Integrand size = 24, antiderivative size = 322 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=-\frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{8 a^9 \sqrt {1-\frac {1}{a^2 x^2}} x^8}+\frac {3 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{7 a^8 \sqrt {1-\frac {1}{a^2 x^2}} x^7}-\frac {8 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{5 a^6 \sqrt {1-\frac {1}{a^2 x^2}} x^5}+\frac {3 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}+\frac {2 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}-\frac {4 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c^4 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:
-1/8*c^4*(c-c/a^2/x^2)^(1/2)/a^9/(1-1/a^2/x^2)^(1/2)/x^8+3/7*c^4*(c-c/a^2/ x^2)^(1/2)/a^8/(1-1/a^2/x^2)^(1/2)/x^7-8/5*c^4*(c-c/a^2/x^2)^(1/2)/a^6/(1- 1/a^2/x^2)^(1/2)/x^5+3/2*c^4*(c-c/a^2/x^2)^(1/2)/a^5/(1-1/a^2/x^2)^(1/2)/x ^4+2*c^4*(c-c/a^2/x^2)^(1/2)/a^4/(1-1/a^2/x^2)^(1/2)/x^3-4*c^4*(c-c/a^2/x^ 2)^(1/2)/a^3/(1-1/a^2/x^2)^(1/2)/x^2+c^4*(c-c/a^2/x^2)^(1/2)*x/(1-1/a^2/x^ 2)^(1/2)-3*c^4*(c-c/a^2/x^2)^(1/2)*ln(x)/a/(1-1/a^2/x^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.30 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{9/2} \left (-\frac {1}{8 a^9 x^8}+\frac {3}{7 a^8 x^7}-\frac {8}{5 a^6 x^5}+\frac {3}{2 a^5 x^4}+\frac {2}{a^4 x^3}-\frac {4}{a^3 x^2}+x-\frac {3 \log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{9/2}} \] Input:
Integrate[(c - c/(a^2*x^2))^(9/2)/E^(3*ArcCoth[a*x]),x]
Output:
((c - c/(a^2*x^2))^(9/2)*(-1/8*1/(a^9*x^8) + 3/(7*a^8*x^7) - 8/(5*a^6*x^5) + 3/(2*a^5*x^4) + 2/(a^4*x^3) - 4/(a^3*x^2) + x - (3*Log[x])/a))/(1 - 1/( a^2*x^2))^(9/2)
Time = 0.79 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.31, 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.
\(\displaystyle \int \left (c-\frac {c}{a^2 x^2}\right )^{9/2} e^{-3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{9/2}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^6 (a x+1)^3}{x^9}dx}{a^9 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} \int \left (a^9-\frac {3 a^8}{x}+\frac {8 a^6}{x^3}-\frac {6 a^5}{x^4}-\frac {6 a^4}{x^5}+\frac {8 a^3}{x^6}-\frac {3 a}{x^8}+\frac {1}{x^9}\right )dx}{a^9 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} \left (a^9 x-3 a^8 \log (x)-\frac {4 a^6}{x^2}+\frac {2 a^5}{x^3}+\frac {3 a^4}{2 x^4}-\frac {8 a^3}{5 x^5}+\frac {3 a}{7 x^7}-\frac {1}{8 x^8}\right )}{a^9 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[(c - c/(a^2*x^2))^(9/2)/E^(3*ArcCoth[a*x]),x]
Output:
(c^4*Sqrt[c - c/(a^2*x^2)]*(-1/8*1/x^8 + (3*a)/(7*x^7) - (8*a^3)/(5*x^5) + (3*a^4)/(2*x^4) + (2*a^5)/x^3 - (4*a^6)/x^2 + a^9*x - 3*a^8*Log[x]))/(a^9 *Sqrt[1 - 1/(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 = 112, normalized size of antiderivative = 0.35
method | result | size |
default | \(-\frac {\left (-280 a^{9} x^{9}+840 a^{8} \ln \left (x \right ) x^{8}+1120 x^{6} a^{6}-560 a^{5} x^{5}-420 a^{4} x^{4}+448 a^{3} x^{3}-120 a x +35\right ) x {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {9}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{280 \left (a x -1\right )^{3} \left (a^{2} x^{2}-1\right )^{3}}\) | \(112\) |
Input:
