Integrand size = 18, antiderivative size = 87 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {2 x}{5 c^5 \sqrt {1-a^2 x^2}}+\frac {1}{5 a c^5 (1-a x)^2 \sqrt {1-a^2 x^2}}+\frac {1}{5 a c^5 (1-a x) \sqrt {1-a^2 x^2}} \] Output:
2/5*x/c^5/(-a^2*x^2+1)^(1/2)+1/5/a/c^5/(-a*x+1)^2/(-a^2*x^2+1)^(1/2)+1/5/a /c^5/(-a*x+1)/(-a^2*x^2+1)^(1/2)
Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {2+a x-4 a^2 x^2+2 a^3 x^3}{5 a c^5 (-1+a x)^2 \sqrt {1-a^2 x^2}} \] Input:
Integrate[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^5),x]
Output:
(2 + a*x - 4*a^2*x^2 + 2*a^3*x^3)/(5*a*c^5*(-1 + a*x)^2*Sqrt[1 - a^2*x^2])
Time = 0.27 (sec) , antiderivative size = 87, 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, 470, 208}
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.
\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle \frac {\int \frac {1}{c^2 (1-a x)^2 \left (1-a^2 x^2\right )^{3/2}}dx}{c^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{(1-a x)^2 \left (1-a^2 x^2\right )^{3/2}}dx}{c^5}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {3}{5} \int \frac {1}{(1-a x) \left (1-a^2 x^2\right )^{3/2}}dx+\frac {1}{5 a (1-a x)^2 \sqrt {1-a^2 x^2}}}{c^5}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {\frac {3}{5} \left (\frac {2}{3} \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {1}{3 a (1-a x) \sqrt {1-a^2 x^2}}\right )+\frac {1}{5 a (1-a x)^2 \sqrt {1-a^2 x^2}}}{c^5}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {3}{5} \left (\frac {2 x}{3 \sqrt {1-a^2 x^2}}+\frac {1}{3 a (1-a x) \sqrt {1-a^2 x^2}}\right )+\frac {1}{5 a (1-a x)^2 \sqrt {1-a^2 x^2}}}{c^5}\) |
Input:
Int[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^5),x]
Output:
(1/(5*a*(1 - a*x)^2*Sqrt[1 - a^2*x^2]) + (3*((2*x)/(3*Sqrt[1 - a^2*x^2]) + 1/(3*a*(1 - a*x)*Sqrt[1 - a^2*x^2])))/5)/c^5
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_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, 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)/(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[((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[(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, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + p + 1, 0] && IntegerQ[2*p]
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 = 0.35 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right )}{5 \left (a x -1\right )^{4} c^{5} a \left (a x +1\right )^{2}}\) | \(56\) |
trager | \(-\frac {\left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \sqrt {-a^{2} x^{2}+1}}{5 c^{5} \left (a x -1\right )^{3} a \left (a x +1\right )}\) | \(56\) |
orering | \(-\frac {\left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \left (a x -1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{5 a \left (a x +1\right )^{2} \left (-a c x +c \right )^{5}}\) | \(60\) |
default | \(\text {Expression too large to display}\) | \(1289\) |
Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/5*(-a^2*x^2+1)^(3/2)*(2*a^3*x^3-4*a^2*x^2+a*x+2)/(a*x-1)^4/c^5/a/(a*x+1) ^2
Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {2 \, a^{4} x^{4} - 4 \, a^{3} x^{3} + 4 \, a x - {\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} - 2}{5 \, {\left (a^{5} c^{5} x^{4} - 2 \, a^{4} c^{5} x^{3} + 2 \, a^{2} c^{5} x - a c^{5}\right )}} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^5,x, algorithm="fricas ")
Output:
1/5*(2*a^4*x^4 - 4*a^3*x^3 + 4*a*x - (2*a^3*x^3 - 4*a^2*x^2 + a*x + 2)*sqr t(-a^2*x^2 + 1) - 2)/(a^5*c^5*x^4 - 2*a^4*c^5*x^3 + 2*a^2*c^5*x - a*c^5)
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx=- \frac {\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{8} x^{8} - 2 a^{7} x^{7} - 2 a^{6} x^{6} + 6 a^{5} x^{5} - 6 a^{3} x^{3} + 2 a^{2} x^{2} + 2 a x - 1}\, dx + \int \left (- \frac {a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{8} x^{8} - 2 a^{7} x^{7} - 2 a^{6} x^{6} + 6 a^{5} x^{5} - 6 a^{3} x^{3} + 2 a^{2} x^{2} + 2 a x - 1}\right )\, dx}{c^{5}} \] Input:
integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a*c*x+c)**5,x)
Output:
-(Integral(sqrt(-a**2*x**2 + 1)/(a**8*x**8 - 2*a**7*x**7 - 2*a**6*x**6 + 6 *a**5*x**5 - 6*a**3*x**3 + 2*a**2*x**2 + 2*a*x - 1), x) + Integral(-a**2*x **2*sqrt(-a**2*x**2 + 1)/(a**8*x**8 - 2*a**7*x**7 - 2*a**6*x**6 + 6*a**5*x **5 - 6*a**3*x**3 + 2*a**2*x**2 + 2*a*x - 1), x))/c**5
\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\int { -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a c x - c\right )}^{5} {\left (a x + 1\right )}^{3}} \,d x } \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^5,x, algorithm="maxima ")
Output:
-integrate((-a^2*x^2 + 1)^(3/2)/((a*c*x - c)^5*(a*x + 1)^3), x)
Result contains complex when optimal does not.
