Integrand size = 27, antiderivative size = 263 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{5 a \sqrt {1-\frac {1}{a^2 x^2}} x^5}+\frac {3 \sqrt {c-\frac {c}{a^2 x^2}}}{4 \sqrt {1-\frac {1}{a^2 x^2}} x^4}-\frac {4 a \sqrt {c-\frac {c}{a^2 x^2}}}{3 \sqrt {1-\frac {1}{a^2 x^2}} x^3}+\frac {2 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}} x^2}-\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}} x}-\frac {4 a^4 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 a^4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1+a x)}{\sqrt {1-\frac {1}{a^2 x^2}}} \]
-1/5*(c-c/a^2/x^2)^(1/2)/a/x^5/(1-1/a^2/x^2)^(1/2)+3/4*(c-c/a^2/x^2)^(1/2) /x^4/(1-1/a^2/x^2)^(1/2)-4/3*a*(c-c/a^2/x^2)^(1/2)/x^3/(1-1/a^2/x^2)^(1/2) +2*a^2*(c-c/a^2/x^2)^(1/2)/x^2/(1-1/a^2/x^2)^(1/2)-4*a^3*(c-c/a^2/x^2)^(1/ 2)/x/(1-1/a^2/x^2)^(1/2)-4*a^4*ln(x)*(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/ 2)+4*a^4*ln(a*x+1)*(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.34 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {1}{5 a x^5}+\frac {3}{4 x^4}-\frac {4 a}{3 x^3}+\frac {2 a^2}{x^2}-\frac {4 a^3}{x}-4 a^4 \log (x)+4 a^4 \log (1+a x)\right )}{\sqrt {1-\frac {1}{a^2 x^2}}} \]
(Sqrt[c - c/(a^2*x^2)]*(-1/5*1/(a*x^5) + 3/(4*x^4) - (4*a)/(3*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 - 1/(a^2*x ^2)]
Time = 0.53 (sec) , antiderivative size = 93, 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 = {6751, 6747, 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.
\(\displaystyle \int \frac {\sqrt {c-\frac {c}{a^2 x^2}} e^{-3 \coth ^{-1}(a x)}}{x^5} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}}}{x^5}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^2}{x^6 (a x+1)}dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \left (\frac {4 a^6}{a x+1}-\frac {4 a^5}{x}+\frac {4 a^4}{x^2}-\frac {4 a^3}{x^3}+\frac {4 a^2}{x^4}-\frac {3 a}{x^5}+\frac {1}{x^6}\right )dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-4 a^5 \log (x)+4 a^5 \log (a x+1)-\frac {4 a^4}{x}+\frac {2 a^3}{x^2}-\frac {4 a^2}{3 x^3}+\frac {3 a}{4 x^4}-\frac {1}{5 x^5}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
(Sqrt[c - c/(a^2*x^2)]*(-1/5*1/x^5 + (3*a)/(4*x^4) - (4*a^2)/(3*x^3) + (2* a^3)/x^2 - (4*a^4)/x - 4*a^5*Log[x] + 4*a^5*Log[1 + a*x]))/(a*Sqrt[1 - 1/( a^2*x^2)])
3.10.27.3.1 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^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[c^p/a^(2*p) Int[(u/x^(2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !Inte gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[c^IntPart[p]*((c + d/x^2)^FracPart[p]/(1 - 1/(a^2*x^2))^FracPart [p]) Int[u*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !IntegerQ[n/2] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.40
method | result | size |
default | \(\frac {\left (240 \ln \left (a x +1\right ) x^{5} a^{5}-240 a^{5} \ln \left (x \right ) x^{5}-240 a^{4} x^{4}+120 a^{3} x^{3}-80 a^{2} x^{2}+45 a x -12\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{60 \left (a x -1\right )^{2} x^{4}}\) | \(106\) |
1/60*(240*ln(a*x+1)*x^5*a^5-240*a^5*ln(x)*x^5-240*a^4*x^4+120*a^3*x^3-80*a ^2*x^2+45*a*x-12)*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*(a*x+1)*((a*x-1)/(a*x+1))^ (3/2)/(a*x-1)^2/x^4
Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.41 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {240 \, a^{6} \sqrt {c} x^{5} \log \left (\frac {2 \, a^{3} c x^{2} + 2 \, a^{2} c x + \sqrt {a^{2} c} {\left (2 \, a x + 1\right )} \sqrt {c} + a c}{a x^{2} + x}\right ) - {\left (240 \, a^{4} x^{4} - 120 \, a^{3} x^{3} + 80 \, a^{2} x^{2} - 45 \, a x + 12\right )} \sqrt {a^{2} c}}{60 \, a^{2} x^{5}} \]
1/60*(240*a^6*sqrt(c)*x^5*log((2*a^3*c*x^2 + 2*a^2*c*x + sqrt(a^2*c)*(2*a* x + 1)*sqrt(c) + a*c)/(a*x^2 + x)) - (240*a^4*x^4 - 120*a^3*x^3 + 80*a^2*x ^2 - 45*a*x + 12)*sqrt(a^2*c))/(a^2*x^5)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\text {Timed out} \]
\[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{5}} \,d x } \]
\[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{x^5} \,d x \]