Integrand size = 24, antiderivative size = 97 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {2 (1+a x)}{7 a \left (c-a^2 c x^2\right )^{7/2}}-\frac {x}{7 c \left (c-a^2 c x^2\right )^{5/2}}-\frac {4 x}{21 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {8 x}{21 c^3 \sqrt {c-a^2 c x^2}} \] Output:
1/7*(-2*a*x-2)/a/(-a^2*c*x^2+c)^(7/2)-1/7*x/c/(-a^2*c*x^2+c)^(5/2)-4/21*x/ c^2/(-a^2*c*x^2+c)^(3/2)-8/21*x/c^3/(-a^2*c*x^2+c)^(1/2)
Time = 0.06 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {\sqrt {1-a^2 x^2} \left (6+9 a x-24 a^2 x^2+4 a^3 x^3+16 a^4 x^4-8 a^5 x^5\right )}{21 a c^3 (1-a x)^{7/2} (1+a x)^{3/2} \sqrt {c-a^2 c x^2}} \] Input:
Integrate[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^(7/2),x]
Output:
-1/21*(Sqrt[1 - a^2*x^2]*(6 + 9*a*x - 24*a^2*x^2 + 4*a^3*x^3 + 16*a^4*x^4 - 8*a^5*x^5))/(a*c^3*(1 - a*x)^(7/2)*(1 + a*x)^(3/2)*Sqrt[c - a^2*c*x^2])
Time = 0.65 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6717, 6691, 457, 209, 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.
\(\displaystyle \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}}dx\) |
\(\Big \downarrow \) 6691 |
\(\displaystyle -c \int \frac {(a x+1)^2}{\left (c-a^2 c x^2\right )^{9/2}}dx\) |
\(\Big \downarrow \) 457 |
\(\displaystyle -c \left (\frac {5 \int \frac {1}{\left (c-a^2 c x^2\right )^{7/2}}dx}{7 c}+\frac {2 (a x+1)}{7 a c \left (c-a^2 c x^2\right )^{7/2}}\right )\) |
\(\Big \downarrow \) 209 |
\(\displaystyle -c \left (\frac {5 \left (\frac {4 \int \frac {1}{\left (c-a^2 c x^2\right )^{5/2}}dx}{5 c}+\frac {x}{5 c \left (c-a^2 c x^2\right )^{5/2}}\right )}{7 c}+\frac {2 (a x+1)}{7 a c \left (c-a^2 c x^2\right )^{7/2}}\right )\) |
\(\Big \downarrow \) 209 |
\(\displaystyle -c \left (\frac {5 \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}+\frac {x}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )}{5 c}+\frac {x}{5 c \left (c-a^2 c x^2\right )^{5/2}}\right )}{7 c}+\frac {2 (a x+1)}{7 a c \left (c-a^2 c x^2\right )^{7/2}}\right )\) |
\(\Big \downarrow \) 208 |
\(\displaystyle -c \left (\frac {5 \left (\frac {4 \left (\frac {2 x}{3 c^2 \sqrt {c-a^2 c x^2}}+\frac {x}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )}{5 c}+\frac {x}{5 c \left (c-a^2 c x^2\right )^{5/2}}\right )}{7 c}+\frac {2 (a x+1)}{7 a c \left (c-a^2 c x^2\right )^{7/2}}\right )\) |
Input:
Int[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^(7/2),x]
Output:
-(c*((2*(1 + a*x))/(7*a*c*(c - a^2*c*x^2)^(7/2)) + (5*(x/(5*c*(c - a^2*c*x ^2)^(5/2)) + (4*(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)))/(7*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 = 64, normalized size of antiderivative = 0.66
method | result | size |
gosper | \(\frac {\left (a x +1\right )^{2} \left (8 a^{5} x^{5}-16 a^{4} x^{4}-4 a^{3} x^{3}+24 a^{2} x^{2}-9 a x -6\right )}{21 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}\) | \(64\) |
orering | \(\frac {\left (a x +1\right )^{2} \left (8 a^{5} x^{5}-16 a^{4} x^{4}-4 a^{3} x^{3}+24 a^{2} x^{2}-9 a x -6\right )}{21 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}\) | \(64\) |
trager | \(\frac {\left (8 a^{5} x^{5}-16 a^{4} x^{4}-4 a^{3} x^{3}+24 a^{2} x^{2}-9 a x -6\right ) \sqrt {-a^{2} c \,x^{2}+c}}{21 c^{4} \left (a x -1\right )^{4} \left (a x +1\right )^{2} a}\) | \(74\) |
default | \(\frac {x}{5 c \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {-a^{2} c \,x^{2}+c}}}{c}+\frac {\frac {2}{7 a c \left (x -\frac {1}{a}\right ) \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {5}{2}}}-\frac {12 a \left (-\frac {-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c}{10 a^{2} c^{2} \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {5}{2}}}+\frac {-\frac {2 \left (-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c \right )}{15 a^{2} c^{2} \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}-\frac {4 \left (-2 \left (x -\frac {1}{a}\right ) a^{2} c -2 a c \right )}{15 a^{2} c^{3} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}}{c}\right )}{7}}{a}\) | \(292\) |
