Integrand size = 27, antiderivative size = 264 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{5 a \sqrt {1-\frac {1}{a^2 x^2}} x^5}+\frac {3 \sqrt {c-\frac {c}{a^2 x^2}}}{4 \sqrt {1-\frac {1}{a^2 x^2}} x^4}+\frac {4 a \sqrt {c-\frac {c}{a^2 x^2}}}{3 \sqrt {1-\frac {1}{a^2 x^2}} x^3}+\frac {2 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}} x}-\frac {4 a^4 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 a^4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1-a x)}{\sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
1/5*(c-c/a^2/x^2)^(1/2)/a/(1-1/a^2/x^2)^(1/2)/x^5+3/4*(c-c/a^2/x^2)^(1/2)/ (1-1/a^2/x^2)^(1/2)/x^4+4/3*a*(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/2)/x^3+ 2*a^2*(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/2)/x^2+4*a^3*(c-c/a^2/x^2)^(1/2 )/(1-1/a^2/x^2)^(1/2)/x-4*a^4*(c-c/a^2/x^2)^(1/2)*ln(x)/(1-1/a^2/x^2)^(1/2 )+4*a^4*(c-c/a^2/x^2)^(1/2)*ln(-a*x+1)/(1-1/a^2/x^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.34 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {1}{5 a x^5}+\frac {3}{4 x^4}+\frac {4 a}{3 x^3}+\frac {2 a^2}{x^2}+\frac {4 a^3}{x}-4 a^4 \log (x)+4 a^4 \log (1-a x)\right )}{\sqrt {1-\frac {1}{a^2 x^2}}} \] Input:
Integrate[(E^(3*ArcCoth[a*x])*Sqrt[c - c/(a^2*x^2)])/x^5,x]
Output:
(Sqrt[c - c/(a^2*x^2)]*(1/(5*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 = 0.91 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.36, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6751, 6747, 25, 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 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int -\frac {(a x+1)^2}{x^6 (1-a x)}dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(a x+1)^2}{x^6 (1-a x)}dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 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 |
\(\displaystyle -\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (4 a^5 \log (x)-4 a^5 \log (1-a x)-\frac {4 a^4}{x}-\frac {2 a^3}{x^2}-\frac {4 a^2}{3 x^3}-\frac {3 a}{4 x^4}-\frac {1}{5 x^5}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(E^(3*ArcCoth[a*x])*Sqrt[c - c/(a^2*x^2)])/x^5,x]
Output:
-((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)]))
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 = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.40
method | result | size |
default | \(\frac {\left (240 \ln \left (a x -1\right ) x^{5} a^{5}-240 a^{5} \ln \left (x \right ) x^{5}+240 a^{4} x^{4}+120 a^{3} x^{3}+80 a^{2} x^{2}+45 a x +12\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x -1\right )}{60 \left (a x +1\right )^{2} x^{4} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(106\) |
Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a^2/x^2)^(1/2)/x^5,x,method=_RETURNVERB OSE)
Output:
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)^2/x^4/((a* x-1)/(a*x+1))^(3/2)
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.41 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {240 \, a^{6} \sqrt {c} x^{5} \log \left (\frac {2 \, a^{3} c x^{2} - 2 \, a^{2} c x - \sqrt {a^{2} c} {\left (2 \, a x - 1\right )} \sqrt {c} + a c}{a x^{2} - x}\right ) + {\left (240 \, a^{4} x^{4} + 120 \, a^{3} x^{3} + 80 \, a^{2} x^{2} + 45 \, a x + 12\right )} \sqrt {a^{2} c}}{60 \, a^{2} x^{5}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a^2/x^2)^(1/2)/x^5,x, algorithm=" fricas")
Output:
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. \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(3/2)*(c-c/a**2/x**2)**(1/2)/x**5,x)
Output:
Timed out
\[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}}}{x^{5} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a^2/x^2)^(1/2)/x^5,x, algorithm=" maxima")
Output:
integrate(sqrt(c - c/(a^2*x^2))/(x^5*((a*x - 1)/(a*x + 1))^(3/2)), x)
Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.33 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {{\left (240 \, a^{5} \log \left ({\left | a x - 1 \right |}\right ) \mathrm {sgn}\left (x\right ) - 240 \, a^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right ) + \frac {240 \, a^{4} x^{4} \mathrm {sgn}\left (x\right ) + 120 \, a^{3} x^{3} \mathrm {sgn}\left (x\right ) + 80 \, a^{2} x^{2} \mathrm {sgn}\left (x\right ) + 45 \, a x \mathrm {sgn}\left (x\right ) + 12 \, \mathrm {sgn}\left (x\right )}{x^{5}}\right )} \sqrt {c} {\left | a \right |}}{60 \, a^{2} \mathrm {sgn}\left (a x + 1\right )} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a^2/x^2)^(1/2)/x^5,x, algorithm=" giac")
Output:
1/60*(240*a^5*log(abs(a*x - 1))*sgn(x) - 240*a^5*log(abs(x))*sgn(x) + (240 *a^4*x^4*sgn(x) + 120*a^3*x^3*sgn(x) + 80*a^2*x^2*sgn(x) + 45*a*x*sgn(x) + 12*sgn(x))/x^5)*sqrt(c)*abs(a)/(a^2*sgn(a*x + 1))
Timed out. \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}}{x^5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \] Input:
int((c - c/(a^2*x^2))^(1/2)/(x^5*((a*x - 1)/(a*x + 1))^(3/2)),x)
Output:
int((c - c/(a^2*x^2))^(1/2)/(x^5*((a*x - 1)/(a*x + 1))^(3/2)), x)
Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.24 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {\sqrt {c}\, \left (240 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}-240 \,\mathrm {log}\left (x \right ) a^{5} x^{5}+240 a^{4} x^{4}+120 a^{3} x^{3}+80 a^{2} x^{2}+45 a x +12\right )}{60 a \,x^{5}} \] Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*(c-c/a^2/x^2)^(1/2)/x^5,x)
Output:
(sqrt(c)*(240*log(a*x - 1)*a**5*x**5 - 240*log(x)*a**5*x**5 + 240*a**4*x** 4 + 120*a**3*x**3 + 80*a**2*x**2 + 45*a*x + 12))/(60*a*x**5)