Integrand size = 12, antiderivative size = 45 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x} \, dx=\frac {4 \sqrt {1-a^2 x^2}}{1+a x}+\arcsin (a x)-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \]
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x} \, dx=\frac {4 \sqrt {1-a^2 x^2}}{1+a x}+\arcsin (a x)+\log (x)-\log \left (1+\sqrt {1-a^2 x^2}\right ) \]
Time = 0.47 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6674, 2351, 564, 25, 243, 73, 221, 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.
\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{x} \, dx\) |
\(\Big \downarrow \) 6674 |
\(\displaystyle \int \frac {(1-a x)^2}{x (a x+1) \sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 2351 |
\(\displaystyle \int \frac {1}{x (a x+1) \sqrt {1-a^2 x^2}}dx+\int \frac {a^2 x-2 a}{(a x+1) \sqrt {1-a^2 x^2}}dx\) |
\(\Big \downarrow \) 564 |
\(\displaystyle -\int -\frac {1}{x \sqrt {1-a^2 x^2}}dx+\int \frac {a^2 x-2 a}{(a x+1) \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {1}{x \sqrt {1-a^2 x^2}}dx+\int \frac {a^2 x-2 a}{(a x+1) \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a x+1}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2+\int \frac {a^2 x-2 a}{(a x+1) \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a x+1}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \int \frac {a^2 x-2 a}{(a x+1) \sqrt {1-a^2 x^2}}dx-\frac {\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}+\frac {\sqrt {1-a^2 x^2}}{a x+1}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \int \frac {a^2 x-2 a}{(a x+1) \sqrt {1-a^2 x^2}}dx-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )+\frac {\sqrt {1-a^2 x^2}}{a x+1}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle a \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )+\frac {4 \sqrt {1-a^2 x^2}}{a x+1}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )+\frac {4 \sqrt {1-a^2 x^2}}{a x+1}+\arcsin (a x)\) |
3.1.55.3.1 Defintions of rubi rules used
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]
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]
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[(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]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 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[(x^m/Sqrt[a + b *x^2])*ExpandToSum[((2^(-n - 1)*(-c)^(m - n - 1))/(d^m*x^m) - (-c + d*x)^(- n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^ 2, 0] && ILtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
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[((Px_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.))/(x_), x_S ymbol] :> Int[PolynomialQuotient[Px, x, x]*(c + d*x)^n*(a + b*x^2)^p, x] + Simp[PolynomialRemainder[Px, x, x] Int[(c + d*x)^n*((a + b*x^2)^p/x), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && PolynomialQ[Px, x]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x )^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/2)*Sqrt[1 - a^2*x^2])), x] / ; FreeQ[{a, c, m}, x] && IntegerQ[(n - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(494\) vs. \(2(41)=82\).
Time = 0.08 (sec) , antiderivative size = 495, normalized size of antiderivative = 11.00
method | result | size |
default | \(\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3}+\sqrt {-a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\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 )-\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 )^{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 )}{a}-\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^{2}}\) | \(495\) |
1/3*(-a^2*x^2+1)^(3/2)+(-a^2*x^2+1)^(1/2)-arctanh(1/(-a^2*x^2+1)^(1/2))-1/ 3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-a*(-1/4*(-2*a^2*(x+1/a)+2*a)/a^2*(-a^ 2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2* (x+1/a)^2+2*a*(x+1/a))^(1/2)))-1/a*(1/a/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1 /a))^(5/2)+3*a*(1/3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+a*(-1/4*(-2*a^2*(x+ 1/a)+2*a)/a^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/2/(a^2)^(1/2)*arctan((a ^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)))))-1/a^2*(-1/a/(x+1/a)^3*( -a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-2*a*(1/a/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*( x+1/a))^(5/2)+3*a*(1/3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+a*(-1/4*(-2*a^2* (x+1/a)+2*a)/a^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/2/(a^2)^(1/2)*arctan ((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))))))
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.82 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x} \, dx=\frac {4 \, a x - 2 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (a x + 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + 4 \, \sqrt {-a^{2} x^{2} + 1} + 4}{a x + 1} \]
(4*a*x - 2*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (a*x + 1)*lo g((sqrt(-a^2*x^2 + 1) - 1)/x) + 4*sqrt(-a^2*x^2 + 1) + 4)/(a*x + 1)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{x \left (a x + 1\right )^{3}}\, dx \]
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} x} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (41) = 82\).
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.91 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x} \, dx=\frac {a \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} - \frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} - \frac {8 \, a}{{\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]
a*arcsin(a*x)*sgn(a)/abs(a) - a*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 8*a/(((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a))
Time = 3.49 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.78 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{x} \, dx=\frac {a\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}-\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )-\frac {4\,a\,\sqrt {1-a^2\,x^2}}{\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]