3.205 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-a c x)^5} \, dx\)

Optimal. Leaf size=129 \[ \frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)}+\frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^2}+\frac {3 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^3}+\frac {\sqrt {1-a^2 x^2}}{7 a c^5 (1-a x)^4} \]

[Out]

1/7*(-a^2*x^2+1)^(1/2)/a/c^5/(-a*x+1)^4+3/35*(-a^2*x^2+1)^(1/2)/a/c^5/(-a*x+1)^3+2/35*(-a^2*x^2+1)^(1/2)/a/c^5
/(-a*x+1)^2+2/35*(-a^2*x^2+1)^(1/2)/a/c^5/(-a*x+1)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6127, 659, 651} \[ \frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)}+\frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^2}+\frac {3 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^3}+\frac {\sqrt {1-a^2 x^2}}{7 a c^5 (1-a x)^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - a*c*x)^5),x]

[Out]

Sqrt[1 - a^2*x^2]/(7*a*c^5*(1 - a*x)^4) + (3*Sqrt[1 - a^2*x^2])/(35*a*c^5*(1 - a*x)^3) + (2*Sqrt[1 - a^2*x^2])
/(35*a*c^5*(1 - a*x)^2) + (2*Sqrt[1 - a^2*x^2])/(35*a*c^5*(1 - a*x))

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

Rule 6127

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[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]

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)}}{(c-a c x)^5} \, dx &=\frac {\int \frac {1}{(c-a c x)^4 \sqrt {1-a^2 x^2}} \, dx}{c}\\ &=\frac {\sqrt {1-a^2 x^2}}{7 a c^5 (1-a x)^4}+\frac {3 \int \frac {1}{(c-a c x)^3 \sqrt {1-a^2 x^2}} \, dx}{7 c^2}\\ &=\frac {\sqrt {1-a^2 x^2}}{7 a c^5 (1-a x)^4}+\frac {3 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^3}+\frac {6 \int \frac {1}{(c-a c x)^2 \sqrt {1-a^2 x^2}} \, dx}{35 c^3}\\ &=\frac {\sqrt {1-a^2 x^2}}{7 a c^5 (1-a x)^4}+\frac {3 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^3}+\frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^2}+\frac {2 \int \frac {1}{(c-a c x) \sqrt {1-a^2 x^2}} \, dx}{35 c^4}\\ &=\frac {\sqrt {1-a^2 x^2}}{7 a c^5 (1-a x)^4}+\frac {3 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^3}+\frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^2}+\frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 51, normalized size = 0.40 \[ -\frac {\sqrt {a x+1} \left (2 a^3 x^3-8 a^2 x^2+13 a x-12\right )}{35 a c^5 (1-a x)^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - a*c*x)^5),x]

[Out]

-1/35*(Sqrt[1 + a*x]*(-12 + 13*a*x - 8*a^2*x^2 + 2*a^3*x^3))/(a*c^5*(1 - a*x)^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 117, normalized size = 0.91 \[ \frac {12 \, a^{4} x^{4} - 48 \, a^{3} x^{3} + 72 \, a^{2} x^{2} - 48 \, a x - {\left (2 \, a^{3} x^{3} - 8 \, a^{2} x^{2} + 13 \, a x - 12\right )} \sqrt {-a^{2} x^{2} + 1} + 12}{35 \, {\left (a^{5} c^{5} x^{4} - 4 \, a^{4} c^{5} x^{3} + 6 \, a^{3} c^{5} x^{2} - 4 \, a^{2} c^{5} x + a c^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x, algorithm="fricas")

[Out]

1/35*(12*a^4*x^4 - 48*a^3*x^3 + 72*a^2*x^2 - 48*a*x - (2*a^3*x^3 - 8*a^2*x^2 + 13*a*x - 12)*sqrt(-a^2*x^2 + 1)
+ 12)/(a^5*c^5*x^4 - 4*a^4*c^5*x^3 + 6*a^3*c^5*x^2 - 4*a^2*c^5*x + a*c^5)

________________________________________________________________________________________

giac [C]  time = 0.21, size = 164, normalized size = 1.27 \[ \frac {1}{280} \, c^{2} {\left (\frac {{\left (5 \, {\left (\frac {2 \, c}{a c x - c} + 1\right )}^{3} \sqrt {-\frac {2 \, c}{a c x - c} - 1} - 21 \, {\left (\frac {2 \, c}{a c x - c} + 1\right )}^{2} \sqrt {-\frac {2 \, c}{a c x - c} - 1} - 35 \, {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} - 35 \, \sqrt {-\frac {2 \, c}{a c x - c} - 1}\right )} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (c)}{a^{2} c^{7}} + \frac {16 i \, \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (c)}{a^{2} c^{7}}\right )} {\left | a \right |} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x, algorithm="giac")

[Out]

1/280*c^2*((5*(2*c/(a*c*x - c) + 1)^3*sqrt(-2*c/(a*c*x - c) - 1) - 21*(2*c/(a*c*x - c) + 1)^2*sqrt(-2*c/(a*c*x
- c) - 1) - 35*(-2*c/(a*c*x - c) - 1)^(3/2) - 35*sqrt(-2*c/(a*c*x - c) - 1))*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)
/(a^2*c^7) + 16*I*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)/(a^2*c^7))*abs(a)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 50, normalized size = 0.39 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 x^{3} a^{3}-8 a^{2} x^{2}+13 a x -12\right )}{35 \left (a x -1\right )^{4} c^{5} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x)

[Out]

-1/35*(-a^2*x^2+1)^(1/2)*(2*a^3*x^3-8*a^2*x^2+13*a*x-12)/(a*x-1)^4/c^5/a

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a c x - c\right )}^{5} {\left (a x + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x, algorithm="maxima")

[Out]

-integrate(sqrt(-a^2*x^2 + 1)/((a*c*x - c)^5*(a*x + 1)), x)

________________________________________________________________________________________

mupad [B]  time = 0.80, size = 167, normalized size = 1.29 \[ -\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {3\,a^4}{35\,c^5\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^3}-\frac {2\,a^4}{35\,c^5\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}+\frac {a^5}{7\,c^5\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^4\,\sqrt {-a^2}}+\frac {2\,a^7}{35\,c^5\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^2\,{\left (-a^2\right )}^{3/2}}\right )}{a^4\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(1/2)/((c - a*c*x)^5*(a*x + 1)),x)

[Out]

-((1 - a^2*x^2)^(1/2)*((3*a^4)/(35*c^5*(x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)^3) - (2*a^4)/(35*c^5*(x*(-a^2)^(1/2)
- (-a^2)^(1/2)/a)) + a^5/(7*c^5*(x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)^4*(-a^2)^(1/2)) + (2*a^7)/(35*c^5*(x*(-a^2)^
(1/2) - (-a^2)^(1/2)/a)^2*(-a^2)^(3/2))))/(a^4*(-a^2)^(1/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{6} x^{6} - 4 a^{5} x^{5} + 5 a^{4} x^{4} - 5 a^{2} x^{2} + 4 a x - 1}\, dx}{c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**5,x)

[Out]

-Integral(sqrt(-a**2*x**2 + 1)/(a**6*x**6 - 4*a**5*x**5 + 5*a**4*x**4 - 5*a**2*x**2 + 4*a*x - 1), x)/c**5

________________________________________________________________________________________