Integrand size = 24, antiderivative size = 74 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {2 (1+a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}-\frac {x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {2 x}{5 c^2 \sqrt {c-a^2 c x^2}} \] Output:
1/5*(-2*a*x-2)/a/(-a^2*c*x^2+c)^(5/2)-1/5*x/c/(-a^2*c*x^2+c)^(3/2)-2/5*x/c ^2/(-a^2*c*x^2+c)^(1/2)
Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {2+a x-4 a^2 x^2+2 a^3 x^3}{5 a c^2 (-1+a x)^2 \sqrt {c-a^2 c x^2}} \] Input:
Integrate[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^(5/2),x]
Output:
-1/5*(2 + a*x - 4*a^2*x^2 + 2*a^3*x^3)/(a*c^2*(-1 + a*x)^2*Sqrt[c - a^2*c* x^2])
Time = 0.60 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6717, 6691, 457, 209, 208}
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.
\(\displaystyle \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 6691 |
\(\displaystyle -c \int \frac {(a x+1)^2}{\left (c-a^2 c x^2\right )^{7/2}}dx\) |
\(\Big \downarrow \) 457 |
\(\displaystyle -c \left (\frac {3 \int \frac {1}{\left (c-a^2 c x^2\right )^{5/2}}dx}{5 c}+\frac {2 (a x+1)}{5 a c \left (c-a^2 c x^2\right )^{5/2}}\right )\) |
\(\Big \downarrow \) 209 |
\(\displaystyle -c \left (\frac {3 \left (\frac {2 \int \frac {1}{\left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}+\frac {x}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )}{5 c}+\frac {2 (a x+1)}{5 a c \left (c-a^2 c x^2\right )^{5/2}}\right )\) |
\(\Big \downarrow \) 208 |
\(\displaystyle -c \left (\frac {3 \left (\frac {2 x}{3 c^2 \sqrt {c-a^2 c x^2}}+\frac {x}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )}{5 c}+\frac {2 (a x+1)}{5 a c \left (c-a^2 c x^2\right )^{5/2}}\right )\) |
Input:
Int[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^(5/2),x]
Output:
-(c*((2*(1 + a*x))/(5*a*c*(c - a^2*c*x^2)^(5/2)) + (3*(x/(3*c*(c - a^2*c*x ^2)^(3/2)) + (2*x)/(3*c^2*Sqrt[c - a^2*c*x^2])))/(5*c)))
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[((c_) + (d_.)*(x_))^2*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*( c + d*x)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d^2*((p + 2)/(b*(p + 1))) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[ b*c^2 + a*d^2, 0] && LtQ[p, -1]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^(n/2) Int[(c + d*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c , d, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0]) && IGtQ[n/ 2, 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 = 47, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(-\frac {\left (a x +1\right )^{2} \left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right )}{5 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\) | \(47\) |
orering | \(-\frac {\left (a x +1\right )^{2} \left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right )}{5 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\) | \(47\) |
trager | \(\frac {\left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \sqrt {-a^{2} c \,x^{2}+c}}{5 c^{3} \left (a x -1\right )^{3} a \left (a x +1\right )}\) | \(57\) |
default | \(\frac {x}{3 c \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {-a^{2} c \,x^{2}+c}}+\frac {\frac {2}{5 a c \left (x -\frac {1}{a}\right ) \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}-\frac {8 a \left (-\frac {-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c}{6 a^{2} c^{2} \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}-\frac {-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c}{3 a^{2} c^{3} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{5}}{a}\) | \(206\) |
Input:
int(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/5*(a*x+1)^2*(2*a^3*x^3-4*a^2*x^2+a*x+2)/a/(-a^2*c*x^2+c)^(5/2)
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 2\right )} \sqrt {-a^{2} c x^{2} + c}}{5 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")
Output:
1/5*(2*a^3*x^3 - 4*a^2*x^2 + a*x + 2)*sqrt(-a^2*c*x^2 + c)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3)
\[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {a x + 1}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \left (a x - 1\right )}\, dx \] Input:
integrate(1/(a*x-1)*(a*x+1)/(-a**2*c*x**2+c)**(5/2),x)
Output:
Integral((a*x + 1)/((-c*(a*x - 1)*(a*x + 1))**(5/2)*(a*x - 1)), x)
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {2}{5 \, {\left ({\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c x - {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a c\right )}} - \frac {2 \, x}{5 \, \sqrt {-a^{2} c x^{2} + c} c^{2}} - \frac {x}{5 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")
Output:
2/5/((-a^2*c*x^2 + c)^(3/2)*a^2*c*x - (-a^2*c*x^2 + c)^(3/2)*a*c) - 2/5*x/ (sqrt(-a^2*c*x^2 + c)*c^2) - 1/5*x/((-a^2*c*x^2 + c)^(3/2)*c)
\[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x - 1\right )}} \,d x } \] Input:
integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")
Output:
integrate((a*x + 1)/((-a^2*c*x^2 + c)^(5/2)*(a*x - 1)), x)
Time = 13.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c-a^2\,c\,x^2}\,\left (2\,a^3\,x^3-4\,a^2\,x^2+a\,x+2\right )}{5\,a\,c^3\,{\left (a\,x-1\right )}^3\,\left (a\,x+1\right )} \] Input:
int((a*x + 1)/((c - a^2*c*x^2)^(5/2)*(a*x - 1)),x)
Output:
((c - a^2*c*x^2)^(1/2)*(a*x - 4*a^2*x^2 + 2*a^3*x^3 + 2))/(5*a*c^3*(a*x - 1)^3*(a*x + 1))
Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.41 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\, a x +\sqrt {-a^{2} x^{2}+1}-4 a^{3} x^{3}+8 a^{2} x^{2}-2 a x -4\right )}{10 \sqrt {-a^{2} x^{2}+1}\, a \,c^{3} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*(sqrt( - a**2*x**2 + 1)*a**2*x**2 - 2*sqrt( - a**2*x**2 + 1)*a*x + sqrt( - a**2*x**2 + 1) - 4*a**3*x**3 + 8*a**2*x**2 - 2*a*x - 4))/(10*sqr t( - a**2*x**2 + 1)*a*c**3*(a**2*x**2 - 2*a*x + 1))