Integrand size = 24, antiderivative size = 301 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=-\frac {\left (c-\frac {c}{a^2 x^2}\right )^{7/2} x}{6 \left (1-a^2 x^2\right )^{7/2}}+\frac {3 a \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^2}{5 \left (1-a^2 x^2\right )^{7/2}}-\frac {a^2 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^3}{4 \left (1-a^2 x^2\right )^{7/2}}-\frac {5 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^4}{3 \left (1-a^2 x^2\right )^{7/2}}+\frac {5 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^5}{2 \left (1-a^2 x^2\right )^{7/2}}+\frac {a^5 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^6}{\left (1-a^2 x^2\right )^{7/2}}-\frac {a^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^8}{\left (1-a^2 x^2\right )^{7/2}}+\frac {3 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 \log (x)}{\left (1-a^2 x^2\right )^{7/2}} \] Output:
-1/6*(c-c/a^2/x^2)^(7/2)*x/(-a^2*x^2+1)^(7/2)+3/5*a*(c-c/a^2/x^2)^(7/2)*x^ 2/(-a^2*x^2+1)^(7/2)-1/4*a^2*(c-c/a^2/x^2)^(7/2)*x^3/(-a^2*x^2+1)^(7/2)-5/ 3*a^3*(c-c/a^2/x^2)^(7/2)*x^4/(-a^2*x^2+1)^(7/2)+5/2*a^4*(c-c/a^2/x^2)^(7/ 2)*x^5/(-a^2*x^2+1)^(7/2)+a^5*(c-c/a^2/x^2)^(7/2)*x^6/(-a^2*x^2+1)^(7/2)-a ^7*(c-c/a^2/x^2)^(7/2)*x^8/(-a^2*x^2+1)^(7/2)+3*a^6*(c-c/a^2/x^2)^(7/2)*x^ 7*ln(x)/(-a^2*x^2+1)^(7/2)
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.33 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {c^3 \sqrt {c-\frac {c}{a^2 x^2}} \left (10-36 a x+15 a^2 x^2+100 a^3 x^3-150 a^4 x^4-60 a^5 x^5+60 a^7 x^7-180 a^6 x^6 \log (x)\right )}{60 a^6 x^5 \sqrt {1-a^2 x^2}} \] Input:
Integrate[(c - c/(a^2*x^2))^(7/2)/E^(3*ArcTanh[a*x]),x]
Output:
(c^3*Sqrt[c - c/(a^2*x^2)]*(10 - 36*a*x + 15*a^2*x^2 + 100*a^3*x^3 - 150*a ^4*x^4 - 60*a^5*x^5 + 60*a^7*x^7 - 180*a^6*x^6*Log[x]))/(60*a^6*x^5*Sqrt[1 - a^2*x^2])
Time = 0.75 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6710, 6700, 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.
\(\displaystyle \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx\) |
\(\Big \downarrow \) 6710 |
\(\displaystyle \frac {x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \int \frac {e^{-3 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^{7/2}}{x^7}dx}{\left (1-a^2 x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \int \frac {(1-a x)^5 (a x+1)^2}{x^7}dx}{\left (1-a^2 x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \int \left (-a^7+\frac {3 a^6}{x}-\frac {a^5}{x^2}-\frac {5 a^4}{x^3}+\frac {5 a^3}{x^4}+\frac {a^2}{x^5}-\frac {3 a}{x^6}+\frac {1}{x^7}\right )dx}{\left (1-a^2 x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \left (a^7 (-x)+3 a^6 \log (x)+\frac {a^5}{x}+\frac {5 a^4}{2 x^2}-\frac {5 a^3}{3 x^3}-\frac {a^2}{4 x^4}+\frac {3 a}{5 x^5}-\frac {1}{6 x^6}\right )}{\left (1-a^2 x^2\right )^{7/2}}\) |
Input:
Int[(c - c/(a^2*x^2))^(7/2)/E^(3*ArcTanh[a*x]),x]
Output:
((c - c/(a^2*x^2))^(7/2)*x^7*(-1/6*1/x^6 + (3*a)/(5*x^5) - a^2/(4*x^4) - ( 5*a^3)/(3*x^3) + (5*a^4)/(2*x^2) + a^5/x - a^7*x + 3*a^6*Log[x]))/(1 - a^2 *x^2)^(7/2)
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^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[x^(2*p)*((c + d/x^2)^p/(1 - a^2*x^2)^p) Int[(u/x^(2*p))*(1 - a ^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !IntegerQ[p] && !IntegerQ[n/2]
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.34
method | result | size |
default | \(-\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {7}{2}} x \left (-60 a^{7} x^{7}+180 a^{6} \ln \left (x \right ) x^{6}+60 a^{5} x^{5}+150 a^{4} x^{4}-100 a^{3} x^{3}-15 a^{2} x^{2}+36 a x -10\right )}{60 \left (a^{2} x^{2}-1\right )^{3} \sqrt {-a^{2} x^{2}+1}}\) | \(102\) |
Input:
int((c-c/a^2/x^2)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x,method=_RETURNVERBO SE)
Output:
-1/60*(c*(a^2*x^2-1)/a^2/x^2)^(7/2)*x/(a^2*x^2-1)^3/(-a^2*x^2+1)^(1/2)*(-6 0*a^7*x^7+180*a^6*ln(x)*x^6+60*a^5*x^5+150*a^4*x^4-100*a^3*x^3-15*a^2*x^2+ 36*a*x-10)
