Integrand size = 25, antiderivative size = 218 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {x \sqrt {1-a^2 x^2}}{a^4 c \sqrt {c-a^2 c x^2}}+\frac {x^2 \sqrt {1-a^2 x^2}}{2 a^3 c \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{2 a^5 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {7 \sqrt {1-a^2 x^2} \log (1-a x)}{4 a^5 c \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (1+a x)}{4 a^5 c \sqrt {c-a^2 c x^2}} \]
x*(-a^2*x^2+1)^(1/2)/a^4/c/(-a^2*c*x^2+c)^(1/2)+1/2*x^2*(-a^2*x^2+1)^(1/2) /a^3/c/(-a^2*c*x^2+c)^(1/2)+1/2*(-a^2*x^2+1)^(1/2)/a^5/c/(-a*x+1)/(-a^2*c* x^2+c)^(1/2)+7/4*ln(-a*x+1)*(-a^2*x^2+1)^(1/2)/a^5/c/(-a^2*c*x^2+c)^(1/2)+ 1/4*ln(a*x+1)*(-a^2*x^2+1)^(1/2)/a^5/c/(-a^2*c*x^2+c)^(1/2)
Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.35 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (2 \left (2 a x+a^2 x^2+\frac {1}{1-a x}\right )+7 \log (1-a x)+\log (1+a x)\right )}{4 a^5 c \sqrt {c-a^2 c x^2}} \]
(Sqrt[1 - a^2*x^2]*(2*(2*a*x + a^2*x^2 + (1 - a*x)^(-1)) + 7*Log[1 - a*x] + Log[1 + a*x]))/(4*a^5*c*Sqrt[c - a^2*c*x^2])
Time = 0.50 (sec) , antiderivative size = 91, 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, 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.
\(\displaystyle \int \frac {x^4 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^4}{(1-a x)^2 (a x+1)}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (\frac {x}{a^3}+\frac {7}{4 a^4 (a x-1)}+\frac {1}{4 a^4 (a x+1)}+\frac {1}{2 a^4 (a x-1)^2}+\frac {1}{a^4}\right )dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {1}{2 a^5 (1-a x)}+\frac {7 \log (1-a x)}{4 a^5}+\frac {\log (a x+1)}{4 a^5}+\frac {x}{a^4}+\frac {x^2}{2 a^3}\right )}{c \sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*(x/a^4 + x^2/(2*a^3) + 1/(2*a^5*(1 - a*x)) + (7*Log[1 - a*x])/(4*a^5) + Log[1 + a*x]/(4*a^5)))/(c*Sqrt[c - a^2*c*x^2])
3.10.67.3.1 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 = 0.17 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.50
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (2 a^{3} x^{3}+2 a^{2} x^{2}+a \ln \left (a x +1\right ) x +7 a \ln \left (a x -1\right ) x -4 a x -\ln \left (a x +1\right )-7 \ln \left (a x -1\right )-2\right )}{4 \left (a^{2} x^{2}-1\right ) c^{2} a^{5} \left (a x -1\right )}\) | \(110\) |
-1/4*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(2*a^3*x^3+2*a^2*x^2+a*ln(a *x+1)*x+7*a*ln(a*x-1)*x-4*a*x-ln(a*x+1)-7*ln(a*x-1)-2)/(a^2*x^2-1)/c^2/a^5 /(a*x-1)
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^4/(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)
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^{4} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
\[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \]
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^4\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {1-a^2\,x^2}} \,d x \]