Integrand size = 25, antiderivative size = 261 \[ \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {2 x \sqrt {1-a^2 x^2}}{a^5 c \sqrt {c-a^2 c x^2}}+\frac {x^2 \sqrt {1-a^2 x^2}}{2 a^4 c \sqrt {c-a^2 c x^2}}+\frac {x^3 \sqrt {1-a^2 x^2}}{3 a^3 c \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{2 a^6 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {9 \sqrt {1-a^2 x^2} \log (1-a x)}{4 a^6 c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (1+a x)}{4 a^6 c \sqrt {c-a^2 c x^2}} \] Output:
2*x*(-a^2*x^2+1)^(1/2)/a^5/c/(-a^2*c*x^2+c)^(1/2)+1/2*x^2*(-a^2*x^2+1)^(1/ 2)/a^4/c/(-a^2*c*x^2+c)^(1/2)+1/3*x^3*(-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^6/c/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+9/4*( -a^2*x^2+1)^(1/2)*ln(-a*x+1)/a^6/c/(-a^2*c*x^2+c)^(1/2)-1/4*(-a^2*x^2+1)^( 1/2)*ln(a*x+1)/a^6/c/(-a^2*c*x^2+c)^(1/2)
Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.33 \[ \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (24 a x+6 a^2 x^2+4 a^3 x^3+\frac {6}{1-a x}+27 \log (1-a x)-3 \log (1+a x)\right )}{12 a^6 c \sqrt {c-a^2 c x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x^5)/(c - a^2*c*x^2)^(3/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(24*a*x + 6*a^2*x^2 + 4*a^3*x^3 + 6/(1 - a*x) + 27*Log[ 1 - a*x] - 3*Log[1 + a*x]))/(12*a^6*c*Sqrt[c - a^2*c*x^2])
Time = 0.51 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.39, 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.
\(\displaystyle \int \frac {x^5 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^5}{(1-a x)^2 (a x+1)}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (\frac {x^2}{a^3}+\frac {x}{a^4}+\frac {9}{4 a^5 (a x-1)}-\frac {1}{4 a^5 (a x+1)}+\frac {1}{2 a^5 (a x-1)^2}+\frac {2}{a^5}\right )dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {1}{2 a^6 (1-a x)}+\frac {9 \log (1-a x)}{4 a^6}-\frac {\log (a x+1)}{4 a^6}+\frac {2 x}{a^5}+\frac {x^2}{2 a^4}+\frac {x^3}{3 a^3}\right )}{c \sqrt {c-a^2 c x^2}}\) |
Input:
Int[(E^ArcTanh[a*x]*x^5)/(c - a^2*c*x^2)^(3/2),x]
Output:
(Sqrt[1 - a^2*x^2]*((2*x)/a^5 + x^2/(2*a^4) + x^3/(3*a^3) + 1/(2*a^6*(1 - a*x)) + (9*Log[1 - a*x])/(4*a^6) - Log[1 + a*x]/(4*a^6)))/(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 = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.41
method | result | size |
default | \(-\frac {\left (-4 a^{4} x^{4}-2 a^{3} x^{3}-18 a^{2} x^{2}+3 \ln \left (a x +1\right ) x a -27 a \ln \left (a x -1\right ) x +24 a x -3 \ln \left (a x +1\right )+27 \ln \left (a x -1\right )+6\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{12 \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right ) c^{2} a^{6}}\) | \(108\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^5/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVE RBOSE)
Output:
-1/12*(-4*a^4*x^4-2*a^3*x^3-18*a^2*x^2+3*ln(a*x+1)*x*a-27*a*ln(a*x-1)*x+24 *a*x-3*ln(a*x+1)+27*ln(a*x-1)+6)/(-a^2*x^2+1)^(1/2)/(a*x-1)*(-c*(a^2*x^2-1 ))^(1/2)/c^2/a^6
\[ \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{5}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^5/(-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^5/(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)
\[ \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^{5} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**5/(-a**2*c*x**2+c)**(3/2),x)
Output:
Integral(x**5*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1 ))**(3/2)), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{5}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^5/(-a^2*c*x^2+c)^(3/2),x, algorithm ="maxima")
Output:
-a*integrate(-x^6/((a^2*c^(3/2)*x^2 - c^(3/2))*(a*x + 1)*(a*x - 1)), x) - 1/2/(a^8*c^(3/2)*x^2 - a^6*c^(3/2)) + log(-a^2*c*x^2 + c)/(a^6*c^(3/2)) - 1/2*sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)/(a^6*c^2)
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.33 \[ \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{6} c^{\frac {3}{2}}} + \frac {9 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a^{6} c^{\frac {3}{2}}} - \frac {1}{2 \, {\left (a x - 1\right )} a^{6} c^{\frac {3}{2}}} + \frac {2 \, a^{6} c^{3} x^{3} + 3 \, a^{5} c^{3} x^{2} + 12 \, a^{4} c^{3} x}{6 \, a^{9} c^{\frac {9}{2}}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^5/(-a^2*c*x^2+c)^(3/2),x, algorithm ="giac")
Output:
-1/4*log(abs(a*x + 1))/(a^6*c^(3/2)) + 9/4*log(abs(a*x - 1))/(a^6*c^(3/2)) - 1/2/((a*x - 1)*a^6*c^(3/2)) + 1/6*(2*a^6*c^3*x^3 + 3*a^5*c^3*x^2 + 12*a ^4*c^3*x)/(a^9*c^(9/2))
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^5\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((x^5*(a*x + 1))/((c - a^2*c*x^2)^(3/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((x^5*(a*x + 1))/((c - a^2*c*x^2)^(3/2)*(1 - a^2*x^2)^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.31 \[ \int \frac {e^{\text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (27 \,\mathrm {log}\left (a x -1\right ) a x -27 \,\mathrm {log}\left (a x -1\right )-3 \,\mathrm {log}\left (a x +1\right ) a x +3 \,\mathrm {log}\left (a x +1\right )+4 a^{4} x^{4}+2 a^{3} x^{3}+18 a^{2} x^{2}-30 a x \right )}{12 a^{6} c^{2} \left (a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^5/(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*(27*log(a*x - 1)*a*x - 27*log(a*x - 1) - 3*log(a*x + 1)*a*x + 3*l og(a*x + 1) + 4*a**4*x**4 + 2*a**3*x**3 + 18*a**2*x**2 - 30*a*x))/(12*a**6 *c**2*(a*x - 1))