Integrand size = 24, antiderivative size = 182 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{6 (1+a x)^3 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \text {arctanh}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]
1/6*a^4*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)^3/(-a^2*c*x^2+c)^(5/2)+1/8*a^4*(1- 1/a^2/x^2)^(5/2)*x^5/(a*x+1)^2/(-a^2*c*x^2+c)^(5/2)+1/8*a^4*(1-1/a^2/x^2)^ (5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)-1/8*a^4*(1-1/a^2/x^2)^(5/2)*x^5*arc tanh(a*x)/(-a^2*c*x^2+c)^(5/2)
Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.39 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-10-9 a x-3 a^2 x^2+3 (1+a x)^3 \text {arctanh}(a x)\right )}{24 c^2 (1+a x)^3 \sqrt {c-a^2 c x^2}} \]
-1/24*(Sqrt[1 - 1/(a^2*x^2)]*x*(-10 - 9*a*x - 3*a^2*x^2 + 3*(1 + a*x)^3*Ar cTanh[a*x]))/(c^2*(1 + a*x)^3*Sqrt[c - a^2*c*x^2])
Time = 0.46 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.50, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6746, 6747, 25, 54, 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.
\(\displaystyle \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int -\frac {1}{(1-a x) (a x+1)^4}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {1}{(1-a x) (a x+1)^4}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \left (\frac {1}{8 (a x+1)^2}+\frac {1}{4 (a x+1)^3}+\frac {1}{2 (a x+1)^4}-\frac {1}{8 \left (a^2 x^2-1\right )}\right )dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (\frac {\text {arctanh}(a x)}{8 a}-\frac {1}{8 a (a x+1)}-\frac {1}{8 a (a x+1)^2}-\frac {1}{6 a (a x+1)^3}\right )}{\left (c-a^2 c x^2\right )^{5/2}}\) |
-((a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(-1/6*1/(a*(1 + a*x)^3) - 1/(8*a*(1 + a *x)^2) - 1/(8*a*(1 + a*x)) + ArcTanh[a*x]/(8*a)))/(c - a^2*c*x^2)^(5/2))
3.7.66.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p) Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && !IntegerQ[p]
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]
Time = 0.52 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 a^{3} \ln \left (a x +1\right ) x^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}+9 a^{2} \ln \left (a x +1\right ) x^{2}-9 a^{2} \ln \left (a x -1\right ) x^{2}-6 a^{2} x^{2}+9 a \ln \left (a x +1\right ) x -9 a \ln \left (a x -1\right ) x -18 a x +3 \ln \left (a x +1\right )-3 \ln \left (a x -1\right )-20\right )}{48 \left (a x +1\right ) \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a}\) | \(169\) |
1/48*((a*x-1)/(a*x+1))^(3/2)/(a*x+1)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(3*a^3 *ln(a*x+1)*x^3-3*a^3*ln(a*x-1)*x^3+9*a^2*ln(a*x+1)*x^2-9*a^2*ln(a*x-1)*x^2 -6*a^2*x^2+9*a*ln(a*x+1)*x-9*a*ln(a*x-1)*x-18*a*x+3*ln(a*x+1)-3*ln(a*x-1)- 20)/(a^2*x^2-1)/c^3/a
Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {3 \, {\left (a^{4} x^{3} + 3 \, a^{3} x^{2} + 3 \, a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) + 2 \, {\left (3 \, a^{2} x^{2} + 9 \, a x + 10\right )} \sqrt {-a^{2} c}}{48 \, {\left (a^{5} c^{3} x^{3} + 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x + a^{2} c^{3}\right )}} \]
-1/48*(3*(a^4*x^3 + 3*a^3*x^2 + 3*a^2*x + a)*sqrt(-c)*log((a^2*c*x^2 + 2*s qrt(-a^2*c)*sqrt(-c)*x + c)/(a^2*x^2 - 1)) + 2*(3*a^2*x^2 + 9*a*x + 10)*sq rt(-a^2*c))/(a^5*c^3*x^3 + 3*a^4*c^3*x^2 + 3*a^3*c^3*x + a^2*c^3)
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{{\left (c-a^2\,c\,x^2\right )}^{5/2}} \,d x \]