Integrand size = 27, antiderivative size = 42 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=4 a \sqrt {c-\frac {c}{a x}}-\frac {2 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c} \] Output:
4*a*(c-c/a/x)^(1/2)-2/3*a*(c-c/a/x)^(3/2)/c
Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.67 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} (1+5 a x)}{3 x} \] Input:
Integrate[(E^(2*ArcCoth[a*x])*Sqrt[c - c/(a*x)])/x^2,x]
Output:
(2*Sqrt[c - c/(a*x)]*(1 + 5*a*x))/(3*x)
Time = 1.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.31, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6717, 6683, 1070, 281, 946, 53, 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.
\(\displaystyle \int \frac {\sqrt {c-\frac {c}{a x}} e^{2 \coth ^{-1}(a x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2}dx\) |
\(\Big \downarrow \) 6683 |
\(\displaystyle -\int \frac {\sqrt {c-\frac {c}{a x}} (a x+1)}{x^2 (1-a x)}dx\) |
\(\Big \downarrow \) 1070 |
\(\displaystyle -\int \frac {\left (a+\frac {1}{x}\right ) \sqrt {c-\frac {c}{a x}}}{\left (\frac {1}{x}-a\right ) x^2}dx\) |
\(\Big \downarrow \) 281 |
\(\displaystyle \frac {c \int \frac {a+\frac {1}{x}}{\sqrt {c-\frac {c}{a x}} x^2}dx}{a}\) |
\(\Big \downarrow \) 946 |
\(\displaystyle -\frac {c \int \frac {a+\frac {1}{x}}{\sqrt {c-\frac {c}{a x}}}d\frac {1}{x}}{a}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {c \int \left (\frac {2 a}{\sqrt {c-\frac {c}{a x}}}-\frac {a \sqrt {c-\frac {c}{a x}}}{c}\right )d\frac {1}{x}}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c \left (\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c^2}-\frac {4 a^2 \sqrt {c-\frac {c}{a x}}}{c}\right )}{a}\) |
Input:
Int[(E^(2*ArcCoth[a*x])*Sqrt[c - c/(a*x)])/x^2,x]
Output:
-((c*((-4*a^2*Sqrt[c - c/(a*x)])/c + (2*a^2*(c - c/(a*x))^(3/2))/(3*c^2))) /a)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ Symbol] :> Simp[(b/d)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] && !(IntegerQ[q] & & SimplerQ[a + b*x^n, c + d*x^n])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m - n + 1, 0]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_.) + (b_.)*(x_)^(n_.))^ (p_.)*((e_) + (f_.)*(x_)^(n_.))^(r_.), x_Symbol] :> Int[x^(m + n*(p + r))*( b + a/x^n)^p*(c + d/x^n)^q*(f + e/x^n)^r, x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && EqQ[mn, -n] && IntegerQ[p] && IntegerQ[r]
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, p}, x] && EqQ[c^2 - a^2*d^2, 0] && !IntegerQ[p] && IntegerQ[n/2] && !G tQ[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 = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.60
method | result | size |
orering | \(\frac {2 \left (5 a x +1\right ) \sqrt {c -\frac {c}{a x}}}{3 x}\) | \(25\) |
gosper | \(\frac {2 \left (5 a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{3 x}\) | \(27\) |
trager | \(\frac {2 \left (5 a x +1\right ) \sqrt {-\frac {-a c x +c}{a x}}}{3 x}\) | \(29\) |
risch | \(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (5 a^{2} x^{2}-4 a x -1\right )}{3 \left (a x -1\right ) x}\) | \(42\) |
default | \(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (6 \sqrt {x \left (a x -1\right )}\, a^{\frac {5}{2}} x^{3}+6 \sqrt {a \,x^{2}-x}\, a^{\frac {5}{2}} x^{3}-12 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x -3 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{2} x^{3}+3 \ln \left (\frac {2 \sqrt {x \left (a x -1\right )}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{2} x^{3}-2 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\right )}{3 x^{2} \sqrt {x \left (a x -1\right )}\, \sqrt {a}}\) | \(173\) |
Input:
int(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^2,x,method=_RETURNVERBOSE)
Output:
2/3*(5*a*x+1)/x*(c-c/a/x)^(1/2)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.67 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {2 \, {\left (5 \, a x + 1\right )} \sqrt {\frac {a c x - c}{a x}}}{3 \, x} \] Input:
integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^2,x, algorithm="fricas")
Output:
2/3*(5*a*x + 1)*sqrt((a*c*x - c)/(a*x))/x
\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{x^{2} \left (a x - 1\right )}\, dx \] Input:
integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)**(1/2)/x**2,x)
Output:
Integral(sqrt(-c*(-1 + 1/(a*x)))*(a*x + 1)/(x**2*(a*x - 1)), x)
\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{{\left (a x - 1\right )} x^{2}} \,d x } \] Input:
integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^2,x, algorithm="maxima")
Output:
integrate((a*x + 1)*sqrt(c - c/(a*x))/((a*x - 1)*x^2), x)
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (36) = 72\).
Time = 0.16 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.88 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {2 \, {\left (3 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{2} a^{2} c + 3 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} a c^{\frac {3}{2}} {\left | a \right |} - a^{2} c^{2}\right )}}{3 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{3} {\left | a \right |} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^2,x, algorithm="giac")
Output:
2/3*(3*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^2*a^2*c + 3*(sqrt(a^2*c)* x - sqrt(a^2*c*x^2 - a*c*x))*a*c^(3/2)*abs(a) - a^2*c^2)/((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^3*abs(a)*sgn(x))
Time = 13.65 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.57 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {2\,\sqrt {c-\frac {c}{a\,x}}\,\left (5\,a\,x+1\right )}{3\,x} \] Input:
int(((c - c/(a*x))^(1/2)*(a*x + 1))/(x^2*(a*x - 1)),x)
Output:
(2*(c - c/(a*x))^(1/2)*(5*a*x + 1))/(3*x)
Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {2 \sqrt {c}\, \left (5 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}\, a x +\sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}-3 a^{2} x^{2}\right )}{3 a \,x^{2}} \] Input:
int(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^2,x)
Output:
(2*sqrt(c)*(5*sqrt(x)*sqrt(a)*sqrt(a*x - 1)*a*x + sqrt(x)*sqrt(a)*sqrt(a*x - 1) - 3*a**2*x**2))/(3*a*x**2)