Integrand size = 12, antiderivative size = 64 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx=\arcsin (a+b x)-\frac {2 (1+a) \text {arctanh}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{\sqrt {1-a^2}} \] Output:
arcsin(b*x+a)-2*(1+a)*arctanh((1-a)^(1/2)*(b*x+a+1)^(1/2)/(1+a)^(1/2)/(-b* x-a+1)^(1/2))/(-a^2+1)^(1/2)
Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.66 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx=\frac {2 \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \sqrt {-1-a} \text {arctanh}\left (\frac {\sqrt {-1-a} \sqrt {1-a-b x}}{\sqrt {-1+a} \sqrt {1+a+b x}}\right )}{\sqrt {-1+a}} \] Input:
Integrate[E^ArcTanh[a + b*x]/x,x]
Output:
(2*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])])/Sqrt[ b] - (2*Sqrt[-1 - a]*ArcTanh[(Sqrt[-1 - a]*Sqrt[1 - a - b*x])/(Sqrt[-1 + a ]*Sqrt[1 + a + b*x])])/Sqrt[-1 + a]
Time = 0.49 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6713, 140, 27, 62, 104, 221, 1090, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx\) |
\(\Big \downarrow \) 6713 |
\(\displaystyle \int \frac {\sqrt {a+b x+1}}{x \sqrt {-a-b x+1}}dx\) |
\(\Big \downarrow \) 140 |
\(\displaystyle b \int \frac {1}{\sqrt {-a-b x+1} \sqrt {a+b x+1}}dx+\int \frac {a+1}{x \sqrt {-a-b x+1} \sqrt {a+b x+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle b \int \frac {1}{\sqrt {-a-b x+1} \sqrt {a+b x+1}}dx+(a+1) \int \frac {1}{x \sqrt {-a-b x+1} \sqrt {a+b x+1}}dx\) |
\(\Big \downarrow \) 62 |
\(\displaystyle b \int \frac {1}{\sqrt {-b^2 x^2-2 a b x+(1-a) (a+1)}}dx+(a+1) \int \frac {1}{x \sqrt {-a-b x+1} \sqrt {a+b x+1}}dx\) |
\(\Big \downarrow \) 104 |
\(\displaystyle b \int \frac {1}{\sqrt {-b^2 x^2-2 a b x+(1-a) (a+1)}}dx+2 (a+1) \int \frac {1}{-a+\frac {(1-a) (a+b x+1)}{-a-b x+1}-1}d\frac {\sqrt {a+b x+1}}{\sqrt {-a-b x+1}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle b \int \frac {1}{\sqrt {-b^2 x^2-2 a b x+(1-a) (a+1)}}dx-\frac {2 (a+1) \text {arctanh}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{\sqrt {1-a^2}}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {1-\frac {\left (-2 x b^2-2 a b\right )^2}{4 b^2}}}d\left (-2 x b^2-2 a b\right )}{2 b}-\frac {2 (a+1) \text {arctanh}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{\sqrt {1-a^2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {2 (a+1) \text {arctanh}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{\sqrt {1-a^2}}-\arcsin \left (\frac {-2 a b-2 b^2 x}{2 b}\right )\) |
Input:
Int[E^ArcTanh[a + b*x]/x,x]
Output:
-ArcSin[(-2*a*b - 2*b^2*x)/(2*b)] - (2*(1 + a)*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x])])/Sqrt[1 - a^2]
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_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Int[ 1/Sqrt[a*c - b*(a - c)*x - b^2*x^2], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]
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[b*d^(m + n)*f^p Int[(a + b*x)^(m - 1)/(c + d*x)^m, x] , x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandToSum[(a + b*x )*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] || !(GtQ[n, 0] || SumSimplerQ[n, -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[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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
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 = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.70
method | result | size |
default | \(\frac {b \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}-\frac {\left (a +1\right ) \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\sqrt {-a^{2}+1}}\) | \(109\) |
Input:
int((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x,x,method=_RETURNVERBOSE)
Output:
b/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+1/b*a)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)) -(a+1)/(-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)
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (54) = 108\).
Time = 0.13 (sec) , antiderivative size = 306, normalized size of antiderivative = 4.78 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx=\left [\frac {1}{2} \, \sqrt {-\frac {a + 1}{a - 1}} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, a^{2} - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{3} + {\left (a^{2} - a\right )} b x - a^{2} - a + 1\right )} \sqrt {-\frac {a + 1}{a - 1}} + 2}{x^{2}}\right ) - \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ), \sqrt {\frac {a + 1}{a - 1}} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {\frac {a + 1}{a - 1}}}{{\left (a + 1\right )} b^{2} x^{2} + a^{3} + 2 \, {\left (a^{2} + a\right )} b x + a^{2} - a - 1}\right ) - \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )\right ] \] Input:
integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x,x, algorithm="fricas")
Output:
[1/2*sqrt(-(a + 1)/(a - 1))*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a) *b*x - 4*a^2 - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^3 + (a^2 - a)*b*x - a^2 - a + 1)*sqrt(-(a + 1)/(a - 1)) + 2)/x^2) - arctan(sqrt(-b^2*x^2 - 2* a*b*x - a^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 - 1)), sqrt((a + 1)/(a - 1))*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt((a + 1)/(a - 1))/((a + 1)*b^2*x^2 + a^3 + 2*(a^2 + a)*b*x + a^2 - a - 1)) - arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a*b*x + a ^2 - 1))]
\[ \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx=\int \frac {a + b x + 1}{x \sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}\, dx \] Input:
integrate((b*x+a+1)/(1-(b*x+a)**2)**(1/2)/x,x)
Output:
Integral((a + b*x + 1)/(x*sqrt(-(a + b*x - 1)*(a + b*x + 1))), x)
Exception generated. \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x,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
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx=-\frac {b \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{{\left | b \right |}} + \frac {2 \, {\left (a b + b\right )} \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left | b \right |}} \] Input:
integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x,x, algorithm="giac")
Output:
-b*arcsin(-b*x - a)*sgn(b)/abs(b) + 2*(a*b + b)*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)*abs(b))
Time = 25.16 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.06 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx=\mathrm {asin}\left (a+b\,x\right )-\frac {\ln \left (\frac {\sqrt {1-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{x}-\frac {a^2-1}{x}-a\,b\right )}{\sqrt {1-a^2}}-\frac {a\,\ln \left (\frac {\sqrt {1-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{x}-\frac {a^2-1}{x}-a\,b\right )}{\sqrt {1-a^2}} \] Input:
int((a + b*x + 1)/(x*(1 - (a + b*x)^2)^(1/2)),x)
Output:
asin(a + b*x) - 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*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)
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x} \, dx=\frac {\mathit {asin} \left (b x +a \right ) a -\mathit {asin} \left (b x +a \right )-2 \sqrt {a^{2}-1}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a -1}{\sqrt {a^{2}-1}}\right )}{a -1} \] Input:
int((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x,x)
Output:
(asin(a + b*x)*a - asin(a + b*x) - 2*sqrt(a**2 - 1)*atan((tan(asin(a + b*x )/2)*a - 1)/sqrt(a**2 - 1)))/(a - 1)