Integrand size = 19, antiderivative size = 79 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx=\frac {2 \sqrt {1-a^2 x^2}}{3 c^2 (1-a x)^2}+\frac {5 \sqrt {1-a^2 x^2}}{3 c^2 (1-a x)}-\frac {\text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c^2} \] Output:
2/3*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)^2+5/3*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)- arctanh((-a^2*x^2+1)^(1/2))/c^2
Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx=\frac {-7-2 a x+5 a^2 x^2-3 (-1+a x) \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{3 c^2 (-1+a x) \sqrt {1-a^2 x^2}} \] Input:
Integrate[E^ArcTanh[a*x]/(x*(c - a*c*x)^2),x]
Output:
(-7 - 2*a*x + 5*a^2*x^2 - 3*(-1 + a*x)*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(3*c^2*(-1 + a*x)*Sqrt[1 - a^2*x^2])
Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {6678, 27, 570, 532, 25, 532, 27, 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.
| \(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx\) |
| \(\Big \downarrow \) 6678 |
| \(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{c^3 x (1-a x)^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {1-a^2 x^2}}{x (1-a x)^3}dx}{c^2}\) |
| \(\Big \downarrow \) 570 |
| \(\displaystyle \frac {\int \frac {(a x+1)^3}{x \left (1-a^2 x^2\right )^{5/2}}dx}{c^2}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {\frac {4 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{3} \int -\frac {5 a x+3}{x \left (1-a^2 x^2\right )^{3/2}}dx}{c^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {5 a x+3}{x \left (1-a^2 x^2\right )^{3/2}}dx+\frac {4 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {5 a x+3}{\sqrt {1-a^2 x^2}}-\int -\frac {3}{x \sqrt {1-a^2 x^2}}dx\right )+\frac {4 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \int \frac {1}{x \sqrt {1-a^2 x^2}}dx+\frac {5 a x+3}{\sqrt {1-a^2 x^2}}\right )+\frac {4 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2+\frac {5 a x+3}{\sqrt {1-a^2 x^2}}\right )+\frac {4 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {5 a x+3}{\sqrt {1-a^2 x^2}}-\frac {3 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}\right )+\frac {4 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {5 a x+3}{\sqrt {1-a^2 x^2}}-3 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )+\frac {4 (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}}{c^2}\) |
Input:
Int[E^ArcTanh[a*x]/(x*(c - a*c*x)^2),x]
Output:
((4*(1 + a*x))/(3*(1 - a^2*x^2)^(3/2)) + ((3 + 5*a*x)/Sqrt[1 - a^2*x^2] - 3*ArcTanh[Sqrt[1 - a^2*x^2]])/3)/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_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(146\) vs. \(2(69)=138\).
Time = 0.23 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.86
| method | result | size |
| default | \(\frac {-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}}{c^{2}}\) | \(147\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c)^2,x,method=_RETURNVERBOSE)
Output:
1/c^2*(-arctanh(1/(-a^2*x^2+1)^(1/2))+2/a*(1/3/a/(x-1/a)^2*(-(x-1/a)^2*a^2 -2*a*(x-1/a))^(1/2)-1/3/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))-1/a/(x -1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx=\frac {7 \, a^{2} x^{2} - 14 \, a x + 3 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} {\left (5 \, a x - 7\right )} + 7}{3 \, {\left (a^{2} c^{2} x^{2} - 2 \, a c^{2} x + c^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c)^2,x, algorithm="fricas")
Output:
1/3*(7*a^2*x^2 - 14*a*x + 3*(a^2*x^2 - 2*a*x + 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - sqrt(-a^2*x^2 + 1)*(5*a*x - 7) + 7)/(a^2*c^2*x^2 - 2*a*c^2*x + c ^2)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx=\frac {\int \frac {a x}{a^{2} x^{3} \sqrt {- a^{2} x^{2} + 1} - 2 a x^{2} \sqrt {- a^{2} x^{2} + 1} + x \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a^{2} x^{3} \sqrt {- a^{2} x^{2} + 1} - 2 a x^{2} \sqrt {- a^{2} x^{2} + 1} + x \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x/(-a*c*x+c)**2,x)
