\(\int \frac {e^{3 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx\) [492]

Optimal result

Integrand size = 22, antiderivative size = 101 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {4 \sqrt {1-a^2 x^2}}{3 a c}+\frac {16 \sqrt {1-a^2 x^2}}{3 a c (1-a x)}-\frac {\left (1-a^2 x^2\right )^{5/2}}{3 a c (1-a x)^4}-\frac {4 \arcsin (a x)}{a c} \] Output:

4/3*(-a^2*x^2+1)^(1/2)/a/c+16/3*(-a^2*x^2+1)^(1/2)/a/c/(-a*x+1)-1/3*(-a^2* 
x^2+1)^(5/2)/a/c/(-a*x+1)^4-4*arcsin(a*x)/a/c
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx=-\frac {(1+a x)^{5/2}+16 \sqrt {2} (-1+a x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1-a x)\right )}{3 a c (1-a x)^{3/2}} \] Input:

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

Output:

-1/3*((1 + a*x)^(5/2) + 16*Sqrt[2]*(-1 + a*x)*Hypergeometric2F1[-3/2, -1/2 
, 1/2, (1 - a*x)/2])/(a*c*(1 - a*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6681, 6678, 571, 463, 455, 223}

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.

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

\(\Big \downarrow \) 6681

\(\displaystyle -\frac {a \int \frac {e^{3 \text {arctanh}(a x)} x}{1-a x}dx}{c}\)

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 571

\(\displaystyle -\frac {a \left (\frac {\left (1-a^2 x^2\right )^{5/2}}{3 a^2 (1-a x)^4}-\frac {4 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(1-a x)^3}dx}{3 a}\right )}{c}\)

\(\Big \downarrow \) 463

\(\displaystyle -\frac {a \left (\frac {\left (1-a^2 x^2\right )^{5/2}}{3 a^2 (1-a x)^4}-\frac {4 \left (\frac {4 \sqrt {1-a^2 x^2}}{a (1-a x)}-\int \frac {a x+3}{\sqrt {1-a^2 x^2}}dx\right )}{3 a}\right )}{c}\)

\(\Big \downarrow \) 455

\(\displaystyle -\frac {a \left (\frac {\left (1-a^2 x^2\right )^{5/2}}{3 a^2 (1-a x)^4}-\frac {4 \left (-3 \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a}+\frac {4 \sqrt {1-a^2 x^2}}{a (1-a x)}\right )}{3 a}\right )}{c}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {a \left (\frac {\left (1-a^2 x^2\right )^{5/2}}{3 a^2 (1-a x)^4}-\frac {4 \left (\frac {\sqrt {1-a^2 x^2}}{a}+\frac {4 \sqrt {1-a^2 x^2}}{a (1-a x)}-\frac {3 \arcsin (a x)}{a}\right )}{3 a}\right )}{c}\)

Input:

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

Output:

-((a*((1 - a^2*x^2)^(5/2)/(3*a^2*(1 - a*x)^4) - (4*(Sqrt[1 - a^2*x^2]/a + 
(4*Sqrt[1 - a^2*x^2])/(a*(1 - a*x)) - (3*ArcSin[a*x])/a))/(3*a)))/c)
 

Defintions of rubi rules used

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 463

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x 
))), x] - Simp[d^(2*n + 2)/b^(n + 1)   Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ 
(2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F 
reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[n, 0] && EqQ[n + p, 
-3/2]
 

rule 571

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

rule 6678

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* 
(x_))^(m_.), x_Symbol] :> Simp[c^n   Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - 
 a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 
0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 
, 0]) && IntegerQ[2*p]
 

rule 6681

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.49

method	result	size
risch	\(-\frac {a^{2} x^{2}-1}{a c \sqrt {-a^{2} x^{2}+1}}-\frac {\left (\frac {4 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {a^{2}}}+\frac {4 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a^{4} \left (x -\frac {1}{a}\right )^{2}}+\frac {20 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a^{3} \left (x -\frac {1}{a}\right )}\right ) a}{c}\)	\(150\)
default	\(\frac {a \left (a^{2} \left (-\frac {x^{2}}{a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {-a^{2} x^{2}+1}}\right )+\frac {8 x}{a \sqrt {-a^{2} x^{2}+1}}+\frac {7}{a^{2} \sqrt {-a^{2} x^{2}+1}}+4 a \left (\frac {x}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}\right )+\frac {\frac {8}{3 a \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}+\frac {8 \left (-2 \left (x -\frac {1}{a}\right ) a^{2}-2 a \right )}{3 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}}{a^{2}}\right )}{c}\)	\(230\)