Input:
int(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/21*(a*x+1)^2*(8*a^5*x^5-16*a^4*x^4-4*a^3*x^3+24*a^2*x^2-9*a*x-6)/a/(-a^2 *c*x^2+c)^(7/2)
Time = 0.20 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.28 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {{\left (8 \, a^{5} x^{5} - 16 \, a^{4} x^{4} - 4 \, a^{3} x^{3} + 24 \, a^{2} x^{2} - 9 \, a x - 6\right )} \sqrt {-a^{2} c x^{2} + c}}{21 \, {\left (a^{7} c^{4} x^{6} - 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} + 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(7/2),x, algorithm="fricas")
Output:
1/21*(8*a^5*x^5 - 16*a^4*x^4 - 4*a^3*x^3 + 24*a^2*x^2 - 9*a*x - 6)*sqrt(-a ^2*c*x^2 + c)/(a^7*c^4*x^6 - 2*a^6*c^4*x^5 - a^5*c^4*x^4 + 4*a^4*c^4*x^3 - a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4)
\[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int \frac {a x + 1}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \left (a x - 1\right )}\, dx \] Input:
integrate(1/(a*x-1)*(a*x+1)/(-a**2*c*x**2+c)**(7/2),x)
Output:
Integral((a*x + 1)/((-c*(a*x - 1)*(a*x + 1))**(7/2)*(a*x - 1)), x)
Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.02 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {2}{7 \, {\left ({\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} a^{2} c x - {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} a c\right )}} - \frac {8 \, x}{21 \, \sqrt {-a^{2} c x^{2} + c} c^{3}} - \frac {4 \, x}{21 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c^{2}} - \frac {x}{7 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} c} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(7/2),x, algorithm="maxima")
Output:
2/7/((-a^2*c*x^2 + c)^(5/2)*a^2*c*x - (-a^2*c*x^2 + c)^(5/2)*a*c) - 8/21*x /(sqrt(-a^2*c*x^2 + c)*c^3) - 4/21*x/((-a^2*c*x^2 + c)^(3/2)*c^2) - 1/7*x/ ((-a^2*c*x^2 + c)^(5/2)*c)
\[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} {\left (a x - 1\right )}} \,d x } \] Input:
integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(7/2),x, algorithm="giac")
Output:
integrate((a*x + 1)/((-a^2*c*x^2 + c)^(7/2)*(a*x - 1)), x)
Time = 13.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.38 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {c-a^2\,c\,x^2}}{14\,a\,c^4\,{\left (a\,x-1\right )}^3}-\frac {\sqrt {c-a^2\,c\,x^2}}{28\,a\,c^4\,{\left (a\,x-1\right )}^4}-\frac {\sqrt {c-a^2\,c\,x^2}\,\left (\frac {11\,x}{42\,c^4}+\frac {5}{28\,a\,c^4}\right )}{{\left (a\,x-1\right )}^2\,{\left (a\,x+1\right )}^2}+\frac {8\,x\,\sqrt {c-a^2\,c\,x^2}}{21\,c^4\,\left (a\,x-1\right )\,\left (a\,x+1\right )} \] Input:
int((a*x + 1)/((c - a^2*c*x^2)^(7/2)*(a*x - 1)),x)
Output:
(c - a^2*c*x^2)^(1/2)/(14*a*c^4*(a*x - 1)^3) - (c - a^2*c*x^2)^(1/2)/(28*a *c^4*(a*x - 1)^4) - ((c - a^2*c*x^2)^(1/2)*((11*x)/(42*c^4) + 5/(28*a*c^4) ))/((a*x - 1)^2*(a*x + 1)^2) + (8*x*(c - a^2*c*x^2)^(1/2))/(21*c^4*(a*x - 1)*(a*x + 1))
Time = 0.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.55 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {c}\, \left (9 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}-18 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+18 \sqrt {-a^{2} x^{2}+1}\, a x -9 \sqrt {-a^{2} x^{2}+1}-16 a^{5} x^{5}+32 a^{4} x^{4}+8 a^{3} x^{3}-48 a^{2} x^{2}+18 a x +12\right )}{42 \sqrt {-a^{2} x^{2}+1}\, a \,c^{4} \left (a^{4} x^{4}-2 a^{3} x^{3}+2 a x -1\right )} \] Input:
int(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(7/2),x)
Output:
(sqrt(c)*(9*sqrt( - a**2*x**2 + 1)*a**4*x**4 - 18*sqrt( - a**2*x**2 + 1)*a **3*x**3 + 18*sqrt( - a**2*x**2 + 1)*a*x - 9*sqrt( - a**2*x**2 + 1) - 16*a **5*x**5 + 32*a**4*x**4 + 8*a**3*x**3 - 48*a**2*x**2 + 18*a*x + 12))/(42*s qrt( - a**2*x**2 + 1)*a*c**4*(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 1))