Integrand size = 24, antiderivative size = 322 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{8 a^9 \sqrt {1-\frac {1}{a^2 x^2}} x^8}+\frac {3 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{7 a^8 \sqrt {1-\frac {1}{a^2 x^2}} x^7}-\frac {8 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{5 a^6 \sqrt {1-\frac {1}{a^2 x^2}} x^5}-\frac {3 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}+\frac {2 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}+\frac {4 c^4 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 c^4 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]
1/8*c^4*(c-c/a^2/x^2)^(1/2)/a^9/x^8/(1-1/a^2/x^2)^(1/2)+3/7*c^4*(c-c/a^2/x ^2)^(1/2)/a^8/x^7/(1-1/a^2/x^2)^(1/2)-8/5*c^4*(c-c/a^2/x^2)^(1/2)/a^6/x^5/ (1-1/a^2/x^2)^(1/2)-3/2*c^4*(c-c/a^2/x^2)^(1/2)/a^5/x^4/(1-1/a^2/x^2)^(1/2 )+2*c^4*(c-c/a^2/x^2)^(1/2)/a^4/x^3/(1-1/a^2/x^2)^(1/2)+4*c^4*(c-c/a^2/x^2 )^(1/2)/a^3/x^2/(1-1/a^2/x^2)^(1/2)+c^4*x*(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2 )^(1/2)+3*c^4*ln(x)*(c-c/a^2/x^2)^(1/2)/a/(1-1/a^2/x^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.30 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{9/2} \left (\frac {1}{8 a^9 x^8}+\frac {3}{7 a^8 x^7}-\frac {8}{5 a^6 x^5}-\frac {3}{2 a^5 x^4}+\frac {2}{a^4 x^3}+\frac {4}{a^3 x^2}+x+\frac {3 \log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{9/2}} \]
((c - c/(a^2*x^2))^(9/2)*(1/(8*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 = 0.42 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.32, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6751, 6747, 25, 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 |
\(\displaystyle \frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} \int -\frac {(1-a x)^3 (a x+1)^6}{x^9}dx}{a^9 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^3 (a x+1)^6}{x^9}dx}{a^9 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 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 |
\(\displaystyle -\frac {c^4 \sqrt {c-\frac {c}{a^2 x^2}} \left (a^9 (-x)-3 a^8 \log (x)-\frac {4 a^6}{x^2}-\frac {2 a^5}{x^3}+\frac {3 a^4}{2 x^4}+\frac {8 a^3}{5 x^5}-\frac {3 a}{7 x^7}-\frac {1}{8 x^8}\right )}{a^9 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
-((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)]))
3.9.46.3.1 Defintions of rubi rules used
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 = 0.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.35
method | result | size |
default | \(\frac {\left (280 a^{9} x^{9}+840 a^{8} \ln \left (x \right ) x^{8}+1120 a^{6} x^{6}+560 a^{5} x^{5}-420 a^{4} x^{4}-448 a^{3} x^{3}+120 a x +35\right ) {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {9}{2}} x}{280 \left (a x +1\right )^{3} \left (a^{2} x^{2}-1\right )^{3} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(112\) |
1/280*(280*a^9*x^9+840*a^8*ln(x)*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)*(c*(a^2*x^2-1)/a^2/x^2)^(9/2)*x/(a*x+1)^3/(a^2*x^2 -1)^3/((a*x-1)/(a*x+1))^(3/2)
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.30 \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\frac {{\left (280 \, a^{9} c^{4} x^{9} + 840 \, a^{8} c^{4} x^{8} \log \left (x\right ) + 1120 \, a^{6} c^{4} x^{6} + 560 \, a^{5} c^{4} x^{5} - 420 \, a^{4} c^{4} x^{4} - 448 \, a^{3} c^{4} x^{3} + 120 \, a c^{4} x + 35 \, c^{4}\right )} \sqrt {a^{2} c}}{280 \, a^{10} x^{8}} \]
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. \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\text {Timed out} \]
\[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\int { \frac {{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {9}{2}}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]
\[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\int { \frac {{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {9}{2}}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int e^{3 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{9/2} \, dx=\int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^{9/2}}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]