int((c-c/a^2/x^2)^(9/2)*((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/280*(-280*a^9*x^9+840*a^8*ln(x)*x^8+1120*x^6*a^6-560*a^5*x^5-420*a^4*x^ 4+448*a^3*x^3-120*a*x+35)*x*(c*(a^2*x^2-1)/a^2/x^2)^(9/2)*((a*x-1)/(a*x+1) )^(3/2)/(a*x-1)^3/(a^2*x^2-1)^3
Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.30 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\frac {{\left (280 \, a^{9} c^{4} x^{9} - 840 \, a^{8} c^{4} x^{8} \log \left (x\right ) - 1120 \, a^{6} c^{4} x^{6} + 560 \, a^{5} c^{4} x^{5} + 420 \, a^{4} c^{4} x^{4} - 448 \, a^{3} c^{4} x^{3} + 120 \, a c^{4} x - 35 \, c^{4}\right )} \sqrt {a^{2} c}}{280 \, a^{10} x^{8}} \] Input:
integrate((c-c/a^2/x^2)^(9/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas ")
Output:
1/280*(280*a^9*c^4*x^9 - 840*a^8*c^4*x^8*log(x) - 1120*a^6*c^4*x^6 + 560*a ^5*c^4*x^5 + 420*a^4*c^4*x^4 - 448*a^3*c^4*x^3 + 120*a*c^4*x - 35*c^4)*sqr t(a^2*c)/(a^10*x^8)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(9/2)*((a*x-1)/(a*x+1))**(3/2),x)
Output:
Timed out
\[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {9}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(9/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima ")
Output:
integrate((c - c/(a^2*x^2))^(9/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.13 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.49 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\frac {1}{280} \, {\left (\frac {280 \, c^{4} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a} - \frac {840 \, c^{4} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{2}} - \frac {1120 \, a^{6} c^{4} x^{6} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 560 \, a^{5} c^{4} x^{5} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 420 \, a^{4} c^{4} x^{4} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + 448 \, a^{3} c^{4} x^{3} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) - 120 \, a c^{4} x \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right ) + 35 \, c^{4} \mathrm {sgn}\left (a x + 1\right ) \mathrm {sgn}\left (x\right )}{a^{10} x^{8}}\right )} \sqrt {c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(9/2)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")
Output:
1/280*(280*c^4*x*sgn(a*x + 1)*sgn(x)/a - 840*c^4*log(abs(x))*sgn(a*x + 1)* sgn(x)/a^2 - (1120*a^6*c^4*x^6*sgn(a*x + 1)*sgn(x) - 560*a^5*c^4*x^5*sgn(a *x + 1)*sgn(x) - 420*a^4*c^4*x^4*sgn(a*x + 1)*sgn(x) + 448*a^3*c^4*x^3*sgn (a*x + 1)*sgn(x) - 120*a*c^4*x*sgn(a*x + 1)*sgn(x) + 35*c^4*sgn(a*x + 1)*s gn(x))/(a^10*x^8))*sqrt(c)*abs(a)
Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\int {\left (c-\frac {c}{a^2\,x^2}\right )}^{9/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \] Input:
int((c - c/(a^2*x^2))^(9/2)*((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int((c - c/(a^2*x^2))^(9/2)*((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.25 \[ \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\frac {\sqrt {c}\, c^{4} \left (-840 \,\mathrm {log}\left (a x \right ) a^{8} x^{8}+280 a^{9} x^{9}-1260 a^{8} x^{8}-1120 a^{6} x^{6}+560 a^{5} x^{5}+420 a^{4} x^{4}-448 a^{3} x^{3}+120 a x -35\right )}{280 a^{9} x^{8}} \] Input:
int((c-c/a^2/x^2)^(9/2)*((a*x-1)/(a*x+1))^(3/2),x)
Output:
(sqrt(c)*c**4*( - 840*log(a*x)*a**8*x**8 + 280*a**9*x**9 - 1260*a**8*x**8 - 1120*a**6*x**6 + 560*a**5*x**5 + 420*a**4*x**4 - 448*a**3*x**3 + 120*a*x - 35))/(280*a**9*x**8)