Integrand size = 27, antiderivative size = 221 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{4 a \sqrt {1-\frac {1}{a^2 x^2}} x^4}+\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}} x^3}-\frac {2 a \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {4 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}} \log (1+a x)}{\sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
-1/4*(c-c/a^2/x^2)^(1/2)/a/(1-1/a^2/x^2)^(1/2)/x^4+(c-c/a^2/x^2)^(1/2)/(1- 1/a^2/x^2)^(1/2)/x^3-2*a*(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/2)/x^2+4*a^2 *(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/2)/x+4*a^3*(c-c/a^2/x^2)^(1/2)*ln(x) /(1-1/a^2/x^2)^(1/2)-4*a^3*(c-c/a^2/x^2)^(1/2)*ln(a*x+1)/(1-1/a^2/x^2)^(1/ 2)
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.34 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {1}{4 a x^4}+\frac {1}{x^3}-\frac {2 a}{x^2}+\frac {4 a^2}{x}+4 a^3 \log (x)-4 a^3 \log (1+a x)\right )}{\sqrt {1-\frac {1}{a^2 x^2}}} \] Input:
Integrate[Sqrt[c - c/(a^2*x^2)]/(E^(3*ArcCoth[a*x])*x^4),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*(-1/4*1/(a*x^4) + x^(-3) - (2*a)/x^2 + (4*a^2)/x + 4*a^3*Log[x] - 4*a^3*Log[1 + a*x]))/Sqrt[1 - 1/(a^2*x^2)]
Time = 0.88 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.36, 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.
\(\displaystyle \int \frac {\sqrt {c-\frac {c}{a^2 x^2}} e^{-3 \coth ^{-1}(a x)}}{x^4} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}}}{x^4}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^2}{x^5 (a x+1)}dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \left (-\frac {4 a^5}{a x+1}+\frac {4 a^4}{x}-\frac {4 a^3}{x^2}+\frac {4 a^2}{x^3}-\frac {3 a}{x^4}+\frac {1}{x^5}\right )dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (4 a^4 \log (x)-4 a^4 \log (a x+1)+\frac {4 a^3}{x}-\frac {2 a^2}{x^2}+\frac {a}{x^3}-\frac {1}{4 x^4}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[Sqrt[c - c/(a^2*x^2)]/(E^(3*ArcCoth[a*x])*x^4),x]
Output:
(Sqrt[c - c/(a^2*x^2)]*(-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]))/(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 = 98, normalized size of antiderivative = 0.44
method | result | size |
default | \(-\frac {\left (16 \ln \left (a x +1\right ) x^{4} a^{4}-16 \ln \left (x \right ) x^{4} a^{4}-16 a^{3} x^{3}+8 a^{2} x^{2}-4 a x +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{4 \left (a x -1\right )^{2} x^{3}}\) | \(98\) |
Input:
int((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x,method=_RETURNVERBOS E)
Output:
-1/4*(16*ln(a*x+1)*x^4*a^4-16*ln(x)*x^4*a^4-16*a^3*x^3+8*a^2*x^2-4*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^ 3
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.45 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\frac {16 \, a^{5} \sqrt {c} x^{4} \log \left (\frac {2 \, a^{3} c x^{2} + 2 \, a^{2} c x - \sqrt {a^{2} c} {\left (2 \, a x + 1\right )} \sqrt {c} + a c}{a x^{2} + x}\right ) + {\left (16 \, a^{3} x^{3} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {a^{2} c}}{4 \, a^{2} x^{4}} \] Input:
integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="fr icas")
Output:
1/4*(16*a^5*sqrt(c)*x^4*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)) + (16*a^3*x^3 - 8*a^2*x^2 + 4*a*x - 1)*sq rt(a^2*c))/(a^2*x^4)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(1/2)*((a*x-1)/(a*x+1))**(3/2)/x**4,x)
Output:
Timed out
\[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="ma xima")
Output:
integrate(sqrt(c - c/(a^2*x^2))*((a*x - 1)/(a*x + 1))^(3/2)/x^4, x)
Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.48 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=-\frac {1}{4} \, {\left (16 \, a^{2} \log \left ({\left | a x + 1 \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 16 \, a^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - \frac {16 \, a^{3} x^{3} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 8 \, a^{2} x^{2} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + 4 \, a x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{2} x^{4}}\right )} \sqrt {c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="gi ac")
Output:
-1/4*(16*a^2*log(abs(a*x + 1))*sgn(a*x + 1)*sgn(x) - 16*a^2*log(abs(x))*sg n(a*x + 1)*sgn(x) - (16*a^3*x^3*sgn(a*x + 1)*sgn(x) - 8*a^2*x^2*sgn(a*x + 1)*sgn(x) + 4*a*x*sgn(a*x + 1)*sgn(x) - sgn(a*x + 1)*sgn(x))/(a^2*x^4))*sq rt(c)*abs(a)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{x^4} \,d x \] Input:
int(((c - c/(a^2*x^2))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x^4,x)
Output:
int(((c - c/(a^2*x^2))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x^4, x)
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.30 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx=\frac {\sqrt {c}\, \left (-16 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+16 \,\mathrm {log}\left (a x \right ) a^{4} x^{4}-4 a^{4} x^{4}+16 a^{3} x^{3}-8 a^{2} x^{2}+4 a x -1\right )}{4 a \,x^{4}} \] Input:
int((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x)
Output:
(sqrt(c)*( - 16*log(a*x + 1)*a**4*x**4 + 16*log(a*x)*a**4*x**4 - 4*a**4*x* *4 + 16*a**3*x**3 - 8*a**2*x**2 + 4*a*x - 1))/(4*a*x**4)