Integrand size = 17, antiderivative size = 64 \[ \int \frac {e^{\text {arctanh}(a x)} x}{c-a c x} \, dx=\frac {\sqrt {1-a^2 x^2}}{a^2 c}+\frac {2 \sqrt {1-a^2 x^2}}{a^2 c (1-a x)}-\frac {2 \arcsin (a x)}{a^2 c} \] Output:
(-a^2*x^2+1)^(1/2)/a^2/c+2*(-a^2*x^2+1)^(1/2)/a^2/c/(-a*x+1)-2*arcsin(a*x) /a^2/c
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\text {arctanh}(a x)} x}{c-a c x} \, dx=\frac {\frac {(3-a x) \sqrt {1+a x}}{\sqrt {1-a x}}+4 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{a^2 c} \] Input:
Integrate[(E^ArcTanh[a*x]*x)/(c - a*c*x),x]
Output:
(((3 - a*x)*Sqrt[1 + a*x])/Sqrt[1 - a*x] + 4*ArcSin[Sqrt[1 - a*x]/Sqrt[2]] )/(a^2*c)
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6678, 27, 563, 455, 223}
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.
\(\displaystyle \int \frac {x e^{\text {arctanh}(a x)}}{c-a c x} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {x \sqrt {1-a^2 x^2}}{c^2 (1-a x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x \sqrt {1-a^2 x^2}}{(1-a x)^2}dx}{c}\) |
\(\Big \downarrow \) 563 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^2 (1-a x)}-\frac {\int \frac {a x+2}{\sqrt {1-a^2 x^2}}dx}{a}}{c}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^2 (1-a x)}-\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{a}}{a}}{c}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\frac {2 \sqrt {1-a^2 x^2}}{a^2 (1-a x)}-\frac {\frac {2 \arcsin (a x)}{a}-\frac {\sqrt {1-a^2 x^2}}{a}}{a}}{c}\) |
Input:
Int[(E^ArcTanh[a*x]*x)/(c - a*c*x),x]
Output:
((2*Sqrt[1 - a^2*x^2])/(a^2*(1 - a*x)) - (-(Sqrt[1 - a^2*x^2]/a) + (2*ArcS in[a*x])/a)/a)/c
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)* b^(n + 2)*(c + d*x))), x] - Simp[d^(2*n - m + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(-n - 1)*(-c)^(m - n - 1) - d^m*x^m*(-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2 , 0] && IGtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
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 = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.48
method | result | size |
default | \(-\frac {-\frac {\sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {2 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {a^{2}}}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{3} \left (x -\frac {1}{a}\right )}}{c}\) | \(95\) |
risch | \(-\frac {a^{2} x^{2}-1}{a^{2} \sqrt {-a^{2} x^{2}+1}\, c}-\frac {\frac {2 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a \sqrt {a^{2}}}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{3} \left (x -\frac {1}{a}\right )}}{c}\) | \(108\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c),x,method=_RETURNVERBOSE)
Output:
-1/c*(-(-a^2*x^2+1)^(1/2)/a^2+2/a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x ^2+1)^(1/2))+2/a^3/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\text {arctanh}(a x)} x}{c-a c x} \, dx=\frac {3 \, a x + 4 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (a x - 3\right )} - 3}{a^{3} c x - a^{2} c} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c),x, algorithm="fricas")
Output:
(3*a*x + 4*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^ 2 + 1)*(a*x - 3) - 3)/(a^3*c*x - a^2*c)
\[ \int \frac {e^{\text {arctanh}(a x)} x}{c-a c x} \, dx=- \frac {\int \frac {x}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{2}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x/(-a*c*x+c),x)
Output:
-(Integral(x/(a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integ ral(a*x**2/(a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x))/c
Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\text {arctanh}(a x)} x}{c-a c x} \, dx=-\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{3} c x - a^{2} c} - \frac {2 \, \arcsin \left (a x\right )}{a^{2} c} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{2} c} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c),x, algorithm="maxima")
Output:
-2*sqrt(-a^2*x^2 + 1)/(a^3*c*x - a^2*c) - 2*arcsin(a*x)/(a^2*c) + sqrt(-a^ 2*x^2 + 1)/(a^2*c)
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.22 \[ \int \frac {e^{\text {arctanh}(a x)} x}{c-a c x} \, dx=-\frac {2 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{a c {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{2} c} + \frac {4}{a c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c),x, algorithm="giac")
Output:
-2*arcsin(a*x)*sgn(a)/(a*c*abs(a)) + sqrt(-a^2*x^2 + 1)/(a^2*c) + 4/(a*c*( (sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a))
Time = 14.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\text {arctanh}(a x)} x}{c-a c x} \, dx=\frac {\sqrt {1-a^2\,x^2}}{a^2\,c}-\frac {2\,\sqrt {1-a^2\,x^2}}{\left (c\,\sqrt {-a^2}-a\,c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a\,c\,\sqrt {-a^2}} \] Input:
int((x*(a*x + 1))/((1 - a^2*x^2)^(1/2)*(c - a*c*x)),x)
Output:
(1 - a^2*x^2)^(1/2)/(a^2*c) - (2*(1 - a^2*x^2)^(1/2))/((c*(-a^2)^(1/2) - a *c*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (2*asinh(x*(-a^2)^(1/2)))/(a*c*(-a^2)^( 1/2))
Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\text {arctanh}(a x)} x}{c-a c x} \, dx=\frac {-2 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-2 \mathit {asin} \left (a x \right ) a x +2 \mathit {asin} \left (a x \right )+\sqrt {-a^{2} x^{2}+1}\, a x -4 \sqrt {-a^{2} x^{2}+1}-a^{2} x^{2}+a x +4}{a^{2} c \left (\sqrt {-a^{2} x^{2}+1}+a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c),x)
Output:
( - 2*sqrt( - a**2*x**2 + 1)*asin(a*x) - 2*asin(a*x)*a*x + 2*asin(a*x) + s qrt( - a**2*x**2 + 1)*a*x - 4*sqrt( - a**2*x**2 + 1) - a**2*x**2 + a*x + 4 )/(a**2*c*(sqrt( - a**2*x**2 + 1) + a*x - 1))