Integrand size = 25, antiderivative size = 155 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx=-\frac {x \sqrt {1-a^2 x^2}}{a^3 \sqrt {c-a^2 c x^2}}-\frac {x^2 \sqrt {1-a^2 x^2}}{2 a^2 \sqrt {c-a^2 c x^2}}-\frac {x^3 \sqrt {1-a^2 x^2}}{3 a \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{a^4 \sqrt {c-a^2 c x^2}} \]
-x*(-a^2*x^2+1)^(1/2)/a^3/(-a^2*c*x^2+c)^(1/2)-1/2*x^2*(-a^2*x^2+1)^(1/2)/ a^2/(-a^2*c*x^2+c)^(1/2)-1/3*x^3*(-a^2*x^2+1)^(1/2)/a/(-a^2*c*x^2+c)^(1/2) -ln(-a*x+1)*(-a^2*x^2+1)^(1/2)/a^4/(-a^2*c*x^2+c)^(1/2)
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.41 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2} \left (a x \left (6+3 a x+2 a^2 x^2\right )+6 \log (1-a x)\right )}{6 a^4 \sqrt {c-a^2 c x^2}} \]
-1/6*(Sqrt[1 - a^2*x^2]*(a*x*(6 + 3*a*x + 2*a^2*x^2) + 6*Log[1 - a*x]))/(a ^4*Sqrt[c - a^2*c*x^2])
Time = 0.45 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.45, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 49, 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.
\(\displaystyle \int \frac {x^3 e^{\text {arctanh}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^3}{\sqrt {1-a^2 x^2}}dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^3}{1-a x}dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {x^2}{a}-\frac {x}{a^2}-\frac {1}{a^3 (a x-1)}-\frac {1}{a^3}\right )dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (-\frac {\log (1-a x)}{a^4}-\frac {x}{a^3}-\frac {x^2}{2 a^2}-\frac {x^3}{3 a}\right )}{\sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*(-(x/a^3) - x^2/(2*a^2) - x^3/(3*a) - Log[1 - a*x]/a^4) )/Sqrt[c - a^2*c*x^2]
3.10.58.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (2 a^{3} x^{3}+3 a^{2} x^{2}+6 a x +6 \ln \left (a x -1\right )\right )}{6 \left (a^{2} x^{2}-1\right ) c \,a^{4}}\) | \(75\) |
1/6*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(2*a^3*x^3+3*a^2*x^2+6*a*x+6 *ln(a*x-1))/(a^2*x^2-1)/c/a^4
Time = 0.30 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.39 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx=\left [\frac {3 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {c} \log \left (\frac {a^{6} c x^{6} - 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} + 4 \, a c x + {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - 2 \, c}{a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1}\right ) + {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 6 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a^{6} c x^{2} - a^{4} c\right )}}, -\frac {6 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c}}{a^{4} c x^{4} - 2 \, a^{3} c x^{3} - a^{2} c x^{2} + 2 \, a c x}\right ) - {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 6 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a^{6} c x^{2} - a^{4} c\right )}}\right ] \]
[1/6*(3*(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 + 3*a^2*x^2 + 6*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2* x^2 + 1))/(a^6*c*x^2 - a^4*c), -1/6*(6*(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 + 3*a^2*x^2 + 6*a*x) *sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c*x^2 - a^4*c)]
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx=\int \frac {x^{3} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{3}}{\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
a*integrate(-x^4/((a*x + 1)*(a*x - 1)), x)/sqrt(c) + 1/2*log(x^2 - 1/a^2)/ (a^4*sqrt(c)) + 1/2*sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)/(a^4*c)
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{3}}{\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\sqrt {c-a^2 c x^2}} \, dx=\int \frac {x^3\,\left (a\,x+1\right )}{\sqrt {c-a^2\,c\,x^2}\,\sqrt {1-a^2\,x^2}} \,d x \]