Integrand size = 22, antiderivative size = 77 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {3 c^3 \sqrt {1-a^2 x^2}}{2 a}+\frac {c^3 \left (1-a^2 x^2\right )^{3/2}}{2 a^3 x^2}-\frac {3 c^3 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{2 a} \] Output:
3/2*c^3*(-a^2*x^2+1)^(1/2)/a+1/2*c^3*(-a^2*x^2+1)^(3/2)/a^3/x^2-3/2*c^3*ar ctanh((-a^2*x^2+1)^(1/2))/a
Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.48 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {c^3 \left (3 a^2 \arcsin (a x)+\frac {1+a^2 x^2-2 a^4 x^4+6 a^2 x^2 \sqrt {1-a^2 x^2} \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{x^2 \sqrt {1-a^2 x^2}}-3 a^2 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{2 a^3} \] Input:
Integrate[E^(3*ArcTanh[a*x])*(c - c/(a*x))^3,x]
Output:
(c^3*(3*a^2*ArcSin[a*x] + (1 + a^2*x^2 - 2*a^4*x^4 + 6*a^2*x^2*Sqrt[1 - a^ 2*x^2]*ArcSin[Sqrt[1 - a*x]/Sqrt[2]])/(x^2*Sqrt[1 - a^2*x^2]) - 3*a^2*ArcT anh[Sqrt[1 - a^2*x^2]]))/(2*a^3)
Time = 0.39 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6681, 6678, 243, 51, 60, 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.
\(\displaystyle \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle -\frac {c^3 \int \frac {e^{3 \text {arctanh}(a x)} (1-a x)^3}{x^3}dx}{a^3}\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle -\frac {c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{x^3}dx}{a^3}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {c^3 \int \frac {\left (1-a^2 x^2\right )^{3/2}}{x^4}dx^2}{2 a^3}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {c^3 \left (-\frac {3}{2} a^2 \int \frac {\sqrt {1-a^2 x^2}}{x^2}dx^2-\frac {\left (1-a^2 x^2\right )^{3/2}}{x^2}\right )}{2 a^3}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {c^3 \left (-\frac {3}{2} a^2 \left (\int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2+2 \sqrt {1-a^2 x^2}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{x^2}\right )}{2 a^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {c^3 \left (-\frac {3}{2} a^2 \left (2 \sqrt {1-a^2 x^2}-\frac {2 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{x^2}\right )}{2 a^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {c^3 \left (-\frac {3}{2} a^2 \left (2 \sqrt {1-a^2 x^2}-2 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )-\frac {\left (1-a^2 x^2\right )^{3/2}}{x^2}\right )}{2 a^3}\) |
Input:
Int[E^(3*ArcTanh[a*x])*(c - c/(a*x))^3,x]
Output:
-1/2*(c^3*(-((1 - a^2*x^2)^(3/2)/x^2) - (3*a^2*(2*Sqrt[1 - a^2*x^2] - 2*Ar cTanh[Sqrt[1 - a^2*x^2]]))/2))/a^3
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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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[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]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol ] :> Simp[d^p Int[u*(1 + c*(x/d))^p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; F reeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]
Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {\left (a^{2} x^{2}-1\right ) c^{3}}{2 x^{2} \sqrt {-a^{2} x^{2}+1}\, a^{3}}-\frac {\left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-2 \sqrt {-a^{2} x^{2}+1}\right ) c^{3}}{2 a}\) | \(72\) |
pseudoelliptic | \(\frac {c^{3} \left (3 \ln \left (\sqrt {-a^{2} x^{2}+1}-1\right ) a^{2} x^{2}-3 \ln \left (1+\sqrt {-a^{2} x^{2}+1}\right ) a^{2} x^{2}+4 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+2 \sqrt {-a^{2} x^{2}+1}\right )}{4 a^{3} x^{2}}\) | \(93\) |
default | \(\frac {c^{3} \left (a^{6} \left (-\frac {x^{2}}{a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {-a^{2} x^{2}+1}}\right )+\frac {1}{2 x^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {3 a^{2} \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{2}-\frac {3 a^{2}}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3}}\) | \(118\) |
meijerg | \(\frac {3 c^{3} \left (\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{2}-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{\sqrt {\pi }\, a}+\frac {c^{3} \left (\frac {\sqrt {\pi }}{2 x^{2} a^{2}}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }\, \left (-20 a^{2} x^{2}+8\right )}{16 a^{2} x^{2}}+\frac {\sqrt {\pi }\, \left (-24 a^{2} x^{2}+8\right )}{16 a^{2} x^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )}{2}\right )}{a \sqrt {\pi }}+\frac {c^{3} \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right )}{4 \sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}+\frac {3 c^{3} \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {\pi }}\) | \(272\) |
Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^3,x,method=_RETURNVERBOSE)
Output:
-1/2*(a^2*x^2-1)/x^2/(-a^2*x^2+1)^(1/2)*c^3/a^3-1/2/a*(3*arctanh(1/(-a^2*x ^2+1)^(1/2))-2*(-a^2*x^2+1)^(1/2))*c^3
Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {3 \, a^{2} c^{3} x^{2} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + 2 \, a^{2} c^{3} x^{2} + {\left (2 \, a^{2} c^{3} x^{2} + c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{2 \, a^{3} x^{2}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^3,x, algorithm="fricas")
Output:
1/2*(3*a^2*c^3*x^2*log((sqrt(-a^2*x^2 + 1) - 1)/x) + 2*a^2*c^3*x^2 + (2*a^ 2*c^3*x^2 + c^3)*sqrt(-a^2*x^2 + 1))/(a^3*x^2)
Time = 7.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.68 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {\begin {cases} - 2 a^{2} \left (- c^{3} \sqrt {- a^{2} x^{2} + 1} - \frac {3 c^{3} \log {\left (\sqrt {- a^{2} x^{2} + 1} - 1 \right )}}{4} + \frac {3 c^{3} \log {\left (\sqrt {- a^{2} x^{2} + 1} + 1 \right )}}{4} + \frac {c^{3}}{4 \left (\sqrt {- a^{2} x^{2} + 1} + 1\right )} + \frac {c^{3}}{4 \left (\sqrt {- a^{2} x^{2} + 1} - 1\right )}\right ) & \text {for}\: a^{2} \neq 0 \\2 a^{2} c^{3} \log {\left (x^{2} \right )} + \frac {c^{3}}{x^{2}} & \text {otherwise} \end {cases}}{2 a^{3}} \] Input:
integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(c-c/a/x)**3,x)
Output:
Piecewise((-2*a**2*(-c**3*sqrt(-a**2*x**2 + 1) - 3*c**3*log(sqrt(-a**2*x** 2 + 1) - 1)/4 + 3*c**3*log(sqrt(-a**2*x**2 + 1) + 1)/4 + c**3/(4*(sqrt(-a* *2*x**2 + 1) + 1)) + c**3/(4*(sqrt(-a**2*x**2 + 1) - 1))), Ne(a**2, 0)), ( 2*a**2*c**3*log(x**2) + c**3/x**2, True))/(2*a**3)
Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (65) = 130\).
Time = 0.11 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.44 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-a^{3} c^{3} {\left (\frac {x^{2}}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {2}{\sqrt {-a^{2} x^{2} + 1} a^{4}}\right )} + \frac {3 \, c^{3} {\left (\frac {1}{\sqrt {-a^{2} x^{2} + 1}} - \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )\right )}}{a} - \frac {3 \, c^{3}}{\sqrt {-a^{2} x^{2} + 1} a} + \frac {{\left (3 \, a^{2} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {3 \, a^{2}}{\sqrt {-a^{2} x^{2} + 1}} + \frac {1}{\sqrt {-a^{2} x^{2} + 1} x^{2}}\right )} c^{3}}{2 \, a^{3}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^3,x, algorithm="maxima")
Output:
-a^3*c^3*(x^2/(sqrt(-a^2*x^2 + 1)*a^2) - 2/(sqrt(-a^2*x^2 + 1)*a^4)) + 3*c ^3*(1/sqrt(-a^2*x^2 + 1) - log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)))/a - 3*c^3/(sqrt(-a^2*x^2 + 1)*a) + 1/2*(3*a^2*log(2*sqrt(-a^2*x^2 + 1)/abs(x ) + 2/abs(x)) - 3*a^2/sqrt(-a^2*x^2 + 1) + 1/(sqrt(-a^2*x^2 + 1)*x^2))*c^3 /a^3
Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {3 \, c^{3} \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - 3 \, c^{3} \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right ) - 4 \, \sqrt {-a^{2} x^{2} + 1} c^{3} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{a^{2} x^{2}}}{4 \, a} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^3,x, algorithm="giac")
Output:
-1/4*(3*c^3*log(sqrt(-a^2*x^2 + 1) + 1) - 3*c^3*log(-sqrt(-a^2*x^2 + 1) + 1) - 4*sqrt(-a^2*x^2 + 1)*c^3 - 2*sqrt(-a^2*x^2 + 1)*c^3/(a^2*x^2))/a
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {c^3\,\sqrt {1-a^2\,x^2}}{a}-\frac {3\,c^3\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{2\,a}+\frac {c^3\,\sqrt {1-a^2\,x^2}}{2\,a^3\,x^2} \] Input:
int(((c - c/(a*x))^3*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)
Output:
(c^3*(1 - a^2*x^2)^(1/2))/a - (3*c^3*atanh((1 - a^2*x^2)^(1/2)))/(2*a) + ( c^3*(1 - a^2*x^2)^(1/2))/(2*a^3*x^2)
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {c^{3} \left (8 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+4 \sqrt {-a^{2} x^{2}+1}+12 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{2} x^{2}-9 a^{2} x^{2}\right )}{8 a^{3} x^{2}} \] Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^3,x)
Output:
(c**3*(8*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 4*sqrt( - a**2*x**2 + 1) + 12* log(tan(asin(a*x)/2))*a**2*x**2 - 9*a**2*x**2))/(8*a**3*x**2)