Integrand size = 25, antiderivative size = 116 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx=-\frac {x \sqrt {1-a^2 x^2}}{a^2 \sqrt {c-a^2 c x^2}}-\frac {x^2 \sqrt {1-a^2 x^2}}{2 a \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{a^3 \sqrt {c-a^2 c x^2}} \] Output:
-x*(-a^2*x^2+1)^(1/2)/a^2/(-a^2*c*x^2+c)^(1/2)-1/2*x^2*(-a^2*x^2+1)^(1/2)/ a/(-a^2*c*x^2+c)^(1/2)-(-a^2*x^2+1)^(1/2)*ln(-a*x+1)/a^3/(-a^2*c*x^2+c)^(1 /2)
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.47 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2} (a x (2+a x)+2 \log (1-a x))}{2 a^3 \sqrt {c-a^2 c x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x^2)/Sqrt[c - a^2*c*x^2],x]
Output:
-1/2*(Sqrt[1 - a^2*x^2]*(a*x*(2 + a*x) + 2*Log[1 - a*x]))/(a^3*Sqrt[c - a^ 2*c*x^2])
Time = 0.74 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.51, 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.
\(\displaystyle \int \frac {x^2 e^{\text {arctanh}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {1-a^2 x^2}}dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^2}{1-a x}dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {x}{a}-\frac {1}{a^2 (a x-1)}-\frac {1}{a^2}\right )dx}{\sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (-\frac {\log (1-a x)}{a^3}-\frac {x}{a^2}-\frac {x^2}{2 a}\right )}{\sqrt {c-a^2 c x^2}}\) |
Input:
Int[(E^ArcTanh[a*x]*x^2)/Sqrt[c - a^2*c*x^2],x]
Output:
(Sqrt[1 - a^2*x^2]*(-(x/a^2) - x^2/(2*a) - Log[1 - a*x]/a^3))/Sqrt[c - a^2 *c*x^2]
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 = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.47
method | result | size |
default | \(-\frac {\left (a^{2} x^{2}+2 a x +2 \ln \left (a x -1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{2 \sqrt {-a^{2} x^{2}+1}\, c \,a^{3}}\) | \(55\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(1/2),x,method=_RETURNVE RBOSE)
Output:
-1/2*(a^2*x^2+2*a*x+2*ln(a*x-1))/(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2) /c/a^3
Time = 0.11 (sec) , antiderivative size = 352, normalized size of antiderivative = 3.03 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx=\left [\frac {{\left (a^{2} x^{2} - 1\right )} \sqrt {c} \log \left (\frac {a^{6} c x^{6} - 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} + 4 \, a c x + {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - 2 \, c}{a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1}\right ) + \sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 2 \, a x\right )} \sqrt {-a^{2} x^{2} + 1}}{2 \, {\left (a^{5} c x^{2} - a^{3} c\right )}}, -\frac {2 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c}}{a^{4} c x^{4} - 2 \, a^{3} c x^{3} - a^{2} c x^{2} + 2 \, a c x}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 2 \, a x\right )} \sqrt {-a^{2} x^{2} + 1}}{2 \, {\left (a^{5} c x^{2} - a^{3} c\right )}}\right ] \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(1/2),x, algorithm ="fricas")
Output:
[1/2*((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)) + sqrt(-a^2*c*x^2 + c)*(a^2*x^2 + 2*a*x)*sqrt(-a^2*x^2 + 1))/(a^5*c *x^2 - a^3*c), -1/2*(2*(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)) - sqrt(-a^2*c*x^2 + c)*(a^2*x^2 + 2*a*x)*sqrt(-a^ 2*x^2 + 1))/(a^5*c*x^2 - a^3*c)]
\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx=\int \frac {x^{2} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**2/(-a**2*c*x**2+c)**(1/2),x)
Output:
Integral(x**2*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*sqrt(-c*(a*x - 1)*(a*x + 1))), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{2}}{\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(1/2),x, algorithm ="maxima")
Output:
integrate((a*x + 1)*x^2/(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)), x)
Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.34 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx=-\frac {a \sqrt {c} x^{2} + 2 \, \sqrt {c} x}{2 \, a^{2} c} - \frac {\log \left ({\left | -a x + 1 \right |}\right )}{a^{3} \sqrt {c}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(1/2),x, algorithm ="giac")
Output:
-1/2*(a*sqrt(c)*x^2 + 2*sqrt(c)*x)/(a^2*c) - log(abs(-a*x + 1))/(a^3*sqrt( c))
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx=\int \frac {x^2\,\left (a\,x+1\right )}{\sqrt {c-a^2\,c\,x^2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((x^2*(a*x + 1))/((c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((x^2*(a*x + 1))/((c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.27 \[ \int \frac {e^{\text {arctanh}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx=\frac {\sqrt {c}\, \left (-2 \,\mathrm {log}\left (a x -1\right )-a^{2} x^{2}-2 a x \right )}{2 a^{3} c} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a^2*c*x^2+c)^(1/2),x)
Output:
(sqrt(c)*( - 2*log(a*x - 1) - a**2*x**2 - 2*a*x))/(2*a**3*c)