Integrand size = 24, antiderivative size = 276 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1-a x)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{8 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}+\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{24 (1+a x)^3 \left (c-a^2 c x^2\right )^{7/2}}+\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1+a x)^2 \left (c-a^2 c x^2\right )^{7/2}}+\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (1+a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7 \text {arctanh}(a x)}{16 \left (c-a^2 c x^2\right )^{7/2}} \] Output:
-1/32*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)^2/(-a^2*c*x^2+c)^(7/2)-1/8*a^6* (1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)/(-a^2*c*x^2+c)^(7/2)+1/24*a^6*(1-1/a^2/x^ 2)^(7/2)*x^7/(a*x+1)^3/(-a^2*c*x^2+c)^(7/2)+3/32*a^6*(1-1/a^2/x^2)^(7/2)*x ^7/(a*x+1)^2/(-a^2*c*x^2+c)^(7/2)+3/16*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(a*x+1) /(-a^2*c*x^2+c)^(7/2)-5/16*a^6*(1-1/a^2/x^2)^(7/2)*x^7*arctanh(a*x)/(-a^2* c*x^2+c)^(7/2)
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.36 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-8+25 a x+25 a^2 x^2-15 a^3 x^3-15 a^4 x^4+15 (-1+a x)^2 (1+a x)^3 \text {arctanh}(a x)\right )}{48 (-1+a x)^2 (c+a c x)^3 \sqrt {c-a^2 c x^2}} \] Input:
Integrate[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^(7/2)),x]
Output:
(Sqrt[1 - 1/(a^2*x^2)]*x*(-8 + 25*a*x + 25*a^2*x^2 - 15*a^3*x^3 - 15*a^4*x ^4 + 15*(-1 + a*x)^2*(1 + a*x)^3*ArcTanh[a*x]))/(48*(-1 + a*x)^2*(c + a*c* x)^3*Sqrt[c - a^2*c*x^2])
Time = 0.84 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.44, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6746, 6747, 25, 54, 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.
| \(\displaystyle \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 6746 |
| \(\displaystyle \frac {x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 6747 |
| \(\displaystyle \frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int -\frac {1}{(1-a x)^3 (a x+1)^4}dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \frac {1}{(1-a x)^3 (a x+1)^4}dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle -\frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \left (\frac {1}{8 (a x-1)^2}+\frac {3}{16 (a x+1)^2}-\frac {1}{16 (a x-1)^3}+\frac {3}{16 (a x+1)^3}+\frac {1}{8 (a x+1)^4}-\frac {5}{16 \left (a^2 x^2-1\right )}\right )dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \left (\frac {5 \text {arctanh}(a x)}{16 a}+\frac {1}{8 a (1-a x)}-\frac {3}{16 a (a x+1)}+\frac {1}{32 a (1-a x)^2}-\frac {3}{32 a (a x+1)^2}-\frac {1}{24 a (a x+1)^3}\right )}{\left (c-a^2 c x^2\right )^{7/2}}\) |
Input:
Int[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^(7/2)),x]
Output:
-((a^7*(1 - 1/(a^2*x^2))^(7/2)*x^7*(1/(32*a*(1 - a*x)^2) + 1/(8*a*(1 - a*x )) - 1/(24*a*(1 + a*x)^3) - 3/(32*a*(1 + a*x)^2) - 3/(16*a*(1 + a*x)) + (5 *ArcTanh[a*x])/(16*a)))/(c - a^2*c*x^2)^(7/2))
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p) Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && !IntegerQ[p]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[c^p/a^(2*p) Int[(u/x^(2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !Inte gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
Time = 0.14 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.87
| method | result | size |
| default | \(\frac {\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (15 \ln \left (a x -1\right ) x^{5} a^{5}-15 \ln \left (a x +1\right ) x^{5} a^{5}+15 \ln \left (a x -1\right ) x^{4} a^{4}-15 \ln \left (a x +1\right ) x^{4} a^{4}+30 a^{4} x^{4}-30 a^{3} \ln \left (a x -1\right ) x^{3}+30 \ln \left (a x +1\right ) x^{3} a^{3}+30 a^{3} x^{3}-30 a^{2} \ln \left (a x -1\right ) x^{2}+30 \ln \left (a x +1\right ) x^{2} a^{2}-50 a^{2} x^{2}+15 a \ln \left (a x -1\right ) x -15 \ln \left (a x +1\right ) x a -50 a x +15 \ln \left (a x -1\right )-15 \ln \left (a x +1\right )+16\right )}{96 \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{4} a \left (a x -1\right )^{2}}\) | \(241\) |
Input:
int(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/96*((a*x-1)/(a*x+1))^(1/2)/(a*x+1)^2*(-c*(a^2*x^2-1))^(1/2)*(15*ln(a*x-1 )*x^5*a^5-15*ln(a*x+1)*x^5*a^5+15*ln(a*x-1)*x^4*a^4-15*ln(a*x+1)*x^4*a^4+3 0*a^4*x^4-30*a^3*ln(a*x-1)*x^3+30*ln(a*x+1)*x^3*a^3+30*a^3*x^3-30*a^2*ln(a *x-1)*x^2+30*ln(a*x+1)*x^2*a^2-50*a^2*x^2+15*a*ln(a*x-1)*x-15*ln(a*x+1)*x* a-50*a*x+15*ln(a*x-1)-15*ln(a*x+1)+16)/(a^2*x^2-1)/c^4/a/(a*x-1)^2
