Integrand size = 24, antiderivative size = 275 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (1+a x)^4 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{12 (1+a x)^3 \left (c-a^2 c x^2\right )^{7/2}}-\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1+a x)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{8 (1+a x) \left (c-a^2 c x^2\right )^{7/2}}+\frac {5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7 \text {arctanh}(a x)}{32 \left (c-a^2 c x^2\right )^{7/2}} \] Output:
1/32*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)/(-a^2*c*x^2+c)^(7/2)-1/16*a^6*(1 -1/a^2/x^2)^(7/2)*x^7/(a*x+1)^4/(-a^2*c*x^2+c)^(7/2)-1/12*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)-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)+5/32*a^6*(1-1/a^2/x^2)^(7/2)*x^7*arctanh(a*x)/(-a^2*c* x^2+c)^(7/2)
Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.36 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (32+15 a x-35 a^2 x^2-45 a^3 x^3-15 a^4 x^4+15 (-1+a x) (1+a x)^4 \text {arctanh}(a x)\right )}{96 c^3 (-1+a x) (1+a x)^4 \sqrt {c-a^2 c x^2}} \] Input:
Integrate[1/(E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^(7/2)),x]
Output:
-1/96*(Sqrt[1 - 1/(a^2*x^2)]*x*(32 + 15*a*x - 35*a^2*x^2 - 45*a^3*x^3 - 15 *a^4*x^4 + 15*(-1 + a*x)*(1 + a*x)^4*ArcTanh[a*x]))/(c^3*(-1 + a*x)*(1 + a *x)^4*Sqrt[c - a^2*c*x^2])
Time = 0.85 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.43, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6746, 6747, 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.
| \(\displaystyle \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 6746 |
| \(\displaystyle \frac {x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 6747 |
| \(\displaystyle \frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \frac {1}{(1-a x)^2 (a x+1)^5}dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \left (\frac {1}{32 (a x-1)^2}+\frac {1}{8 (a x+1)^2}+\frac {3}{16 (a x+1)^3}+\frac {1}{4 (a x+1)^4}+\frac {1}{4 (a x+1)^5}-\frac {5}{32 \left (a^2 x^2-1\right )}\right )dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \left (\frac {5 \text {arctanh}(a x)}{32 a}+\frac {1}{32 a (1-a x)}-\frac {1}{8 a (a x+1)}-\frac {3}{32 a (a x+1)^2}-\frac {1}{12 a (a x+1)^3}-\frac {1}{16 a (a x+1)^4}\right )}{\left (c-a^2 c x^2\right )^{7/2}}\) |
Input:
Int[1/(E^(3*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)) - 1/(16*a*(1 + a*x)^4 ) - 1/(12*a*(1 + a*x)^3) - 3/(32*a*(1 + a*x)^2) - 1/(8*a*(1 + a*x)) + (5*A rcTanh[a*x])/(32*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 = 0.88
| method | result | size |
| default | \(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (15 \ln \left (a x +1\right ) x^{5} a^{5}-15 \ln \left (a x -1\right ) x^{5} a^{5}+45 \ln \left (a x +1\right ) x^{4} a^{4}-45 \ln \left (a x -1\right ) x^{4} a^{4}-30 a^{4} x^{4}+30 \ln \left (a x +1\right ) x^{3} a^{3}-30 a^{3} \ln \left (a x -1\right ) x^{3}-90 a^{3} x^{3}-30 \ln \left (a x +1\right ) x^{2} a^{2}+30 a^{2} \ln \left (a x -1\right ) x^{2}-70 a^{2} x^{2}-45 \ln \left (a x +1\right ) x a +45 a \ln \left (a x -1\right ) x +30 a x -15 \ln \left (a x +1\right )+15 \ln \left (a x -1\right )+64\right )}{192 \left (a x -1\right )^{2} \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{4} a}\) | \(241\) |
Input:
int(((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/192*((a*x-1)/(a*x+1))^(3/2)/(a*x-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+45*ln(a*x+1)*x^4*a^4-45*ln(a*x-1 )*x^4*a^4-30*a^4*x^4+30*ln(a*x+1)*x^3*a^3-30*a^3*ln(a*x-1)*x^3-90*a^3*x^3- 30*ln(a*x+1)*x^2*a^2+30*a^2*ln(a*x-1)*x^2-70*a^2*x^2-45*ln(a*x+1)*x*a+45*a *ln(a*x-1)*x+30*a*x-15*ln(a*x+1)+15*ln(a*x-1)+64)/(a^2*x^2-1)/c^4/a
Time = 0.10 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {15 \, {\left (a^{6} x^{5} + 3 \, a^{5} x^{4} + 2 \, a^{4} x^{3} - 2 \, a^{3} x^{2} - 3 \, a^{2} x - a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) + 2 \, {\left (15 \, a^{4} x^{4} + 45 \, a^{3} x^{3} + 35 \, a^{2} x^{2} - 15 \, a x - 32\right )} \sqrt {-a^{2} c}}{192 \, {\left (a^{7} c^{4} x^{5} + 3 \, a^{6} c^{4} x^{4} + 2 \, a^{5} c^{4} x^{3} - 2 \, a^{4} c^{4} x^{2} - 3 \, a^{3} c^{4} x - a^{2} c^{4}\right )}} \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="frica s")
Output:
