Integrand size = 27, antiderivative size = 96 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=4 a^3 \sqrt {c-\frac {c}{a x}}-\frac {10 a^3 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+\frac {8 a^3 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}-\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{7 c^3} \] Output:
4*a^3*(c-c/a/x)^(1/2)-10/3*a^3*(c-c/a/x)^(3/2)/c+8/5*a^3*(c-c/a/x)^(5/2)/c ^2-2/7*a^3*(c-c/a/x)^(7/2)/c^3
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.46 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} \left (15+39 a x+52 a^2 x^2+104 a^3 x^3\right )}{105 x^3} \] Input:
Integrate[(E^(2*ArcCoth[a*x])*Sqrt[c - c/(a*x)])/x^4,x]
Output:
(2*Sqrt[c - c/(a*x)]*(15 + 39*a*x + 52*a^2*x^2 + 104*a^3*x^3))/(105*x^3)
Time = 1.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.09, 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, 948, 86, 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.
\(\displaystyle \int \frac {\sqrt {c-\frac {c}{a x}} e^{2 \coth ^{-1}(a x)}}{x^4} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4}dx\) |
\(\Big \downarrow \) 6683 |
\(\displaystyle -\int \frac {\sqrt {c-\frac {c}{a x}} (a x+1)}{x^4 (1-a x)}dx\) |
\(\Big \downarrow \) 1070 |
\(\displaystyle -\int \frac {\left (a+\frac {1}{x}\right ) \sqrt {c-\frac {c}{a x}}}{\left (\frac {1}{x}-a\right ) x^4}dx\) |
\(\Big \downarrow \) 281 |
\(\displaystyle \frac {c \int \frac {a+\frac {1}{x}}{\sqrt {c-\frac {c}{a x}} x^4}dx}{a}\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {c \int \frac {a+\frac {1}{x}}{\sqrt {c-\frac {c}{a x}} x^2}d\frac {1}{x}}{a}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {c \int \left (-\frac {\left (c-\frac {c}{a x}\right )^{5/2} a^3}{c^3}+\frac {4 \left (c-\frac {c}{a x}\right )^{3/2} a^3}{c^2}-\frac {5 \sqrt {c-\frac {c}{a x}} a^3}{c}+\frac {2 a^3}{\sqrt {c-\frac {c}{a x}}}\right )d\frac {1}{x}}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c \left (\frac {2 a^4 \left (c-\frac {c}{a x}\right )^{7/2}}{7 c^4}-\frac {8 a^4 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^3}+\frac {10 a^4 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c^2}-\frac {4 a^4 \sqrt {c-\frac {c}{a x}}}{c}\right )}{a}\) |
Input:
Int[(E^(2*ArcCoth[a*x])*Sqrt[c - c/(a*x)])/x^4,x]
Output:
-((c*((-4*a^4*Sqrt[c - c/(a*x)])/c + (10*a^4*(c - c/(a*x))^(3/2))/(3*c^2) - (8*a^4*(c - c/(a*x))^(5/2))/(5*c^3) + (2*a^4*(c - c/(a*x))^(7/2))/(7*c^4 )))/a)
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]
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 = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.43
method | result | size |
orering | \(\frac {2 \left (104 a^{3} x^{3}+52 a^{2} x^{2}+39 a x +15\right ) \sqrt {c -\frac {c}{a x}}}{105 x^{3}}\) | \(41\) |
gosper | \(\frac {2 \left (104 a^{3} x^{3}+52 a^{2} x^{2}+39 a x +15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{105 x^{3}}\) | \(43\) |
trager | \(\frac {2 \left (104 a^{3} x^{3}+52 a^{2} x^{2}+39 a x +15\right ) \sqrt {-\frac {-a c x +c}{a x}}}{105 x^{3}}\) | \(45\) |
risch | \(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (104 a^{4} x^{4}-52 a^{3} x^{3}-13 a^{2} x^{2}-24 a x -15\right )}{105 \left (a x -1\right ) x^{3}}\) | \(58\) |
default | \(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (210 \sqrt {x \left (a x -1\right )}\, a^{\frac {9}{2}} x^{5}+210 \sqrt {a \,x^{2}-x}\, a^{\frac {9}{2}} x^{5}-420 \left (a \,x^{2}-x \right )^{\frac {3}{2}} a^{\frac {7}{2}} x^{3}-105 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{4} x^{5}+105 \ln \left (\frac {2 \sqrt {x \left (a x -1\right )}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{4} x^{5}-212 a^{\frac {5}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{2}-108 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x -30 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\right )}{105 x^{4} \sqrt {x \left (a x -1\right )}\, \sqrt {a}}\) | \(211\) |
