Integrand size = 25, antiderivative size = 176 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {x \sqrt {1-a^2 x^2}}{a^3 c \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{2 a^4 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \log (1-a x)}{4 a^4 c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (1+a x)}{4 a^4 c \sqrt {c-a^2 c x^2}} \] Output:
x*(-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^4 /c/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+5/4*(-a^2*x^2+1)^(1/2)*ln(-a*x+1)/a^4/c/( -a^2*c*x^2+c)^(1/2)-1/4*(-a^2*x^2+1)^(1/2)*ln(a*x+1)/a^4/c/(-a^2*c*x^2+c)^ (1/2)
Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.40 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (4 a x+\frac {2}{1-a x}+5 \log (1-a x)-\log (1+a x)\right )}{4 a^4 c \sqrt {c-a^2 c x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x^3)/(c - a^2*c*x^2)^(3/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(4*a*x + 2/(1 - a*x) + 5*Log[1 - a*x] - Log[1 + a*x]))/ (4*a^4*c*Sqrt[c - a^2*c*x^2])
Time = 0.81 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.46, 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.
\(\displaystyle \int \frac {x^3 e^{\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^3}{(1-a x)^2 (a x+1)}dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {1}{4 a^3 (a x+1)}+\frac {1}{a^3}+\frac {5}{4 a^3 (a x-1)}+\frac {1}{2 a^3 (a x-1)^2}\right )dx}{c \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {1}{2 a^4 (1-a x)}+\frac {5 \log (1-a x)}{4 a^4}-\frac {\log (a x+1)}{4 a^4}+\frac {x}{a^3}\right )}{c \sqrt {c-a^2 c x^2}}\) |
Input:
Int[(E^ArcTanh[a*x]*x^3)/(c - a^2*c*x^2)^(3/2),x]
Output:
(Sqrt[1 - a^2*x^2]*(x/a^3 + 1/(2*a^4*(1 - a*x)) + (5*Log[1 - a*x])/(4*a^4) - Log[1 + a*x]/(4*a^4)))/(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 = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.52
method | result | size |
default | \(-\frac {\left (-4 a^{2} x^{2}+\ln \left (a x +1\right ) x a -5 a \ln \left (a x -1\right ) x +4 a x -\ln \left (a x +1\right )+5 \ln \left (a x -1\right )+2\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{4 \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right ) c^{2} a^{4}}\) | \(91\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVE RBOSE)
Output:
-1/4*(-4*a^2*x^2+ln(a*x+1)*x*a-5*a*ln(a*x-1)*x+4*a*x-ln(a*x+1)+5*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^4
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{3}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-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^3/(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)
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^{3} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**3/(-a**2*c*x**2+c)**(3/2),x)
Output:
Integral(x**3*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1 ))**(3/2)), x)
\[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (a x + 1\right )} x^{3}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(3/2),x, algorithm ="maxima")
Output:
-a*integrate(-x^4/((a^2*c^(3/2)*x^2 - c^(3/2))*(a*x + 1)*(a*x - 1)), x) - 1/2/(a^6*c^(3/2)*x^2 - a^4*c^(3/2)) + 1/2*log(-a^2*c*x^2 + c)/(a^4*c^(3/2) )
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.31 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {x}{a^{3} c^{\frac {3}{2}}} - \frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a^{4} c^{\frac {3}{2}}} + \frac {5 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a^{4} c^{\frac {3}{2}}} - \frac {1}{2 \, {\left (a x - 1\right )} a^{4} c^{\frac {3}{2}}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(3/2),x, algorithm ="giac")
Output:
x/(a^3*c^(3/2)) - 1/4*log(abs(a*x + 1))/(a^4*c^(3/2)) + 5/4*log(abs(a*x - 1))/(a^4*c^(3/2)) - 1/2/((a*x - 1)*a^4*c^(3/2))
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^3\,\left (a\,x+1\right )}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((x^3*(a*x + 1))/((c - a^2*c*x^2)^(3/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((x^3*(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 = 64, normalized size of antiderivative = 0.36 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (5 \,\mathrm {log}\left (a x -1\right ) a x -5 \,\mathrm {log}\left (a x -1\right )-\mathrm {log}\left (a x +1\right ) a x +\mathrm {log}\left (a x +1\right )+4 a^{2} x^{2}-6 a x \right )}{4 a^{4} c^{2} \left (a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*(5*log(a*x - 1)*a*x - 5*log(a*x - 1) - log(a*x + 1)*a*x + log(a*x + 1) + 4*a**2*x**2 - 6*a*x))/(4*a**4*c**2*(a*x - 1))