Integrand size = 27, antiderivative size = 101 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=-\frac {\sqrt {c-a^2 c x^2}}{3 x^3}-\frac {a \sqrt {c-a^2 c x^2}}{x^2}-\frac {5 a^2 \sqrt {c-a^2 c x^2}}{3 x}-a^3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \] Output:
-1/3*(-a^2*c*x^2+c)^(1/2)/x^3-a*(-a^2*c*x^2+c)^(1/2)/x^2-5/3*a^2*(-a^2*c*x ^2+c)^(1/2)/x-a^3*c^(1/2)*arctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=-\frac {\left (1+3 a x+5 a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 x^3}+a^3 \sqrt {c} \log (x)-a^3 \sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right ) \] Input:
Integrate[(E^(2*ArcTanh[a*x])*Sqrt[c - a^2*c*x^2])/x^4,x]
Output:
-1/3*((1 + 3*a*x + 5*a^2*x^2)*Sqrt[c - a^2*c*x^2])/x^3 + a^3*Sqrt[c]*Log[x ] - a^3*Sqrt[c]*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2]]
Time = 0.68 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {6701, 540, 25, 27, 539, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx\) |
\(\Big \downarrow \) 6701 |
\(\displaystyle c \int \frac {(a x+1)^2}{x^4 \sqrt {c-a^2 c x^2}}dx\) |
\(\Big \downarrow \) 540 |
\(\displaystyle c \left (-\frac {\int -\frac {a c (5 a x+6)}{x^3 \sqrt {c-a^2 c x^2}}dx}{3 c}-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle c \left (\frac {\int \frac {a c (5 a x+6)}{x^3 \sqrt {c-a^2 c x^2}}dx}{3 c}-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (\frac {1}{3} a \int \frac {5 a x+6}{x^3 \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 539 |
\(\displaystyle c \left (\frac {1}{3} a \left (-\frac {\int -\frac {2 a c (3 a x+5)}{x^2 \sqrt {c-a^2 c x^2}}dx}{2 c}-\frac {3 \sqrt {c-a^2 c x^2}}{c x^2}\right )-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle c \left (\frac {1}{3} a \left (a \int \frac {3 a x+5}{x^2 \sqrt {c-a^2 c x^2}}dx-\frac {3 \sqrt {c-a^2 c x^2}}{c x^2}\right )-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 534 |
\(\displaystyle c \left (\frac {1}{3} a \left (a \left (3 a \int \frac {1}{x \sqrt {c-a^2 c x^2}}dx-\frac {5 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {3 \sqrt {c-a^2 c x^2}}{c x^2}\right )-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle c \left (\frac {1}{3} a \left (a \left (\frac {3}{2} a \int \frac {1}{x^2 \sqrt {c-a^2 c x^2}}dx^2-\frac {5 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {3 \sqrt {c-a^2 c x^2}}{c x^2}\right )-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle c \left (\frac {1}{3} a \left (a \left (-\frac {3 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2 c}}d\sqrt {c-a^2 c x^2}}{a c}-\frac {5 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {3 \sqrt {c-a^2 c x^2}}{c x^2}\right )-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle c \left (\frac {1}{3} a \left (a \left (-\frac {3 a \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {5 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {3 \sqrt {c-a^2 c x^2}}{c x^2}\right )-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\) |
Input:
Int[(E^(2*ArcTanh[a*x])*Sqrt[c - a^2*c*x^2])/x^4,x]
Output:
c*(-1/3*Sqrt[c - a^2*c*x^2]/(c*x^3) + (a*((-3*Sqrt[c - a^2*c*x^2])/(c*x^2) + a*((-5*Sqrt[c - a^2*c*x^2])/(c*x) - (3*a*ArcTanh[Sqrt[c - a^2*c*x^2]/Sq rt[c]])/Sqrt[c])))/3)
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_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*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*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
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 = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {\left (5 a^{4} x^{4}+3 a^{3} x^{3}-4 a^{2} x^{2}-3 a x -1\right ) c}{3 x^{3} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-a^{3} \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\) | \(87\) |
default | \(-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3 c \,x^{3}}+2 a \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}-\frac {a^{2} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )}{2}\right )+2 a^{2} \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{c x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )\right )+2 a^{3} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )-2 a^{3} \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}-\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )\) | \(317\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x,method=_RETURNVERBOS E)
