Integrand size = 27, antiderivative size = 188 \[ \int e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=-\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} x^2}{a^2 \sqrt {1-a^2 x^2}}-\frac {2 \sqrt {c-\frac {c}{a^2 x^2}} x^3}{a \sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^4}{\sqrt {1-a^2 x^2}}-\frac {a \sqrt {c-\frac {c}{a^2 x^2}} x^5}{4 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} x \log (1-a x)}{a^3 \sqrt {1-a^2 x^2}} \] Output:
-4*(c-c/a^2/x^2)^(1/2)*x^2/a^2/(-a^2*x^2+1)^(1/2)-2*(c-c/a^2/x^2)^(1/2)*x^ 3/a/(-a^2*x^2+1)^(1/2)-(c-c/a^2/x^2)^(1/2)*x^4/(-a^2*x^2+1)^(1/2)-1/4*a*(c -c/a^2/x^2)^(1/2)*x^5/(-a^2*x^2+1)^(1/2)-4*(c-c/a^2/x^2)^(1/2)*x*ln(-a*x+1 )/a^3/(-a^2*x^2+1)^(1/2)
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.38 \[ \int e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left (-\frac {4 x}{a^2}-\frac {2 x^2}{a}-x^3-\frac {a x^4}{4}-\frac {4 \log (1-a x)}{a^3}\right )}{\sqrt {1-a^2 x^2}} \] Input:
Integrate[E^(3*ArcTanh[a*x])*Sqrt[c - c/(a^2*x^2)]*x^3,x]
Output:
(Sqrt[c - c/(a^2*x^2)]*x*((-4*x)/a^2 - (2*x^2)/a - x^3 - (a*x^4)/4 - (4*Lo g[1 - a*x])/a^3))/Sqrt[1 - a^2*x^2]
Time = 0.51 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.38, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6710, 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.
\(\displaystyle \int x^3 e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} \, dx\) |
\(\Big \downarrow \) 6710 |
\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int e^{3 \text {arctanh}(a x)} x^2 \sqrt {1-a^2 x^2}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {x^2 (a x+1)^2}{1-a x}dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \left (-a x^3-3 x^2-\frac {4 x}{a}-\frac {4}{a^2 (a x-1)}-\frac {4}{a^2}\right )dx}{\sqrt {1-a^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {4 \log (1-a x)}{a^3}-\frac {4 x}{a^2}-\frac {a x^4}{4}-\frac {2 x^2}{a}-x^3\right )}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[E^(3*ArcTanh[a*x])*Sqrt[c - c/(a^2*x^2)]*x^3,x]
Output:
(Sqrt[c - c/(a^2*x^2)]*x*((-4*x)/a^2 - (2*x^2)/a - x^3 - (a*x^4)/4 - (4*Lo g[1 - a*x])/a^3))/Sqrt[1 - a^2*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_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[x^(2*p)*((c + d/x^2)^p/(1 - a^2*x^2)^p) Int[(u/x^(2*p))*(1 - a ^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !IntegerQ[p] && !IntegerQ[n/2]
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.39
method | result | size |
default | \(-\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \left (a^{4} x^{4}+4 a^{3} x^{3}+8 a^{2} x^{2}+16 a x +16 \ln \left (a x -1\right )\right )}{4 \sqrt {-a^{2} x^{2}+1}\, a^{3}}\) | \(74\) |
Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)*x^3,x,method=_RETURNV ERBOSE)
Output:
-1/4*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x*(a^4*x^4+4*a^3*x^3+8*a^2*x^2+16*a*x+1 6*ln(a*x-1))/(-a^2*x^2+1)^(1/2)/a^3
Time = 0.12 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.23 \[ \int e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\left [\frac {8 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \log \left (\frac {a^{6} c x^{6} - 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} + 4 \, a c x - {\left (a^{5} x^{5} - 4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} - 4 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 2 \, c}{a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1}\right ) + {\left (a^{5} x^{5} + 4 \, a^{4} x^{4} + 8 \, a^{3} x^{3} + 16 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{6} x^{2} - a^{4}\right )}}, \frac {16 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {c} \arctan \left (\frac {{\left (a^{2} x^{2} - 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{3} c x^{3} - 2 \, a^{2} c x^{2} - a c x + 2 \, c}\right ) + {\left (a^{5} x^{5} + 4 \, a^{4} x^{4} + 8 \, a^{3} x^{3} + 16 \, a^{2} x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{6} x^{2} - a^{4}\right )}}\right ] \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)*x^3,x, algorith m="fricas")
