Integrand size = 27, antiderivative size = 184 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=-\frac {4 x \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-a^2 x^2}}+\frac {2 x^2 \sqrt {c-a^2 c x^2}}{a \sqrt {1-a^2 x^2}}-\frac {x^3 \sqrt {c-a^2 c x^2}}{\sqrt {1-a^2 x^2}}+\frac {a x^4 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-a^2 x^2}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1+a x)}{a^3 \sqrt {1-a^2 x^2}} \]
-4*x*(-a^2*c*x^2+c)^(1/2)/a^2/(-a^2*x^2+1)^(1/2)+2*x^2*(-a^2*c*x^2+c)^(1/2 )/a/(-a^2*x^2+1)^(1/2)-x^3*(-a^2*c*x^2+c)^(1/2)/(-a^2*x^2+1)^(1/2)+1/4*a*x ^4*(-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^3/(-a^2*x^2+1)^(1/2)
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.38 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (-\frac {4 x}{a^2}+\frac {2 x^2}{a}-x^3+\frac {a x^4}{4}+\frac {4 \log (1+a x)}{a^3}\right )}{\sqrt {1-a^2 x^2}} \]
(Sqrt[c - a^2*c*x^2]*((-4*x)/a^2 + (2*x^2)/a - x^3 + (a*x^4)/4 + (4*Log[1 + a*x])/a^3))/Sqrt[1 - a^2*x^2]
Time = 0.48 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.38, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6703, 6700, 99, 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.
\(\displaystyle \int x^2 e^{-3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int e^{-3 \text {arctanh}(a x)} x^2 \sqrt {1-a^2 x^2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {x^2 (1-a x)^2}{a x+1}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \left (a x^3-3 x^2+\frac {4 x}{a}+\frac {4}{a^2 (a x+1)}-\frac {4}{a^2}\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \left (\frac {4 \log (a x+1)}{a^3}-\frac {4 x}{a^2}+\frac {a x^4}{4}+\frac {2 x^2}{a}-x^3\right )}{\sqrt {1-a^2 x^2}}\) |
(Sqrt[c - a^2*c*x^2]*((-4*x)/a^2 + (2*x^2)/a - x^3 + (a*x^4)/4 + (4*Log[1 + a*x])/a^3))/Sqrt[1 - a^2*x^2]
3.13.63.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.43
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a^{4} x^{4}-4 a^{3} x^{3}+8 a^{2} x^{2}-16 a x +16 \ln \left (a x +1\right )\right )}{4 \left (a^{2} x^{2}-1\right ) a^{3}}\) | \(79\) |
-1/4*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(a^4*x^4-4*a^3*x^3+8*a^2*x^ 2-16*a*x+16*ln(a*x+1))/(a^2*x^2-1)/a^3
Time = 0.28 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.08 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\left [\frac {8 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {c} \log \left (\frac {a^{6} c x^{6} + 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} - 4 \, a c x - {\left (a^{4} x^{4} + 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + 4 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - 2 \, c}{a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1}\right ) - {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 16 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{4 \, {\left (a^{5} x^{2} - a^{3}\right )}}, \frac {16 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c}}{a^{4} c x^{4} + 2 \, a^{3} c x^{3} - a^{2} c x^{2} - 2 \, a c x}\right ) - {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 16 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{4 \, {\left (a^{5} x^{2} - a^{3}\right )}}\right ] \]
[1/4*(8*(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)) - (a^4*x^4 - 4*a^3*x^3 + 8*a^2*x^2 - 16*a*x)*sqrt(-a^2*c*x^2 + c) *sqrt(-a^2*x^2 + 1))/(a^5*x^2 - a^3), 1/4*(16*(a^2*x^2 - 1)*sqrt(-c)*arcta n(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)) - (a^4*x^4 - 4*a^3*x^3 + 8 *a^2*x^2 - 16*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^5*x^2 - a^3 )]
\[ \int e^{-3 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int \frac {x^{2} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{\left (a x + 1\right )^{3}}\, dx \]
\[ \int e^{-3 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{{\left (a x + 1\right )}^{3}} \,d x } \]
\[ \int e^{-3 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{{\left (a x + 1\right )}^{3}} \,d x } \]
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int \frac {x^2\,\sqrt {c-a^2\,c\,x^2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \]