\(\int \frac {e^{n \text {arctanh}(a x)} x}{\sqrt {c-a^2 c x^2}} \, dx\) [1374]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 169 \[ \int \frac {e^{n \text {arctanh}(a x)} x}{\sqrt {c-a^2 c x^2}} \, dx=-\frac {(1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1+n}{2}} \sqrt {1-a^2 x^2}}{a^2 \sqrt {c-a^2 c x^2}}-\frac {2^{\frac {1+n}{2}} n (1-a x)^{\frac {1-n}{2}} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-a x)\right )}{a^2 (1-n) \sqrt {c-a^2 c x^2}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.15 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int \frac {e^{n \text {arctanh}(a x)} x}{\sqrt {c-a^2 c x^2}} \, dx=\frac {(1-a x)^{\frac {1}{2}-\frac {n}{2}} \sqrt {1-a^2 x^2} \left (-\left ((-1+n) (1+a x)^{\frac {1+n}{2}}\right )+2^{\frac {3+n}{2}} n \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-\frac {n}{2},\frac {1}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},\frac {1}{2}-\frac {a x}{2}\right )\right )}{a^2 \left (-1+n^2\right ) \sqrt {c-a^2 c x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 88, 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.

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

\(\Big \downarrow \) 6703

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 88

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

(Sqrt[1 - a^2*x^2]*(-(((1 - a*x)^((1 - n)/2)*(1 + a*x)^((1 + n)/2))/(a^2*( 
1 + n))) - (2^((3 + n)/2)*n*(1 - a*x)^((1 - n)/2)*Hypergeometric2F1[(-1 - 
n)/2, (1 - n)/2, (3 - n)/2, (1 - a*x)/2])/(a^2*(1 - n)*(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 88
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], 
 x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimpl 
erQ[p, 1]
 

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 [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

int((x*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(1/2),x)
 

Output:

int((x*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(1/2), x)
 

Reduce [F]

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

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

Output:

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