3.2.42 \(\int e^{3 \text {arctanh}(a x)} x^m \, dx\) [142]

3.2.42.1 Optimal result

Integrand size = 12, antiderivative size = 151 \[ \int e^{3 \text {arctanh}(a x)} x^m \, dx=-\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}-\frac {a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{2+m}+\frac {4 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}+\frac {4 a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{2+m} \]

output

-3*x^(1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)-a*x^(2+m) 
*hypergeom([1/2, 1+1/2*m],[2+1/2*m],a^2*x^2)/(2+m)+4*x^(1+m)*hypergeom([3/ 
2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)+4*a*x^(2+m)*hypergeom([3/2, 1+1/2 
*m],[2+1/2*m],a^2*x^2)/(2+m)
 

3.2.42.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.61 \[ \int e^{3 \text {arctanh}(a x)} x^m \, dx=\frac {x^{1+m} \sqrt {-1-a x} \sqrt {1-a x} \left (\operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {1}{2},2+m,-a x,a x\right )-2 \operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {3}{2},2+m,-a x,a x\right )\right )}{(1+m) \sqrt {-1+a x} \sqrt {1+a x}} \]

input

Integrate[E^(3*ArcTanh[a*x])*x^m,x]
 

output

(x^(1 + m)*Sqrt[-1 - a*x]*Sqrt[1 - a*x]*(AppellF1[1 + m, -1/2, 1/2, 2 + m, 
 -(a*x), a*x] - 2*AppellF1[1 + m, -1/2, 3/2, 2 + m, -(a*x), a*x]))/((1 + m 
)*Sqrt[-1 + a*x]*Sqrt[1 + a*x])
 

3.2.42.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6674, 2355, 557, 278, 583, 557, 278}

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.

\(\displaystyle \int x^m e^{3 \text {arctanh}(a x)} \, dx\)

\(\Big \downarrow \) 6674

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

\(\Big \downarrow \) 2355

\(\displaystyle \int \frac {x^m (-a x-3)}{\sqrt {1-a^2 x^2}}dx+4 \int \frac {x^m}{(1-a x) \sqrt {1-a^2 x^2}}dx\)

\(\Big \downarrow \) 557

\(\displaystyle -a \int \frac {x^{m+1}}{\sqrt {1-a^2 x^2}}dx-3 \int \frac {x^m}{\sqrt {1-a^2 x^2}}dx+4 \int \frac {x^m}{(1-a x) \sqrt {1-a^2 x^2}}dx\)

\(\Big \downarrow \) 278

\(\displaystyle 4 \int \frac {x^m}{(1-a x) \sqrt {1-a^2 x^2}}dx-\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}\)

\(\Big \downarrow \) 583

\(\displaystyle 4 \int \frac {x^m (a x+1)}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}\)

\(\Big \downarrow \) 557

\(\displaystyle 4 \left (a \int \frac {x^{m+1}}{\left (1-a^2 x^2\right )^{3/2}}dx+\int \frac {x^m}{\left (1-a^2 x^2\right )^{3/2}}dx\right )-\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}+4 \left (\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}\right )\)

input

Int[E^(3*ArcTanh[a*x])*x^m,x]
 

output

(-3*x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(1 + 
m) - (a*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/( 
2 + m) + 4*((x^(1 + m)*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, a^2*x^ 
2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/2, a^ 
2*x^2])/(2 + m))
 

3.2.42.3.1 Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 583

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, 0]
 

rule 2355

Int[(Px_)*((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2) 
^(p_.), x_Symbol] :> Int[PolynomialQuotient[Px, c + d*x, x]*(e*x)^m*(c + d* 
x)^(n + 1)*(a + b*x^2)^p, x] + Simp[PolynomialRemainder[Px, c + d*x, x]   I 
nt[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p} 
, x] && PolynomialQ[Px, x] && LtQ[n, 0]
 

rule 6674

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x 
)^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/2)*Sqrt[1 - a^2*x^2])), x] / 
; FreeQ[{a, c, m}, x] && IntegerQ[(n - 1)/2]
 

3.2.42.4 Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.92

method	result	size
meijerg	\(\frac {x^{1+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {1}{2}+\frac {m}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{1+m}+\frac {a^{3} x^{4+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, 2+\frac {m}{2}\right ], \left [3+\frac {m}{2}\right ], a^{2} x^{2}\right )}{4+m}+\frac {3 a^{2} x^{3+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {3}{2}+\frac {m}{2}\right ], \left [\frac {5}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{3+m}+\frac {3 a \,x^{2+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, 1+\frac {m}{2}\right ], \left [2+\frac {m}{2}\right ], a^{2} x^{2}\right )}{2+m}\)	\(139\)

input

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m,x,method=_RETURNVERBOSE)
 

output

x^(1+m)*hypergeom([3/2,1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)+a^3/(4+m)*x^( 
4+m)*hypergeom([3/2,2+1/2*m],[3+1/2*m],a^2*x^2)+3*a^2/(3+m)*x^(3+m)*hyperg 
eom([3/2,3/2+1/2*m],[5/2+1/2*m],a^2*x^2)+3*a*x^(2+m)*hypergeom([3/2,1+1/2* 
m],[2+1/2*m],a^2*x^2)/(2+m)
 

3.2.42.5 Fricas [F]

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

input

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m,x, algorithm="fricas")
 

output

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

3.2.42.6 Sympy [F]

\[ \int e^{3 \text {arctanh}(a x)} x^m \, dx=\int \frac {x^{m} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

input

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*x**m,x)
 

output

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

3.2.42.7 Maxima [F]

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

input

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m,x, algorithm="maxima")
 

output

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

3.2.42.8 Giac [F(-2)]

Exception generated. \[ \int e^{3 \text {arctanh}(a x)} x^m \, dx=\text {Exception raised: TypeError} \]

input

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^m,x, algorithm="giac")
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

3.2.42.9 Mupad [F(-1)]

Timed out. \[ \int e^{3 \text {arctanh}(a x)} x^m \, dx=\int \frac {x^m\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

input

int((x^m*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)
 

output

int((x^m*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2), x)