Integrand size = 24, antiderivative size = 183 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{4 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {3 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \text {arctanh}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \] Output:
1/8*a^4*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)/(-a^2*c*x^2+c)^(5/2)-1/8*a^4*(1-1 /a^2/x^2)^(5/2)*x^5/(a*x+1)^2/(-a^2*c*x^2+c)^(5/2)-1/4*a^4*(1-1/a^2/x^2)^( 5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)+3/8*a^4*(1-1/a^2/x^2)^(5/2)*x^5*arct anh(a*x)/(-a^2*c*x^2+c)^(5/2)
Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.44 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (2-3 a x-3 a^2 x^2+3 (-1+a x) (1+a x)^2 \text {arctanh}(a x)\right )}{8 (-1+a x) (c+a c x)^2 \sqrt {c-a^2 c x^2}} \] Input:
Integrate[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^(5/2)),x]
Output:
(Sqrt[1 - 1/(a^2*x^2)]*x*(2 - 3*a*x - 3*a^2*x^2 + 3*(-1 + a*x)*(1 + a*x)^2 *ArcTanh[a*x]))/(8*(-1 + a*x)*(c + a*c*x)^2*Sqrt[c - a^2*c*x^2])
Time = 0.77 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.50, 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.
\(\displaystyle \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {1}{(1-a x)^2 (a x+1)^3}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \left (\frac {1}{8 (a x-1)^2}+\frac {1}{4 (a x+1)^2}+\frac {1}{4 (a x+1)^3}-\frac {3}{8 \left (a^2 x^2-1\right )}\right )dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (\frac {3 \text {arctanh}(a x)}{8 a}+\frac {1}{8 a (1-a x)}-\frac {1}{4 a (a x+1)}-\frac {1}{8 a (a x+1)^2}\right )}{\left (c-a^2 c x^2\right )^{5/2}}\) |
Input:
Int[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^(5/2)),x]
Output:
(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(1/(8*a*(1 - a*x)) - 1/(8*a*(1 + a*x)^2) - 1/(4*a*(1 + a*x)) + (3*ArcTanh[a*x])/(8*a)))/(c - a^2*c*x^2)^(5/2)
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 = 169, normalized size of antiderivative = 0.92
method | result | size |
default | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 \ln \left (a x +1\right ) x^{3} a^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}+3 \ln \left (a x +1\right ) x^{2} a^{2}-3 a^{2} \ln \left (a x -1\right ) x^{2}-6 a^{2} x^{2}-3 \ln \left (a x +1\right ) x a +3 a \ln \left (a x -1\right ) x -6 a x -3 \ln \left (a x +1\right )+3 \ln \left (a x -1\right )+4\right )}{16 \left (a x +1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a \left (a x -1\right )}\) | \(169\) |
Input:
int(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/16*((a*x-1)/(a*x+1))^(1/2)/(a*x+1)*(-c*(a^2*x^2-1))^(1/2)*(3*ln(a*x+1)* x^3*a^3-3*a^3*ln(a*x-1)*x^3+3*ln(a*x+1)*x^2*a^2-3*a^2*ln(a*x-1)*x^2-6*a^2* x^2-3*ln(a*x+1)*x*a+3*a*ln(a*x-1)*x-6*a*x-3*ln(a*x+1)+3*ln(a*x-1)+4)/(a^2* x^2-1)/c^3/a/(a*x-1)
Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {3 \, {\left (a^{4} x^{3} + a^{3} x^{2} - a^{2} x - a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, {\left (3 \, a^{2} x^{2} + 3 \, a x - 2\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{5} c^{3} x^{3} + a^{4} c^{3} x^{2} - a^{3} c^{3} x - a^{2} c^{3}\right )}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="frica s")
Output:
-1/16*(3*(a^4*x^3 + 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*(3*a^2*x^2 + 3*a*x - 2)*sqrt(-a ^2*c))/(a^5*c^3*x^3 + a^4*c^3*x^2 - a^3*c^3*x - a^2*c^3)
Timed out. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(((a*x-1)/(a*x+1))**(1/2)/(-a**2*c*x**2+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxim a")
Output:
integrate(sqrt((a*x - 1)/(a*x + 1))/(-a^2*c*x^2 + c)^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.44 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (\frac {3 \, \log \left ({\left | a x + 1 \right |}\right )}{a c^{2}} - \frac {3 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{2}} - \frac {2 \, {\left (3 \, a^{2} x^{2} + 3 \, a x - 2\right )}}{{\left (a x + 1\right )}^{2} {\left (a x - 1\right )} a c^{2}}\right )} \mathrm {sgn}\left (a x + 1\right )}{16 \, \sqrt {-c}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac" )
Output:
1/16*(3*log(abs(a*x + 1))/(a*c^2) - 3*log(abs(a*x - 1))/(a*c^2) - 2*(3*a^2 *x^2 + 3*a*x - 2)/((a*x + 1)^2*(a*x - 1)*a*c^2))*sgn(a*x + 1)/sqrt(-c)
Timed out. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{{\left (c-a^2\,c\,x^2\right )}^{5/2}} \,d x \] Input:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a^2*c*x^2)^(5/2),x)
Output:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a^2*c*x^2)^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.36 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, i \left (6 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}+6 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}-6 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x -6 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )+6 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}+6 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}-6 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x -6 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )-12 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}-12 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}+12 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x +12 \,\mathrm {log}\left (\sqrt {-a x +1}\right )+3 a^{3} x^{3}-9 a^{2} x^{2}-15 a x +5\right )}{32 a \,c^{3} \left (a^{3} x^{3}+a^{2} x^{2}-a x -1\right )} \] Input:
int(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*i*(6*log(sqrt( - a*x + 1) - sqrt(2))*a**3*x**3 + 6*log(sqrt( - a* x + 1) - sqrt(2))*a**2*x**2 - 6*log(sqrt( - a*x + 1) - sqrt(2))*a*x - 6*lo g(sqrt( - a*x + 1) - sqrt(2)) + 6*log(sqrt( - a*x + 1) + sqrt(2))*a**3*x** 3 + 6*log(sqrt( - a*x + 1) + sqrt(2))*a**2*x**2 - 6*log(sqrt( - a*x + 1) + sqrt(2))*a*x - 6*log(sqrt( - a*x + 1) + sqrt(2)) - 12*log(sqrt( - a*x + 1 ))*a**3*x**3 - 12*log(sqrt( - a*x + 1))*a**2*x**2 + 12*log(sqrt( - a*x + 1 ))*a*x + 12*log(sqrt( - a*x + 1)) + 3*a**3*x**3 - 9*a**2*x**2 - 15*a*x + 5 ))/(32*a*c**3*(a**3*x**3 + a**2*x**2 - a*x - 1))