Integrand size = 24, antiderivative size = 74 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {2 (1+a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}+\frac {x}{5 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 x}{5 c^2 \sqrt {c-a^2 c x^2}} \]
2/5*(a*x+1)/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 = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {2+a x-4 a^2 x^2+2 a^3 x^3}{5 a c^2 (-1+a x)^2 \sqrt {c-a^2 c x^2}} \]
Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle c \left (\frac {3 \left (\frac {2 x}{3 c^2 \sqrt {c-a^2 c x^2}}+\frac {x}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )}{5 c}+\frac {2 (a x+1)}{5 a c \left (c-a^2 c x^2\right )^{5/2}}\right )\) |
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))
3.12.23.3.1 Defintions of rubi rules used
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]
Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(\frac {\left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \left (a x +1\right )^{2}}{5 \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} a}\) | \(47\) |
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 | \(-\frac {x}{3 c \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {2 x}{3 c^{2} \sqrt {-a^{2} c \,x^{2}+c}}-\frac {2 \left (\frac {1}{5 a c \left (x -\frac {1}{a}\right ) \left (-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {-2 a^{2} c \left (x -\frac {1}{a}\right )-2 a c}{6 a^{2} c^{2} \left (-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}-\frac {-2 a^{2} c \left (x -\frac {1}{a}\right )-2 a c}{3 a^{2} c^{3} \sqrt {-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c}}\right )}{5}\right )}{a}\) | \(206\) |
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 2\right )} \sqrt {-a^{2} c x^{2} + c}}{5 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \]
-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)
\[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=- \int \frac {a x}{a^{5} c^{2} x^{5} \sqrt {- a^{2} c x^{2} + c} - a^{4} c^{2} x^{4} \sqrt {- a^{2} c x^{2} + c} - 2 a^{3} c^{2} x^{3} \sqrt {- a^{2} c x^{2} + c} + 2 a^{2} c^{2} x^{2} \sqrt {- a^{2} c x^{2} + c} + a c^{2} x \sqrt {- a^{2} c x^{2} + c} - c^{2} \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {1}{a^{5} c^{2} x^{5} \sqrt {- a^{2} c x^{2} + c} - a^{4} c^{2} x^{4} \sqrt {- a^{2} c x^{2} + c} - 2 a^{3} c^{2} x^{3} \sqrt {- a^{2} c x^{2} + c} + 2 a^{2} c^{2} x^{2} \sqrt {- a^{2} c x^{2} + c} + a c^{2} x \sqrt {- a^{2} c x^{2} + c} - c^{2} \sqrt {- a^{2} c x^{2} + c}}\, dx \]
-Integral(a*x/(a**5*c**2*x**5*sqrt(-a**2*c*x**2 + c) - a**4*c**2*x**4*sqrt (-a**2*c*x**2 + c) - 2*a**3*c**2*x**3*sqrt(-a**2*c*x**2 + c) + 2*a**2*c**2 *x**2*sqrt(-a**2*c*x**2 + c) + a*c**2*x*sqrt(-a**2*c*x**2 + c) - c**2*sqrt (-a**2*c*x**2 + c)), x) - Integral(1/(a**5*c**2*x**5*sqrt(-a**2*c*x**2 + c ) - a**4*c**2*x**4*sqrt(-a**2*c*x**2 + c) - 2*a**3*c**2*x**3*sqrt(-a**2*c* x**2 + c) + 2*a**2*c**2*x**2*sqrt(-a**2*c*x**2 + c) + a*c**2*x*sqrt(-a**2* c*x**2 + c) - c**2*sqrt(-a**2*c*x**2 + c)), x)
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (62) = 124\).
Time = 0.24 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.95 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {1}{5} \, a {\left (\frac {a}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{4} c x + {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{3} c} - \frac {a}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{4} c x - {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{3} c} - \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{3} c x + {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c} - \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{3} c x - {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c} + \frac {2 \, x}{\sqrt {-a^{2} c x^{2} + c} a c^{2}} + \frac {x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a c}\right )} \]
1/5*a*(a/((-a^2*c*x^2 + c)^(3/2)*a^4*c*x + (-a^2*c*x^2 + c)^(3/2)*a^3*c) - a/((-a^2*c*x^2 + c)^(3/2)*a^4*c*x - (-a^2*c*x^2 + c)^(3/2)*a^3*c) - 1/((- a^2*c*x^2 + c)^(3/2)*a^3*c*x + (-a^2*c*x^2 + c)^(3/2)*a^2*c) - 1/((-a^2*c* x^2 + c)^(3/2)*a^3*c*x - (-a^2*c*x^2 + c)^(3/2)*a^2*c) + 2*x/(sqrt(-a^2*c* x^2 + c)*a*c^2) + x/((-a^2*c*x^2 + c)^(3/2)*a*c))
\[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { -\frac {{\left (a x + 1\right )}^{2}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a^{2} x^{2} - 1\right )}} \,d x } \]
Time = 3.84 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c-a^2\,c\,x^2}\,\left (2\,a^3\,x^3-4\,a^2\,x^2+a\,x+2\right )}{5\,a\,c^3\,{\left (a\,x-1\right )}^3\,\left (a\,x+1\right )} \]