Integrand size = 27, antiderivative size = 220 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{4 x^3 \sqrt {1-a^2 x^2}}+\frac {a \sqrt {c-\frac {c}{a^2 x^2}}}{x^2 \sqrt {1-a^2 x^2}}-\frac {2 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{x \sqrt {1-a^2 x^2}}+\frac {4 a^4 \sqrt {c-\frac {c}{a^2 x^2}} x \log (x)}{\sqrt {1-a^2 x^2}}-\frac {4 a^4 \sqrt {c-\frac {c}{a^2 x^2}} x \log (1+a x)}{\sqrt {1-a^2 x^2}} \] Output:
4*a^3*(c-c/a^2/x^2)^(1/2)/(-a^2*x^2+1)^(1/2)-1/4*(c-c/a^2/x^2)^(1/2)/x^3/( -a^2*x^2+1)^(1/2)+a*(c-c/a^2/x^2)^(1/2)/x^2/(-a^2*x^2+1)^(1/2)-2*a^2*(c-c/ a^2/x^2)^(1/2)/x/(-a^2*x^2+1)^(1/2)+4*a^4*(c-c/a^2/x^2)^(1/2)*x*ln(x)/(-a^ 2*x^2+1)^(1/2)-4*a^4*(c-c/a^2/x^2)^(1/2)*x*ln(a*x+1)/(-a^2*x^2+1)^(1/2)
Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.35 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left (-\frac {1}{4 x^4}+\frac {a}{x^3}-\frac {2 a^2}{x^2}+\frac {4 a^3}{x}+4 a^4 \log (x)-4 a^4 \log (1+a x)\right )}{\sqrt {1-a^2 x^2}} \] Input:
Integrate[Sqrt[c - c/(a^2*x^2)]/(E^(3*ArcTanh[a*x])*x^4),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*x*(-1/4*1/x^4 + a/x^3 - (2*a^2)/x^2 + (4*a^3)/x + 4 *a^4*Log[x] - 4*a^4*Log[1 + a*x]))/Sqrt[1 - a^2*x^2]
Time = 0.56 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.35, 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.
| \(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx\) |
| \(\Big \downarrow \) 6710 |
| \(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {1-a^2 x^2}}{x^5}dx}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 6700 |
| \(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^2}{x^5 (a x+1)}dx}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \left (-\frac {4 a^5}{a x+1}+\frac {4 a^4}{x}-\frac {4 a^3}{x^2}+\frac {4 a^2}{x^3}-\frac {3 a}{x^4}+\frac {1}{x^5}\right )dx}{\sqrt {1-a^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (4 a^4 \log (x)-4 a^4 \log (a x+1)+\frac {4 a^3}{x}-\frac {2 a^2}{x^2}+\frac {a}{x^3}-\frac {1}{4 x^4}\right )}{\sqrt {1-a^2 x^2}}\) |
Input:
Int[Sqrt[c - c/(a^2*x^2)]/(E^(3*ArcTanh[a*x])*x^4),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*x*(-1/4*1/x^4 + a/x^3 - (2*a^2)/x^2 + (4*a^3)/x + 4 *a^4*Log[x] - 4*a^4*Log[1 + a*x]))/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 = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.38
| method | result | size |
| default | \(\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (16 \ln \left (x \right ) x^{4} a^{4}-16 \ln \left (a x +1\right ) x^{4} a^{4}+16 a^{3} x^{3}-8 a^{2} x^{2}+4 a x -1\right )}{4 x^{3} \sqrt {-a^{2} x^{2}+1}}\) | \(83\) |
Input:
int((c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x,method=_RETURNV ERBOSE)
Output:
1/4*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)/x^3*(16*ln(x)*x^4*a^4-16*ln(a*x+1)*x^4*a ^4+16*a^3*x^3-8*a^2*x^2+4*a*x-1)/(-a^2*x^2+1)^(1/2)
Time = 0.14 (sec) , antiderivative size = 551, normalized size of antiderivative = 2.50 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\left [\frac {8 \, {\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt {-c} \log \left (\frac {4 \, a^{5} c x^{5} + {\left (2 \, a^{6} + 4 \, a^{5} + 6 \, a^{4} + 4 \, a^{3} + a^{2}\right )} c x^{6} + {\left (4 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} - 4 \, a - 1\right )} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, a c x - {\left (4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} - {\left (4 \, a^{4} + 6 \, a^{3} + 4 \, a^{2} + a\right )} x^{5} + 4 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{4} x^{6} + 2 \, a^{3} x^{5} - 2 \, a x^{3} - x^{2}}\right ) - {\left (16 \, a^{3} x^{3} - {\left (16 \, a^{3} - 8 \, a^{2} + 4 \, a - 1\right )} x^{4} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{2} x^{5} - x^{3}\right )}}, -\frac {16 \, {\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt {c} \arctan \left (-\frac {{\left (2 \, a^{2} x^{2} + {\left (2 \, a^{3} + 2 \, a^{2} + a\right )} x^{3} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + a^{2}\right )} c x^{4} + {\left (a^{2} + 2 \, a + 1\right )} c x^{2} - 2 \, a c x - c}\right ) + {\left (16 \, a^{3} x^{3} - {\left (16 \, a^{3} - 8 \, a^{2} + 4 \, a - 1\right )} x^{4} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{2} x^{5} - x^{3}\right )}}\right ] \] Input:
integrate((c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x, algorith m="fricas")
Output:
[1/4*(8*(a^5*x^5 - a^3*x^3)*sqrt(-c)*log((4*a^5*c*x^5 + (2*a^6 + 4*a^5 + 6 *a^4 + 4*a^3 + a^2)*c*x^6 + (4*a^4 - 4*a^3 - 6*a^2 - 4*a - 1)*c*x^4 - 5*a^ 2*c*x^2 - 4*a*c*x - (4*a^4*x^4 + 6*a^3*x^3 - (4*a^4 + 6*a^3 + 4*a^2 + a)*x ^5 + 4*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^ 2*x^2)) - c)/(a^4*x^6 + 2*a^3*x^5 - 2*a*x^3 - x^2)) - (16*a^3*x^3 - (16*a^ 3 - 8*a^2 + 4*a - 1)*x^4 - 8*a^2*x^2 + 4*a*x - 1)*sqrt(-a^2*x^2 + 1)*sqrt( (a^2*c*x^2 - c)/(a^2*x^2)))/(a^2*x^5 - x^3), -1/4*(16*(a^5*x^5 - a^3*x^3)* sqrt(c)*arctan(-(2*a^2*x^2 + (2*a^3 + 2*a^2 + a)*x^3 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(2*a^3*c*x^3 - (2*a^3 + a^2)* c*x^4 + (a^2 + 2*a + 1)*c*x^2 - 2*a*c*x - c)) + (16*a^3*x^3 - (16*a^3 - 8* a^2 + 4*a - 1)*x^4 - 8*a^2*x^2 + 4*a*x - 1)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c *x^2 - c)/(a^2*x^2)))/(a^2*x^5 - x^3)]
\[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )}}{x^{4} \left (a x + 1\right )^{3}}\, dx \] Input:
integrate((c-c/a**2/x**2)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/x**4,x)
Output:
Integral((-(a*x - 1)*(a*x + 1))**(3/2)*sqrt(-c*(-1 + 1/(a*x))*(1 + 1/(a*x) ))/(x**4*(a*x + 1)**3), x)
\[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{{\left (a x + 1\right )}^{3} x^{4}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x, algorith m="maxima")
Output:
integrate((-a^2*x^2 + 1)^(3/2)*sqrt(c - c/(a^2*x^2))/((a*x + 1)^3*x^4), x)
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.33 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=-\frac {1}{4} \, {\left (16 \, a^{2} \log \left ({\left | a x + 1 \right |}\right ) \mathrm {sgn}\left (x\right ) - 16 \, a^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right ) - \frac {16 \, a^{3} x^{3} \mathrm {sgn}\left (x\right ) - 8 \, a^{2} x^{2} \mathrm {sgn}\left (x\right ) + 4 \, a x \mathrm {sgn}\left (x\right ) - \mathrm {sgn}\left (x\right )}{a^{2} x^{4}}\right )} \sqrt {-c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x, algorith m="giac")
Output:
-1/4*(16*a^2*log(abs(a*x + 1))*sgn(x) - 16*a^2*log(abs(x))*sgn(x) - (16*a^ 3*x^3*sgn(x) - 8*a^2*x^2*sgn(x) + 4*a*x*sgn(x) - sgn(x))/(a^2*x^4))*sqrt(- c)*abs(a)
Timed out. \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (1-a^2\,x^2\right )}^{3/2}}{x^4\,{\left (a\,x+1\right )}^3} \,d x \] Input:
int(((c - c/(a^2*x^2))^(1/2)*(1 - a^2*x^2)^(3/2))/(x^4*(a*x + 1)^3),x)
Output:
int(((c - c/(a^2*x^2))^(1/2)*(1 - a^2*x^2)^(3/2))/(x^4*(a*x + 1)^3), x)
Time = 0.15 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.26 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\frac {\sqrt {c}\, i \left (16 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}-16 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-16 a^{3} x^{3}+8 a^{2} x^{2}-4 a x +1\right )}{4 a \,x^{4}} \] Input:
int((c-c/a^2/x^2)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^4,x)
Output:
(sqrt(c)*i*(16*log(a*x + 1)*a**4*x**4 - 16*log(x)*a**4*x**4 - 16*a**3*x**3 + 8*a**2*x**2 - 4*a*x + 1))/(4*a*x**4)