Integrand size = 27, antiderivative size = 146 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{2 a \sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {3 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {4 a \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 a \sqrt {c-\frac {c}{a^2 x^2}} \log (1+a x)}{\sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
-1/2*(c-c/a^2/x^2)^(1/2)/a/(1-1/a^2/x^2)^(1/2)/x^2+3*(c-c/a^2/x^2)^(1/2)/( 1-1/a^2/x^2)^(1/2)/x+4*a*(c-c/a^2/x^2)^(1/2)*ln(x)/(1-1/a^2/x^2)^(1/2)-4*a *(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 = 60, normalized size of antiderivative = 0.41 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {1}{2 a x^2}+\frac {3}{x}+4 a \log (x)-4 a \log (1+a x)\right )}{\sqrt {1-\frac {1}{a^2 x^2}}} \] Input:
Integrate[Sqrt[c - c/(a^2*x^2)]/(E^(3*ArcCoth[a*x])*x^2),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*(-1/2*1/(a*x^2) + 3/x + 4*a*Log[x] - 4*a*Log[1 + a* x]))/Sqrt[1 - 1/(a^2*x^2)]
Time = 0.87 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.45, 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.
\(\displaystyle \int \frac {\sqrt {c-\frac {c}{a^2 x^2}} e^{-3 \coth ^{-1}(a x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}}}{x^2}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^2}{x^3 (a x+1)}dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \left (-\frac {4 a^3}{a x+1}+\frac {4 a^2}{x}-\frac {3 a}{x^2}+\frac {1}{x^3}\right )dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (4 a^2 \log (x)-4 a^2 \log (a x+1)+\frac {3 a}{x}-\frac {1}{2 x^2}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[Sqrt[c - c/(a^2*x^2)]/(E^(3*ArcCoth[a*x])*x^2),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*(-1/2*1/x^2 + (3*a)/x + 4*a^2*Log[x] - 4*a^2*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 = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.56
method | result | size |
default | \(-\frac {\left (8 \ln \left (a x +1\right ) x^{2} a^{2}-8 a^{2} \ln \left (x \right ) x^{2}-6 a x +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{2 \left (a x -1\right )^{2} x}\) | \(82\) |
Input:
int((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^2,x,method=_RETURNVERBOS E)
Output:
-1/2*(8*ln(a*x+1)*x^2*a^2-8*a^2*ln(x)*x^2-6*a*x+1)*(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
Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.57 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=\frac {8 \, a^{3} \sqrt {c} x^{2} \log \left (\frac {2 \, a^{3} c x^{2} + 2 \, a^{2} c x - \sqrt {a^{2} c} {\left (2 \, a x + 1\right )} \sqrt {c} + a c}{a x^{2} + x}\right ) + \sqrt {a^{2} c} {\left (6 \, a x - 1\right )}}{2 \, a^{2} x^{2}} \] Input:
integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^2,x, algorithm="fr icas")
Output:
1/2*(8*a^3*sqrt(c)*x^2*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)) + sqrt(a^2*c)*(6*a*x - 1))/(a^2*x^2)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(1/2)*((a*x-1)/(a*x+1))**(3/2)/x**2,x)
Output:
Timed out
\[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{2}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^2,x, algorithm="ma xima")
Output:
integrate(sqrt(c - c/(a^2*x^2))*((a*x - 1)/(a*x + 1))^(3/2)/x^2, x)
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.47 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=-\frac {1}{2} \, {\left (8 \, \log \left ({\left | a x + 1 \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 8 \, \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - \frac {6 \, a x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{2} x^{2}}\right )} \sqrt {c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^2,x, algorithm="gi ac")
Output:
-1/2*(8*log(abs(a*x + 1))*sgn(a*x + 1)*sgn(x) - 8*log(abs(x))*sgn(a*x + 1) *sgn(x) - (6*a*x*sgn(a*x + 1)*sgn(x) - sgn(a*x + 1)*sgn(x))/(a^2*x^2))*sqr t(c)*abs(a)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=\int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{x^2} \,d x \] Input:
int(((c - c/(a^2*x^2))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x^2,x)
Output:
int(((c - c/(a^2*x^2))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x^2, x)
Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.34 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=\frac {\sqrt {c}\, \left (-8 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+8 \,\mathrm {log}\left (a x \right ) a^{2} x^{2}-3 a^{2} x^{2}+6 a x -1\right )}{2 a \,x^{2}} \] Input:
int((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^2,x)
Output:
(sqrt(c)*( - 8*log(a*x + 1)*a**2*x**2 + 8*log(a*x)*a**2*x**2 - 3*a**2*x**2 + 6*a*x - 1))/(2*a*x**2)