\(\int e^{3 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx\) [456]

Optimal result

Integrand size = 24, antiderivative size = 169 \[ \int e^{3 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\frac {8 (e x)^{1+m} \sqrt {c+a c x}}{e \sqrt {1-a x} \sqrt {1+a x}}+\frac {2 (e x)^{1+m} \sqrt {1-a x} \sqrt {c+a c x}}{e (3+2 m) \sqrt {1+a x}}+\frac {2 \left (23+40 m+16 m^2\right ) (a x)^{-m} (e x)^m \sqrt {1-a x} \sqrt {c+a c x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1-a x\right )}{a (3+2 m) \sqrt {1+a x}} \] Output:

8*(e*x)^(1+m)*(a*c*x+c)^(1/2)/e/(-a*x+1)^(1/2)/(a*x+1)^(1/2)+2*(e*x)^(1+m) 
*(-a*x+1)^(1/2)*(a*c*x+c)^(1/2)/e/(3+2*m)/(a*x+1)^(1/2)+2*(16*m^2+40*m+23) 
*(e*x)^m*(-a*x+1)^(1/2)*(a*c*x+c)^(1/2)*hypergeom([1/2, -m],[3/2],-a*x+1)/ 
a/(3+2*m)/((a*x)^m)/(a*x+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.59 \[ \int e^{3 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=-\frac {c x (e x)^m \sqrt {1+a x} \left (-2 (1+m) (13+8 m-a x)+\left (23+40 m+16 m^2\right ) \sqrt {1-a x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,a x\right )\right )}{(1+m) (3+2 m) \sqrt {1-a x} \sqrt {c+a c x}} \] Input:

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

Output:

-((c*x*(e*x)^m*Sqrt[1 + a*x]*(-2*(1 + m)*(13 + 8*m - a*x) + (23 + 40*m + 1 
6*m^2)*Sqrt[1 - a*x]*Hypergeometric2F1[1/2, 1 + m, 2 + m, a*x]))/((1 + m)* 
(3 + 2*m)*Sqrt[1 - a*x]*Sqrt[c + a*c*x]))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.70, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6680, 37, 100, 27, 90, 74}

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.

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

\(\Big \downarrow \) 6680

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

\(\Big \downarrow \) 37

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {\sqrt {a c x+c} \left (\frac {8 (e x)^{m+1}}{e \sqrt {1-a x}}-\frac {2 \int \frac {a^2 e (e x)^m (8 m+a x+7)}{2 \sqrt {1-a x}}dx}{a^2 e}\right )}{\sqrt {a x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a c x+c} \left (\frac {8 (e x)^{m+1}}{e \sqrt {1-a x}}-\int \frac {(e x)^m (8 m+a x+7)}{\sqrt {1-a x}}dx\right )}{\sqrt {a x+1}}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\sqrt {a c x+c} \left (-\frac {\left (16 m^2+40 m+23\right ) \int \frac {(e x)^m}{\sqrt {1-a x}}dx}{2 m+3}+\frac {2 \sqrt {1-a x} (e x)^{m+1}}{e (2 m+3)}+\frac {8 (e x)^{m+1}}{e \sqrt {1-a x}}\right )}{\sqrt {a x+1}}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {\sqrt {a c x+c} \left (-\frac {\left (16 m^2+40 m+23\right ) (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},m+1,m+2,a x\right )}{e (m+1) (2 m+3)}+\frac {2 \sqrt {1-a x} (e x)^{m+1}}{e (2 m+3)}+\frac {8 (e x)^{m+1}}{e \sqrt {1-a x}}\right )}{\sqrt {a x+1}}\)

Input:

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

Output:

(Sqrt[c + a*c*x]*((8*(e*x)^(1 + m))/(e*Sqrt[1 - a*x]) + (2*(e*x)^(1 + m)*S 
qrt[1 - a*x])/(e*(3 + 2*m)) - ((23 + 40*m + 16*m^2)*(e*x)^(1 + m)*Hypergeo 
metric2F1[1/2, 1 + m, 2 + m, a*x])/(e*(1 + m)*(3 + 2*m))))/Sqrt[1 + a*x]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 37

Int[(u_.)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> S 
imp[(a + b*x)^m/(c + d*x)^m   Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, 
b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !SimplerQ[a + b*x, c + d*x]
 

rule 74

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

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 6680

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

Maple [F]

\[\int \frac {\left (a x +1\right )^{3} \left (e x \right )^{m} \sqrt {a c x +c}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int e^{3 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\text {too large to display} \] Input:

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

Output:

(2*e**m*sqrt(c)*(4*x**m*sqrt( - a*x + 1)*a**2*m**2*x**2 - x**m*sqrt( - a*x 
 + 1)*a**2*x**2 + 12*x**m*sqrt( - a*x + 1)*a*m**2*x + 14*x**m*sqrt( - a*x 
+ 1)*a*m*x - 10*x**m*sqrt( - a*x + 1)*a*x + 16*x**m*sqrt( - a*x + 1)*m**2 
+ 40*x**m*sqrt( - a*x + 1)*m + 23*x**m*sqrt( - a*x + 1) + 32*int((x**m*sqr 
t( - a*x + 1))/(8*a**2*m**3*x**3 + 12*a**2*m**2*x**3 - 2*a**2*m*x**3 - 3*a 
**2*x**3 - 16*a*m**3*x**2 - 24*a*m**2*x**2 + 4*a*m*x**2 + 6*a*x**2 + 8*m** 
3*x + 12*m**2*x - 2*m*x - 3*x),x)*a*m**6*x + 144*int((x**m*sqrt( - a*x + 1 
))/(8*a**2*m**3*x**3 + 12*a**2*m**2*x**3 - 2*a**2*m*x**3 - 3*a**2*x**3 - 1 
6*a*m**3*x**2 - 24*a*m**2*x**2 + 4*a*m*x**2 + 6*a*x**2 + 8*m**3*x + 12*m** 
2*x - 2*m*x - 3*x),x)*a*m**5*x + 200*int((x**m*sqrt( - a*x + 1))/(8*a**2*m 
**3*x**3 + 12*a**2*m**2*x**3 - 2*a**2*m*x**3 - 3*a**2*x**3 - 16*a*m**3*x** 
2 - 24*a*m**2*x**2 + 4*a*m*x**2 + 6*a*x**2 + 8*m**3*x + 12*m**2*x - 2*m*x 
- 3*x),x)*a*m**4*x + 60*int((x**m*sqrt( - a*x + 1))/(8*a**2*m**3*x**3 + 12 
*a**2*m**2*x**3 - 2*a**2*m*x**3 - 3*a**2*x**3 - 16*a*m**3*x**2 - 24*a*m**2 
*x**2 + 4*a*m*x**2 + 6*a*x**2 + 8*m**3*x + 12*m**2*x - 2*m*x - 3*x),x)*a*m 
**3*x - 52*int((x**m*sqrt( - a*x + 1))/(8*a**2*m**3*x**3 + 12*a**2*m**2*x* 
*3 - 2*a**2*m*x**3 - 3*a**2*x**3 - 16*a*m**3*x**2 - 24*a*m**2*x**2 + 4*a*m 
*x**2 + 6*a*x**2 + 8*m**3*x + 12*m**2*x - 2*m*x - 3*x),x)*a*m**2*x - 24*in 
t((x**m*sqrt( - a*x + 1))/(8*a**2*m**3*x**3 + 12*a**2*m**2*x**3 - 2*a**2*m 
*x**3 - 3*a**2*x**3 - 16*a*m**3*x**2 - 24*a*m**2*x**2 + 4*a*m*x**2 + 6*...