Integrand size = 20, antiderivative size = 50 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c \sqrt {1-a^2 x^2}}{a}-\frac {2 c \arcsin (a x)}{a}+\frac {c \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{a} \]
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.20 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c \left (\sqrt {1-a^2 x^2}-\arcsin (a x)+2 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )+\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{a} \]
(c*(Sqrt[1 - a^2*x^2] - ArcSin[a*x] + 2*ArcSin[Sqrt[1 - a*x]/Sqrt[2]] + Ar cTanh[Sqrt[1 - a^2*x^2]]))/a
Time = 0.36 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {6681, 6678, 570, 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.
\(\displaystyle \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle -\frac {c \int \frac {e^{3 \text {arctanh}(a x)} (1-a x)}{x}dx}{a}\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle -\frac {c \int \frac {\left (1-a^2 x^2\right )^{3/2}}{x (1-a x)^2}dx}{a}\) |
\(\Big \downarrow \) 570 |
\(\displaystyle -\frac {c \int \frac {(a x+1)^2}{x \sqrt {1-a^2 x^2}}dx}{a}\) |
\(\Big \downarrow \) 541 |
\(\displaystyle -\frac {c \left (-\frac {\int -\frac {a^2 (2 a x+1)}{x \sqrt {1-a^2 x^2}}dx}{a^2}-\sqrt {1-a^2 x^2}\right )}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {c \left (\frac {\int \frac {a^2 (2 a x+1)}{x \sqrt {1-a^2 x^2}}dx}{a^2}-\sqrt {1-a^2 x^2}\right )}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {c \left (\int \frac {2 a x+1}{x \sqrt {1-a^2 x^2}}dx-\sqrt {1-a^2 x^2}\right )}{a}\) |
\(\Big \downarrow \) 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 |
\(\displaystyle -\frac {c \left (-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\sqrt {1-a^2 x^2}+2 \arcsin (a x)\right )}{a}\) |
3.5.69.3.1 Defintions of rubi rules used
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_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
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. \(144\) vs. \(2(46)=92\).
Time = 0.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.90
method | result | size |
default | \(\frac {c \left (-\frac {2 a x}{\sqrt {-a^{2} x^{2}+1}}+a^{4} \left (-\frac {x^{2}}{a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {-a^{2} x^{2}+1}}\right )-\frac {1}{\sqrt {-a^{2} x^{2}+1}}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+2 a^{3} \left (\frac {x}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}\right )\right )}{a}\) | \(145\) |
meijerg | \(-\frac {2 c x}{\sqrt {-a^{2} x^{2}+1}}-\frac {c \left (-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{2}\right )}{\sqrt {\pi }\, a}+\frac {c \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right )}{4 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}-\frac {2 c \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{\sqrt {\pi }\, \sqrt {-a^{2}}}\) | \(193\) |
c/a*(-2*a/(-a^2*x^2+1)^(1/2)*x+a^4*(-x^2/a^2/(-a^2*x^2+1)^(1/2)+2/a^4/(-a^ 2*x^2+1)^(1/2))-1/(-a^2*x^2+1)^(1/2)+arctanh(1/(-a^2*x^2+1)^(1/2))+2*a^3*( x/a^2/(-a^2*x^2+1)^(1/2)-1/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+ 1)^(1/2))))
Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.32 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {4 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - c \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1} c}{a} \]
(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
Time = 5.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.04 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=- a c \left (\begin {cases} - \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {for}\: a^{2} \neq 0 \\\frac {x^{2}}{2} & \text {otherwise} \end {cases}\right ) - 2 c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{\sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\x & \text {otherwise} \end {cases}\right ) - \frac {c \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right )}{a} \]
-a*c*Piecewise((-sqrt(-a**2*x**2 + 1)/a**2, Ne(a**2, 0)), (x**2/2, True)) - 2*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
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (46) = 92\).
Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.80 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=-a^{3} c {\left (\frac {x^{2}}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {2}{\sqrt {-a^{2} x^{2} + 1} a^{4}}\right )} + 2 \, a^{2} c {\left (\frac {x}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {\arcsin \left (a x\right )}{a^{3}}\right )} - \frac {2 \, c x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {c {\left (\frac {1}{\sqrt {-a^{2} x^{2} + 1}} - \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )\right )}}{a} \]
-a^3*c*(x^2/(sqrt(-a^2*x^2 + 1)*a^2) - 2/(sqrt(-a^2*x^2 + 1)*a^4)) + 2*a^2 *c*(x/(sqrt(-a^2*x^2 + 1)*a^2) - arcsin(a*x)/a^3) - 2*c*x/sqrt(-a^2*x^2 + 1) - c*(1/sqrt(-a^2*x^2 + 1) - log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) )/a
Time = 0.31 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.36 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=-\frac {2 \, c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} + \frac {c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} \]
-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 = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.12 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c\,\sqrt {1-a^2\,x^2}}{a}+\frac {c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{a}-\frac {2\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}} \]