\(\int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx\) [475]

Optimal result

Integrand size = 24, antiderivative size = 53 \[ \int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\frac {(e x)^{1+m} \sqrt {c+a c x} \operatorname {AppellF1}\left (1+m,2,-\frac {5}{2},2+m,a x,-a x\right )}{e (1+m) \sqrt {1+a x}} \] Output:

(e*x)^(1+m)*(a*c*x+c)^(1/2)*AppellF1(1+m,2,-5/2,2+m,a*x,-a*x)/e/(1+m)/(a*x 
+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.68 \[ \int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=-\frac {x (e x)^m \sqrt {c+a c x} \left (4 \operatorname {AppellF1}\left (1+m,-\frac {1}{2},1,2+m,-a x,a x\right )-4 \operatorname {AppellF1}\left (1+m,2,-\frac {1}{2},2+m,a x,-a x\right )-\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1+m,2+m,-a x\right )\right )}{(1+m) \sqrt {1+a x}} \] Input:

Integrate[E^(4*ArcTanh[a*x])*(e*x)^m*Sqrt[c + a*c*x],x]
 

Output:

-((x*(e*x)^m*Sqrt[c + a*c*x]*(4*AppellF1[1 + m, -1/2, 1, 2 + m, -(a*x), a* 
x] - 4*AppellF1[1 + m, 2, -1/2, 2 + m, a*x, -(a*x)] - Hypergeometric2F1[-1 
/2, 1 + m, 2 + m, -(a*x)]))/((1 + m)*Sqrt[1 + a*x]))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6680, 35, 152, 150}

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.

\(\displaystyle \int e^{4 \text {arctanh}(a x)} \sqrt {a c x+c} (e x)^m \, dx\)

\(\Big \downarrow \) 6680

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

\(\Big \downarrow \) 35

\(\displaystyle \frac {\int \frac {(e x)^m (a x c+c)^{5/2}}{(1-a x)^2}dx}{c^2}\)

\(\Big \downarrow \) 152

\(\displaystyle \frac {\sqrt {a c x+c} \int \frac {(e x)^m (a x+1)^{5/2}}{(1-a x)^2}dx}{\sqrt {a x+1}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\sqrt {a c x+c} (e x)^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {5}{2},2,m+2,-a x,a x\right )}{e (m+1) \sqrt {a x+1}}\)

Input:

Int[E^(4*ArcTanh[a*x])*(e*x)^m*Sqrt[c + a*c*x],x]
 

Output:

((e*x)^(1 + m)*Sqrt[c + a*c*x]*AppellF1[1 + m, -5/2, 2, 2 + m, -(a*x), a*x 
])/(e*(1 + m)*Sqrt[1 + a*x])
 

Defintions of rubi rules used

rule 35

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

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 152

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) 
Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, 
 n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]
 

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]

\[\int \frac {\left (a x +1\right )^{4} \left (e x \right )^{m} \sqrt {a c x +c}}{\left (-a^{2} x^{2}+1\right )^{2}}d x\]

Input:

int((a*x+1)^4/(-a^2*x^2+1)^2*(e*x)^m*(a*c*x+c)^(1/2),x)
 

Output:

int((a*x+1)^4/(-a^2*x^2+1)^2*(e*x)^m*(a*c*x+c)^(1/2),x)
 

Fricas [F]

\[ \int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\int { \frac {\sqrt {a c x + c} {\left (a x + 1\right )}^{4} \left (e x\right )^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}} \,d x } \] Input:

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(e*x)^m*(a*c*x+c)^(1/2),x, algorithm="f 
ricas")
 

Output:

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

Sympy [F]

\[ \int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\int \frac {\sqrt {c \left (a x + 1\right )} \left (e x\right )^{m} \left (a x + 1\right )^{2}}{\left (a x - 1\right )^{2}}\, dx \] Input:

integrate((a*x+1)**4/(-a**2*x**2+1)**2*(e*x)**m*(a*c*x+c)**(1/2),x)
 

Output:

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

Maxima [F]

\[ \int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\int { \frac {\sqrt {a c x + c} {\left (a x + 1\right )}^{4} \left (e x\right )^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}} \,d x } \] Input:

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(e*x)^m*(a*c*x+c)^(1/2),x, algorithm="m 
axima")
 

Output:

integrate(sqrt(a*c*x + c)*(a*x + 1)^4*(e*x)^m/(a^2*x^2 - 1)^2, x)
 

