Integrand size = 12, antiderivative size = 98 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx=-\frac {\sqrt {1-a-b x} \sqrt {1+a+b x}}{(1-a) x}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{(1-a) \sqrt {1-a^2}} \] Output:
-(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/(1-a)/x-2*b*arctanh((1-a)^(1/2)*(b*x+a+1 )^(1/2)/(1+a)^(1/2)/(-b*x-a+1)^(1/2))/(1-a)/(-a^2+1)^(1/2)
Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx=\frac {\sqrt {-((-1+a+b x) (1+a+b x))}}{(-1+a) x}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {-1-a} \sqrt {1-a-b x}}{\sqrt {-1+a} \sqrt {1+a+b x}}\right )}{\sqrt {-1-a} (-1+a)^{3/2}} \] Input:
Integrate[E^ArcTanh[a + b*x]/x^2,x]
Output:
Sqrt[-((-1 + a + b*x)*(1 + a + b*x))]/((-1 + a)*x) - (2*b*ArcTanh[(Sqrt[-1 - a]*Sqrt[1 - a - b*x])/(Sqrt[-1 + a]*Sqrt[1 + a + b*x])])/(Sqrt[-1 - a]* (-1 + a)^(3/2))
Time = 0.43 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6713, 105, 104, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 6713 |
\(\displaystyle \int \frac {\sqrt {a+b x+1}}{x^2 \sqrt {-a-b x+1}}dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {b \int \frac {1}{x \sqrt {-a-b x+1} \sqrt {a+b x+1}}dx}{1-a}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{(1-a) x}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {2 b \int \frac {1}{-a+\frac {(1-a) (a+b x+1)}{-a-b x+1}-1}d\frac {\sqrt {a+b x+1}}{\sqrt {-a-b x+1}}}{1-a}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{(1-a) x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 b \text {arctanh}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \sqrt {1-a^2}}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{(1-a) x}\) |
Input:
Int[E^ArcTanh[a + b*x]/x^2,x]
Output:
-((Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/((1 - a)*x)) - (2*b*ArcTanh[(Sqrt[ 1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x])])/((1 - a)*Sqrt[ 1 - a^2])
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
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[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.) , x_Symbol] :> Int[(d + e*x)^m*((1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^( n/2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.19
method | result | size |
risch | \(-\frac {b^{2} x^{2}+2 a b x +a^{2}-1}{\left (-1+a \right ) x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {b \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-1+a \right ) \sqrt {-a^{2}+1}}\) | \(117\) |
default | \(-\frac {b \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\sqrt {-a^{2}+1}}+\left (a +1\right ) \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{\left (-a^{2}+1\right ) x}-\frac {a b \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )\) | \(168\) |
Input:
int((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-1/(-1+a)*(b^2*x^2+2*a*b*x+a^2-1)/x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+b/(-1+a )/(-a^2+1)^(1/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a ^2+1)^(1/2))/x)
Time = 0.12 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.88 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx=\left [-\frac {\sqrt {-a^{2} + 1} b x \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{2 \, {\left (a^{3} - a^{2} - a + 1\right )} x}, -\frac {\sqrt {a^{2} - 1} b x \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{{\left (a^{3} - a^{2} - a + 1\right )} x}\right ] \] Input:
integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^2,x, algorithm="fricas")
Output:
[-1/2*(sqrt(-a^2 + 1)*b*x*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b *x - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1) - 4*a^2 + 2)/x^2) - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^2 - 1))/((a^3 - a^2 - a + 1)*x), -(sqrt(a^2 - 1)*b*x*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a ^2 + 1)*(a*b*x + a^2 - 1)*sqrt(a^2 - 1)/((a^2 - 1)*b^2*x^2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1)) - sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^2 - 1))/((a ^3 - a^2 - a + 1)*x)]
\[ \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx=\int \frac {a + b x + 1}{x^{2} \sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}\, dx \] Input:
integrate((b*x+a+1)/(1-(b*x+a)**2)**(1/2)/x**2,x)
Output:
Integral((a + b*x + 1)/(x**2*sqrt(-(a + b*x - 1)*(a + b*x + 1))), x)
Exception generated. \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^2,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for more details)Is
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (79) = 158\).
Time = 0.13 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.31 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx=-\frac {2 \, b^{2} \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left (a {\left | b \right |} - {\left | b \right |}\right )}} + \frac {2 \, {\left (a b^{2} - \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} b^{2}}{b^{2} x + a b}\right )}}{{\left (a^{2} {\left | b \right |} - a {\left | b \right |}\right )} {\left (\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a}{{\left (b^{2} x + a b\right )}^{2}} + a - \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}}{b^{2} x + a b}\right )}} \] Input:
integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^2,x, algorithm="giac")
Output:
-2*b^2*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - 1)/sqrt(a^2 - 1))/(sqrt(a^2 - 1)*(a*abs(b) - abs(b))) + 2*(a*b^2 - (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*b^2/(b^2*x + a*b))/((a^2*a bs(b) - a*abs(b))*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a/(b^ 2*x + a*b)^2 + a - 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2* x + a*b)))
Time = 24.45 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.46 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx=\frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{x\,\left (a^2-1\right )}-\frac {b\,\ln \left (\frac {\sqrt {1-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{x}-\frac {a^2-1}{x}-a\,b\right )}{\sqrt {1-a^2}}+\frac {a^2\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a-1}{\sqrt {1-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}\right )}{{\left (1-a^2\right )}^{3/2}}+\frac {a\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{x\,\left (a^2-1\right )}+\frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a-1}{\sqrt {1-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}\right )}{{\left (1-a^2\right )}^{3/2}} \] Input:
int((a + b*x + 1)/(x^2*(1 - (a + b*x)^2)^(1/2)),x)
Output:
(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/2)/(x*(a^2 - 1)) - (b*log(((1 - a^2)^(1/2 )*(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/2))/x - (a^2 - 1)/x - a*b))/(1 - a^2)^( 1/2) + (a^2*b*atanh((a^2 + a*b*x - 1)/((1 - a^2)^(1/2)*(1 - b^2*x^2 - 2*a* b*x - a^2)^(1/2))))/(1 - a^2)^(3/2) + (a*(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/ 2))/(x*(a^2 - 1)) + (a*b*atanh((a^2 + a*b*x - 1)/((1 - a^2)^(1/2)*(1 - b^2 *x^2 - 2*a*b*x - a^2)^(1/2))))/(1 - a^2)^(3/2)
Time = 0.15 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x^2} \, dx=\frac {2 \sqrt {a^{2}-1}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a -1}{\sqrt {a^{2}-1}}\right ) b x +\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{2}-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x \left (a^{3}-a^{2}-a +1\right )} \] Input:
int((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^2,x)
Output:
(2*sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a - 1)/sqrt(a**2 - 1))*b*x + sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*a**2 - sqrt( - a**2 - 2*a*b*x - b* *2*x**2 + 1))/(x*(a**3 - a**2 - a + 1))