Integrand size = 17, antiderivative size = 46 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx=-\frac {c \sqrt {1-a^2 x^2}}{2 x^2}+\frac {1}{2} a^2 c \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \] Output:
-1/2*c*(-a^2*x^2+1)^(1/2)/x^2+1/2*a^2*c*arctanh((-a^2*x^2+1)^(1/2))
Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.46 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx=\frac {c \left (-1+a^2 x^2+a^2 x^2 \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{2 x^2 \sqrt {1-a^2 x^2}} \] Input:
Integrate[(E^ArcTanh[a*x]*(c - a*c*x))/x^3,x]
Output:
(c*(-1 + a^2*x^2 + a^2*x^2*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]]))/ (2*x^2*Sqrt[1 - a^2*x^2])
Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6678, 243, 51, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{x^3}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} c \int \frac {\sqrt {1-a^2 x^2}}{x^4}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} c \left (-\frac {1}{2} a^2 \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\frac {\sqrt {1-a^2 x^2}}{x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} c \left (\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}-\frac {\sqrt {1-a^2 x^2}}{x^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} c \left (a^2 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{x^2}\right )\) |
Input:
Int[(E^ArcTanh[a*x]*(c - a*c*x))/x^3,x]
Output:
(c*(-(Sqrt[1 - a^2*x^2]/x^2) + a^2*ArcTanh[Sqrt[1 - a^2*x^2]]))/2
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* (x_))^(m_.), x_Symbol] :> Simp[c^n Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 , 0]) && IntegerQ[2*p]
Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87
method | result | size |
default | \(-c \left (-\frac {a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}\right )\) | \(40\) |
risch | \(\frac {\left (a^{2} x^{2}-1\right ) c}{2 x^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right ) c}{2}\) | \(48\) |
pseudoelliptic | \(-\frac {c \left (\ln \left (\sqrt {-a^{2} x^{2}+1}-1\right ) a^{2} x^{2}-\ln \left (1+\sqrt {-a^{2} x^{2}+1}\right ) a^{2} x^{2}+2 \sqrt {-a^{2} x^{2}+1}\right )}{4 x^{2}}\) | \(67\) |
meijerg | \(-\frac {a^{2} c \left (\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{2 \sqrt {\pi }}-\frac {a^{2} c \left (\frac {\sqrt {\pi }}{x^{2} a^{2}}-\frac {\left (1-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{2}-\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right )}{8 a^{2} x^{2}}+\frac {\sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}}{a^{2} x^{2}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{2 \sqrt {\pi }}\) | \(158\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x,method=_RETURNVERBOSE)
Output:
-c*(-1/2*a^2*arctanh(1/(-a^2*x^2+1)^(1/2))+1/2*(-a^2*x^2+1)^(1/2)/x^2)
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx=-\frac {a^{2} c x^{2} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1} c}{2 \, x^{2}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="fricas")
Output:
-1/2*(a^2*c*x^2*log((sqrt(-a^2*x^2 + 1) - 1)/x) + sqrt(-a^2*x^2 + 1)*c)/x^ 2
Time = 3.48 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.89 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx=\frac {\begin {cases} - 2 a^{2} \left (\frac {c \log {\left (\sqrt {- a^{2} x^{2} + 1} - 1 \right )}}{4} - \frac {c \log {\left (\sqrt {- a^{2} x^{2} + 1} + 1 \right )}}{4} - \frac {c}{4 \left (\sqrt {- a^{2} x^{2} + 1} + 1\right )} - \frac {c}{4 \left (\sqrt {- a^{2} x^{2} + 1} - 1\right )}\right ) & \text {for}\: a^{2} \neq 0 \\- \frac {c}{x^{2}} & \text {otherwise} \end {cases}}{2} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)/x**3,x)
Output:
Piecewise((-2*a**2*(c*log(sqrt(-a**2*x**2 + 1) - 1)/4 - c*log(sqrt(-a**2*x **2 + 1) + 1)/4 - c/(4*(sqrt(-a**2*x**2 + 1) + 1)) - c/(4*(sqrt(-a**2*x**2 + 1) - 1))), Ne(a**2, 0)), (-c/x**2, True))/2
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx=\frac {1}{2} \, a^{2} c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {\sqrt {-a^{2} x^{2} + 1} c}{2 \, x^{2}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="maxima")
Output:
1/2*a^2*c*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) - 1/2*sqrt(-a^2*x^2 + 1)*c/x^2
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx=\frac {1}{4} \, {\left (c \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - c \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right ) - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c}{a^{2} x^{2}}\right )} a^{2} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="giac")
Output:
1/4*(c*log(sqrt(-a^2*x^2 + 1) + 1) - c*log(-sqrt(-a^2*x^2 + 1) + 1) - 2*sq rt(-a^2*x^2 + 1)*c/(a^2*x^2))*a^2
Time = 14.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx=\frac {a^2\,c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{2}-\frac {c\,\sqrt {1-a^2\,x^2}}{2\,x^2} \] Input:
int(((c - a*c*x)*(a*x + 1))/(x^3*(1 - a^2*x^2)^(1/2)),x)
Output:
(a^2*c*atanh((1 - a^2*x^2)^(1/2)))/2 - (c*(1 - a^2*x^2)^(1/2))/(2*x^2)
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\text {arctanh}(a x)} (c-a c x)}{x^3} \, dx=-\frac {c \left (\sqrt {-a^{2} x^{2}+1}+\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{2} x^{2}\right )}{2 x^{2}} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x)
Output:
( - c*(sqrt( - a**2*x**2 + 1) + log(tan(asin(a*x)/2))*a**2*x**2))/(2*x**2)