Giac [F]

\[ \int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\int { \frac {\sqrt {a c x + c} {\left (a x + 1\right )}^{4} \left (e x\right )^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}} \,d x } \] Input:

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(e*x)^m*(a*c*x+c)^(1/2),x, algorithm="g 
iac")
 

Output:

integrate(sqrt(a*c*x + c)*(a*x + 1)^4*(e*x)^m/(a^2*x^2 - 1)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\int \frac {{\left (e\,x\right )}^m\,\sqrt {c+a\,c\,x}\,{\left (a\,x+1\right )}^4}{{\left (a^2\,x^2-1\right )}^2} \,d x \] Input:

int(((e*x)^m*(c + a*c*x)^(1/2)*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)
 

Output:

int(((e*x)^m*(c + a*c*x)^(1/2)*(a*x + 1)^4)/(a^2*x^2 - 1)^2, x)
 

Reduce [F]

\[ \int e^{4 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\text {too large to display} \] Input:

int((a*x+1)^4/(-a^2*x^2+1)^2*(e*x)^m*(a*c*x+c)^(1/2),x)
 

Output:

(2*e**m*sqrt(c)*(6*x**m*sqrt(a*x + 1)*a**2*m*x**2 + 3*x**m*sqrt(a*x + 1)*a 
**2*x**2 + 18*x**m*sqrt(a*x + 1)*a*m*x + 36*x**m*sqrt(a*x + 1)*a*x - 16*x* 
*m*sqrt(a*x + 1)*m**2 - 44*x**m*sqrt(a*x + 1)*m - 27*x**m*sqrt(a*x + 1) + 
64*int((x**m*sqrt(a*x + 1)*x)/(4*a**3*m**2*x**3 + 8*a**3*m*x**3 + 3*a**3*x 
**3 - 4*a**2*m**2*x**2 - 8*a**2*m*x**2 - 3*a**2*x**2 - 4*a*m**2*x - 8*a*m* 
x - 3*a*x + 4*m**2 + 8*m + 3),x)*a**3*m**5*x + 368*int((x**m*sqrt(a*x + 1) 
*x)/(4*a**3*m**2*x**3 + 8*a**3*m*x**3 + 3*a**3*x**3 - 4*a**2*m**2*x**2 - 8 
*a**2*m*x**2 - 3*a**2*x**2 - 4*a*m**2*x - 8*a*m*x - 3*a*x + 4*m**2 + 8*m + 
 3),x)*a**3*m**4*x + 860*int((x**m*sqrt(a*x + 1)*x)/(4*a**3*m**2*x**3 + 8* 
a**3*m*x**3 + 3*a**3*x**3 - 4*a**2*m**2*x**2 - 8*a**2*m*x**2 - 3*a**2*x**2 
 - 4*a*m**2*x - 8*a*m*x - 3*a*x + 4*m**2 + 8*m + 3),x)*a**3*m**3*x + 1084* 
int((x**m*sqrt(a*x + 1)*x)/(4*a**3*m**2*x**3 + 8*a**3*m*x**3 + 3*a**3*x**3 
 - 4*a**2*m**2*x**2 - 8*a**2*m*x**2 - 3*a**2*x**2 - 4*a*m**2*x - 8*a*m*x - 
 3*a*x + 4*m**2 + 8*m + 3),x)*a**3*m**2*x + 729*int((x**m*sqrt(a*x + 1)*x) 
/(4*a**3*m**2*x**3 + 8*a**3*m*x**3 + 3*a**3*x**3 - 4*a**2*m**2*x**2 - 8*a* 
*2*m*x**2 - 3*a**2*x**2 - 4*a*m**2*x - 8*a*m*x - 3*a*x + 4*m**2 + 8*m + 3) 
,x)*a**3*m*x + 180*int((x**m*sqrt(a*x + 1)*x)/(4*a**3*m**2*x**3 + 8*a**3*m 
*x**3 + 3*a**3*x**3 - 4*a**2*m**2*x**2 - 8*a**2*m*x**2 - 3*a**2*x**2 - 4*a 
*m**2*x - 8*a*m*x - 3*a*x + 4*m**2 + 8*m + 3),x)*a**3*x - 64*int((x**m*sqr 
t(a*x + 1)*x)/(4*a**3*m**2*x**3 + 8*a**3*m*x**3 + 3*a**3*x**3 - 4*a**2*...