-1/192*(15*(a^6*x^5 + 3*a^5*x^4 + 2*a^4*x^3 - 2*a^3*x^2 - 3*a^2*x - a)*sqr t(-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 + 45*a^3*x^3 + 35*a^2*x^2 - 15*a*x - 32)*sqrt(-a^2*c))/(a^7*c^4 *x^5 + 3*a^6*c^4*x^4 + 2*a^5*c^4*x^3 - 2*a^4*c^4*x^2 - 3*a^3*c^4*x - a^2*c ^4)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\text {Timed out} \] Input:
integrate(((a*x-1)/(a*x+1))**(3/2)/(-a**2*c*x**2+c)**(7/2),x)
Output:
Timed out
\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="maxim a")
Output:
integrate(((a*x - 1)/(a*x + 1))^(3/2)/(-a^2*c*x^2 + c)^(7/2), x)
Time = 0.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.48 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {\frac {15 \, \log \left ({\left | a x + 1 \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{a c^{3}} - \frac {15 \, \log \left ({\left | a x - 1 \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{a c^{3}} - \frac {2 \, {\left (15 \, a^{4} x^{4} \mathrm {sgn}\left (a x + 1\right ) + 45 \, a^{3} x^{3} \mathrm {sgn}\left (a x + 1\right ) + 35 \, a^{2} x^{2} \mathrm {sgn}\left (a x + 1\right ) - 15 \, a x \mathrm {sgn}\left (a x + 1\right ) - 32 \, \mathrm {sgn}\left (a x + 1\right )\right )}}{{\left (a x + 1\right )}^{4} {\left (a x - 1\right )} a c^{3}}}{192 \, \sqrt {-c}} \] Input:
integrate(((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="giac" )
Output:
-1/192*(15*log(abs(a*x + 1))*sgn(a*x + 1)/(a*c^3) - 15*log(abs(a*x - 1))*s gn(a*x + 1)/(a*c^3) - 2*(15*a^4*x^4*sgn(a*x + 1) + 45*a^3*x^3*sgn(a*x + 1) + 35*a^2*x^2*sgn(a*x + 1) - 15*a*x*sgn(a*x + 1) - 32*sgn(a*x + 1))/((a*x + 1)^4*(a*x - 1)*a*c^3))/sqrt(-c)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{{\left (c-a^2\,c\,x^2\right )}^{7/2}} \,d x \] Input:
int(((a*x - 1)/(a*x + 1))^(3/2)/(c - a^2*c*x^2)^(7/2),x)
Output:
int(((a*x - 1)/(a*x + 1))^(3/2)/(c - a^2*c*x^2)^(7/2), x)
Time = 0.15 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.43 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {c}\, i \left (-60 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{5} x^{5}-180 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{4} x^{4}-120 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}+120 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}+180 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +60 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )-60 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{5} x^{5}-180 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{4} x^{4}-120 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}+120 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}+180 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x +60 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )+120 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{5} x^{5}+360 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{4} x^{4}+240 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}-240 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}-360 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x -120 \,\mathrm {log}\left (\sqrt {-a x +1}\right )-15 a^{5} x^{5}+75 a^{4} x^{4}+330 a^{3} x^{3}+310 a^{2} x^{2}-75 a x -241\right )}{768 a \,c^{4} \left (a^{5} x^{5}+3 a^{4} x^{4}+2 a^{3} x^{3}-2 a^{2} x^{2}-3 a x -1\right )} \] Input:
int(((a*x-1)/(a*x+1))^(3/2)/(-a^2*c*x^2+c)^(7/2),x)
Output:
(sqrt(c)*i*( - 60*log(sqrt( - a*x + 1) - sqrt(2))*a**5*x**5 - 180*log(sqrt ( - a*x + 1) - sqrt(2))*a**4*x**4 - 120*log(sqrt( - a*x + 1) - sqrt(2))*a* *3*x**3 + 120*log(sqrt( - a*x + 1) - sqrt(2))*a**2*x**2 + 180*log(sqrt( - a*x + 1) - sqrt(2))*a*x + 60*log(sqrt( - a*x + 1) - sqrt(2)) - 60*log(sqrt ( - a*x + 1) + sqrt(2))*a**5*x**5 - 180*log(sqrt( - a*x + 1) + sqrt(2))*a* *4*x**4 - 120*log(sqrt( - a*x + 1) + sqrt(2))*a**3*x**3 + 120*log(sqrt( - a*x + 1) + sqrt(2))*a**2*x**2 + 180*log(sqrt( - a*x + 1) + sqrt(2))*a*x + 60*log(sqrt( - a*x + 1) + sqrt(2)) + 120*log(sqrt( - a*x + 1))*a**5*x**5 + 360*log(sqrt( - a*x + 1))*a**4*x**4 + 240*log(sqrt( - a*x + 1))*a**3*x**3 - 240*log(sqrt( - a*x + 1))*a**2*x**2 - 360*log(sqrt( - a*x + 1))*a*x - 1 20*log(sqrt( - a*x + 1)) - 15*a**5*x**5 + 75*a**4*x**4 + 330*a**3*x**3 + 3 10*a**2*x**2 - 75*a*x - 241))/(768*a*c**4*(a**5*x**5 + 3*a**4*x**4 + 2*a** 3*x**3 - 2*a**2*x**2 - 3*a*x - 1))