Integrand size = 18, antiderivative size = 58 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c (1-a x) \sqrt {1-a^2 x^2}}{a^2 x}+\frac {c \arcsin (a x)}{a}+\frac {c \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{a} \]
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c \left ((1-a x) \sqrt {1-a^2 x^2}+a x \arcsin (a x)+a x \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{a^2 x} \]
(c*((1 - a*x)*Sqrt[1 - a^2*x^2] + a*x*ArcSin[a*x] + a*x*ArcTanh[Sqrt[1 - a ^2*x^2]]))/(a^2*x)
Time = 0.36 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6707, 6698, 536, 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.
\(\displaystyle \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle -\frac {c \int \frac {e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )}{x^2}dx}{a^2}\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle -\frac {c \int \frac {(a x+1) \sqrt {1-a^2 x^2}}{x^2}dx}{a^2}\) |
\(\Big \downarrow \) 536 |
\(\displaystyle -\frac {c \left (\int \frac {a-a^2 x}{x \sqrt {1-a^2 x^2}}dx-\frac {(1-a x) \sqrt {1-a^2 x^2}}{x}\right )}{a^2}\) |
\(\Big \downarrow \) 538 |
\(\displaystyle -\frac {c \left (a^2 \left (-\int \frac {1}{\sqrt {1-a^2 x^2}}dx\right )+a \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\frac {(1-a x) \sqrt {1-a^2 x^2}}{x}\right )}{a^2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {c \left (a \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2} (1-a x)}{x}-a \arcsin (a x)\right )}{a^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {c \left (\frac {1}{2} a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\frac {\sqrt {1-a^2 x^2} (1-a x)}{x}-a \arcsin (a x)\right )}{a^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {c \left (-\frac {\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a}-\frac {\sqrt {1-a^2 x^2} (1-a x)}{x}-a \arcsin (a x)\right )}{a^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {c \left (-a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sqrt {1-a^2 x^2} (1-a x)}{x}-a \arcsin (a x)\right )}{a^2}\) |
-((c*(-(((1 - a*x)*Sqrt[1 - a^2*x^2])/x) - a*ArcSin[a*x] - a*ArcTanh[Sqrt[ 1 - a^2*x^2]]))/a^2)
3.7.31.3.1 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[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_)^2, x_Symbol] :> S imp[(-(2*c*p - d*x))*((a + b*x^2)^p/(2*p*x)), x] + Int[(a*d + 2*b*c*p*x)*(( a + b*x^2)^(p - 1)/x), x] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && Integer Q[2*p]
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[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0]) && IGtQ[(n + 1)/2, 0] && !IntegerQ[p - n/2]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.41
method | result | size |
default | \(\frac {c \left (\frac {a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-\sqrt {-a^{2} x^{2}+1}\, a +\frac {\sqrt {-a^{2} x^{2}+1}}{x}+a \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{a^{2}}\) | \(82\) |
risch | \(-\frac {\left (a^{2} x^{2}-1\right ) c}{x \sqrt {-a^{2} x^{2}+1}\, a^{2}}+\frac {\left (\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-\sqrt {-a^{2} x^{2}+1}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right ) c}{a}\) | \(92\) |
meijerg | \(-\frac {c \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{2 a \sqrt {\pi }}-\frac {c \left (-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )+\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }\right )}{2 a \sqrt {\pi }}+\frac {c \arcsin \left (a x \right )}{a}+\frac {c \sqrt {-a^{2} x^{2}+1}}{a^{2} x}\) | \(114\) |
c/a^2*(a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-(-a^2*x^2+ 1)^(1/2)*a+(-a^2*x^2+1)^(1/2)/x+a*arctanh(1/(-a^2*x^2+1)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.45 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {2 \, a c x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + a c x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + a c x + \sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )}}{a^{2} x} \]
-(2*a*c*x*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + a*c*x*log((sqrt(-a^2*x^ 2 + 1) - 1)/x) + a*c*x + sqrt(-a^2*x^2 + 1)*(a*c*x - c))/(a^2*x)
Time = 2.62 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.47 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=a c \left (\begin {cases} - \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {for}\: a^{2} \neq 0 \\\frac {x^{2}}{2} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{\sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\x & \text {otherwise} \end {cases}\right ) - \frac {c \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right )}{a} - \frac {c \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right )}{a^{2}} \]
a*c*Piecewise((-sqrt(-a**2*x**2 + 1)/a**2, Ne(a**2, 0)), (x**2/2, True)) + c*Piecewise((log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/sqrt(-a* *2), Ne(a**2, 0)), (x, True)) - c*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x **2) > 1), (I*asin(1/(a*x)), True))/a - c*Piecewise((-I*sqrt(a**2*x**2 - 1 )/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True))/a**2
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.36 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c \arcsin \left (a x\right )}{a} + \frac {c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )}{a} - \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} + \frac {\sqrt {-a^{2} x^{2} + 1} c}{a^{2} x} \]
c*arcsin(a*x)/a + c*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x))/a - sqrt(- a^2*x^2 + 1)*c/a + sqrt(-a^2*x^2 + 1)*c/(a^2*x)
Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (54) = 108\).
Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.21 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {a^{2} c x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} + \frac {c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} + \frac {c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c}{2 \, a^{2} x {\left | a \right |}} \]
-1/2*a^2*c*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*abs(a)) + 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 + 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c/ (a^2*x*abs(a))
Time = 3.57 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.31 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{a}-\frac {c\,\sqrt {1-a^2\,x^2}}{a}+\frac {c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {c\,\sqrt {1-a^2\,x^2}}{a^2\,x} \]