Integrand size = 17, antiderivative size = 65 \[ \int \frac {e^{\text {arctanh}(a x)} x}{(c-a c x)^3} \, dx=\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^2 c^3 (1-a x)^4}-\frac {4 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 c^3 (1-a x)^3} \] Output:
1/5*(-a^2*x^2+1)^(3/2)/a^2/c^3/(-a*x+1)^4-4/15*(-a^2*x^2+1)^(3/2)/a^2/c^3/ (-a*x+1)^3
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.54 \[ \int \frac {e^{\text {arctanh}(a x)} x}{(c-a c x)^3} \, dx=\frac {(1+a x)^{3/2} (-1+4 a x)}{15 a^2 c^3 (1-a x)^{5/2}} \] Input:
Integrate[(E^ArcTanh[a*x]*x)/(c - a*c*x)^3,x]
Output:
((1 + a*x)^(3/2)*(-1 + 4*a*x))/(15*a^2*c^3*(1 - a*x)^(5/2))
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {6678, 27, 571, 460}
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.
\(\displaystyle \int \frac {x e^{\text {arctanh}(a x)}}{(c-a c x)^3} \, dx\) |
\(\Big \downarrow \) 6678 |
\(\displaystyle c \int \frac {x \sqrt {1-a^2 x^2}}{c^4 (1-a x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x \sqrt {1-a^2 x^2}}{(1-a x)^4}dx}{c^3}\) |
\(\Big \downarrow \) 571 |
\(\displaystyle \frac {\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^2 (1-a x)^4}-\frac {4 \int \frac {\sqrt {1-a^2 x^2}}{(1-a x)^3}dx}{5 a}}{c^3}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^2 (1-a x)^4}-\frac {4 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 (1-a x)^3}}{c^3}\) |
Input:
Int[(E^ArcTanh[a*x]*x)/(c - a*c*x)^3,x]
Output:
((1 - a^2*x^2)^(3/2)/(5*a^2*(1 - a*x)^4) - (4*(1 - a^2*x^2)^(3/2))/(15*a^2 *(1 - a*x)^3))/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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*(n + p + 1))), x] + Simp[n/(2*d* (n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ((LtQ[n, -1] && !IGtQ[n + p + 1, 0]) || (LtQ[n, 0] && LtQ[p, -1]) || EqQ[n + 2*p + 2, 0]) && NeQ[n + p + 1, 0]
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 = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(\frac {\left (4 a x -1\right ) \left (a x +1\right )^{2}}{15 \left (a x -1\right )^{2} c^{3} \sqrt {-a^{2} x^{2}+1}\, a^{2}}\) | \(41\) |
trager | \(-\frac {\left (4 a^{2} x^{2}+3 a x -1\right ) \sqrt {-a^{2} x^{2}+1}}{15 c^{3} \left (a x -1\right )^{3} a^{2}}\) | \(42\) |
orering | \(-\frac {\left (4 a x -1\right ) \left (a x -1\right ) \left (a x +1\right )^{2}}{15 a^{2} \sqrt {-a^{2} x^{2}+1}\, \left (-a c x +c \right )^{3}}\) | \(45\) |
default | \(-\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{3} \left (x -\frac {1}{a}\right )}+\frac {\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {4 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a^{4}}+\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}}{a^{3}}}{c^{3}}\) | \(262\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c)^3,x,method=_RETURNVERBOSE)
Output:
1/15*(4*a*x-1)*(a*x+1)^2/(a*x-1)^2/c^3/(-a^2*x^2+1)^(1/2)/a^2
Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\text {arctanh}(a x)} x}{(c-a c x)^3} \, dx=-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x + {\left (4 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1} - 1}{15 \, {\left (a^{5} c^{3} x^{3} - 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x - a^{2} c^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c)^3,x, algorithm="fricas")
Output:
-1/15*(a^3*x^3 - 3*a^2*x^2 + 3*a*x + (4*a^2*x^2 + 3*a*x - 1)*sqrt(-a^2*x^2 + 1) - 1)/(a^5*c^3*x^3 - 3*a^4*c^3*x^2 + 3*a^3*c^3*x - a^2*c^3)
\[ \int \frac {e^{\text {arctanh}(a x)} x}{(c-a c x)^3} \, dx=- \frac {\int \frac {x}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{2}}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x/(-a*c*x+c)**3,x)
Output:
-(Integral(x/(a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + 3*a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral( a*x**2/(a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + 3*a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x))/c**3
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (55) = 110\).
