Integrand size = 22, antiderivative size = 146 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=-\frac {\left (c-\frac {c}{a^2 x^2}\right )^{3/2} x}{2 \left (1-a^2 x^2\right )^{3/2}}-\frac {a \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^2}{\left (1-a^2 x^2\right )^{3/2}}-\frac {a^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^4}{\left (1-a^2 x^2\right )^{3/2}}-\frac {a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3 \log (x)}{\left (1-a^2 x^2\right )^{3/2}} \] Output:
-1/2*(c-c/a^2/x^2)^(3/2)*x/(-a^2*x^2+1)^(3/2)-a*(c-c/a^2/x^2)^(3/2)*x^2/(- a^2*x^2+1)^(3/2)-a^3*(c-c/a^2/x^2)^(3/2)*x^4/(-a^2*x^2+1)^(3/2)-a^2*(c-c/a ^2/x^2)^(3/2)*x^3*ln(x)/(-a^2*x^2+1)^(3/2)
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.49 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {c \sqrt {c-\frac {c}{a^2 x^2}} \left (1+2 a x+3 a^2 x^2+2 a^3 x^3+2 a^2 x^2 \log (x)\right )}{2 a^2 x \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^ArcTanh[a*x]*(c - c/(a^2*x^2))^(3/2),x]
Output:
(c*Sqrt[c - c/(a^2*x^2)]*(1 + 2*a*x + 3*a^2*x^2 + 2*a^3*x^3 + 2*a^2*x^2*Lo g[x]))/(2*a^2*x*Sqrt[1 - a^2*x^2])
Time = 0.44 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.41, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6710, 6700, 84, 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.
\(\displaystyle \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 6710 |
\(\displaystyle \frac {x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \int \frac {e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )^{3/2}}{x^3}dx}{\left (1-a^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \int \frac {(1-a x) (a x+1)^2}{x^3}dx}{\left (1-a^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle \frac {x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \int \left (-a^3-\frac {a^2}{x}+\frac {a}{x^2}+\frac {1}{x^3}\right )dx}{\left (1-a^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \left (a^3 (-x)-a^2 \log (x)-\frac {a}{x}-\frac {1}{2 x^2}\right )}{\left (1-a^2 x^2\right )^{3/2}}\) |
Input:
Int[E^ArcTanh[a*x]*(c - c/(a^2*x^2))^(3/2),x]
Output:
((c - c/(a^2*x^2))^(3/2)*x^3*(-1/2*1/x^2 - a/x - a^3*x - a^2*Log[x]))/(1 - a^2*x^2)^(3/2)
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
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 = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {3}{2}} x \left (2 a^{3} x^{3}+2 a^{2} \ln \left (x \right ) x^{2}+2 a x +1\right )}{2 \left (a^{2} x^{2}-1\right ) \sqrt {-a^{2} x^{2}+1}}\) | \(70\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a^2/x^2)^(3/2),x,method=_RETURNVERBOSE )
Output:
1/2*(c*(a^2*x^2-1)/a^2/x^2)^(3/2)*x/(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)*(2*a^3* x^3+2*a^2*ln(x)*x^2+2*a*x+1)
Time = 0.12 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.57 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\left [\frac {{\left (a^{3} c x^{3} - a c x\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} + {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) - {\left (2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + 2 \, a + 1\right )} c x^{2} + 2 \, a c x + c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{3} - a^{2} x\right )}}, -\frac {2 \, {\left (a^{3} c x^{3} - a c x\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} - a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} + {\left (a^{2} - 1\right )} c x^{2} - c}\right ) + {\left (2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + 2 \, a + 1\right )} c x^{2} + 2 \, a c x + c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{3} - a^{2} x\right )}}\right ] \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a^2/x^2)^(3/2),x, algorithm="fri cas")
Output:
[1/2*((a^3*c*x^3 - a*c*x)*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)) - (2*a^3*c*x^3 - (2*a^3 + 2*a + 1)*c*x^2 + 2*a*c*x + c )*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^4*x^3 - a^2*x), - 1/2*(2*(a^3*c*x^3 - a*c*x)*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) ) + (2*a^3*c*x^3 - (2*a^3 + 2*a + 1)*c*x^2 + 2*a*c*x + c)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^4*x^3 - a^2*x)]
\[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\int \frac {\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {3}{2}} \left (a x + 1\right )}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(c-c/a**2/x**2)**(3/2),x)
Output:
Integral((-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))**(3/2)*(a*x + 1)/sqrt(-(a*x - 1 )*(a*x + 1)), x)
\[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\int { \frac {{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {3}{2}}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a^2/x^2)^(3/2),x, algorithm="max ima")
Output:
integrate((a*x + 1)*(c - c/(a^2*x^2))^(3/2)/sqrt(-a^2*x^2 + 1), x)
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.34 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {1}{2} \, {\left (\frac {2 \, c x \mathrm {sgn}\left (x\right )}{a} + \frac {2 \, c \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{2}} + \frac {2 \, a c x \mathrm {sgn}\left (x\right ) + c \mathrm {sgn}\left (x\right )}{a^{4} x^{2}}\right )} \sqrt {-c} {\left | a \right |} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a^2/x^2)^(3/2),x, algorithm="gia c")
Output:
1/2*(2*c*x*sgn(x)/a + 2*c*log(abs(x))*sgn(x)/a^2 + (2*a*c*x*sgn(x) + c*sgn (x))/(a^4*x^2))*sqrt(-c)*abs(a)
Timed out. \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^{3/2}\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \] Input:
int(((c - c/(a^2*x^2))^(3/2)*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
int(((c - c/(a^2*x^2))^(3/2)*(a*x + 1))/(1 - a^2*x^2)^(1/2), x)
Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.25 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {\sqrt {c}\, c i \left (2 \,\mathrm {log}\left (x \right ) a^{2} x^{2}+2 a^{3} x^{3}+2 a x +1\right )}{2 a^{3} x^{2}} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a^2/x^2)^(3/2),x)
Output:
(sqrt(c)*c*i*(2*log(x)*a**2*x**2 + 2*a**3*x**3 + 2*a*x + 1))/(2*a**3*x**2)