Output:
(Integral(a*x/(a**2*x**3*sqrt(-a**2*x**2 + 1) - 2*a*x**2*sqrt(-a**2*x**2 + 1) + x*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a**2*x**3*sqrt(-a**2*x**2 + 1) - 2*a*x**2*sqrt(-a**2*x**2 + 1) + x*sqrt(-a**2*x**2 + 1)), x))/c**2
\[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx=\int { \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )}^{2} x} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c)^2,x, algorithm="maxima")
Output:
integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(a*c*x - c)^2*x), x)
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 245, normalized size of antiderivative = 3.10 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx=\frac {{\left (\frac {{\left (3 \, \log \left (2\right ) - 6 \, \log \left (i + 1\right ) + 10 i\right )} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )}{c} + \frac {6 \, \log \left (\sqrt {-\frac {2 \, c}{a c x - c} - 1} + 1\right )}{c \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )} - \frac {6 \, \log \left ({\left | \sqrt {-\frac {2 \, c}{a c x - c} - 1} - 1 \right |}\right )}{c \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )} - \frac {2 \, {\left (c^{2} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{2} \mathrm {sgn}\left (a\right )^{2} \mathrm {sgn}\left (c\right )^{2} + 6 \, c^{2} \sqrt {-\frac {2 \, c}{a c x - c} - 1} \mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{2} \mathrm {sgn}\left (a\right )^{2} \mathrm {sgn}\left (c\right )^{2}\right )}}{c^{3} \mathrm {sgn}\left (\frac {1}{a c x - c}\right )^{3} \mathrm {sgn}\left (a\right )^{3} \mathrm {sgn}\left (c\right )^{3}}\right )} a}{6 \, c {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c)^2,x, algorithm="giac")
Output:
1/6*((3*log(2) - 6*log(I + 1) + 10*I)*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)/c + 6*log(sqrt(-2*c/(a*c*x - c) - 1) + 1)/(c*sgn(1/(a*c*x - c))*sgn(a)*sgn(c) ) - 6*log(abs(sqrt(-2*c/(a*c*x - c) - 1) - 1))/(c*sgn(1/(a*c*x - c))*sgn(a )*sgn(c)) - 2*(c^2*(-2*c/(a*c*x - c) - 1)^(3/2)*sgn(1/(a*c*x - c))^2*sgn(a )^2*sgn(c)^2 + 6*c^2*sqrt(-2*c/(a*c*x - c) - 1)*sgn(1/(a*c*x - c))^2*sgn(a )^2*sgn(c)^2)/(c^3*sgn(1/(a*c*x - c))^3*sgn(a)^3*sgn(c)^3))*a/(c*abs(a))
Time = 14.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.51 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx=\frac {2\,a^2\,\sqrt {1-a^2\,x^2}}{3\,\left (a^4\,c^2\,x^2-2\,a^3\,c^2\,x+a^2\,c^2\right )}-\frac {\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{c^2}+\frac {5\,a\,\sqrt {1-a^2\,x^2}}{3\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )} \] Input:
int((a*x + 1)/(x*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^2),x)
Output:
(2*a^2*(1 - a^2*x^2)^(1/2))/(3*(a^2*c^2 - 2*a^3*c^2*x + a^4*c^2*x^2)) - at anh((1 - a^2*x^2)^(1/2))/c^2 + (5*a*(1 - a^2*x^2)^(1/2))/(3*(-a^2)^(1/2)*( c^2*x*(-a^2)^(1/2) - (c^2*(-a^2)^(1/2))/a))
Time = 0.15 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\text {arctanh}(a x)}}{x (c-a c x)^2} \, dx=\frac {3 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-9 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+9 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-3 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right )-6 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}+6 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-8}{3 c^{2} \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c)^2,x)
Output:
(3*log(tan(asin(a*x)/2))*tan(asin(a*x)/2)**3 - 9*log(tan(asin(a*x)/2))*tan (asin(a*x)/2)**2 + 9*log(tan(asin(a*x)/2))*tan(asin(a*x)/2) - 3*log(tan(as in(a*x)/2)) - 6*tan(asin(a*x)/2)**3 + 6*tan(asin(a*x)/2) - 8)/(3*c**2*(tan (asin(a*x)/2)**3 - 3*tan(asin(a*x)/2)**2 + 3*tan(asin(a*x)/2) - 1))