Output:
[1/4*(8*(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^5*x^5 - 4*a^4*x^4 + 6*a^3*x^3 - 4*a^2*x^2)*sq rt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - 2*c)/(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)) + (a^5*x^5 + 4*a^4*x^4 + 8*a^3*x^3 + 16*a^2*x^2) *sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^6*x^2 - a^4), 1/4* (16*(a^2*x^2 - 1)*sqrt(c)*arctan((a^2*x^2 - 2*a*x + 2)*sqrt(-a^2*x^2 + 1)* sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^3*c*x^3 - 2*a^2*c*x^2 - a*c*x + 2*c)) + (a^5*x^5 + 4*a^4*x^4 + 8*a^3*x^3 + 16*a^2*x^2)*sqrt(-a^2*x^2 + 1) *sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^6*x^2 - a^4)]
\[ \int e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\int \frac {x^{3} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(c-c/a**2/x**2)**(1/2)*x**3,x)
Output:
Integral(x**3*sqrt(-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))*(a*x + 1)**3/(-(a*x - 1)*(a*x + 1))**(3/2), x)
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.04 \[ \int e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {1}{4} \, a^{3} {\left (\frac {i \, a^{2} \sqrt {c} x^{4} + 2 i \, \sqrt {c} x^{2}}{a^{5}} + \frac {2 i \, \sqrt {c} \log \left (a x + 1\right )}{a^{7}} + \frac {2 i \, \sqrt {c} \log \left (a x - 1\right )}{a^{7}}\right )} + \frac {1}{2} \, a^{2} {\left (\frac {2 \, {\left (i \, a^{2} \sqrt {c} x^{3} + 3 i \, \sqrt {c} x\right )}}{a^{5}} - \frac {3 i \, \sqrt {c} \log \left (a x + 1\right )}{a^{6}} + \frac {3 i \, \sqrt {c} \log \left (a x - 1\right )}{a^{6}}\right )} - \frac {3}{2} \, a {\left (-\frac {i \, \sqrt {c} x^{2}}{a^{3}} - \frac {i \, \sqrt {c} \log \left (a x + 1\right )}{a^{5}} - \frac {i \, \sqrt {c} \log \left (a x - 1\right )}{a^{5}}\right )} + \frac {i \, \sqrt {c} x}{a^{3}} - \frac {i \, \sqrt {c} \log \left (a x + 1\right )}{2 \, a^{4}} + \frac {i \, \sqrt {c} \log \left (a x - 1\right )}{2 \, a^{4}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)*x^3,x, algorith m="maxima")
Output:
1/4*a^3*((I*a^2*sqrt(c)*x^4 + 2*I*sqrt(c)*x^2)/a^5 + 2*I*sqrt(c)*log(a*x + 1)/a^7 + 2*I*sqrt(c)*log(a*x - 1)/a^7) + 1/2*a^2*(2*(I*a^2*sqrt(c)*x^3 + 3*I*sqrt(c)*x)/a^5 - 3*I*sqrt(c)*log(a*x + 1)/a^6 + 3*I*sqrt(c)*log(a*x - 1)/a^6) - 3/2*a*(-I*sqrt(c)*x^2/a^3 - I*sqrt(c)*log(a*x + 1)/a^5 - I*sqrt( c)*log(a*x - 1)/a^5) + I*sqrt(c)*x/a^3 - 1/2*I*sqrt(c)*log(a*x + 1)/a^4 + 1/2*I*sqrt(c)*log(a*x - 1)/a^4
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.35 \[ \int e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=-\frac {1}{4} \, \sqrt {-c} {\left (\frac {16 \, \log \left ({\left | a x - 1 \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{5}} + \frac {a^{11} x^{4} \mathrm {sgn}\left (x\right ) + 4 \, a^{10} x^{3} \mathrm {sgn}\left (x\right ) + 8 \, a^{9} x^{2} \mathrm {sgn}\left (x\right ) + 16 \, a^{8} x \mathrm {sgn}\left (x\right )}{a^{12}}\right )} {\left | a \right |} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)*x^3,x, algorith m="giac")
Output:
-1/4*sqrt(-c)*(16*log(abs(a*x - 1))*sgn(x)/a^5 + (a^11*x^4*sgn(x) + 4*a^10 *x^3*sgn(x) + 8*a^9*x^2*sgn(x) + 16*a^8*x*sgn(x))/a^12)*abs(a)
Timed out. \[ \int e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\int \frac {x^3\,\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \] Input:
int((x^3*(c - c/(a^2*x^2))^(1/2)*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)
Output:
int((x^3*(c - c/(a^2*x^2))^(1/2)*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2), x)
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.24 \[ \int e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx=\frac {\sqrt {c}\, i \left (-16 \,\mathrm {log}\left (a x -1\right )-a^{4} x^{4}-4 a^{3} x^{3}-8 a^{2} x^{2}-16 a x \right )}{4 a^{4}} \] Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a^2/x^2)^(1/2)*x^3,x)
Output:
(sqrt(c)*i*( - 16*log(a*x - 1) - a**4*x**4 - 4*a**3*x**3 - 8*a**2*x**2 - 1 6*a*x))/(4*a**4)