Integrand size = 22, antiderivative size = 277 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{24 (1-a x)^3 \left (c-a^2 c x^2\right )^{7/2}}+\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1-a x)^2 \left (c-a^2 c x^2\right )^{7/2}}+\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{16 (1-a x) \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{32 (1+a x)^2 \left (c-a^2 c x^2\right )^{7/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}{8 (1+a x) \left (c-a^2 c x^2\right )^{7/2}}+\frac {5 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7 \text {arctanh}(a x)}{16 \left (c-a^2 c x^2\right )^{7/2}} \]
1/24*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)^3/(-a^2*c*x^2+c)^(7/2)+3/32*a^6* (1-1/a^2/x^2)^(7/2)*x^7/(-a*x+1)^2/(-a^2*c*x^2+c)^(7/2)+3/16*a^6*(1-1/a^2/ x^2)^(7/2)*x^7/(-a*x+1)/(-a^2*c*x^2+c)^(7/2)-1/32*a^6*(1-1/a^2/x^2)^(7/2)* x^7/(a*x+1)^2/(-a^2*c*x^2+c)^(7/2)-1/8*a^6*(1-1/a^2/x^2)^(7/2)*x^7/(a*x+1) /(-a^2*c*x^2+c)^(7/2)+5/16*a^6*(1-1/a^2/x^2)^(7/2)*x^7*arctanh(a*x)/(-a^2* c*x^2+c)^(7/2)
Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.36 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-8-25 a x+25 a^2 x^2+15 a^3 x^3-15 a^4 x^4+15 (-1+a x)^3 (1+a x)^2 \text {arctanh}(a x)\right )}{48 c^3 (-1+a x)^3 (1+a x)^2 \sqrt {c-a^2 c x^2}} \]
-1/48*(Sqrt[1 - 1/(a^2*x^2)]*x*(-8 - 25*a*x + 25*a^2*x^2 + 15*a^3*x^3 - 15 *a^4*x^4 + 15*(-1 + a*x)^3*(1 + a*x)^2*ArcTanh[a*x]))/(c^3*(-1 + a*x)^3*(1 + a*x)^2*Sqrt[c - a^2*c*x^2])
Time = 0.48 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.44, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6746, 6747, 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.
\(\displaystyle \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2} x^7}dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \frac {1}{(1-a x)^4 (a x+1)^3}dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \int \left (\frac {3}{16 (a x-1)^2}+\frac {1}{8 (a x+1)^2}-\frac {3}{16 (a x-1)^3}+\frac {1}{16 (a x+1)^3}+\frac {1}{8 (a x-1)^4}-\frac {5}{16 \left (a^2 x^2-1\right )}\right )dx}{\left (c-a^2 c x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^7 x^7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2} \left (\frac {5 \text {arctanh}(a x)}{16 a}+\frac {3}{16 a (1-a x)}-\frac {1}{8 a (a x+1)}+\frac {3}{32 a (1-a x)^2}-\frac {1}{32 a (a x+1)^2}+\frac {1}{24 a (1-a x)^3}\right )}{\left (c-a^2 c x^2\right )^{7/2}}\) |
(a^7*(1 - 1/(a^2*x^2))^(7/2)*x^7*(1/(24*a*(1 - a*x)^3) + 3/(32*a*(1 - a*x) ^2) + 3/(16*a*(1 - a*x)) - 1/(32*a*(1 + a*x)^2) - 1/(8*a*(1 + a*x)) + (5*A rcTanh[a*x])/(16*a)))/(c - a^2*c*x^2)^(7/2)
3.7.22.3.1 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 = 0.54 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (15 \ln \left (a x +1\right ) x^{5} a^{5}-15 \ln \left (a x -1\right ) x^{5} a^{5}-15 \ln \left (a x +1\right ) x^{4} a^{4}+15 \ln \left (a x -1\right ) x^{4} a^{4}-30 a^{4} x^{4}-30 a^{3} \ln \left (a x +1\right ) x^{3}+30 a^{3} \ln \left (a x -1\right ) x^{3}+30 a^{3} x^{3}+30 a^{2} \ln \left (a x +1\right ) x^{2}-30 a^{2} \ln \left (a x -1\right ) x^{2}+50 a^{2} x^{2}+15 a \ln \left (a x +1\right ) x -15 a \ln \left (a x -1\right ) x -50 a x -15 \ln \left (a x +1\right )+15 \ln \left (a x -1\right )-16\right )}{96 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{4} a \left (a x +1\right )^{2}}\) | \(241\) |
1/96/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)^2*(-c*(a^2*x^2-1))^(1/2)*(15*ln(a*x+1 )*x^5*a^5-15*ln(a*x-1)*x^5*a^5-15*ln(a*x+1)*x^4*a^4+15*ln(a*x-1)*x^4*a^4-3 0*a^4*x^4-30*a^3*ln(a*x+1)*x^3+30*a^3*ln(a*x-1)*x^3+30*a^3*x^3+30*a^2*ln(a *x+1)*x^2-30*a^2*ln(a*x-1)*x^2+50*a^2*x^2+15*a*ln(a*x+1)*x-15*a*ln(a*x-1)* x-50*a*x-15*ln(a*x+1)+15*ln(a*x-1)-16)/(a^2*x^2-1)/c^4/a/(a*x+1)^2
Time = 0.27 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {15 \, {\left (a^{6} x^{5} - a^{5} x^{4} - 2 \, a^{4} x^{3} + 2 \, a^{3} x^{2} + a^{2} x - a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) + 2 \, {\left (15 \, a^{4} x^{4} - 15 \, a^{3} x^{3} - 25 \, a^{2} x^{2} + 25 \, a x + 8\right )} \sqrt {-a^{2} c}}{96 \, {\left (a^{7} c^{4} x^{5} - a^{6} c^{4} x^{4} - 2 \, a^{5} c^{4} x^{3} + 2 \, a^{4} c^{4} x^{2} + a^{3} c^{4} x - a^{2} c^{4}\right )}} \]
-1/96*(15*(a^6*x^5 - a^5*x^4 - 2*a^4*x^3 + 2*a^3*x^2 + a^2*x - a)*sqrt(-c) *log((a^2*c*x^2 + 2*sqrt(-a^2*c)*sqrt(-c)*x + c)/(a^2*x^2 - 1)) + 2*(15*a^ 4*x^4 - 15*a^3*x^3 - 25*a^2*x^2 + 25*a*x + 8)*sqrt(-a^2*c))/(a^7*c^4*x^5 - a^6*c^4*x^4 - 2*a^5*c^4*x^3 + 2*a^4*c^4*x^2 + a^3*c^4*x - a^2*c^4)
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]
\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int \frac {1}{{\left (c-a^2\,c\,x^2\right )}^{7/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]