Integrand size = 20, antiderivative size = 50 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c \sqrt {1-a^2 x^2}}{a}+\frac {2 c \arcsin (a x)}{a}+\frac {c \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{a} \] Output:
c*(-a^2*x^2+1)^(1/2)/a+2*c*arcsin(a*x)/a+c*arctanh((-a^2*x^2+1)^(1/2))/a
Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.60 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c \left (\frac {1}{\sqrt {1-a^2 x^2}}-\frac {a^2 x^2}{\sqrt {1-a^2 x^2}}+\arcsin (a x)-2 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )+\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{a} \] Input:
Integrate[(c - c/(a*x))/E^ArcTanh[a*x],x]
Output:
(c*(1/Sqrt[1 - a^2*x^2] - (a^2*x^2)/Sqrt[1 - a^2*x^2] + ArcSin[a*x] - 2*Ar cSin[Sqrt[1 - a*x]/Sqrt[2]] + ArcTanh[Sqrt[1 - a^2*x^2]]))/a
Time = 0.61 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6681, 6678, 541, 25, 27, 538, 223, 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.
\(\displaystyle \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle -\frac {c \int \frac {e^{-\text {arctanh}(a x)} (1-a x)}{x}dx}{a}\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle -\frac {c \int \frac {(1-a x)^2}{x \sqrt {1-a^2 x^2}}dx}{a}\) |
\(\Big \downarrow \) 541 |
\(\displaystyle -\frac {c \left (-\frac {\int -\frac {a^2 (1-2 a x)}{x \sqrt {1-a^2 x^2}}dx}{a^2}-\sqrt {1-a^2 x^2}\right )}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {c \left (\frac {\int \frac {a^2 (1-2 a x)}{x \sqrt {1-a^2 x^2}}dx}{a^2}-\sqrt {1-a^2 x^2}\right )}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {c \left (\int \frac {1-2 a x}{x \sqrt {1-a^2 x^2}}dx-\sqrt {1-a^2 x^2}\right )}{a}\) |
\(\Big \downarrow \) 538 |
\(\displaystyle -\frac {c \left (-2 a \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\sqrt {1-a^2 x^2}\right )}{a}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {c \left (\int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\sqrt {1-a^2 x^2}-2 \arcsin (a x)\right )}{a}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {c \left (\frac {1}{2} \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\sqrt {1-a^2 x^2}-2 \arcsin (a x)\right )}{a}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {c \left (-\frac {\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}-\sqrt {1-a^2 x^2}-2 \arcsin (a x)\right )}{a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {c \left (-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\sqrt {1-a^2 x^2}-2 \arcsin (a x)\right )}{a}\) |
Input:
Int[(c - c/(a*x))/E^ArcTanh[a*x],x]
Output:
-((c*(-Sqrt[1 - a^2*x^2] - 2*ArcSin[a*x] - ArcTanh[Sqrt[1 - a^2*x^2]]))/a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x ] + Simp[1/(b*(m + n + 2*p + 1)) Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) *x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol ] :> Simp[d^p Int[u*(1 + c*(x/d))^p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; F reeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(97\) vs. \(2(46)=92\).
Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.96
method | result | size |
default | \(\frac {c \left (-\sqrt {-a^{2} x^{2}+1}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+2 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {2 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{a}\) | \(98\) |
Input:
int((c-c/a/x)/(a*x+1)*(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
c/a*(-(-a^2*x^2+1)^(1/2)+arctanh(1/(-a^2*x^2+1)^(1/2))+2*(-a^2*(x+1/a)^2+2 *a*(x+1/a))^(1/2)+2*a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a *(x+1/a))^(1/2)))
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.34 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=-\frac {4 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + c \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} c}{a} \] Input:
integrate((c-c/a/x)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")
Output:
-(4*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + c*log((sqrt(-a^2*x^2 + 1) - 1)/x) - sqrt(-a^2*x^2 + 1)*c)/a
\[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c \left (\int \left (- \frac {\sqrt {- a^{2} x^{2} + 1}}{a x^{2} + x}\right )\, dx + \int \frac {a x \sqrt {- a^{2} x^{2} + 1}}{a x^{2} + x}\, dx\right )}{a} \] Input:
integrate((c-c/a/x)/(a*x+1)*(-a**2*x**2+1)**(1/2),x)
Output:
c*(Integral(-sqrt(-a**2*x**2 + 1)/(a*x**2 + x), x) + Integral(a*x*sqrt(-a* *2*x**2 + 1)/(a*x**2 + x), x))/a
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.40 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=a c {\left (\frac {\arcsin \left (a x\right )}{a^{2}} + \frac {\log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )}{a^{2}}\right )} + c {\left (\frac {\arcsin \left (a x\right )}{a} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a}\right )} \] Input:
integrate((c-c/a/x)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="maxima")
Output:
a*c*(arcsin(a*x)/a^2 + log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x))/a^2) + c*(arcsin(a*x)/a + sqrt(-a^2*x^2 + 1)/a)
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.36 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {2 \, c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} + \frac {c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} \] Input:
integrate((c-c/a/x)/(a*x+1)*(-a^2*x^2+1)^(1/2),x, algorithm="giac")
Output:
2*c*arcsin(a*x)*sgn(a)/abs(a) + c*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + sqrt(-a^2*x^2 + 1)*c/a
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.12 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c\,\sqrt {1-a^2\,x^2}}{a}+\frac {c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{a}+\frac {2\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}} \] Input:
int(((c - c/(a*x))*(1 - a^2*x^2)^(1/2))/(a*x + 1),x)
Output:
(c*(1 - a^2*x^2)^(1/2))/a + (c*atanh((1 - a^2*x^2)^(1/2)))/a + (2*c*asinh( x*(-a^2)^(1/2)))/(-a^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c \left (2 \mathit {asin} \left (a x \right )+\sqrt {-a^{2} x^{2}+1}-\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right )-1\right )}{a} \] Input:
int((c-c/a/x)/(a*x+1)*(-a^2*x^2+1)^(1/2),x)
Output:
(c*(2*asin(a*x) + sqrt( - a**2*x**2 + 1) - log(tan(asin(a*x)/2)) - 1))/a