\(\int \frac {e^{2 \text {arctanh}(a x)}}{x^2 (c-a^2 c x^2)^{3/2}} \, dx\) [1158]

Optimal result

Integrand size = 27, antiderivative size = 112 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {2 a (1+a x)}{3 \left (c-a^2 c x^2\right )^{3/2}}+\frac {7+6 a x}{3 c x \sqrt {c-a^2 c x^2}}-\frac {10 \sqrt {c-a^2 c x^2}}{3 c^2 x}-\frac {2 a \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{c^{3/2}} \] Output:

2/3*a*(a*x+1)/(-a^2*c*x^2+c)^(3/2)+1/3*(6*a*x+7)/c/x/(-a^2*c*x^2+c)^(1/2)- 
10/3*(-a^2*c*x^2+c)^(1/2)/c^2/x-2*a*arctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))/ 
c^(3/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\left (-3+14 a x-10 a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 c^2 x (-1+a x)^2}+\frac {2 a \log (x)}{c^{3/2}}-\frac {2 a \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right )}{c^{3/2}} \] Input:

Integrate[E^(2*ArcTanh[a*x])/(x^2*(c - a^2*c*x^2)^(3/2)),x]
 

Output:

((-3 + 14*a*x - 10*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(3*c^2*x*(-1 + a*x)^2) + 
(2*a*Log[x])/c^(3/2) - (2*a*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2]])/c^(3/2)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6701, 532, 25, 2336, 27, 534, 243, 73, 221}

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.

\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 532

\(\displaystyle c \left (\frac {2 a (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\int -\frac {4 a^2 x^2+6 a x+3}{x^2 \left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {\int \frac {4 a^2 x^2+6 a x+3}{x^2 \left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}+\frac {2 a (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 2336

\(\displaystyle c \left (\frac {\frac {a (7 a x+6)}{c \sqrt {c-a^2 c x^2}}-\frac {\int -\frac {3 (2 a x+1)}{x^2 \sqrt {c-a^2 c x^2}}dx}{c}}{3 c}+\frac {2 a (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {\frac {3 \int \frac {2 a x+1}{x^2 \sqrt {c-a^2 c x^2}}dx}{c}+\frac {a (7 a x+6)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 534

\(\displaystyle c \left (\frac {\frac {3 \left (2 a \int \frac {1}{x \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{c x}\right )}{c}+\frac {a (7 a x+6)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle c \left (\frac {\frac {3 \left (a \int \frac {1}{x^2 \sqrt {c-a^2 c x^2}}dx^2-\frac {\sqrt {c-a^2 c x^2}}{c x}\right )}{c}+\frac {a (7 a x+6)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c \left (\frac {\frac {3 \left (-\frac {2 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2 c}}d\sqrt {c-a^2 c x^2}}{a c}-\frac {\sqrt {c-a^2 c x^2}}{c x}\right )}{c}+\frac {a (7 a x+6)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c \left (\frac {\frac {3 \left (-\frac {2 a \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a^2 c x^2}}{c x}\right )}{c}+\frac {a (7 a x+6)}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 a (a x+1)}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\)

Input:

Int[E^(2*ArcTanh[a*x])/(x^2*(c - a^2*c*x^2)^(3/2)),x]
 

Output:

c*((2*a*(1 + a*x))/(3*c*(c - a^2*c*x^2)^(3/2)) + ((a*(6 + 7*a*x))/(c*Sqrt[ 
c - a^2*c*x^2]) + (3*(-(Sqrt[c - a^2*c*x^2]/(c*x)) - (2*a*ArcTanh[Sqrt[c - 
 a^2*c*x^2]/Sqrt[c]])/Sqrt[c]))/c)/(3*c))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 6701

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.42

method	result	size
risch	\(\frac {a^{2} x^{2}-1}{x \sqrt {-c \left (a^{2} x^{2}-1\right )}\, c}+\frac {-\frac {2 a \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{\sqrt {c}}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{3 a c \left (x -\frac {1}{a}\right )^{2}}-\frac {7 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{3 c \left (x -\frac {1}{a}\right )}}{c}\)	\(159\)
default	\(-\frac {1}{c x \sqrt {-a^{2} c \,x^{2}+c}}+\frac {2 a^{2} x}{c \sqrt {-a^{2} c \,x^{2}+c}}+2 a \left (\frac {1}{c \sqrt {-a^{2} c \,x^{2}+c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{c^{\frac {3}{2}}}\right )-2 a \left (\frac {1}{3 a c \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}+\frac {-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c}{3 a \,c^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )\)	\(203\)

Input:

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

Output:

