Integrand size = 27, antiderivative size = 138 \[ \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{2 a^3 c (1+a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{4 a^3 c \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2} \log (1+a x)}{4 a^3 c \sqrt {c-a^2 c x^2}} \] Output:
-1/2*(-a^2*x^2+1)^(1/2)/a^3/c/(a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/4*(-a^2*x^2+1 )^(1/2)*ln(-a*x+1)/a^3/c/(-a^2*c*x^2+c)^(1/2)-3/4*(-a^2*x^2+1)^(1/2)*ln(a* x+1)/a^3/c/(-a^2*c*x^2+c)^(1/2)
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.52 \[ \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1-a^2 x^2} (2+(1+a x) \log (1-a x)+3 (1+a x) \log (1+a x))}{4 a^3 (c+a c x) \sqrt {c-a^2 c x^2}} \] Input:
Integrate[x^2/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^(3/2)),x]
Output:
-1/4*(Sqrt[1 - a^2*x^2]*(2 + (1 + a*x)*Log[1 - a*x] + 3*(1 + a*x)*Log[1 + a*x]))/(a^3*(c + a*c*x)*Sqrt[c - a^2*c*x^2])
Time = 0.54 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54, 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.
\(\displaystyle \int \frac {x^2 e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^2}{(1-a x) (a x+1)^2}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {3}{4 a^2 (a x+1)}+\frac {1}{2 a^2 (a x+1)^2}-\frac {1}{4 a^2 (a x-1)}\right )dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (-\frac {1}{2 a^3 (a x+1)}-\frac {\log (1-a x)}{4 a^3}-\frac {3 \log (a x+1)}{4 a^3}\right )}{c \sqrt {c-a^2 c x^2}}\) |
Input:
Int[x^2/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^(3/2)),x]
Output:
(Sqrt[1 - a^2*x^2]*(-1/2*1/(a^3*(1 + a*x)) - Log[1 - a*x]/(4*a^3) - (3*Log [1 + a*x])/(4*a^3)))/(c*Sqrt[c - a^2*c*x^2])
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 = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.56
method | result | size |
default | \(-\frac {\left (3 \ln \left (a x +1\right ) x a +a \ln \left (a x -1\right ) x +3 \ln \left (a x +1\right )+\ln \left (a x -1\right )+2\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{4 \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) c^{2} a^{3}}\) | \(77\) |
Input:
int(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVE RBOSE)
Output:
-1/4*(3*ln(a*x+1)*x*a+a*ln(a*x-1)*x+3*ln(a*x+1)+ln(a*x-1)+2)/(-a^2*x^2+1)^ (1/2)/(a*x+1)*(-c*(a^2*x^2-1))^(1/2)/c^2/a^3
\[ \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a x + 1\right )}} \,d x } \] Input:
integrate(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm ="fricas")
Output:
integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^2/(a^5*c^2*x^5 + a^4*c^ 2*x^4 - 2*a^3*c^2*x^3 - 2*a^2*c^2*x^2 + a*c^2*x + c^2), x)
\[ \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (a x + 1\right )}\, dx \] Input:
integrate(x**2/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a**2*c*x**2+c)**(3/2),x)
Output:
Integral(x**2*sqrt(-(a*x - 1)*(a*x + 1))/((-c*(a*x - 1)*(a*x + 1))**(3/2)* (a*x + 1)), x)
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.38 \[ \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c}}{2 \, {\left (a^{4} c^{2} x + a^{3} c^{2}\right )}} - \frac {3 \, \log \left (a x + 1\right )}{4 \, a^{3} c^{\frac {3}{2}}} - \frac {\log \left (a x - 1\right )}{4 \, a^{3} c^{\frac {3}{2}}} \] Input:
integrate(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm ="maxima")
Output:
-1/2*sqrt(c)/(a^4*c^2*x + a^3*c^2) - 3/4*log(a*x + 1)/(a^3*c^(3/2)) - 1/4* log(a*x - 1)/(a^3*c^(3/2))
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.34 \[ \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {3 \, \log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{3} c^{\frac {3}{2}}} - \frac {\log \left ({\left | -a x + 1 \right |}\right )}{4 \, a^{3} c^{\frac {3}{2}}} - \frac {1}{2 \, {\left (a x + 1\right )} a^{3} c^{\frac {3}{2}}} \] Input:
integrate(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm ="giac")
Output:
-3/4*log(abs(a*x + 1))/(a^3*c^(3/2)) - 1/4*log(abs(-a*x + 1))/(a^3*c^(3/2) ) - 1/2/((a*x + 1)*a^3*c^(3/2))
Timed out. \[ \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^2\,\sqrt {1-a^2\,x^2}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\left (a\,x+1\right )} \,d x \] Input:
int((x^2*(1 - a^2*x^2)^(1/2))/((c - a^2*c*x^2)^(3/2)*(a*x + 1)),x)
Output:
int((x^2*(1 - a^2*x^2)^(1/2))/((c - a^2*c*x^2)^(3/2)*(a*x + 1)), x)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.42 \[ \int \frac {e^{-\text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (-\mathrm {log}\left (a x -1\right ) a x -\mathrm {log}\left (a x -1\right )-3 \,\mathrm {log}\left (a x +1\right ) a x -3 \,\mathrm {log}\left (a x +1\right )+2 a x \right )}{4 a^{3} c^{2} \left (a x +1\right )} \] Input:
int(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*( - log(a*x - 1)*a*x - log(a*x - 1) - 3*log(a*x + 1)*a*x - 3*log( a*x + 1) + 2*a*x))/(4*a**3*c**2*(a*x + 1))