Integrand size = 27, antiderivative size = 194 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {\sqrt {c-a^2 c x^2}}{3 a \sqrt {1-\frac {1}{a^2 x^2}} x^4}+\frac {3 \sqrt {c-a^2 c x^2}}{2 \sqrt {1-\frac {1}{a^2 x^2}} x^3}+\frac {4 a \sqrt {c-a^2 c x^2}}{\sqrt {1-\frac {1}{a^2 x^2}} x^2}-\frac {4 a^2 \sqrt {c-a^2 c x^2} \log (x)}{\sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {4 a^2 \sqrt {c-a^2 c x^2} \log (1-a x)}{\sqrt {1-\frac {1}{a^2 x^2}} x} \] Output:
1/3*(-a^2*c*x^2+c)^(1/2)/a/(1-1/a^2/x^2)^(1/2)/x^4+3/2*(-a^2*c*x^2+c)^(1/2 )/(1-1/a^2/x^2)^(1/2)/x^3+4*a*(-a^2*c*x^2+c)^(1/2)/(1-1/a^2/x^2)^(1/2)/x^2 -4*a^2*(-a^2*c*x^2+c)^(1/2)*ln(x)/(1-1/a^2/x^2)^(1/2)/x+4*a^2*(-a^2*c*x^2+ c)^(1/2)*ln(-a*x+1)/(1-1/a^2/x^2)^(1/2)/x
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.39 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (\frac {1}{3 a x^3}+\frac {3}{2 x^2}+\frac {4 a}{x}-4 a^2 \log (x)+4 a^2 \log (1-a x)\right )}{\sqrt {1-\frac {1}{a^2 x^2}} x} \] Input:
Integrate[(E^(3*ArcCoth[a*x])*Sqrt[c - a^2*c*x^2])/x^4,x]
Output:
(Sqrt[c - a^2*c*x^2]*(1/(3*a*x^3) + 3/(2*x^2) + (4*a)/x - 4*a^2*Log[x] + 4 *a^2*Log[1 - a*x]))/(Sqrt[1 - 1/(a^2*x^2)]*x)
Time = 0.84 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.41, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6746, 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.
\(\displaystyle \int \frac {\sqrt {c-a^2 c x^2} e^{3 \coth ^{-1}(a x)}}{x^4} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}}}{x^3}dx}{x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int -\frac {(a x+1)^2}{x^4 (1-a x)}dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \int \frac {(a x+1)^2}{x^4 (1-a x)}dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \int \left (-\frac {4 a^4}{a x-1}+\frac {4 a^3}{x}+\frac {4 a^2}{x^2}+\frac {3 a}{x^3}+\frac {1}{x^4}\right )dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \left (4 a^3 \log (x)-4 a^3 \log (1-a x)-\frac {4 a^2}{x}-\frac {3 a}{2 x^2}-\frac {1}{3 x^3}\right )}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(E^(3*ArcCoth[a*x])*Sqrt[c - a^2*c*x^2])/x^4,x]
Output:
-((Sqrt[c - a^2*c*x^2]*(-1/3*1/x^3 - (3*a)/(2*x^2) - (4*a^2)/x + 4*a^3*Log [x] - 4*a^3*Log[1 - a*x]))/(a*Sqrt[1 - 1/(a^2*x^2)]*x))
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_Symbo l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p) Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && !IntegerQ[p]
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]
Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.44
method | result | size |
default | \(\frac {\left (24 a^{3} \ln \left (a x -1\right ) x^{3}-24 \ln \left (x \right ) x^{3} a^{3}+24 a^{2} x^{2}+9 a x +2\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x -1\right )}{6 x^{3} \left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(85\) |
Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(1/2)/x^4,x,method=_RETURNVER BOSE)
Output:
1/6*(24*a^3*ln(a*x-1)*x^3-24*ln(x)*x^3*a^3+24*a^2*x^2+9*a*x+2)*(-c*(a^2*x^ 2-1))^(1/2)*(a*x-1)/x^3/(a*x+1)^2/((a*x-1)/(a*x+1))^(3/2)
Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.51 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {24 \, a^{4} \sqrt {-c} x^{3} \log \left (\frac {2 \, a^{3} c x^{2} - 2 \, a^{2} c x + \sqrt {-a^{2} c} {\left (2 \, a x - 1\right )} \sqrt {-c} + a c}{a x^{2} - x}\right ) + {\left (24 \, a^{2} x^{2} + 9 \, a x + 2\right )} \sqrt {-a^{2} c}}{6 \, a x^{3}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm= "fricas")
Output:
1/6*(24*a^4*sqrt(-c)*x^3*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)) + (24*a^2*x^2 + 9*a*x + 2)*sqrt(-a^2*c ))/(a*x^3)
Timed out. \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(3/2)*(-a**2*c*x**2+c)**(1/2)/x**4,x)
Output:
Timed out
\[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c}}{x^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm= "maxima")
Output:
integrate(sqrt(-a^2*c*x^2 + c)/(x^4*((a*x - 1)/(a*x + 1))^(3/2)), x)
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.28 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {{\left (24 \, a^{3} \log \left ({\left | a x - 1 \right |}\right ) - 24 \, a^{3} \log \left ({\left | x \right |}\right ) + \frac {24 \, a^{2} x^{2} + 9 \, a x + 2}{x^{3}}\right )} \sqrt {-c}}{6 \, \mathrm {sgn}\left (a x + 1\right )} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm= "giac")
Output:
1/6*(24*a^3*log(abs(a*x - 1)) - 24*a^3*log(abs(x)) + (24*a^2*x^2 + 9*a*x + 2)/x^3)*sqrt(-c)/sgn(a*x + 1)
Timed out. \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\int \frac {\sqrt {c-a^2\,c\,x^2}}{x^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \] Input:
int((c - a^2*c*x^2)^(1/2)/(x^4*((a*x - 1)/(a*x + 1))^(3/2)),x)
Output:
int((c - a^2*c*x^2)^(1/2)/(x^4*((a*x - 1)/(a*x + 1))^(3/2)), x)
Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.42 \[ \int \frac {e^{3 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {\sqrt {c}\, i \left (24 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{3} x^{3}+24 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{3} x^{3}-48 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}+8 a^{3} x^{3}-24 a^{2} x^{2}-9 a x -2\right )}{6 x^{3}} \] Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^(1/2)/x^4,x)
Output:
(sqrt(c)*i*(24*log(sqrt( - a*x + 1) - 1)*a**3*x**3 + 24*log(sqrt( - a*x + 1) + 1)*a**3*x**3 - 48*log(sqrt( - a*x + 1))*a**3*x**3 + 8*a**3*x**3 - 24* a**2*x**2 - 9*a*x - 2))/(6*x**3)