Integrand size = 23, antiderivative size = 42 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx=\frac {\sqrt {c-a c x}}{x}+3 a \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \] Output:
(-a*c*x+c)^(1/2)/x+3*a*c^(1/2)*arctanh((-a*c*x+c)^(1/2)/c^(1/2))
Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx=\frac {\sqrt {c-a c x}}{x}+3 a \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right ) \] Input:
Integrate[(E^(2*ArcCoth[a*x])*Sqrt[c - a*c*x])/x^2,x]
Output:
Sqrt[c - a*c*x]/x + 3*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]]
Time = 0.68 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6717, 6680, 35, 87, 73, 221}
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.
\(\displaystyle \int \frac {\sqrt {c-a c x} e^{2 \coth ^{-1}(a x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^2}dx\) |
\(\Big \downarrow \) 6680 |
\(\displaystyle -\int \frac {(a x+1) \sqrt {c-a c x}}{x^2 (1-a x)}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle -c \int \frac {a x+1}{x^2 \sqrt {c-a c x}}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -c \left (\frac {3}{2} a \int \frac {1}{x \sqrt {c-a c x}}dx-\frac {\sqrt {c-a c x}}{c x}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -c \left (-\frac {3 \int \frac {1}{\frac {1}{a}-\frac {c-a c x}{a c}}d\sqrt {c-a c x}}{c}-\frac {\sqrt {c-a c x}}{c x}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -c \left (-\frac {3 a \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a c x}}{c x}\right )\) |
Input:
Int[(E^(2*ArcCoth[a*x])*Sqrt[c - a*c*x])/x^2,x]
Output:
-(c*(-(Sqrt[c - a*c*x]/(c*x)) - (3*a*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/Sqr t[c]))
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Int[u*(c + d*x)^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c , d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && !(IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2) Int[ u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.02
method | result | size |
risch | \(-\frac {\left (a x -1\right ) c}{x \sqrt {-c \left (a x -1\right )}}+3 a \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )\) | \(43\) |
pseudoelliptic | \(\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}}{\sqrt {c}}\right ) a c x +\sqrt {-c \left (a x -1\right )}\, \sqrt {c}}{x \sqrt {c}}\) | \(43\) |
derivativedivides | \(-2 c a \left (-\frac {\sqrt {-a c x +c}}{2 a c x}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )\) | \(45\) |
default | \(2 c a \left (\frac {\sqrt {-a c x +c}}{2 a c x}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )\) | \(45\) |
Input:
int(1/(a*x-1)*(a*x+1)*(-a*c*x+c)^(1/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-(a*x-1)/x/(-c*(a*x-1))^(1/2)*c+3*a*c^(1/2)*arctanh((-a*c*x+c)^(1/2)/c^(1/ 2))
Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.40 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx=\left [\frac {3 \, a \sqrt {c} x \log \left (\frac {a c x - 2 \, \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{x}\right ) + 2 \, \sqrt {-a c x + c}}{2 \, x}, \frac {3 \, a \sqrt {-c} x \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + \sqrt {-a c x + c}}{x}\right ] \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a*c*x+c)^(1/2)/x^2,x, algorithm="fricas")
Output:
[1/2*(3*a*sqrt(c)*x*log((a*c*x - 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x) + 2* sqrt(-a*c*x + c))/x, (3*a*sqrt(-c)*x*arctan(sqrt(-a*c*x + c)*sqrt(-c)/(a*c *x - c)) + sqrt(-a*c*x + c))/x]
\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx=\int \frac {\sqrt {- c \left (a x - 1\right )} \left (a x + 1\right )}{x^{2} \left (a x - 1\right )}\, dx \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a*c*x+c)**(1/2)/x**2,x)
Output:
Integral(sqrt(-c*(a*x - 1))*(a*x + 1)/(x**2*(a*x - 1)), x)
Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.48 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx=-\frac {1}{2} \, a c {\left (\frac {3 \, \log \left (\frac {\sqrt {-a c x + c} - \sqrt {c}}{\sqrt {-a c x + c} + \sqrt {c}}\right )}{\sqrt {c}} - \frac {2 \, \sqrt {-a c x + c}}{a c x}\right )} \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a*c*x+c)^(1/2)/x^2,x, algorithm="maxima")
Output:
-1/2*a*c*(3*log((sqrt(-a*c*x + c) - sqrt(c))/(sqrt(-a*c*x + c) + sqrt(c))) /sqrt(c) - 2*sqrt(-a*c*x + c)/(a*c*x))
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx=-\frac {\frac {3 \, a^{2} c \arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {\sqrt {-a c x + c} a}{x}}{a} \] Input:
integrate(1/(a*x-1)*(a*x+1)*(-a*c*x+c)^(1/2)/x^2,x, algorithm="giac")
Output:
-(3*a^2*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - sqrt(-a*c*x + c)*a/ x)/a
Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx=\frac {\sqrt {c-a\,c\,x}}{x}+3\,a\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {c-a\,c\,x}}{\sqrt {c}}\right ) \] Input:
int(((c - a*c*x)^(1/2)*(a*x + 1))/(x^2*(a*x - 1)),x)
Output:
(c - a*c*x)^(1/2)/x + 3*a*c^(1/2)*atanh((c - a*c*x)^(1/2)/c^(1/2))
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx=\frac {\sqrt {c}\, \left (2 \sqrt {-a x +1}-3 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a x +3 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a x \right )}{2 x} \] Input:
int(1/(a*x-1)*(a*x+1)*(-a*c*x+c)^(1/2)/x^2,x)
Output:
(sqrt(c)*(2*sqrt( - a*x + 1) - 3*log(sqrt( - a*x + 1) - 1)*a*x + 3*log(sqr t( - a*x + 1) + 1)*a*x))/(2*x)