\(\int e^{3 \text {arctanh}(a x)} (c x)^m \, dx\) [136]

Optimal result

Integrand size = 14, antiderivative size = 127 \[ \int e^{3 \text {arctanh}(a x)} (c x)^m \, dx=\frac {4 (c x)^{1+m} (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {(3+4 m) (c x)^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{c (1+m)}-\frac {a (5+4 m) (c x)^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{c^2 (2+m)} \] Output:

4*(c*x)^(1+m)*(a*x+1)/c/(-a^2*x^2+1)^(1/2)-(3+4*m)*(c*x)^(1+m)*hypergeom([ 
1/2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/c/(1+m)-a*(5+4*m)*(c*x)^(2+m)*hyperge 
om([1/2, 1+1/2*m],[2+1/2*m],a^2*x^2)/c^2/(2+m)
 

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 = 93, normalized size of antiderivative = 0.73 \[ \int e^{3 \text {arctanh}(a x)} (c x)^m \, dx=\frac {x (c x)^m \sqrt {-1-a x} \sqrt {1-a x} \left (\operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {1}{2},2+m,-a x,a x\right )-2 \operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {3}{2},2+m,-a x,a x\right )\right )}{(1+m) \sqrt {-1+a x} \sqrt {1+a x}} \] Input:

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

Output:

(x*(c*x)^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])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.35, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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.

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

\(\Big \downarrow \) 6674

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

\(\Big \downarrow \) 2355

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 278

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

\(\Big \downarrow \) 583

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

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]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.17

method	result	size
meijerg	\(\frac {\left (x c \right )^{m} x \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {m}{2}+\frac {1}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{1+m}+\frac {a^{3} \left (x c \right )^{m} x^{4} \operatorname {hypergeom}\left (\left [\frac {3}{2}, 2+\frac {m}{2}\right ], \left [\frac {m}{2}+3\right ], a^{2} x^{2}\right )}{4+m}+\frac {3 \left (x c \right )^{m} a^{2} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {3}{2}+\frac {m}{2}\right ], \left [\frac {5}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{3+m}+\frac {3 a \left (x c \right )^{m} x^{2} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {m}{2}+1\right ], \left [2+\frac {m}{2}\right ], a^{2} x^{2}\right )}{2+m}\)	\(149\)

Input:

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

Output:

(x*c)^m/(1+m)*x*hypergeom([3/2,1/2*m+1/2],[3/2+1/2*m],a^2*x^2)+a^3*(x*c)^m 
/(4+m)*x^4*hypergeom([3/2,2+1/2*m],[1/2*m+3],a^2*x^2)+3*(x*c)^m*a^2/(3+m)* 
x^3*hypergeom([3/2,3/2+1/2*m],[5/2+1/2*m],a^2*x^2)+3*a*(x*c)^m/(2+m)*x^2*h 
ypergeom([3/2,1/2*m+1],[2+1/2*m],a^2*x^2)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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