Integrand size = 25, antiderivative size = 84 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{5 x^2 \sqrt {1-a x}}+\frac {4 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{15 x \sqrt {1-a x}} \] Output:
-2/5*(c-c/a/x)^(1/2)*(a*x+1)^(3/2)/x^2/(-a*x+1)^(1/2)+4/15*a*(c-c/a/x)^(1/ 2)*(a*x+1)^(3/2)/x/(-a*x+1)^(1/2)
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2} (-3+2 a x)}{15 x^2 \sqrt {1-a x}} \] Input:
Integrate[(E^ArcTanh[a*x]*Sqrt[c - c/(a*x)])/x^3,x]
Output:
(2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2)*(-3 + 2*a*x))/(15*x^2*Sqrt[1 - a*x])
Time = 0.75 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6684, 6678, 516, 55, 48}
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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\) |
\(\Big \downarrow \) 6684 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {e^{\text {arctanh}(a x)} \sqrt {1-a x}}{x^{7/2}}dx}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {\sqrt {1-a^2 x^2}}{x^{7/2} \sqrt {1-a x}}dx}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 516 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {\sqrt {a x+1}}{x^{7/2}}dx}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {2}{5} a \int \frac {\sqrt {a x+1}}{x^{5/2}}dx-\frac {2 (a x+1)^{3/2}}{5 x^{5/2}}\right )}{\sqrt {1-a x}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\sqrt {x} \left (\frac {4 a (a x+1)^{3/2}}{15 x^{3/2}}-\frac {2 (a x+1)^{3/2}}{5 x^{5/2}}\right ) \sqrt {c-\frac {c}{a x}}}{\sqrt {1-a x}}\) |
Input:
Int[(E^ArcTanh[a*x]*Sqrt[c - c/(a*x)])/x^3,x]
Output:
(Sqrt[c - c/(a*x)]*Sqrt[x]*((-2*(1 + a*x)^(3/2))/(5*x^(5/2)) + (4*a*(1 + a *x)^(3/2))/(15*x^(3/2))))/Sqrt[1 - a*x]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; Free Q[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Simp[x^p*((c + d/x)^p/(1 + c*(x/d))^p) Int[u*(1 + c*(x/d))^p*(E^(n*Ar cTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*d^2, 0] && !IntegerQ[p]
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.52
method | result | size |
orering | \(\frac {2 \left (2 a x -3\right ) \left (a x +1\right )^{2} \sqrt {c -\frac {c}{a x}}}{15 x^{2} \sqrt {-a^{2} x^{2}+1}}\) | \(44\) |
gosper | \(\frac {2 \left (a x +1\right )^{2} \left (2 a x -3\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{15 x^{2} \sqrt {-a^{2} x^{2}+1}}\) | \(46\) |
default | \(-\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (2 a x -3\right )}{15 x^{2} \left (a x -1\right )}\) | \(51\) |
risch | \(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}\, \left (2 a^{3} x^{3}+a^{2} x^{2}-4 a x -3\right )}{15 \sqrt {-a^{2} x^{2}+1}\, x^{2} \sqrt {-\left (a x +1\right ) a c x}}\) | \(89\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^3,x,method=_RETURNVERBOSE )
Output:
2/15*(2*a*x-3)/x^2*(a*x+1)^2/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=-\frac {2 \, {\left (2 \, a^{2} x^{2} - a x - 3\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{15 \, {\left (a x^{3} - x^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^3,x, algorithm="fri cas")
Output:
-2/15*(2*a^2*x^2 - a*x - 3)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x))/(a* x^3 - x^2)
\[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(c-c/a/x)**(1/2)/x**3,x)
Output:
Integral(sqrt(-c*(-1 + 1/(a*x)))*(a*x + 1)/(x**3*sqrt(-(a*x - 1)*(a*x + 1) )), x)
\[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{3}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^3,x, algorithm="max ima")
Output:
integrate((a*x + 1)*sqrt(c - c/(a*x))/(sqrt(-a^2*x^2 + 1)*x^3), x)
\[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{3}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^3,x, algorithm="gia c")
Output:
integrate((a*x + 1)*sqrt(c - c/(a*x))/(sqrt(-a^2*x^2 + 1)*x^3), x)
Time = 22.82 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.62 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=-\frac {\sqrt {c-\frac {c}{a\,x}}\,\left (-\frac {4\,a^3\,x^3}{15}-\frac {2\,a^2\,x^2}{15}+\frac {8\,a\,x}{15}+\frac {2}{5}\right )}{x^2\,\sqrt {1-a^2\,x^2}} \] Input:
int(((c - c/(a*x))^(1/2)*(a*x + 1))/(x^3*(1 - a^2*x^2)^(1/2)),x)
Output:
-((c - c/(a*x))^(1/2)*((8*a*x)/15 - (2*a^2*x^2)/15 - (4*a^3*x^3)/15 + 2/5) )/(x^2*(1 - a^2*x^2)^(1/2))
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.42 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \sqrt {x}\, \sqrt {c}\, \sqrt {a}\, \sqrt {a x +1}\, i \left (2 a^{2} x^{2}-a x -3\right )}{15 a \,x^{3}} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^3,x)
Output:
(2*sqrt(x)*sqrt(c)*sqrt(a)*sqrt(a*x + 1)*i*(2*a**2*x**2 - a*x - 3))/(15*a* x**3)