Time = 0.11 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.03 \[ \int \frac {e^{\text {arctanh}(a x)} x}{(c-a c x)^3} \, dx=-\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{5 \, {\left (a^{5} c^{3} x^{3} - 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x - a^{2} c^{3}\right )}} - \frac {11 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{4} c^{3} x^{2} - 2 \, a^{3} c^{3} x + a^{2} c^{3}\right )}} - \frac {4 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{3} c^{3} x - a^{2} c^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c)^3,x, algorithm="maxima")
Output:
-2/5*sqrt(-a^2*x^2 + 1)/(a^5*c^3*x^3 - 3*a^4*c^3*x^2 + 3*a^3*c^3*x - a^2*c ^3) - 11/15*sqrt(-a^2*x^2 + 1)/(a^4*c^3*x^2 - 2*a^3*c^3*x + a^2*c^3) - 4/1 5*sqrt(-a^2*x^2 + 1)/(a^3*c^3*x - a^2*c^3)
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (55) = 110\).
Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.86 \[ \int \frac {e^{\text {arctanh}(a x)} x}{(c-a c x)^3} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} + \frac {5 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - 1\right )}}{15 \, a c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c)^3,x, algorithm="giac")
Output:
2/15*(5*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 5*(sqrt(-a^2*x^2 + 1)*ab s(a) + a)^2/(a^4*x^2) + 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 1 )/(a*c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))
Time = 14.01 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.20 \[ \int \frac {e^{\text {arctanh}(a x)} x}{(c-a c x)^3} \, dx=-\frac {360\,a^6\,c^3\,\sqrt {1-a^2\,x^2}-600\,a^6\,c^3\,{\left (1-a^2\,x^2\right )}^{3/2}+225\,a^6\,c^3\,{\left (1-a^2\,x^2\right )}^{5/2}+360\,a^7\,c^3\,x\,\sqrt {1-a^2\,x^2}-420\,a^7\,c^3\,x\,{\left (1-a^2\,x^2\right )}^{3/2}+60\,a^7\,c^3\,x\,{\left (1-a^2\,x^2\right )}^{5/2}}{225\,a^8\,c^6\,{\left (a^2\,x^2-1\right )}^3} \] Input:
int((x*(a*x + 1))/((1 - a^2*x^2)^(1/2)*(c - a*c*x)^3),x)
Output:
-(360*a^6*c^3*(1 - a^2*x^2)^(1/2) - 600*a^6*c^3*(1 - a^2*x^2)^(3/2) + 225* a^6*c^3*(1 - a^2*x^2)^(5/2) + 360*a^7*c^3*x*(1 - a^2*x^2)^(1/2) - 420*a^7* c^3*x*(1 - a^2*x^2)^(3/2) + 60*a^7*c^3*x*(1 - a^2*x^2)^(5/2))/(225*a^8*c^6 *(a^2*x^2 - 1)^3)
Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\text {arctanh}(a x)} x}{(c-a c x)^3} \, dx=\frac {-2 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-\frac {2 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}}{3}-\frac {2 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )}{3}+\frac {2}{15}}{a^{2} c^{3} \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{5}-5 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}+10 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-10 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+5 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x/(-a*c*x+c)^3,x)
Output:
(2*( - 15*tan(asin(a*x)/2)**3 - 5*tan(asin(a*x)/2)**2 - 5*tan(asin(a*x)/2) + 1))/(15*a**2*c**3*(tan(asin(a*x)/2)**5 - 5*tan(asin(a*x)/2)**4 + 10*tan (asin(a*x)/2)**3 - 10*tan(asin(a*x)/2)**2 + 5*tan(asin(a*x)/2) - 1))