Input:
int(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^4,x,method=_RETURNVERBOSE)
Output:
2/105*(104*a^3*x^3+52*a^2*x^2+39*a*x+15)/x^3*(c-c/a/x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.46 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \, {\left (104 \, a^{3} x^{3} + 52 \, a^{2} x^{2} + 39 \, a x + 15\right )} \sqrt {\frac {a c x - c}{a x}}}{105 \, x^{3}} \] Input:
integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^4,x, algorithm="fricas")
Output:
2/105*(104*a^3*x^3 + 52*a^2*x^2 + 39*a*x + 15)*sqrt((a*c*x - c)/(a*x))/x^3
\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{x^{4} \left (a x - 1\right )}\, dx \] Input:
integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)**(1/2)/x**4,x)
Output:
Integral(sqrt(-c*(-1 + 1/(a*x)))*(a*x + 1)/(x**4*(a*x - 1)), x)
\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{{\left (a x - 1\right )} x^{4}} \,d x } \] Input:
integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^4,x, algorithm="maxima")
Output:
integrate((a*x + 1)*sqrt(c - c/(a*x))/((a*x - 1)*x^4), x)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (82) = 164\).
Time = 0.24 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.11 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \, {\left (140 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{4} a^{4} c^{2} + 105 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{3} a^{3} c^{\frac {5}{2}} {\left | a \right |} - 231 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{2} a^{4} c^{3} + 105 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} a^{3} c^{\frac {7}{2}} {\left | a \right |} - 15 \, a^{4} c^{4}\right )}}{105 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{7} {\left | a \right |} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^4,x, algorithm="giac")
Output:
2/105*(140*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^4*a^4*c^2 + 105*(sqrt (a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^3*a^3*c^(5/2)*abs(a) - 231*(sqrt(a^2* c)*x - sqrt(a^2*c*x^2 - a*c*x))^2*a^4*c^3 + 105*(sqrt(a^2*c)*x - sqrt(a^2* c*x^2 - a*c*x))*a^3*c^(7/2)*abs(a) - 15*a^4*c^4)/((sqrt(a^2*c)*x - sqrt(a^ 2*c*x^2 - a*c*x))^7*abs(a)*sgn(x))
Time = 13.59 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {208\,a^3\,\sqrt {c-\frac {c}{a\,x}}}{105}+\frac {2\,\sqrt {c-\frac {c}{a\,x}}}{7\,x^3}+\frac {26\,a\,\sqrt {c-\frac {c}{a\,x}}}{35\,x^2}+\frac {104\,a^2\,\sqrt {c-\frac {c}{a\,x}}}{105\,x} \] Input:
int(((c - c/(a*x))^(1/2)*(a*x + 1))/(x^4*(a*x - 1)),x)
Output:
(208*a^3*(c - c/(a*x))^(1/2))/105 + (2*(c - c/(a*x))^(1/2))/(7*x^3) + (26* a*(c - c/(a*x))^(1/2))/(35*x^2) + (104*a^2*(c - c/(a*x))^(1/2))/(105*x)
Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \sqrt {c}\, \left (104 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}\, a^{3} x^{3}+52 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}\, a^{2} x^{2}+39 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}\, a x +15 \sqrt {x}\, \sqrt {a}\, \sqrt {a x -1}-104 a^{4} x^{4}\right )}{105 a \,x^{4}} \] Input:
int(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^4,x)
Output:
(2*sqrt(c)*(104*sqrt(x)*sqrt(a)*sqrt(a*x - 1)*a**3*x**3 + 52*sqrt(x)*sqrt( a)*sqrt(a*x - 1)*a**2*x**2 + 39*sqrt(x)*sqrt(a)*sqrt(a*x - 1)*a*x + 15*sqr t(x)*sqrt(a)*sqrt(a*x - 1) - 104*a**4*x**4))/(105*a*x**4)