Input:

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

Output:

-1/a/c*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)-(4/a/(a^2)^(1/2)*arctan((a^2)^(1/2)* 
x/(-a^2*x^2+1)^(1/2))+4/3/a^4/(x-1/a)^2*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2) 
+20/3/a^3/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))*a/c
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {19 \, a^{2} x^{2} - 38 \, a x + 24 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{2} x^{2} - 26 \, a x + 19\right )} \sqrt {-a^{2} x^{2} + 1} + 19}{3 \, {\left (a^{3} c x^{2} - 2 \, a^{2} c x + a c\right )}} \] Input:

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

Output:

1/3*(19*a^2*x^2 - 38*a*x + 24*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^2 
+ 1) - 1)/(a*x)) + (3*a^2*x^2 - 26*a*x + 19)*sqrt(-a^2*x^2 + 1) + 19)/(a^3 
*c*x^2 - 2*a^2*c*x + a*c)
 

Sympy [F]

\[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {a \left (\int \frac {x}{- a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {3 a x^{2}}{- a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {3 a^{2} x^{3}}{- a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a^{3} x^{4}}{- a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx\right )}{c} \] Input:

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

Output:

a*(Integral(x/(-a**3*x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x**2 
 + 1) + a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral(3* 
a*x**2/(-a**3*x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x**2 + 1) + 
 a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral(3*a**2*x* 
*3/(-a**3*x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x**2 + 1) + a*x 
*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral(a**3*x**4/(-a 
**3*x**3*sqrt(-a**2*x**2 + 1) + a**2*x**2*sqrt(-a**2*x**2 + 1) + a*x*sqrt( 
-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x))/c
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.25 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx=-\frac {4 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{c {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a c} - \frac {8 \, {\left (\frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} - 4\right )}}{3 \, c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{3} {\left | a \right |}} \] Input:

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

Output:

-4*arcsin(a*x)*sgn(a)/(c*abs(a)) + sqrt(-a^2*x^2 + 1)/(a*c) - 8/3*(9*(sqrt 
(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/( 
a^4*x^2) - 4)/(c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^3*abs(a))
 

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.28 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {\sqrt {1-a^2\,x^2}}{a\,c}-\frac {20\,\sqrt {1-a^2\,x^2}}{3\,\left (\frac {c\,\sqrt {-a^2}}{a}-c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c\,\sqrt {-a^2}}-\frac {4\,a\,\sqrt {1-a^2\,x^2}}{3\,\left (c\,a^4\,x^2-2\,c\,a^3\,x+c\,a^2\right )} \] Input:

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

Output:

(1 - a^2*x^2)^(1/2)/(a*c) - (20*(1 - a^2*x^2)^(1/2))/(3*((c*(-a^2)^(1/2))/ 
a - c*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (4*asinh(x*(-a^2)^(1/2)))/(c*(-a^2)^ 
(1/2)) - (4*a*(1 - a^2*x^2)^(1/2))/(3*(a^2*c + a^4*c*x^2 - 2*a^3*c*x))
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.79 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {-12 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x +12 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-12 \mathit {asin} \left (a x \right ) a^{2} x^{2}+24 \mathit {asin} \left (a x \right ) a x -12 \mathit {asin} \left (a x \right )+3 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-15 \sqrt {-a^{2} x^{2}+1}\, a x +8 \sqrt {-a^{2} x^{2}+1}-3 a^{3} x^{3}+34 a^{2} x^{2}-15 a x -8}{3 a c \left (\sqrt {-a^{2} x^{2}+1}\, a x -\sqrt {-a^{2} x^{2}+1}+a^{2} x^{2}-2 a x +1\right )} \] Input:

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

Output:

( - 12*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x + 12*sqrt( - a**2*x**2 + 1)*as 
in(a*x) - 12*asin(a*x)*a**2*x**2 + 24*asin(a*x)*a*x - 12*asin(a*x) + 3*sqr 
t( - a**2*x**2 + 1)*a**2*x**2 - 15*sqrt( - a**2*x**2 + 1)*a*x + 8*sqrt( - 
a**2*x**2 + 1) - 3*a**3*x**3 + 34*a**2*x**2 - 15*a*x - 8)/(3*a*c*(sqrt( - 
a**2*x**2 + 1)*a*x - sqrt( - a**2*x**2 + 1) + a**2*x**2 - 2*a*x + 1))