3.1261 \(\int \frac{e^{-3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx\)

Optimal. Leaf size=138 \[ \frac{16 x}{63 c^4 \sqrt{1-a^2 x^2}}+\frac{8 x}{63 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (a x+1) \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (a x+1)^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{1}{9 a c^4 (a x+1)^3 \left (1-a^2 x^2\right )^{3/2}} \]

[Out]

(8*x)/(63*c^4*(1 - a^2*x^2)^(3/2)) - 1/(9*a*c^4*(1 + a*x)^3*(1 - a^2*x^2)^(3/2)) - 2/(21*a*c^4*(1 + a*x)^2*(1
- a^2*x^2)^(3/2)) - 2/(21*a*c^4*(1 + a*x)*(1 - a^2*x^2)^(3/2)) + (16*x)/(63*c^4*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.0930441, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6139, 655, 659, 192, 191} \[ \frac{16 x}{63 c^4 \sqrt{1-a^2 x^2}}+\frac{8 x}{63 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (a x+1) \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (a x+1)^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{1}{9 a c^4 (a x+1)^3 \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^4),x]

[Out]

(8*x)/(63*c^4*(1 - a^2*x^2)^(3/2)) - 1/(9*a*c^4*(1 + a*x)^3*(1 - a^2*x^2)^(3/2)) - 2/(21*a*c^4*(1 + a*x)^2*(1
- a^2*x^2)^(3/2)) - 2/(21*a*c^4*(1 + a*x)*(1 - a^2*x^2)^(3/2)) + (16*x)/(63*c^4*Sqrt[1 - a^2*x^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]

Rule 655

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx &=\frac{\int \frac{(1-a x)^3}{\left (1-a^2 x^2\right )^{11/2}} \, dx}{c^4}\\ &=\frac{\int \frac{1}{(1+a x)^3 \left (1-a^2 x^2\right )^{5/2}} \, dx}{c^4}\\ &=-\frac{1}{9 a c^4 (1+a x)^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{2 \int \frac{1}{(1+a x)^2 \left (1-a^2 x^2\right )^{5/2}} \, dx}{3 c^4}\\ &=-\frac{1}{9 a c^4 (1+a x)^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (1+a x)^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{10 \int \frac{1}{(1+a x) \left (1-a^2 x^2\right )^{5/2}} \, dx}{21 c^4}\\ &=-\frac{1}{9 a c^4 (1+a x)^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (1+a x)^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (1+a x) \left (1-a^2 x^2\right )^{3/2}}+\frac{8 \int \frac{1}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{21 c^4}\\ &=\frac{8 x}{63 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{1}{9 a c^4 (1+a x)^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (1+a x)^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (1+a x) \left (1-a^2 x^2\right )^{3/2}}+\frac{16 \int \frac{1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{63 c^4}\\ &=\frac{8 x}{63 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{1}{9 a c^4 (1+a x)^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (1+a x)^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2}{21 a c^4 (1+a x) \left (1-a^2 x^2\right )^{3/2}}+\frac{16 x}{63 c^4 \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0323553, size = 75, normalized size = 0.54 \[ -\frac{16 a^6 x^6+48 a^5 x^5+24 a^4 x^4-56 a^3 x^3-66 a^2 x^2-6 a x+19}{63 a c^4 (1-a x)^{3/2} (a x+1)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^4),x]

[Out]

-(19 - 6*a*x - 66*a^2*x^2 - 56*a^3*x^3 + 24*a^4*x^4 + 48*a^5*x^5 + 16*a^6*x^6)/(63*a*c^4*(1 - a*x)^(3/2)*(1 +
a*x)^(9/2))

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Maple [A]  time = 0.03, size = 74, normalized size = 0.5 \begin{align*} -{\frac{16\,{x}^{6}{a}^{6}+48\,{x}^{5}{a}^{5}+24\,{x}^{4}{a}^{4}-56\,{x}^{3}{a}^{3}-66\,{a}^{2}{x}^{2}-6\,ax+19}{63\, \left ( ax+1 \right ) ^{3}{c}^{4}a} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^4,x)

[Out]

-1/63/(-a^2*x^2+1)^(3/2)*(16*a^6*x^6+48*a^5*x^5+24*a^4*x^4-56*a^3*x^3-66*a^2*x^2-6*a*x+19)/(a*x+1)^3/c^4/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{4}{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^4,x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((a^2*c*x^2 - c)^4*(a*x + 1)^3), x)

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Fricas [A]  time = 3.07722, size = 419, normalized size = 3.04 \begin{align*} -\frac{19 \, a^{7} x^{7} + 57 \, a^{6} x^{6} + 19 \, a^{5} x^{5} - 95 \, a^{4} x^{4} - 95 \, a^{3} x^{3} + 19 \, a^{2} x^{2} + 57 \, a x +{\left (16 \, a^{6} x^{6} + 48 \, a^{5} x^{5} + 24 \, a^{4} x^{4} - 56 \, a^{3} x^{3} - 66 \, a^{2} x^{2} - 6 \, a x + 19\right )} \sqrt{-a^{2} x^{2} + 1} + 19}{63 \,{\left (a^{8} c^{4} x^{7} + 3 \, a^{7} c^{4} x^{6} + a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} - 5 \, a^{4} c^{4} x^{3} + a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x + a c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^4,x, algorithm="fricas")

[Out]

-1/63*(19*a^7*x^7 + 57*a^6*x^6 + 19*a^5*x^5 - 95*a^4*x^4 - 95*a^3*x^3 + 19*a^2*x^2 + 57*a*x + (16*a^6*x^6 + 48
*a^5*x^5 + 24*a^4*x^4 - 56*a^3*x^3 - 66*a^2*x^2 - 6*a*x + 19)*sqrt(-a^2*x^2 + 1) + 19)/(a^8*c^4*x^7 + 3*a^7*c^
4*x^6 + a^6*c^4*x^5 - 5*a^5*c^4*x^4 - 5*a^4*c^4*x^3 + a^3*c^4*x^2 + 3*a^2*c^4*x + a*c^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a**2*c*x**2+c)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{4}{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^4,x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((a^2*c*x^2 - c)^4*(a*x + 1)^3), x)