Integrand size = 24, antiderivative size = 322 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=-\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{6 a^7 \sqrt {1-\frac {1}{a^2 x^2}} x^6}+\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{5 a^6 \sqrt {1-\frac {1}{a^2 x^2}} x^5}+\frac {3 c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}-\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}-\frac {3 c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {3 c^3 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]
-1/6*c^3*(c-c/a^2/x^2)^(1/2)/a^7/x^6/(1-1/a^2/x^2)^(1/2)+1/5*c^3*(c-c/a^2/ x^2)^(1/2)/a^6/x^5/(1-1/a^2/x^2)^(1/2)+3/4*c^3*(c-c/a^2/x^2)^(1/2)/a^5/x^4 /(1-1/a^2/x^2)^(1/2)-c^3*(c-c/a^2/x^2)^(1/2)/a^4/x^3/(1-1/a^2/x^2)^(1/2)-3 /2*c^3*(c-c/a^2/x^2)^(1/2)/a^3/x^2/(1-1/a^2/x^2)^(1/2)+3*c^3*(c-c/a^2/x^2) ^(1/2)/a^2/x/(1-1/a^2/x^2)^(1/2)+c^3*x*(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^( 1/2)-c^3*ln(x)*(c-c/a^2/x^2)^(1/2)/a/(1-1/a^2/x^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.30 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{7/2} \left (-\frac {1}{6 a^7 x^6}+\frac {1}{5 a^6 x^5}+\frac {3}{4 a^5 x^4}-\frac {1}{a^4 x^3}-\frac {3}{2 a^3 x^2}+\frac {3}{a^2 x}+x-\frac {\log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2}} \]
((c - c/(a^2*x^2))^(7/2)*(-1/6*1/(a^7*x^6) + 1/(5*a^6*x^5) + 3/(4*a^5*x^4) - 1/(a^4*x^3) - 3/(2*a^3*x^2) + 3/(a^2*x) + x - Log[x]/a))/(1 - 1/(a^2*x^ 2))^(7/2)
Time = 0.42 (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.
\(\displaystyle \int \left (c-\frac {c}{a^2 x^2}\right )^{7/2} e^{-\coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6751 |
\(\displaystyle \frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \int e^{-\coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{7/2}dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(1-a x)^4 (a x+1)^3}{x^7}dx}{a^7 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \int \left (a^7-\frac {a^6}{x}-\frac {3 a^5}{x^2}+\frac {3 a^4}{x^3}+\frac {3 a^3}{x^4}-\frac {3 a^2}{x^5}-\frac {a}{x^6}+\frac {1}{x^7}\right )dx}{a^7 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \left (a^7 x-a^6 \log (x)+\frac {3 a^5}{x}-\frac {3 a^4}{2 x^2}-\frac {a^3}{x^3}+\frac {3 a^2}{4 x^4}+\frac {a}{5 x^5}-\frac {1}{6 x^6}\right )}{a^7 \sqrt {1-\frac {1}{a^2 x^2}}}\) |
(c^3*Sqrt[c - c/(a^2*x^2)]*(-1/6*1/x^6 + a/(5*x^5) + (3*a^2)/(4*x^4) - a^3 /x^3 - (3*a^4)/(2*x^2) + (3*a^5)/x + a^7*x - a^6*Log[x]))/(a^7*Sqrt[1 - 1/ (a^2*x^2)])
3.9.55.3.1 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 = 0.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.35
method | result | size |
default | \(-\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {7}{2}} \sqrt {\frac {a x -1}{a x +1}}\, x \left (-60 a^{7} x^{7}+60 a^{6} \ln \left (x \right ) x^{6}-180 a^{5} x^{5}+90 a^{4} x^{4}+60 a^{3} x^{3}-45 a^{2} x^{2}-12 a x +10\right )}{60 \left (a x -1\right ) \left (a^{2} x^{2}-1\right )^{3}}\) | \(112\) |
-1/60*(c*(a^2*x^2-1)/a^2/x^2)^(7/2)*((a*x-1)/(a*x+1))^(1/2)*x*(-60*a^7*x^7 +60*a^6*ln(x)*x^6-180*a^5*x^5+90*a^4*x^4+60*a^3*x^3-45*a^2*x^2-12*a*x+10)/ (a*x-1)/(a^2*x^2-1)^3
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.30 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {{\left (60 \, a^{7} c^{3} x^{7} - 60 \, a^{6} c^{3} x^{6} \log \left (x\right ) + 180 \, a^{5} c^{3} x^{5} - 90 \, a^{4} c^{3} x^{4} - 60 \, a^{3} c^{3} x^{3} + 45 \, a^{2} c^{3} x^{2} + 12 \, a c^{3} x - 10 \, c^{3}\right )} \sqrt {a^{2} c}}{60 \, a^{8} x^{6}} \]
1/60*(60*a^7*c^3*x^7 - 60*a^6*c^3*x^6*log(x) + 180*a^5*c^3*x^5 - 90*a^4*c^ 3*x^4 - 60*a^3*c^3*x^3 + 45*a^2*c^3*x^2 + 12*a*c^3*x - 10*c^3)*sqrt(a^2*c) /(a^8*x^6)
Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\text {Timed out} \]
\[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}} \sqrt {\frac {a x - 1}{a x + 1}} \,d x } \]
\[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}} \sqrt {\frac {a x - 1}{a x + 1}} \,d x } \]
Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\int {\left (c-\frac {c}{a^2\,x^2}\right )}^{7/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}} \,d x \]