Integrand size = 18, antiderivative size = 97 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx=\frac {\sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)^3}+\frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)^2}+\frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)} \] Output:
1/5*(-a^2*x^2+1)^(1/2)/a/c^4/(-a*x+1)^3+2/15*(-a^2*x^2+1)^(1/2)/a/c^4/(-a* x+1)^2+2/15*(-a^2*x^2+1)^(1/2)/a/c^4/(-a*x+1)
Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.44 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx=\frac {\sqrt {1+a x} \left (7-6 a x+2 a^2 x^2\right )}{15 a c^4 (1-a x)^{5/2}} \] Input:
Integrate[1/(E^ArcTanh[a*x]*(c - a*c*x)^4),x]
Output:
(Sqrt[1 + a*x]*(7 - 6*a*x + 2*a^2*x^2))/(15*a*c^4*(1 - a*x)^(5/2))
Time = 0.44 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6677, 27, 461, 461, 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.
\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle \frac {\int \frac {1}{c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}dx}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{(1-a x)^3 \sqrt {1-a^2 x^2}}dx}{c^4}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {2}{5} \int \frac {1}{(1-a x)^2 \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{5 a (1-a x)^3}}{c^4}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {2}{5} \left (\frac {1}{3} \int \frac {1}{(1-a x) \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)^2}\right )+\frac {\sqrt {1-a^2 x^2}}{5 a (1-a x)^3}}{c^4}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {\frac {2}{5} \left (\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)}+\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)^2}\right )+\frac {\sqrt {1-a^2 x^2}}{5 a (1-a x)^3}}{c^4}\) |
Input:
Int[1/(E^ArcTanh[a*x]*(c - a*c*x)^4),x]
Output:
(Sqrt[1 - a^2*x^2]/(5*a*(1 - a*x)^3) + (2*(Sqrt[1 - a^2*x^2]/(3*a*(1 - a*x )^2) + Sqrt[1 - a^2*x^2]/(3*a*(1 - a*x))))/5)/c^4
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(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] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[c^n Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.43
method | result | size |
gosper | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 a^{2} x^{2}-6 a x +7\right )}{15 \left (a x -1\right )^{3} a \,c^{4}}\) | \(42\) |
trager | \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 a^{2} x^{2}-6 a x +7\right )}{15 \left (a x -1\right )^{3} a \,c^{4}}\) | \(42\) |
orering | \(-\frac {\left (2 a^{2} x^{2}-6 a x +7\right ) \left (a x -1\right ) \sqrt {-a^{2} x^{2}+1}}{15 a \left (-a c x +c \right )^{4}}\) | \(46\) |
default | \(\frac {\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{5 a \left (x -\frac {1}{a}\right )^{4}}-\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{15 \left (x -\frac {1}{a}\right )^{3}}}{2 a^{4}}-\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{12 a^{4} \left (x -\frac {1}{a}\right )^{3}}+\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a \left (x -\frac {1}{a}\right )^{2}}+a \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{8 a^{2}}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{16 a}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{16 a}}{c^{4}}\) | \(394\) |
Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x,method=_RETURNVERBOSE)
Output:
-1/15*(-a^2*x^2+1)^(1/2)*(2*a^2*x^2-6*a*x+7)/(a*x-1)^3/a/c^4
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx=\frac {7 \, a^{3} x^{3} - 21 \, a^{2} x^{2} + 21 \, a x - {\left (2 \, a^{2} x^{2} - 6 \, a x + 7\right )} \sqrt {-a^{2} x^{2} + 1} - 7}{15 \, {\left (a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x - a c^{4}\right )}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x, algorithm="fricas")
Output:
1/15*(7*a^3*x^3 - 21*a^2*x^2 + 21*a*x - (2*a^2*x^2 - 6*a*x + 7)*sqrt(-a^2* x^2 + 1) - 7)/(a^4*c^4*x^3 - 3*a^3*c^4*x^2 + 3*a^2*c^4*x - a*c^4)
\[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx=\frac {\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\, dx}{c^{4}} \] Input:
integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**4,x)
Output:
Integral(sqrt(-a**2*x**2 + 1)/(a**5*x**5 - 3*a**4*x**4 + 2*a**3*x**3 + 2*a **2*x**2 - 3*a*x + 1), x)/c**4
\[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a c x - c\right )}^{4} {\left (a x + 1\right )}} \,d x } \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x, algorithm="maxima")
Output:
integrate(sqrt(-a^2*x^2 + 1)/((a*c*x - c)^4*(a*x + 1)), x)
Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.49 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx=-\frac {2 \, {\left (\frac {20 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {40 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {30 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 7\right )}}{15 \, c^{4} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \] Input:
integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x, algorithm="giac")
Output:
-2/15*(20*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 40*(sqrt(-a^2*x^2 + 1) *abs(a) + a)^2/(a^4*x^2) + 30*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 7)/(c^4*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))
Time = 0.06 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.31 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx=\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2\,a^3}{15\,c^4\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}-\frac {a^3}{5\,c^4\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^3}+\frac {2\,a^4}{15\,c^4\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^2\,\sqrt {-a^2}}\right )}{a^3\,\sqrt {-a^2}} \] Input:
int((1 - a^2*x^2)^(1/2)/((c - a*c*x)^4*(a*x + 1)),x)
Output:
((1 - a^2*x^2)^(1/2)*((2*a^3)/(15*c^4*(x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)) - a^3/(5*c^4*(x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)^3) + (2*a^4)/(15*c^4*(x*(-a^ 2)^(1/2) - (-a^2)^(1/2)/a)^2*(-a^2)^(1/2))))/(a^3*(-a^2)^(1/2))
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^4} \, dx=\frac {-\frac {2 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}}{3}+\frac {2 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )}{3}-\frac {8}{15}}{a \,c^{4} \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{5}-5 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}+10 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-10 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+5 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:
int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x)
Output:
(2*( - 3*tan(asin(a*x)/2)**5 - 10*tan(asin(a*x)/2)**2 + 5*tan(asin(a*x)/2) - 4))/(15*a*c**4*(tan(asin(a*x)/2)**5 - 5*tan(asin(a*x)/2)**4 + 10*tan(as in(a*x)/2)**3 - 10*tan(asin(a*x)/2)**2 + 5*tan(asin(a*x)/2) - 1))