Time = 0.15 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.00 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {1}{40} \, {\left (a {\left (\frac {5}{a^{3} c^{7} \sqrt {-\frac {2 \, c}{a c x - c} - 1}} - \frac {a^{12} c^{28} {\left (\frac {2 \, c}{a c x - c} + 1\right )}^{2} \sqrt {-\frac {2 \, c}{a c x - c} - 1} + 5 \, a^{12} c^{28} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} + 15 \, a^{12} c^{28} \sqrt {-\frac {2 \, c}{a c x - c} - 1}}{a^{15} c^{35}}\right )} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right ) + \frac {16 i \, \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )}{a^{2} c^{7}}\right )} c^{2} {\left | a \right |} \] Input:
integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^5,x, algorithm="giac")
Output:
1/40*(a*(5/(a^3*c^7*sqrt(-2*c/(a*c*x - c) - 1)) - (a^12*c^28*(2*c/(a*c*x - c) + 1)^2*sqrt(-2*c/(a*c*x - c) - 1) + 5*a^12*c^28*(-2*c/(a*c*x - c) - 1) ^(3/2) + 15*a^12*c^28*sqrt(-2*c/(a*c*x - c) - 1))/(a^15*c^35))*sgn(1/(a*c* x - c))*sgn(a)*sgn(c) + 16*I*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)/(a^2*c^7))*c ^2*abs(a)
Time = 13.90 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.68 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {3\,a\,\sqrt {1-a^2\,x^2}}{20\,\left (a^4\,c^5\,x^2-2\,a^3\,c^5\,x+a^2\,c^5\right )}+\frac {\sqrt {1-a^2\,x^2}}{8\,\sqrt {-a^2}\,\left (c^5\,x\,\sqrt {-a^2}+\frac {c^5\,\sqrt {-a^2}}{a}\right )}+\frac {11\,\sqrt {1-a^2\,x^2}}{40\,\sqrt {-a^2}\,\left (c^5\,x\,\sqrt {-a^2}-\frac {c^5\,\sqrt {-a^2}}{a}\right )}+\frac {\sqrt {1-a^2\,x^2}}{10\,\sqrt {-a^2}\,\left (3\,c^5\,x\,\sqrt {-a^2}-\frac {c^5\,\sqrt {-a^2}}{a}+a^2\,c^5\,x^3\,\sqrt {-a^2}-3\,a\,c^5\,x^2\,\sqrt {-a^2}\right )} \] Input:
int((1 - a^2*x^2)^(3/2)/((c - a*c*x)^5*(a*x + 1)^3),x)
Output:
(3*a*(1 - a^2*x^2)^(1/2))/(20*(a^2*c^5 - 2*a^3*c^5*x + a^4*c^5*x^2)) + (1 - a^2*x^2)^(1/2)/(8*(-a^2)^(1/2)*(c^5*x*(-a^2)^(1/2) + (c^5*(-a^2)^(1/2))/ a)) + (11*(1 - a^2*x^2)^(1/2))/(40*(-a^2)^(1/2)*(c^5*x*(-a^2)^(1/2) - (c^5 *(-a^2)^(1/2))/a)) + (1 - a^2*x^2)^(1/2)/(10*(-a^2)^(1/2)*(3*c^5*x*(-a^2)^ (1/2) - (c^5*(-a^2)^(1/2))/a + a^2*c^5*x^3*(-a^2)^(1/2) - 3*a*c^5*x^2*(-a^ 2)^(1/2)))
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {-\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+2 \sqrt {-a^{2} x^{2}+1}\, a x -\sqrt {-a^{2} x^{2}+1}+4 a^{3} x^{3}-8 a^{2} x^{2}+2 a x +4}{10 \sqrt {-a^{2} x^{2}+1}\, a \,c^{5} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^5,x)
Output:
( - sqrt( - a**2*x**2 + 1)*a**2*x**2 + 2*sqrt( - a**2*x**2 + 1)*a*x - sqrt ( - a**2*x**2 + 1) + 4*a**3*x**3 - 8*a**2*x**2 + 2*a*x + 4)/(10*sqrt( - a* *2*x**2 + 1)*a*c**5*(a**2*x**2 - 2*a*x + 1))