Output:
1/3*(5*a^4*x^4+3*a^3*x^3-4*a^2*x^2-3*a*x-1)/x^3/(-c*(a^2*x^2-1))^(1/2)*c-a ^3*c^(1/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)
Time = 0.13 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.52 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\left [\frac {3 \, a^{3} \sqrt {c} x^{3} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (5 \, a^{2} x^{2} + 3 \, a x + 1\right )}}{6 \, x^{3}}, \frac {3 \, a^{3} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (5 \, a^{2} x^{2} + 3 \, a x + 1\right )}}{3 \, x^{3}}\right ] \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="fr icas")
Output:
[1/6*(3*a^3*sqrt(c)*x^3*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)*(5*a^2*x^2 + 3*a*x + 1))/x^3, 1/3*(3*a ^3*sqrt(-c)*x^3*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/c) - sqrt(-a^2*c*x^2 + c)*(5*a^2*x^2 + 3*a*x + 1))/x^3]
\[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=- \int \frac {\sqrt {- a^{2} c x^{2} + c}}{a x^{5} - x^{4}}\, dx - \int \frac {a x \sqrt {- a^{2} c x^{2} + c}}{a x^{5} - x^{4}}\, dx \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*(-a**2*c*x**2+c)**(1/2)/x**4,x)
Output:
-Integral(sqrt(-a**2*c*x**2 + c)/(a*x**5 - x**4), x) - Integral(a*x*sqrt(- a**2*c*x**2 + c)/(a*x**5 - x**4), x)
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.39 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=-\frac {\sqrt {a x + 1} \sqrt {-a x + 1} a^{2} \sqrt {c}}{x} + \frac {a^{4} c^{\frac {3}{2}} \log \left (\frac {\sqrt {-a^{2} c x^{2} + c} - \sqrt {c}}{\sqrt {-a^{2} c x^{2} + c} + \sqrt {c}}\right ) - \frac {2 \, \sqrt {-a^{2} c x^{2} + c} a^{2} c}{x^{2}}}{2 \, a c} - \frac {{\left (2 \, a^{2} \sqrt {c} x^{2} + \sqrt {c}\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{3 \, x^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="ma xima")
Output:
-sqrt(a*x + 1)*sqrt(-a*x + 1)*a^2*sqrt(c)/x + 1/2*(a^4*c^(3/2)*log((sqrt(- a^2*c*x^2 + c) - sqrt(c))/(sqrt(-a^2*c*x^2 + c) + sqrt(c))) - 2*sqrt(-a^2* c*x^2 + c)*a^2*c/x^2)/(a*c) - 1/3*(2*a^2*sqrt(c)*x^2 + sqrt(c))*sqrt(a*x + 1)*sqrt(-a*x + 1)/x^3
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (85) = 170\).
Time = 0.14 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.48 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {2 \, a^{3} c \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {2 \, {\left (3 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{5} a^{3} c - 3 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{4} a^{2} \sqrt {-c} c {\left | a \right |} + 12 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} a^{2} \sqrt {-c} c^{2} {\left | a \right |} - 3 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )} a^{3} c^{3} - 5 \, a^{2} \sqrt {-c} c^{3} {\left | a \right |}\right )}}{3 \, {\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="gi ac")
Output:
2*a^3*c*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) - 2/3*(3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^5*a^3*c - 3*(sqrt(-a^2*c )*x - sqrt(-a^2*c*x^2 + c))^4*a^2*sqrt(-c)*c*abs(a) + 12*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2*a^2*sqrt(-c)*c^2*abs(a) - 3*(sqrt(-a^2*c)*x - sqr t(-a^2*c*x^2 + c))*a^3*c^3 - 5*a^2*sqrt(-c)*c^3*abs(a))/((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c)^3
Timed out. \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=-\int \frac {\sqrt {c-a^2\,c\,x^2}\,{\left (a\,x+1\right )}^2}{x^4\,\left (a^2\,x^2-1\right )} \,d x \] Input:
int(-((c - a^2*c*x^2)^(1/2)*(a*x + 1)^2)/(x^4*(a^2*x^2 - 1)),x)
Output:
-int(((c - a^2*c*x^2)^(1/2)*(a*x + 1)^2)/(x^4*(a^2*x^2 - 1)), x)
Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.70 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {\sqrt {c}\, \left (-5 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-3 \sqrt {-a^{2} x^{2}+1}\, a x -\sqrt {-a^{2} x^{2}+1}+3 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{3} x^{3}\right )}{3 x^{3}} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x)
Output:
(sqrt(c)*( - 5*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 3*sqrt( - a**2*x**2 + 1) *a*x - sqrt( - a**2*x**2 + 1) + 3*log(tan(asin(a*x)/2))*a**3*x**3))/(3*x** 3)