Integrand size = 24, antiderivative size = 263 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^2}-\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)}+\frac {7 \sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {23 \sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \] Output:
(1-1/a^2/x^2)^(1/2)*x/c^2/(c-c/a^2/x^2)^(1/2)+1/8*(1-1/a^2/x^2)^(1/2)/a/c^ 2/(c-c/a^2/x^2)^(1/2)/(-a*x+1)+1/8*(1-1/a^2/x^2)^(1/2)/a/c^2/(c-c/a^2/x^2) ^(1/2)/(a*x+1)^2-(1-1/a^2/x^2)^(1/2)/a/c^2/(c-c/a^2/x^2)^(1/2)/(a*x+1)+7/1 6*(1-1/a^2/x^2)^(1/2)*ln(-a*x+1)/a/c^2/(c-c/a^2/x^2)^(1/2)-23/16*(1-1/a^2/ x^2)^(1/2)*ln(a*x+1)/a/c^2/(c-c/a^2/x^2)^(1/2)
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.36 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (2 \left (8 x+\frac {1}{a (1+a x)^2}+\frac {1}{a-a^2 x}-\frac {8}{a+a^2 x}\right )+\frac {7 \log (1-a x)}{a}-\frac {23 \log (1+a x)}{a}\right )}{16 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \] Input:
Integrate[1/(E^ArcCoth[a*x]*(c - c/(a^2*x^2))^(5/2)),x]
Output:
((1 - 1/(a^2*x^2))^(5/2)*(2*(8*x + 1/(a*(1 + a*x)^2) + (a - a^2*x)^(-1) - 8/(a + a^2*x)) + (7*Log[1 - a*x])/a - (23*Log[1 + a*x])/a))/(16*(c - c/(a^ 2*x^2))^(5/2))
Time = 0.74 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.42, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6751, 6747, 99, 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.
\(\displaystyle \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {\sqrt {1-\frac {1}{a^2 x^2}} \int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}}dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \int \frac {x^5}{(1-a x)^2 (a x+1)^3}dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \int \left (-\frac {23}{16 a^5 (a x+1)}+\frac {1}{a^5 (a x+1)^2}-\frac {1}{4 a^5 (a x+1)^3}+\frac {1}{a^5}+\frac {7}{16 a^5 (a x-1)}+\frac {1}{8 a^5 (a x-1)^2}\right )dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^5 \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {1}{8 a^6 (1-a x)}-\frac {1}{a^6 (a x+1)}+\frac {1}{8 a^6 (a x+1)^2}+\frac {7 \log (1-a x)}{16 a^6}-\frac {23 \log (a x+1)}{16 a^6}+\frac {x}{a^5}\right )}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\) |
Input:
Int[1/(E^ArcCoth[a*x]*(c - c/(a^2*x^2))^(5/2)),x]
Output:
(a^5*Sqrt[1 - 1/(a^2*x^2)]*(x/a^5 + 1/(8*a^6*(1 - a*x)) + 1/(8*a^6*(1 + a* x)^2) - 1/(a^6*(1 + a*x)) + (7*Log[1 - a*x])/(16*a^6) - (23*Log[1 + a*x])/ (16*a^6)))/(c^2*Sqrt[c - c/(a^2*x^2)])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[c^IntPart[p]*((c + d/x^2)^FracPart[p]/(1 - 1/(a^2*x^2))^FracPart [p]) Int[u*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !IntegerQ[n/2] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.08 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.67
method | result | size |
default | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \left (a x -1\right ) \left (-16 a^{4} x^{4}+23 \ln \left (a x +1\right ) x^{3} a^{3}-7 a^{3} \ln \left (a x -1\right ) x^{3}-16 a^{3} x^{3}+23 \ln \left (a x +1\right ) x^{2} a^{2}-7 a^{2} \ln \left (a x -1\right ) x^{2}+34 a^{2} x^{2}-23 \ln \left (a x +1\right ) x a +7 a \ln \left (a x -1\right ) x +18 a x -23 \ln \left (a x +1\right )+7 \ln \left (a x -1\right )-12\right )}{16 a^{6} x^{5} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\) | \(175\) |
Input:
int(((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/16*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)*(a*x-1)*(-16*a^4*x^4+23*ln(a*x+1)*x^ 3*a^3-7*a^3*ln(a*x-1)*x^3-16*a^3*x^3+23*ln(a*x+1)*x^2*a^2-7*a^2*ln(a*x-1)* x^2+34*a^2*x^2-23*ln(a*x+1)*x*a+7*a*ln(a*x-1)*x+18*a*x-23*ln(a*x+1)+7*ln(a *x-1)-12)/a^6/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)
Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.51 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {{\left (16 \, a^{4} x^{4} + 16 \, a^{3} x^{3} - 34 \, a^{2} x^{2} - 18 \, a x - 23 \, {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \log \left (a x + 1\right ) + 7 \, {\left (a^{3} x^{3} + a^{2} x^{2} - a x - 1\right )} \log \left (a x - 1\right ) + 12\right )} \sqrt {a^{2} c}}{16 \, {\left (a^{5} c^{3} x^{3} + a^{4} c^{3} x^{2} - a^{3} c^{3} x - a^{2} c^{3}\right )}} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="fricas ")
Output:
1/16*(16*a^4*x^4 + 16*a^3*x^3 - 34*a^2*x^2 - 18*a*x - 23*(a^3*x^3 + a^2*x^ 2 - a*x - 1)*log(a*x + 1) + 7*(a^3*x^3 + a^2*x^2 - a*x - 1)*log(a*x - 1) + 12)*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. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(((a*x-1)/(a*x+1))**(1/2)/(c-c/a**2/x**2)**(5/2),x)
Output:
Timed out
\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="maxima ")
Output:
integrate(sqrt((a*x - 1)/(a*x + 1))/(c - c/(a^2*x^2))^(5/2), x)
Exception generated. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}} \,d x \] Input:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - c/(a^2*x^2))^(5/2),x)
Output:
int(((a*x - 1)/(a*x + 1))^(1/2)/(c - c/(a^2*x^2))^(5/2), x)
Time = 0.15 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.56 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (7 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+7 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-7 \,\mathrm {log}\left (a x -1\right ) a x -7 \,\mathrm {log}\left (a x -1\right )-23 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-23 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+23 \,\mathrm {log}\left (a x +1\right ) a x +23 \,\mathrm {log}\left (a x +1\right )+16 a^{4} x^{4}+50 a^{3} x^{3}-52 a x -22\right )}{16 a \,c^{3} \left (a^{3} x^{3}+a^{2} x^{2}-a x -1\right )} \] Input:
int(((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(5/2),x)
Output:
(sqrt(c)*(7*log(a*x - 1)*a**3*x**3 + 7*log(a*x - 1)*a**2*x**2 - 7*log(a*x - 1)*a*x - 7*log(a*x - 1) - 23*log(a*x + 1)*a**3*x**3 - 23*log(a*x + 1)*a* *2*x**2 + 23*log(a*x + 1)*a*x + 23*log(a*x + 1) + 16*a**4*x**4 + 50*a**3*x **3 - 52*a*x - 22))/(16*a*c**3*(a**3*x**3 + a**2*x**2 - a*x - 1))