Time = 0.10 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {15 \, {\left (a^{6} x^{5} + a^{5} x^{4} - 2 \, a^{4} x^{3} - 2 \, a^{3} x^{2} + a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, {\left (15 \, a^{4} x^{4} + 15 \, a^{3} x^{3} - 25 \, a^{2} x^{2} - 25 \, a x + 8\right )} \sqrt {-a^{2} c}}{96 \, {\left (a^{7} c^{4} x^{5} + a^{6} c^{4} x^{4} - 2 \, a^{5} c^{4} x^{3} - 2 \, a^{4} c^{4} x^{2} + a^{3} c^{4} x + a^{2} c^{4}\right )}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="frica s")
Output:
-1/96*(15*(a^6*x^5 + a^5*x^4 - 2*a^4*x^3 - 2*a^3*x^2 + a^2*x + a)*sqrt(-c) *log((a^2*c*x^2 - 2*sqrt(-a^2*c)*sqrt(-c)*x + c)/(a^2*x^2 - 1)) - 2*(15*a^ 4*x^4 + 15*a^3*x^3 - 25*a^2*x^2 - 25*a*x + 8)*sqrt(-a^2*c))/(a^7*c^4*x^5 + a^6*c^4*x^4 - 2*a^5*c^4*x^3 - 2*a^4*c^4*x^2 + a^3*c^4*x + a^2*c^4)
Timed out. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\text {Timed out} \] Input:
integrate(((a*x-1)/(a*x+1))**(1/2)/(-a**2*c*x**2+c)**(7/2),x)
Output:
Timed out
\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="maxim a")
Output:
integrate(sqrt((a*x - 1)/(a*x + 1))/(-a^2*c*x^2 + c)^(7/2), x)
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.35 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {{\left (\frac {15 \, \log \left ({\left | a x + 1 \right |}\right )}{a c^{3}} - \frac {15 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{3}} - \frac {2 \, {\left (15 \, a^{4} x^{4} + 15 \, a^{3} x^{3} - 25 \, a^{2} x^{2} - 25 \, a x + 8\right )}}{{\left (a x + 1\right )}^{3} {\left (a x - 1\right )}^{2} a c^{3}}\right )} \mathrm {sgn}\left (a x + 1\right )}{96 \, \sqrt {-c}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="giac" )
Output:
1/96*(15*log(abs(a*x + 1))/(a*c^3) - 15*log(abs(a*x - 1))/(a*c^3) - 2*(15* a^4*x^4 + 15*a^3*x^3 - 25*a^2*x^2 - 25*a*x + 8)/((a*x + 1)^3*(a*x - 1)^2*a *c^3))*sgn(a*x + 1)/sqrt(-c)
Timed out. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int \frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{{\left (c-a^2\,c\,x^2\right )}^{7/2}} \,d x \] Input:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a^2*c*x^2)^(7/2),x)
Output:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a^2*c*x^2)^(7/2), x)
Time = 0.17 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {c}\, i \left (15 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{5} x^{5}+15 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{4} x^{4}-30 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}-30 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}+15 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +15 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )+15 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{5} x^{5}+15 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{4} x^{4}-30 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}-30 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}+15 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x +15 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )-30 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{5} x^{5}-30 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{4} x^{4}+60 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}+60 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}-30 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x -30 \,\mathrm {log}\left (\sqrt {-a x +1}\right )+5 a^{5} x^{5}-25 a^{4} x^{4}-40 a^{3} x^{3}+40 a^{2} x^{2}+55 a x -11\right )}{96 a \,c^{4} \left (a^{5} x^{5}+a^{4} x^{4}-2 a^{3} x^{3}-2 a^{2} x^{2}+a x +1\right )} \] Input:
int(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(7/2),x)
Output:
(sqrt(c)*i*(15*log(sqrt( - a*x + 1) - sqrt(2))*a**5*x**5 + 15*log(sqrt( - a*x + 1) - sqrt(2))*a**4*x**4 - 30*log(sqrt( - a*x + 1) - sqrt(2))*a**3*x* *3 - 30*log(sqrt( - a*x + 1) - sqrt(2))*a**2*x**2 + 15*log(sqrt( - a*x + 1 ) - sqrt(2))*a*x + 15*log(sqrt( - a*x + 1) - sqrt(2)) + 15*log(sqrt( - a*x + 1) + sqrt(2))*a**5*x**5 + 15*log(sqrt( - a*x + 1) + sqrt(2))*a**4*x**4 - 30*log(sqrt( - a*x + 1) + sqrt(2))*a**3*x**3 - 30*log(sqrt( - a*x + 1) + sqrt(2))*a**2*x**2 + 15*log(sqrt( - a*x + 1) + sqrt(2))*a*x + 15*log(sqrt ( - a*x + 1) + sqrt(2)) - 30*log(sqrt( - a*x + 1))*a**5*x**5 - 30*log(sqrt ( - a*x + 1))*a**4*x**4 + 60*log(sqrt( - a*x + 1))*a**3*x**3 + 60*log(sqrt ( - a*x + 1))*a**2*x**2 - 30*log(sqrt( - a*x + 1))*a*x - 30*log(sqrt( - a* x + 1)) + 5*a**5*x**5 - 25*a**4*x**4 - 40*a**3*x**3 + 40*a**2*x**2 + 55*a* x - 11))/(96*a*c**4*(a**5*x**5 + a**4*x**4 - 2*a**3*x**3 - 2*a**2*x**2 + a *x + 1))