Integrand size = 24, antiderivative size = 183 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1+a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{4 a c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \]
1/8*(-a^2*x^2+1)^(1/2)/a/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/8*(-a^2*x^2+1 )^(1/2)/a/c^2/(a*x+1)^2/(-a^2*c*x^2+c)^(1/2)-1/4*(-a^2*x^2+1)^(1/2)/a/c^2/ (a*x+1)/(-a^2*c*x^2+c)^(1/2)+3/8*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a/c^2/(-a ^2*c*x^2+c)^(1/2)
Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.45 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (2-3 a x-3 a^2 x^2+3 (-1+a x) (1+a x)^2 \text {arctanh}(a x)\right )}{8 a (-1+a x) (c+a c x)^2 \sqrt {c-a^2 c x^2}} \]
(Sqrt[1 - a^2*x^2]*(2 - 3*a*x - 3*a^2*x^2 + 3*(-1 + a*x)*(1 + a*x)^2*ArcTa nh[a*x]))/(8*a*(-1 + a*x)*(c + a*c*x)^2*Sqrt[c - a^2*c*x^2])
Time = 0.38 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6693, 6690, 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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{-\text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {1}{(1-a x)^2 (a x+1)^3}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{8 (a x-1)^2}+\frac {1}{4 (a x+1)^2}+\frac {1}{4 (a x+1)^3}-\frac {3}{8 \left (a^2 x^2-1\right )}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {3 \text {arctanh}(a x)}{8 a}+\frac {1}{8 a (1-a x)}-\frac {1}{4 a (a x+1)}-\frac {1}{8 a (a x+1)^2}\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*(1/(8*a*(1 - a*x)) - 1/(8*a*(1 + a*x)^2) - 1/(4*a*(1 + a*x)) + (3*ArcTanh[a*x])/(8*a)))/(c^2*Sqrt[c - a^2*c*x^2])
3.13.13.3.1 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^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p Int[(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a , c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPart[p]) Int [(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0])
Time = 0.32 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 a^{3} \ln \left (a x +1\right ) x^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}+3 a^{2} \ln \left (a x +1\right ) x^{2}-3 a^{2} \ln \left (a x -1\right ) x^{2}-6 a^{2} x^{2}-3 a \ln \left (a x +1\right ) x +3 a \ln \left (a x -1\right ) x -6 a x -3 \ln \left (a x +1\right )+3 \ln \left (a x -1\right )+4\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} a \left (a x +1\right )^{2} \left (a x -1\right )}\) | \(166\) |
-1/16*(-a^2*x^2+1)^(1/2)*(-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+3*a^2*ln(a*x+1)*x^2-3*a^2*ln(a*x-1)*x^2-6*a^2*x^2-3*a*ln(a* x+1)*x+3*a*ln(a*x-1)*x-6*a*x-3*ln(a*x+1)+3*ln(a*x-1)+4)/(a^2*x^2-1)/c^3/a/ (a*x+1)^2/(a*x-1)
Time = 0.28 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.46 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{5} x^{5} + a^{4} x^{4} - 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} + a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) - 4 \, {\left (2 \, a^{3} x^{3} - a^{2} x^{2} - 5 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{32 \, {\left (a^{6} c^{3} x^{5} + a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} + a^{2} c^{3} x + a c^{3}\right )}}, \frac {3 \, {\left (a^{5} x^{5} + a^{4} x^{4} - 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} + a x + 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) - 2 \, {\left (2 \, a^{3} x^{3} - a^{2} x^{2} - 5 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{16 \, {\left (a^{6} c^{3} x^{5} + a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} + a^{2} c^{3} x + a c^{3}\right )}}\right ] \]
[1/32*(3*(a^5*x^5 + a^4*x^4 - 2*a^3*x^3 - 2*a^2*x^2 + a*x + 1)*sqrt(c)*log (-(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a^2*c*x^2 - 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x ^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)) - 4*(2*a^3*x^3 - a^2*x^2 - 5*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 + a^5*c^3*x^4 - 2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 + a^2*c^3 *x + a*c^3), 1/16*(3*(a^5*x^5 + a^4*x^4 - 2*a^3*x^3 - 2*a^2*x^2 + a*x + 1) *sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*sqrt(-c)*x/(a ^4*c*x^4 - c)) - 2*(2*a^3*x^3 - a^2*x^2 - 5*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt (-a^2*x^2 + 1))/(a^6*c^3*x^5 + a^5*c^3*x^4 - 2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 + a^2*c^3*x + a*c^3)]
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.45 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {3 \, a^{2} x^{2} + 3 \, a x - 2}{8 \, {\left (a^{4} c^{\frac {5}{2}} x^{3} + a^{3} c^{\frac {5}{2}} x^{2} - a^{2} c^{\frac {5}{2}} x - a c^{\frac {5}{2}}\right )}} + \frac {3 \, \log \left (a x + 1\right )}{16 \, a c^{\frac {5}{2}}} - \frac {3 \, \log \left (a x - 1\right )}{16 \, a c^{\frac {5}{2}}} \]
-1/8*(3*a^2*x^2 + 3*a*x - 2)/(a^4*c^(5/2)*x^3 + a^3*c^(5/2)*x^2 - a^2*c^(5 /2)*x - a*c^(5/2)) + 3/16*log(a*x + 1)/(a*c^(5/2)) - 3/16*log(a*x - 1)/(a* c^(5/2))
\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\left (a\,x+1\right )} \,d x \]