Integrand size = 24, antiderivative size = 185 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{6 a c^2 (1-a x)^3 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \]
1/6*(-a^2*x^2+1)^(1/2)/a/c^2/(-a*x+1)^3/(-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)+1/8*(-a^2*x^2+1)^(1/2)/a/c ^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+1/8*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a/c^2 /(-a^2*c*x^2+c)^(1/2)
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.39 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-10+9 a x-3 a^2 x^2+3 (-1+a x)^3 \text {arctanh}(a x)\right )}{24 a c^2 (-1+a x)^3 \sqrt {c-a^2 c x^2}} \]
(Sqrt[1 - a^2*x^2]*(-10 + 9*a*x - 3*a^2*x^2 + 3*(-1 + a*x)^3*ArcTanh[a*x]) )/(24*a*c^2*(-1 + a*x)^3*Sqrt[c - a^2*c*x^2])
Time = 0.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.49, 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.
\(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6693 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{3 \text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {1}{(1-a x)^4 (a x+1)}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{8 (a x-1)^2}-\frac {1}{4 (a x-1)^3}+\frac {1}{2 (a x-1)^4}-\frac {1}{8 \left (a^2 x^2-1\right )}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {\text {arctanh}(a x)}{8 a}+\frac {1}{8 a (1-a x)}+\frac {1}{8 a (1-a x)^2}+\frac {1}{6 a (1-a x)^3}\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*(1/(6*a*(1 - a*x)^3) + 1/(8*a*(1 - a*x)^2) + 1/(8*a*(1 - a*x)) + ArcTanh[a*x]/(8*a)))/(c^2*Sqrt[c - a^2*c*x^2])
3.12.71.3.1 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 = 0.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 a^{3} \ln \left (a x +1\right ) x^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}-9 a^{2} \ln \left (a x +1\right ) x^{2}+9 a^{2} \ln \left (a x -1\right ) x^{2}-6 a^{2} x^{2}+9 a \ln \left (a x +1\right ) x -9 a \ln \left (a x -1\right ) x +18 a x -3 \ln \left (a x +1\right )+3 \ln \left (a x -1\right )-20\right )}{48 \left (a^{2} x^{2}-1\right ) c^{3} a \left (a x -1\right )^{3}}\) | \(159\) |
-1/48*(-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^3-9*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/(a*x-1)^3
Time = 0.30 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.48 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{5} x^{5} - 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 3 \, a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) + 4 \, {\left (10 \, a^{3} x^{3} - 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{96 \, {\left (a^{6} c^{3} x^{5} - 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x + a c^{3}\right )}}, \frac {3 \, {\left (a^{5} x^{5} - 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 3 \, a x + 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) + 2 \, {\left (10 \, a^{3} x^{3} - 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{48 \, {\left (a^{6} c^{3} x^{5} - 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x + a c^{3}\right )}}\right ] \]
[1/96*(3*(a^5*x^5 - 3*a^4*x^4 + 2*a^3*x^3 + 2*a^2*x^2 - 3*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*(10*a^3*x^3 - 27*a^2*x^2 + 21*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt( -a^2*x^2 + 1))/(a^6*c^3*x^5 - 3*a^5*c^3*x^4 + 2*a^4*c^3*x^3 + 2*a^3*c^3*x^ 2 - 3*a^2*c^3*x + a*c^3), 1/48*(3*(a^5*x^5 - 3*a^4*x^4 + 2*a^3*x^3 + 2*a^2 *x^2 - 3*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*(10*a^3*x^3 - 27*a^2*x^2 + 21*a*x)*sqr t(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 - 3*a^5*c^3*x^4 + 2*a^4 *c^3*x^3 + 2*a^3*c^3*x^2 - 3*a^2*c^3*x + a*c^3)]
\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {{\left (a\,x+1\right )}^3}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]