3.1199 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx\)

Optimal. Leaf size=97 \[ \frac {16 x}{35 c^4 \sqrt {1-a^2 x^2}}+\frac {8 x}{35 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac {1-a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}} \]

[Out]

1/7*(a*x-1)/a/c^4/(-a^2*x^2+1)^(7/2)+6/35*x/c^4/(-a^2*x^2+1)^(5/2)+8/35*x/c^4/(-a^2*x^2+1)^(3/2)+16/35*x/c^4/(
-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6139, 639, 192, 191} \[ \frac {16 x}{35 c^4 \sqrt {1-a^2 x^2}}+\frac {8 x}{35 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac {1-a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^4),x]

[Out]

-(1 - a*x)/(7*a*c^4*(1 - a^2*x^2)^(7/2)) + (6*x)/(35*c^4*(1 - a^2*x^2)^(5/2)) + (8*x)/(35*c^4*(1 - a^2*x^2)^(3
/2)) + (16*x)/(35*c^4*Sqrt[1 - a^2*x^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 6139

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a^2*x^2)^(p + n/
2)/(1 - a*x)^n, x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a^2*c + d, 0] && IntegerQ[p] && ILtQ[(n - 1)/2, 0] &&
!IntegerQ[p - n/2]

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx &=\frac {\int \frac {1-a x}{\left (1-a^2 x^2\right )^{9/2}} \, dx}{c^4}\\ &=-\frac {1-a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 \int \frac {1}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{7 c^4}\\ &=-\frac {1-a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {24 \int \frac {1}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{35 c^4}\\ &=-\frac {1-a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x}{35 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {16 \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{35 c^4}\\ &=-\frac {1-a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x}{35 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {16 x}{35 c^4 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 75, normalized size = 0.77 \[ \frac {16 a^6 x^6+16 a^5 x^5-40 a^4 x^4-40 a^3 x^3+30 a^2 x^2+30 a x-5}{35 a c^4 (1-a x)^{5/2} (a x+1)^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - a^2*c*x^2)^4),x]

[Out]

(-5 + 30*a*x + 30*a^2*x^2 - 40*a^3*x^3 - 40*a^4*x^4 + 16*a^5*x^5 + 16*a^6*x^6)/(35*a*c^4*(1 - a*x)^(5/2)*(1 +
a*x)^(7/2))

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fricas [B]  time = 0.59, size = 197, normalized size = 2.03 \[ -\frac {5 \, a^{7} x^{7} + 5 \, a^{6} x^{6} - 15 \, a^{5} x^{5} - 15 \, a^{4} x^{4} + 15 \, a^{3} x^{3} + 15 \, a^{2} x^{2} - 5 \, a x + {\left (16 \, a^{6} x^{6} + 16 \, a^{5} x^{5} - 40 \, a^{4} x^{4} - 40 \, a^{3} x^{3} + 30 \, a^{2} x^{2} + 30 \, a x - 5\right )} \sqrt {-a^{2} x^{2} + 1} - 5}{35 \, {\left (a^{8} c^{4} x^{7} + a^{7} c^{4} x^{6} - 3 \, a^{6} c^{4} x^{5} - 3 \, a^{5} c^{4} x^{4} + 3 \, a^{4} c^{4} x^{3} + 3 \, a^{3} c^{4} x^{2} - a^{2} c^{4} x - a c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^4,x, algorithm="fricas")

[Out]

-1/35*(5*a^7*x^7 + 5*a^6*x^6 - 15*a^5*x^5 - 15*a^4*x^4 + 15*a^3*x^3 + 15*a^2*x^2 - 5*a*x + (16*a^6*x^6 + 16*a^
5*x^5 - 40*a^4*x^4 - 40*a^3*x^3 + 30*a^2*x^2 + 30*a*x - 5)*sqrt(-a^2*x^2 + 1) - 5)/(a^8*c^4*x^7 + a^7*c^4*x^6
- 3*a^6*c^4*x^5 - 3*a^5*c^4*x^4 + 3*a^4*c^4*x^3 + 3*a^3*c^4*x^2 - a^2*c^4*x - a*c^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a^{2} c x^{2} - c\right )}^{4} {\left (a x + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^4,x, algorithm="giac")

[Out]

integrate(sqrt(-a^2*x^2 + 1)/((a^2*c*x^2 - c)^4*(a*x + 1)), x)

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maple [A]  time = 0.03, size = 74, normalized size = 0.76 \[ \frac {16 x^{6} a^{6}+16 x^{5} a^{5}-40 x^{4} a^{4}-40 x^{3} a^{3}+30 a^{2} x^{2}+30 a x -5}{35 \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}} \left (a x +1\right ) c^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^4,x)

[Out]

1/35/(-a^2*x^2+1)^(5/2)*(16*a^6*x^6+16*a^5*x^5-40*a^4*x^4-40*a^3*x^3+30*a^2*x^2+30*a*x-5)/(a*x+1)/c^4/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a^{2} c x^{2} - c\right )}^{4} {\left (a x + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*x^2 + 1)/((a^2*c*x^2 - c)^4*(a*x + 1)), x)

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mupad [B]  time = 1.18, size = 145, normalized size = 1.49 \[ \frac {\sqrt {1-a^2\,x^2}\,\left (\frac {8\,x}{35\,c^4}+\frac {1}{56\,a\,c^4}\right )}{{\left (a\,x-1\right )}^2\,{\left (a\,x+1\right )}^2}-\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {17\,x}{70\,c^4}-\frac {1}{7\,a\,c^4}\right )}{{\left (a\,x-1\right )}^3\,{\left (a\,x+1\right )}^3}-\frac {\sqrt {1-a^2\,x^2}}{56\,a\,c^4\,{\left (a\,x+1\right )}^4}-\frac {16\,x\,\sqrt {1-a^2\,x^2}}{35\,c^4\,\left (a\,x-1\right )\,\left (a\,x+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(1/2)/((c - a^2*c*x^2)^4*(a*x + 1)),x)

[Out]

((1 - a^2*x^2)^(1/2)*((8*x)/(35*c^4) + 1/(56*a*c^4)))/((a*x - 1)^2*(a*x + 1)^2) - ((1 - a^2*x^2)^(1/2)*((17*x)
/(70*c^4) - 1/(7*a*c^4)))/((a*x - 1)^3*(a*x + 1)^3) - (1 - a^2*x^2)^(1/2)/(56*a*c^4*(a*x + 1)^4) - (16*x*(1 -
a^2*x^2)^(1/2))/(35*c^4*(a*x - 1)*(a*x + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{- a^{7} x^{7} \sqrt {- a^{2} x^{2} + 1} - a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 3 a^{5} x^{5} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 3 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a**2*c*x**2+c)**4,x)

[Out]

Integral(1/(-a**7*x**7*sqrt(-a**2*x**2 + 1) - a**6*x**6*sqrt(-a**2*x**2 + 1) + 3*a**5*x**5*sqrt(-a**2*x**2 + 1
) + 3*a**4*x**4*sqrt(-a**2*x**2 + 1) - 3*a**3*x**3*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2 + 1) + a
*x*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x)/c**4

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