(a^2*x^2-1)/x/(-c*(a^2*x^2-1))^(1/2)/c+(-2*a/c^(1/2)*ln((2*c+2*c^(1/2)*(-a 
^2*c*x^2+c)^(1/2))/x)+1/3/a/c/(x-1/a)^2*(-(x-1/a)^2*a^2*c-2*(x-1/a)*a*c)^( 
1/2)-7/3/c/(x-1/a)*(-(x-1/a)^2*a^2*c-2*(x-1/a)*a*c)^(1/2))/c
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.04 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a^{3} x^{3} - 2 \, a^{2} x^{2} + a x\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (10 \, a^{2} x^{2} - 14 \, a x + 3\right )}}{3 \, {\left (a^{2} c^{2} x^{3} - 2 \, a c^{2} x^{2} + c^{2} x\right )}}, \frac {6 \, {\left (a^{3} x^{3} - 2 \, a^{2} x^{2} + a x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (10 \, a^{2} x^{2} - 14 \, a x + 3\right )}}{3 \, {\left (a^{2} c^{2} x^{3} - 2 \, a c^{2} x^{2} + c^{2} x\right )}}\right ] \] Input:

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

Output:

[1/3*(3*(a^3*x^3 - 2*a^2*x^2 + a*x)*sqrt(c)*log(-(a^2*c*x^2 + 2*sqrt(-a^2* 
c*x^2 + c)*sqrt(c) - 2*c)/x^2) - sqrt(-a^2*c*x^2 + c)*(10*a^2*x^2 - 14*a*x 
 + 3))/(a^2*c^2*x^3 - 2*a*c^2*x^2 + c^2*x), 1/3*(6*(a^3*x^3 - 2*a^2*x^2 + 
a*x)*sqrt(-c)*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/c) - sqrt(-a^2*c*x^2 + 
c)*(10*a^2*x^2 - 14*a*x + 3))/(a^2*c^2*x^3 - 2*a*c^2*x^2 + c^2*x)]
 

Sympy [F]

\[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=- \int \frac {a x}{- a^{3} c x^{5} \sqrt {- a^{2} c x^{2} + c} + a^{2} c x^{4} \sqrt {- a^{2} c x^{2} + c} + a c x^{3} \sqrt {- a^{2} c x^{2} + c} - c x^{2} \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {1}{- a^{3} c x^{5} \sqrt {- a^{2} c x^{2} + c} + a^{2} c x^{4} \sqrt {- a^{2} c x^{2} + c} + a c x^{3} \sqrt {- a^{2} c x^{2} + c} - c x^{2} \sqrt {- a^{2} c x^{2} + c}}\, dx \] Input:

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

Output:

-Integral(a*x/(-a**3*c*x**5*sqrt(-a**2*c*x**2 + c) + a**2*c*x**4*sqrt(-a** 
2*c*x**2 + c) + a*c*x**3*sqrt(-a**2*c*x**2 + c) - c*x**2*sqrt(-a**2*c*x**2 
 + c)), x) - Integral(1/(-a**3*c*x**5*sqrt(-a**2*c*x**2 + c) + a**2*c*x**4 
*sqrt(-a**2*c*x**2 + c) + a*c*x**3*sqrt(-a**2*c*x**2 + c) - c*x**2*sqrt(-a 
**2*c*x**2 + c)), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=-2 \, a^{2} \sqrt {-c} c {\left (\frac {2 \, {\left | a \right |} \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{a^{3} c^{3}} - \frac {1}{{\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )} a^{2} c^{2}}\right )} {\left | a \right |} \] Input:

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

Output:

-2*a^2*sqrt(-c)*c*(2*abs(a)*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c) 
)/sqrt(-c))/(a^3*c^3) - 1/(((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c) 
*a^2*c^2))*abs(a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.76 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (-10 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+14 \sqrt {-a^{2} x^{2}+1}\, a x -3 \sqrt {-a^{2} x^{2}+1}+3 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}-1\right ) a^{3} x^{3}-6 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}-1\right ) a^{2} x^{2}+3 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}-1\right ) a x -3 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}+1\right ) a^{3} x^{3}+6 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}+1\right ) a^{2} x^{2}-3 \,\mathrm {log}\left (\sqrt {-a^{2} x^{2}+1}+1\right ) a x \right )}{3 c^{2} x \left (a^{2} x^{2}-2 a x +1\right )} \] Input:

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

Output:

(sqrt(c)*( - 10*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 14*sqrt( - a**2*x**2 + 
1)*a*x - 3*sqrt( - a**2*x**2 + 1) + 3*log(sqrt( - a**2*x**2 + 1) - 1)*a**3 
*x**3 - 6*log(sqrt( - a**2*x**2 + 1) - 1)*a**2*x**2 + 3*log(sqrt( - a**2*x 
**2 + 1) - 1)*a*x - 3*log(sqrt( - a**2*x**2 + 1) + 1)*a**3*x**3 + 6*log(sq 
rt( - a**2*x**2 + 1) + 1)*a**2*x**2 - 3*log(sqrt( - a**2*x**2 + 1) + 1)*a* 
x))/(3*c**2*x*(a**2*x**2 - 2*a*x + 1))