Integrand size = 24, antiderivative size = 235 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}-\frac {2 c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c^2 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
-1/4*c^2*(c-c/a^2/x^2)^(1/2)/a^5/(1-1/a^2/x^2)^(1/2)/x^4+c^2*(c-c/a^2/x^2) ^(1/2)/a^4/(1-1/a^2/x^2)^(1/2)/x^3-c^2*(c-c/a^2/x^2)^(1/2)/a^3/(1-1/a^2/x^ 2)^(1/2)/x^2-2*c^2*(c-c/a^2/x^2)^(1/2)/a^2/(1-1/a^2/x^2)^(1/2)/x+c^2*(c-c/ a^2/x^2)^(1/2)*x/(1-1/a^2/x^2)^(1/2)-3*c^2*(c-c/a^2/x^2)^(1/2)*ln(x)/a/(1- 1/a^2/x^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.34 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} \left (-\frac {5}{4 a}-\frac {1}{4 a^5 x^4}+\frac {1}{a^4 x^3}-\frac {1}{a^3 x^2}-\frac {2}{a^2 x}+x-\frac {3 \log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}} \] Input:
Integrate[(c - c/(a^2*x^2))^(5/2)/E^(3*ArcCoth[a*x]),x]
Output:
((c - c/(a^2*x^2))^(5/2)*(-5/(4*a) - 1/(4*a^5*x^4) + 1/(a^4*x^3) - 1/(a^3* x^2) - 2/(a^2*x) + x - (3*Log[x])/a))/(1 - 1/(a^2*x^2))^(5/2)
Time = 0.68 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6751, 6747, 84, 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.
\(\displaystyle \int \left (c-\frac {c}{a^2 x^2}\right )^{5/2} e^{-3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{5/2}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^4 (a x+1)}{x^5}dx}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \int \left (a^5-\frac {3 a^4}{x}+\frac {2 a^3}{x^2}+\frac {2 a^2}{x^3}-\frac {3 a}{x^4}+\frac {1}{x^5}\right )dx}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \left (a^5 x-3 a^4 \log (x)-\frac {2 a^3}{x}-\frac {a^2}{x^2}+\frac {a}{x^3}-\frac {1}{4 x^4}\right )}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(c - c/(a^2*x^2))^(5/2)/E^(3*ArcCoth[a*x]),x]
Output:
(c^2*Sqrt[c - c/(a^2*x^2)]*(-1/4*1/x^4 + a/x^3 - a^2/x^2 - (2*a^3)/x + a^5 *x - 3*a^4*Log[x]))/(a^5*Sqrt[1 - 1/(a^2*x^2)])
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
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 = 96, normalized size of antiderivative = 0.41
method | result | size |
default | \(-\frac {\left (-4 a^{5} x^{5}+12 \ln \left (x \right ) x^{4} a^{4}+8 a^{3} x^{3}+4 a^{2} x^{2}-4 a x +1\right ) x {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{4 \left (a x -1\right )^{3} \left (a^{2} x^{2}-1\right )}\) | \(96\) |
Input:
int((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/4*(-4*a^5*x^5+12*ln(x)*x^4*a^4+8*a^3*x^3+4*a^2*x^2-4*a*x+1)*x*(c*(a^2*x ^2-1)/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(3/2)/(a*x-1)^3/(a^2*x^2-1)
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.31 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {{\left (4 \, a^{5} c^{2} x^{5} - 12 \, a^{4} c^{2} x^{4} \log \left (x\right ) - 8 \, a^{3} c^{2} x^{3} - 4 \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x - c^{2}\right )} \sqrt {a^{2} c}}{4 \, a^{6} x^{4}} \] Input:
integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas ")
Output:
1/4*(4*a^5*c^2*x^5 - 12*a^4*c^2*x^4*log(x) - 8*a^3*c^2*x^3 - 4*a^2*c^2*x^2 + 4*a*c^2*x - c^2)*sqrt(a^2*c)/(a^6*x^4)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(5/2)*((a*x-1)/(a*x+1))**(3/2),x)
Output:
Timed out
\[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima ")
Output:
integrate((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.50 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {1}{4} \, {\left (\frac {4 \, c^{2} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a} - \frac {12 \, c^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{2}} - \frac {8 \, a^{3} c^{2} x^{3} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + 4 \, a^{2} c^{2} x^{2} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 4 \, a c^{2} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + c^{2} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{6} x^{4}}\right )} \sqrt {c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")
Output:
1/4*(4*c^2*x*sgn(a*x + 1)*sgn(x)/a - 12*c^2*log(abs(x))*sgn(a*x + 1)*sgn(x )/a^2 - (8*a^3*c^2*x^3*sgn(a*x + 1)*sgn(x) + 4*a^2*c^2*x^2*sgn(a*x + 1)*sg n(x) - 4*a*c^2*x*sgn(a*x + 1)*sgn(x) + c^2*sgn(a*x + 1)*sgn(x))/(a^6*x^4)) *sqrt(c)*abs(a)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int {\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \] Input:
int((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.27 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} \left (-12 \,\mathrm {log}\left (a x \right ) a^{4} x^{4}+4 a^{5} x^{5}-8 a^{4} x^{4}-8 a^{3} x^{3}-4 a^{2} x^{2}+4 a x -1\right )}{4 a^{5} x^{4}} \] Input:
int((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(3/2),x)
Output:
(sqrt(c)*c**2*( - 12*log(a*x)*a**4*x**4 + 4*a**5*x**5 - 8*a**4*x**4 - 8*a* *3*x**3 - 4*a**2*x**2 + 4*a*x - 1))/(4*a**5*x**4)