Time = 0.16 (sec) , antiderivative size = 546, normalized size of antiderivative = 1.81 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\left [\frac {90 \, {\left (a^{7} c^{3} x^{7} - a^{5} c^{3} x^{5}\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} - {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) - {\left (60 \, a^{7} c^{3} x^{7} - 60 \, a^{5} c^{3} x^{5} - 150 \, a^{4} c^{3} x^{4} - {\left (60 \, a^{7} - 60 \, a^{5} - 150 \, a^{4} + 100 \, a^{3} + 15 \, a^{2} - 36 \, a + 10\right )} c^{3} x^{6} + 100 \, a^{3} c^{3} x^{3} + 15 \, a^{2} c^{3} x^{2} - 36 \, a c^{3} x + 10 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{60 \, {\left (a^{8} x^{7} - a^{6} x^{5}\right )}}, \frac {180 \, {\left (a^{7} c^{3} x^{7} - a^{5} c^{3} x^{5}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} - a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} + {\left (a^{2} - 1\right )} c x^{2} - c}\right ) - {\left (60 \, a^{7} c^{3} x^{7} - 60 \, a^{5} c^{3} x^{5} - 150 \, a^{4} c^{3} x^{4} - {\left (60 \, a^{7} - 60 \, a^{5} - 150 \, a^{4} + 100 \, a^{3} + 15 \, a^{2} - 36 \, a + 10\right )} c^{3} x^{6} + 100 \, a^{3} c^{3} x^{3} + 15 \, a^{2} c^{3} x^{2} - 36 \, a c^{3} x + 10 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{60 \, {\left (a^{8} x^{7} - a^{6} x^{5}\right )}}\right ] \] Input:
integrate((c-c/a^2/x^2)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="f ricas")
Output:
[1/60*(90*(a^7*c^3*x^7 - a^5*c^3*x^5)*sqrt(-c)*log((a^2*c*x^6 + a^2*c*x^2 - c*x^4 - (a*x^5 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/( a^2*x^2)) - c)/(a^2*x^4 - x^2)) - (60*a^7*c^3*x^7 - 60*a^5*c^3*x^5 - 150*a ^4*c^3*x^4 - (60*a^7 - 60*a^5 - 150*a^4 + 100*a^3 + 15*a^2 - 36*a + 10)*c^ 3*x^6 + 100*a^3*c^3*x^3 + 15*a^2*c^3*x^2 - 36*a*c^3*x + 10*c^3)*sqrt(-a^2* x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^8*x^7 - a^6*x^5), 1/60*(180*( a^7*c^3*x^7 - a^5*c^3*x^5)*sqrt(c)*arctan(sqrt(-a^2*x^2 + 1)*(a*x^3 - a*x) *sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^4 + (a^2 - 1)*c*x^2 - c) ) - (60*a^7*c^3*x^7 - 60*a^5*c^3*x^5 - 150*a^4*c^3*x^4 - (60*a^7 - 60*a^5 - 150*a^4 + 100*a^3 + 15*a^2 - 36*a + 10)*c^3*x^6 + 100*a^3*c^3*x^3 + 15*a ^2*c^3*x^2 - 36*a*c^3*x + 10*c^3)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/ (a^2*x^2)))/(a^8*x^7 - a^6*x^5)]
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\text {Timed out} \] Input:
integrate((c-c/a**2/x**2)**(7/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)
Output:
Timed out
\[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {7}{2}}}{{\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate((c-c/a^2/x^2)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="m axima")
Output:
integrate((-a^2*x^2 + 1)^(3/2)*(c - c/(a^2*x^2))^(7/2)/(a*x + 1)^3, x)
Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.37 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {1}{60} \, {\left (\frac {60 \, c^{3} x \mathrm {sgn}\left (x\right )}{a} - \frac {180 \, c^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{2}} - \frac {60 \, a^{5} c^{3} x^{5} \mathrm {sgn}\left (x\right ) + 150 \, a^{4} c^{3} x^{4} \mathrm {sgn}\left (x\right ) - 100 \, a^{3} c^{3} x^{3} \mathrm {sgn}\left (x\right ) - 15 \, a^{2} c^{3} x^{2} \mathrm {sgn}\left (x\right ) + 36 \, a c^{3} x \mathrm {sgn}\left (x\right ) - 10 \, c^{3} \mathrm {sgn}\left (x\right )}{a^{8} x^{6}}\right )} \sqrt {-c} {\left | a \right |} \] Input:
integrate((c-c/a^2/x^2)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="g iac")
Output:
1/60*(60*c^3*x*sgn(x)/a - 180*c^3*log(abs(x))*sgn(x)/a^2 - (60*a^5*c^3*x^5 *sgn(x) + 150*a^4*c^3*x^4*sgn(x) - 100*a^3*c^3*x^3*sgn(x) - 15*a^2*c^3*x^2 *sgn(x) + 36*a*c^3*x*sgn(x) - 10*c^3*sgn(x))/(a^8*x^6))*sqrt(-c)*abs(a)
Timed out. \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^{7/2}\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \] Input:
int(((c - c/(a^2*x^2))^(7/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3,x)
Output:
int(((c - c/(a^2*x^2))^(7/2)*(1 - a^2*x^2)^(3/2))/(a*x + 1)^3, x)
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.23 \[ \int e^{-3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{7/2} \, dx=\frac {\sqrt {c}\, c^{3} i \left (180 \,\mathrm {log}\left (x \right ) a^{6} x^{6}-60 a^{7} x^{7}+60 a^{5} x^{5}+150 a^{4} x^{4}-100 a^{3} x^{3}-15 a^{2} x^{2}+36 a x -10\right )}{60 a^{7} x^{6}} \] Input:
int((c-c/a^2/x^2)^(7/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)
Output:
(sqrt(c)*c**3*i*(180*log(x)*a**6*x**6 - 60*a**7*x**7 + 60*a**5*x**5 + 150* a**4*x**4 - 100*a**3*x**3 - 15*a**2*x**2 + 36*a*x - 10))/(60*a**7*x**6)