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 = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=-\frac {(1+4 a x) \sqrt {c-a^2 c x^2}}{2 x^2}+\frac {3}{2} a^2 \sqrt {c} \log (x)-\frac {3}{2} a^2 \sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right ) \] Input:
Integrate[(E^(2*ArcTanh[a*x])*Sqrt[c - a^2*c*x^2])/x^3,x]
Output:
-1/2*((1 + 4*a*x)*Sqrt[c - a^2*c*x^2])/x^2 + (3*a^2*Sqrt[c]*Log[x])/2 - (3 *a^2*Sqrt[c]*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2]])/2
Time = 0.40 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6701, 540, 25, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx\) |
\(\Big \downarrow \) 6701 |
\(\displaystyle c \int \frac {(a x+1)^2}{x^3 \sqrt {c-a^2 c x^2}}dx\) |
\(\Big \downarrow \) 540 |
\(\displaystyle c \left (-\frac {\int -\frac {a c (3 a x+4)}{x^2 \sqrt {c-a^2 c x^2}}dx}{2 c}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle c \left (\frac {\int \frac {a c (3 a x+4)}{x^2 \sqrt {c-a^2 c x^2}}dx}{2 c}-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (\frac {1}{2} a \int \frac {3 a x+4}{x^2 \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{2 c x^2}\right )\) |
\(\Big \downarrow \) 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[(E^(2*ArcTanh[a*x])*Sqrt[c - a^2*c*x^2])/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[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]) && IGtQ[n/2, 0]
Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\left (4 a^{3} x^{3}+a^{2} x^{2}-4 a x -1\right ) c}{2 x^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {3 a^{2} \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{2}\) | \(78\) |
default | \(-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}+\frac {3 a^{2} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )}{2}+2 a \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{c x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )\right )-2 a^{2} \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}-\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )\) | \(239\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/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 = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.77 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\left [\frac {3 \, a^{2} \sqrt {c} x^{2} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (4 \, a x + 1\right )}}{4 \, x^{2}}, \frac {3 \, a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (4 \, a x + 1\right )}}{2 \, x^{2}}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/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]
\[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=- \int \frac {\sqrt {- a^{2} c x^{2} + c}}{a x^{4} - x^{3}}\, dx - \int \frac {a x \sqrt {- a^{2} c x^{2} + c}}{a x^{4} - x^{3}}\, dx \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(-a**2*c*x**2+c)**(1/2)/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)
\[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\int { -\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a x + 1\right )}^{2}}{{\left (a^{2} x^{2} - 1\right )} x^{3}} \,d x } \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/x^3,x, algorithm="ma xima")
Output:
-integrate(sqrt(-a^2*c*x^2 + c)*(a*x + 1)^2/((a^2*x^2 - 1)*x^3), x)
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (64) = 128\).
Time = 0.16 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.58 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=\frac {3 \, a^{2} c \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {{\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{3} a^{2} c - 4 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} a \sqrt {-c} c {\left | a \right |} + {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )} a^{2} c^{2} + 4 \, a \sqrt {-c} c^{2} {\left | a \right |}}{{\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/x^3,x, algorithm="gi ac")
Output:
3*a^2*c*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) - ((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^3*a^2*c - 4*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2*a*sqrt(-c)*c*abs(a) + (sqrt(-a^2*c)*x - sqrt(-a^2* c*x^2 + c))*a^2*c^2 + 4*a*sqrt(-c)*c^2*abs(a))/((sqrt(-a^2*c)*x - sqrt(-a^ 2*c*x^2 + c))^2 - c)^2
Timed out. \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx=-\int \frac {\sqrt {c-a^2\,c\,x^2}\,{\left (a\,x+1\right )}^2}{x^3\,\left (a^2\,x^2-1\right )} \,d x \] Input:
int(-((c - a^2*c*x^2)^(1/2)*(a*x + 1)^2)/(x^3*(a^2*x^2 - 1)),x)
Output:
-int(((c - a^2*c*x^2)^(1/2)*(a*x + 1)^2)/(x^3*(a^2*x^2 - 1)), 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*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/x^3,x)
Output:
(sqrt(c)*( - 4*sqrt( - a**2*x**2 + 1)*a*x - sqrt( - a**2*x**2 + 1) + 3*log (tan(asin(a*x)/2))*a**2*x**2))/(2*x**2)