Integrand size = 27, antiderivative size = 78 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=-\frac {\sqrt {c-a^2 c x^2}}{2 x^2}+\frac {2 a \sqrt {c-a^2 c x^2}}{x}-\frac {3}{2} a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \] Output:
-1/2*(-a^2*c*x^2+c)^(1/2)/x^2+2*a*(-a^2*c*x^2+c)^(1/2)/x-3/2*a^2*c^(1/2)*a rctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))
Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\frac {1}{2} \left (\frac {(-1+4 a x) \sqrt {c-a^2 c x^2}}{x^2}+3 a^2 \sqrt {c} \log (x)-3 a^2 \sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right )\right ) \] Input:
Integrate[Sqrt[c - a^2*c*x^2]/(E^(2*ArcTanh[a*x])*x^3),x]
Output:
(((-1 + 4*a*x)*Sqrt[c - a^2*c*x^2])/x^2 + 3*a^2*Sqrt[c]*Log[x] - 3*a^2*Sqr t[c]*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2]])/2
Time = 0.62 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6702, 540, 27, 534, 243, 73, 221}
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.
\(\displaystyle \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx\) |
\(\Big \downarrow \) 6702 |
\(\displaystyle c \int \frac {(1-a x)^2}{x^3 \sqrt {c-a^2 c x^2}}dx\) |
\(\Big \downarrow \) 540 |
\(\displaystyle c \left (-\frac {\int \frac {a c (4-3 a x)}{x^2 \sqrt {c-a^2 c x^2}}dx}{2 c}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (-\frac {1}{2} a \int \frac {4-3 a x}{x^2 \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 534 |
\(\displaystyle c \left (-\frac {1}{2} a \left (-3 a \int \frac {1}{x \sqrt {c-a^2 c x^2}}dx-\frac {4 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle c \left (-\frac {1}{2} a \left (-\frac {3}{2} a \int \frac {1}{x^2 \sqrt {c-a^2 c x^2}}dx^2-\frac {4 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle c \left (-\frac {1}{2} a \left (\frac {3 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2 c}}d\sqrt {c-a^2 c x^2}}{a c}-\frac {4 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle c \left (-\frac {1}{2} a \left (\frac {3 a \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {4 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
Input:
Int[Sqrt[c - a^2*c*x^2]/(E^(2*ArcTanh[a*x])*x^3),x]
Output:
c*(-1/2*Sqrt[c - a^2*c*x^2]/(c*x^2) - (a*((-4*Sqrt[c - a^2*c*x^2])/(c*x) + (3*a*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]])/Sqrt[c]))/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) , x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[1/c^(n/2) Int[x^m*((c + d*x^2)^(p + n/2)/(1 - a*x)^n), x] , x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && !(IntegerQ[p] || GtQ[c, 0]) && ILtQ[n/2, 0]
Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01
method | result | size |
risch | \(-\frac {\left (4 a^{3} x^{3}-a^{2} x^{2}-4 a x +1\right ) c}{2 x^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {3 a^{2} \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{2}\) | \(79\) |
default | \(-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}+\frac {3 a^{2} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )}{2}-2 a \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{c x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )\right )-2 a^{2} \left (\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}+\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )\) | \(230\) |
Input:
int((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^3,x,method=_RETURNVERBOS E)
Output:
-1/2*(4*a^3*x^3-a^2*x^2-4*a*x+1)/x^2/(-c*(a^2*x^2-1))^(1/2)*c-3/2*a^2*c^(1 /2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)
Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.76 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\left [\frac {3 \, a^{2} \sqrt {c} x^{2} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (4 \, a x - 1\right )}}{4 \, x^{2}}, \frac {3 \, a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) + \sqrt {-a^{2} c x^{2} + c} {\left (4 \, a x - 1\right )}}{2 \, x^{2}}\right ] \] Input:
integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^3,x, algorithm="fr icas")
Output:
[1/4*(3*a^2*sqrt(c)*x^2*log(-(a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) + 2*sqrt(-a^2*c*x^2 + c)*(4*a*x - 1))/x^2, 1/2*(3*a^2*sqrt(-c)* x^2*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/c) + sqrt(-a^2*c*x^2 + c)*(4*a*x - 1))/x^2]
\[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=- \int \left (- \frac {\sqrt {- a^{2} c x^{2} + c}}{a x^{4} + x^{3}}\right )\, dx - \int \frac {a x \sqrt {- a^{2} c x^{2} + c}}{a x^{4} + x^{3}}\, dx \] Input:
integrate((-a**2*c*x**2+c)**(1/2)/(a*x+1)**2*(-a**2*x**2+1)/x**3,x)
Output:
-Integral(-sqrt(-a**2*c*x**2 + c)/(a*x**4 + x**3), x) - Integral(a*x*sqrt( -a**2*c*x**2 + c)/(a*x**4 + x**3), x)
\[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\int { -\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 1\right )}}{{\left (a x + 1\right )}^{2} x^{3}} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^3,x, algorithm="ma xima")
Output:
-integrate(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 - 1)/((a*x + 1)^2*x^3), x)
Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (64) = 128\).
Time = 0.17 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.95 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\frac {1}{4} \, {\left (\frac {12 \, a c \arctan \left (\frac {\sqrt {-c + \frac {2 \, c}{a x + 1}}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{\sqrt {-c}} - \frac {{\left (3 \, \pi a c - 8 \, a c\right )} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{\sqrt {-c}} + \frac {3 \, a c^{2} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right ) - 5 \, a c {\left (-c + \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{{\left (c - \frac {c}{a x + 1}\right )}^{2}}\right )} {\left | a \right |} \] Input:
integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^3,x, algorithm="gi ac")
Output:
1/4*(12*a*c*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(-c))*sgn(1/(a*x + 1))*sgn (a)/sqrt(-c) - (3*pi*a*c - 8*a*c)*sgn(1/(a*x + 1))*sgn(a)/sqrt(-c) + (3*a* c^2*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) - 5*a*c*(-c + 2*c/(a* x + 1))^(3/2)*sgn(1/(a*x + 1))*sgn(a))/(c - c/(a*x + 1))^2)*abs(a)
Timed out. \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=-\int \frac {\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )}{x^3\,{\left (a\,x+1\right )}^2} \,d x \] Input:
int(-((c - a^2*c*x^2)^(1/2)*(a^2*x^2 - 1))/(x^3*(a*x + 1)^2),x)
Output:
-int(((c - a^2*c*x^2)^(1/2)*(a^2*x^2 - 1))/(x^3*(a*x + 1)^2), x)
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\frac {\sqrt {c}\, \left (4 \sqrt {-a^{2} x^{2}+1}\, a x -\sqrt {-a^{2} x^{2}+1}+3 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{2} x^{2}\right )}{2 x^{2}} \] Input:
int((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^3,x)
Output:
(sqrt(c)*(4*sqrt( - a**2*x**2 + 1)*a*x - sqrt( - a**2*x**2 + 1) + 3*log(ta n(asin(a*x)/2))*a**2*x**2))/(2*x**2)