 + a*x)^(n/2) + 2^( 
(1 + n)/2)*(1 + n)*(-1 + a*x)*Sqrt[1 + a*x]*Hypergeometric2F1[(1 - n)/2, ( 
1 - n)/2, (3 - n)/2, (1 - a*x)/2]))/(a^3*c*(-1 + n)*(1 + n)*Sqrt[1 + a*x]* 
Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6697, 6693, 6690, 79}

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.

\(\displaystyle \int \frac {x^2 e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 6697

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

\(\Big \downarrow \) 6693

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

\(\Big \downarrow \) 6690

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

\(\Big \downarrow \) 79

\(\displaystyle \frac {2^{\frac {n+1}{2}} \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-a x)\right )}{a^3 c (1-n) \sqrt {c-a^2 c x^2}}-\frac {(n-a x) e^{n \text {arctanh}(a x)}}{a^3 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}}\)

Input:

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

Output:

-((E^(n*ArcTanh[a*x])*(n - a*x))/(a^3*c*(1 - n^2)*Sqrt[c - a^2*c*x^2])) + 
(2^((1 + n)/2)*(1 - a*x)^((1 - n)/2)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[( 
1 - n)/2, (1 - n)/2, (3 - n)/2, (1 - a*x)/2])/(a^3*c*(1 - n)*Sqrt[c - a^2* 
c*x^2])
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 6690

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

rule 6693

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

rule 6697

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^2*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[(-(n + 2*(p + 1)*a*x))*(c + d*x^2)^(p + 1)*(E^(n*ArcTanh[a*x])/( 
a*d*(n^2 - 4*(p + 1)^2))), x] + Simp[(n^2 + 2*(p + 1))/(d*(n^2 - 4*(p + 1)^ 
2))   Int[(c + d*x^2)^(p + 1)*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, 
 n}, x] && EqQ[a^2*c + d, 0] && LtQ[p, -1] &&  !IntegerQ[n] && NeQ[n^2 - 4* 
(p + 1)^2, 0] && IntegerQ[2*p]
 

Maple [F]

\[\int \frac {{\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )} x^{2}}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}d x\]

Input:

int(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x)
 

Output:

int(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x)
 

Fricas [F]

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

integrate(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="frica 
s")
 

Output:

integral(sqrt(-a^2*c*x^2 + c)*x^2*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(a^4*c^2* 
x^4 - 2*a^2*c^2*x^2 + c^2), x)
 

Sympy [F]

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

integrate(exp(n*atanh(a*x))*x**2/(-a**2*c*x**2+c)**(3/2),x)
 

Output:

Integral(x**2*exp(n*atanh(a*x))/(-c*(a*x - 1)*(a*x + 1))**(3/2), x)
 

Maxima [F]

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

integrate(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxim 
a")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

int((x^2*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(3/2),x)
 

Output:

int((x^2*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(3/2), x)
 

Reduce [F]

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

int(exp(n*atanh(a*x))*x^2/(-a^2*c*x^2+c)^(3/2),x)
 

Output:

( - int((e**(atanh(a*x)*n)*x**2)/(sqrt( - a**2*x**2 + 1)*a**2*x**2 - sqrt( 
 - a**2*x**2 + 1)),x))/(sqrt(c)*c)