Integrand size = 8, antiderivative size = 54 \[ \int e^{-3 \text {arctanh}(a x)} \, dx=-\frac {4 (1-a x)}{a \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a}-\frac {3 \arcsin (a x)}{a} \] Output:
(4*a*x-4)/a/(-a^2*x^2+1)^(1/2)-(-a^2*x^2+1)^(1/2)/a-3*arcsin(a*x)/a
Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.72 \[ \int e^{-3 \text {arctanh}(a x)} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-1-\frac {4}{1+a x}\right )}{a}-\frac {3 \arcsin (a x)}{a} \] Input:
Integrate[E^(-3*ArcTanh[a*x]),x]
Output:
(Sqrt[1 - a^2*x^2]*(-1 - 4/(1 + a*x)))/a - (3*ArcSin[a*x])/a
Time = 0.42 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6673, 711, 25, 27, 671, 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.
\(\displaystyle \int e^{-3 \text {arctanh}(a x)} \, dx\) |
\(\Big \downarrow \) 6673 |
\(\displaystyle \int \frac {(1-a x)^2}{(a x+1) \sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 711 |
\(\displaystyle -\frac {\int -\frac {a^4 (1-3 a x)}{(a x+1) \sqrt {1-a^2 x^2}}dx}{a^4}-\frac {\sqrt {1-a^2 x^2}}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a^4 (1-3 a x)}{(a x+1) \sqrt {1-a^2 x^2}}dx}{a^4}-\frac {\sqrt {1-a^2 x^2}}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1-3 a x}{(a x+1) \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{a}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle -3 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{a}-\frac {4 \sqrt {1-a^2 x^2}}{a (a x+1)}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {\sqrt {1-a^2 x^2}}{a}-\frac {4 \sqrt {1-a^2 x^2}}{a (a x+1)}-\frac {3 \arcsin (a x)}{a}\) |
Input:
Int[E^(-3*ArcTanh[a*x]),x]
Output:
-(Sqrt[1 - a^2*x^2]/a) - (4*Sqrt[1 - a^2*x^2])/(a*(1 + a*x)) - (3*ArcSin[a *x])/a
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m + p + 1)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (c_.)*(x_ )^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + c*x^2)^(p + 1) /(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x) ^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - 2*e*g^n*(m + p + n)*(d + e*x)^(n - 2)*(a*e - c*d*x), x], x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && Eq Q[c*d^2 + a*e^2, 0] && IGtQ[n, 0] && NeQ[m + n + 2*p + 1, 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.)), x_Symbol] :> Int[(1 + a*x)^((n + 1)/2)/( (1 - a*x)^((n - 1)/2)*Sqrt[1 - a^2*x^2]), x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2]
Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.65
method | result | size |
risch | \(\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-\frac {4 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a^{2} \left (x +\frac {1}{a}\right )}\) | \(89\) |
default | \(\frac {-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{3}}-2 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{3}}\) | \(191\) |
Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/a*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)-3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^ 2*x^2+1)^(1/2))-4/a^2/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.19 \[ \int e^{-3 \text {arctanh}(a x)} \, dx=-\frac {5 \, a x - 6 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x + 5\right )} + 5}{a^{2} x + a} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")
Output:
-(5*a*x - 6*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x ^2 + 1)*(a*x + 5) + 5)/(a^2*x + a)
\[ \int e^{-3 \text {arctanh}(a x)} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (a x + 1\right )^{3}}\, dx \] Input:
integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)
Output:
Integral((-(a*x - 1)*(a*x + 1))**(3/2)/(a*x + 1)**3, x)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17 \[ \int e^{-3 \text {arctanh}(a x)} \, dx=\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{3} x^{2} + 2 \, a^{2} x + a} - \frac {3 \, \arcsin \left (a x\right )}{a} - \frac {6 \, \sqrt {-a^{2} x^{2} + 1}}{a^{2} x + a} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")
Output:
(-a^2*x^2 + 1)^(3/2)/(a^3*x^2 + 2*a^2*x + a) - 3*arcsin(a*x)/a - 6*sqrt(-a ^2*x^2 + 1)/(a^2*x + a)
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.19 \[ \int e^{-3 \text {arctanh}(a x)} \, dx=-\frac {3 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a} + \frac {8}{{\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")
Output:
-3*arcsin(a*x)*sgn(a)/abs(a) - sqrt(-a^2*x^2 + 1)/a + 8/(((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.50 \[ \int e^{-3 \text {arctanh}(a x)} \, dx=\frac {4\,\sqrt {1-a^2\,x^2}}{\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}} \] Input:
int((1 - a^2*x^2)^(3/2)/(a*x + 1)^3,x)
Output:
(4*(1 - a^2*x^2)^(1/2))/((x*(-a^2)^(1/2) + (-a^2)^(1/2)/a)*(-a^2)^(1/2)) - (1 - a^2*x^2)^(1/2)/a - (3*asinh(x*(-a^2)^(1/2)))/(-a^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.72 \[ \int e^{-3 \text {arctanh}(a x)} \, dx=\frac {-3 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )+3 \mathit {asin} \left (a x \right ) a x +3 \mathit {asin} \left (a x \right )+\sqrt {-a^{2} x^{2}+1}\, a x +8 \sqrt {-a^{2} x^{2}+1}+a^{2} x^{2}+a x -8}{a \left (\sqrt {-a^{2} x^{2}+1}-a x -1\right )} \] Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
Output:
( - 3*sqrt( - a**2*x**2 + 1)*asin(a*x) + 3*asin(a*x)*a*x + 3*asin(a*x) + s qrt( - a**2*x**2 + 1)*a*x + 8*sqrt( - a**2*x**2 + 1) + a**2*x**2 + a*x - 8 )/(a*(sqrt( - a**2*x**2 + 1) - a*x - 1))