Integrand size = 25, antiderivative size = 149 \[ \int e^{-3 \text {arctanh}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {4 x \sqrt {c-a^2 c x^2}}{a \sqrt {1-a^2 x^2}}-\frac {3 x^2 \sqrt {c-a^2 c x^2}}{2 \sqrt {1-a^2 x^2}}+\frac {a x^3 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {c-a^2 c x^2} \log (1+a x)}{a^2 \sqrt {1-a^2 x^2}} \]
4*x*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-3/2*x^2*(-a^2*c*x^2+c)^(1/2) /(-a^2*x^2+1)^(1/2)+1/3*a*x^3*(-a^2*c*x^2+c)^(1/2)/(-a^2*x^2+1)^(1/2)-4*ln (a*x+1)*(-a^2*c*x^2+c)^(1/2)/a^2/(-a^2*x^2+1)^(1/2)
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.42 \[ \int e^{-3 \text {arctanh}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (\frac {4 x}{a}-\frac {3 x^2}{2}+\frac {a x^3}{3}-\frac {4 \log (1+a x)}{a^2}\right )}{\sqrt {1-a^2 x^2}} \]
(Sqrt[c - a^2*c*x^2]*((4*x)/a - (3*x^2)/2 + (a*x^3)/3 - (4*Log[1 + a*x])/a ^2))/Sqrt[1 - a^2*x^2]
Time = 0.41 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.42, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 86, 2009}
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.
\(\displaystyle \int x e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int e^{-3 \text {arctanh}(a x)} x \sqrt {1-a^2 x^2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {x (1-a x)^2}{a x+1}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \left (a x^2-3 x+\frac {4}{a}-\frac {4}{a (a x+1)}\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (-\frac {4 \log (a x+1)}{a^2}+\frac {a x^3}{3}+\frac {4 x}{a}-\frac {3 x^2}{2}\right )}{\sqrt {1-a^2 x^2}}\) |
(Sqrt[c - a^2*c*x^2]*((4*x)/a - (3*x^2)/2 + (a*x^3)/3 - (4*Log[1 + a*x])/a ^2))/Sqrt[1 - a^2*x^2]
3.13.64.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar t[p]) Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0]) && !I ntegerQ[n/2]
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (-2 a^{3} x^{3}+9 a^{2} x^{2}-24 a x +24 \ln \left (a x +1\right )\right )}{6 \left (a^{2} x^{2}-1\right ) a^{2}}\) | \(72\) |
1/6*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(-2*a^3*x^3+9*a^2*x^2-24*a*x +24*ln(a*x+1))/(a^2*x^2-1)/a^2
Time = 0.27 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.46 \[ \int e^{-3 \text {arctanh}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\left [\frac {12 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {c} \log \left (\frac {a^{6} c x^{6} + 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} - 4 \, a c x + {\left (a^{4} x^{4} + 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + 4 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - 2 \, c}{a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1}\right ) - {\left (2 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 24 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a^{4} x^{2} - a^{2}\right )}}, -\frac {24 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c}}{a^{4} c x^{4} + 2 \, a^{3} c x^{3} - a^{2} c x^{2} - 2 \, a c x}\right ) + {\left (2 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 24 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a^{4} x^{2} - a^{2}\right )}}\right ] \]
[1/6*(12*(a^2*x^2 - 1)*sqrt(c)*log((a^6*c*x^6 + 4*a^5*c*x^5 + 5*a^4*c*x^4 - 4*a^2*c*x^2 - 4*a*c*x + (a^4*x^4 + 4*a^3*x^3 + 6*a^2*x^2 + 4*a*x)*sqrt(- a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - 2*c)/(a^4*x^4 + 2*a^3*x^3 - 2* a*x - 1)) - (2*a^3*x^3 - 9*a^2*x^2 + 24*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^ 2*x^2 + 1))/(a^4*x^2 - a^2), -1/6*(24*(a^2*x^2 - 1)*sqrt(-c)*arctan(sqrt(- a^2*c*x^2 + c)*(a^2*x^2 + 2*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt(-c)/(a^4*c*x^ 4 + 2*a^3*c*x^3 - a^2*c*x^2 - 2*a*c*x)) + (2*a^3*x^3 - 9*a^2*x^2 + 24*a*x) *sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^4*x^2 - a^2)]
\[ \int e^{-3 \text {arctanh}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int \frac {x \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{\left (a x + 1\right )^{3}}\, dx \]
\[ \int e^{-3 \text {arctanh}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{{\left (a x + 1\right )}^{3}} \,d x } \]
\[ \int e^{-3 \text {arctanh}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{{\left (a x + 1\right )}^{3}} \,d x } \]
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int \frac {x\,\sqrt {c-a^2\,c\,x^2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \]