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

Optimal. Leaf size=159 \[ \frac{(1-a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}-\frac{38 (1-a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{137 (1-a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{245-181 a x}{35 a c^3 \sqrt{1-a^2 x^2}}-\frac{3 \sin ^{-1}(a x)}{a c^3} \]

[Out]

(1 - a*x)^3/(7*a*c^3*(1 - a^2*x^2)^(7/2)) - (38*(1 - a*x)^2)/(35*a*c^3*(1 - a^2*x^2)^(5/2)) + (137*(1 - a*x))/
(35*a*c^3*(1 - a^2*x^2)^(3/2)) - (245 - 181*a*x)/(35*a*c^3*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(a*c^3) - (3
*ArcSin[a*x])/(a*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.450863, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6157, 6149, 1635, 1814, 641, 216} \[ \frac{(1-a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}-\frac{38 (1-a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{137 (1-a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{245-181 a x}{35 a c^3 \sqrt{1-a^2 x^2}}-\frac{3 \sin ^{-1}(a x)}{a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - a*x)^3/(7*a*c^3*(1 - a^2*x^2)^(7/2)) - (38*(1 - a*x)^2)/(35*a*c^3*(1 - a^2*x^2)^(5/2)) + (137*(1 - a*x))/
(35*a*c^3*(1 - a^2*x^2)^(3/2)) - (245 - 181*a*x)/(35*a*c^3*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(a*c^3) - (3
*ArcSin[a*x])/(a*c^3)

Rule 6157

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

Rule 6149

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 641

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^3} \, dx &=-\frac{a^6 \int \frac{e^{-3 \tanh ^{-1}(a x)} x^6}{\left (1-a^2 x^2\right )^3} \, dx}{c^3}\\ &=-\frac{a^6 \int \frac{x^6 (1-a x)^3}{\left (1-a^2 x^2\right )^{9/2}} \, dx}{c^3}\\ &=\frac{(1-a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac{a^6 \int \frac{(1-a x)^2 \left (\frac{3}{a^6}-\frac{7 x}{a^5}+\frac{7 x^2}{a^4}-\frac{7 x^3}{a^3}+\frac{7 x^4}{a^2}-\frac{7 x^5}{a}\right )}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{7 c^3}\\ &=\frac{(1-a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}-\frac{38 (1-a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{a^6 \int \frac{(1-a x) \left (\frac{61}{a^6}-\frac{140 x}{a^5}+\frac{105 x^2}{a^4}-\frac{70 x^3}{a^3}+\frac{35 x^4}{a^2}\right )}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{35 c^3}\\ &=\frac{(1-a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}-\frac{38 (1-a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{137 (1-a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^6 \int \frac{\frac{228}{a^6}-\frac{630 x}{a^5}+\frac{315 x^2}{a^4}-\frac{105 x^3}{a^3}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{105 c^3}\\ &=\frac{(1-a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}-\frac{38 (1-a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{137 (1-a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{245-181 a x}{35 a c^3 \sqrt{1-a^2 x^2}}-\frac{a^6 \int \frac{\frac{315}{a^6}-\frac{105 x}{a^5}}{\sqrt{1-a^2 x^2}} \, dx}{105 c^3}\\ &=\frac{(1-a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}-\frac{38 (1-a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{137 (1-a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{245-181 a x}{35 a c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^3}\\ &=\frac{(1-a x)^3}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}-\frac{38 (1-a x)^2}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{137 (1-a x)}{35 a c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{245-181 a x}{35 a c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{3 \sin ^{-1}(a x)}{a c^3}\\ \end{align*}

Mathematica [A]  time = 0.126437, size = 94, normalized size = 0.59 \[ \frac{35 a^5 x^5+286 a^4 x^4+368 a^3 x^3-125 a^2 x^2-105 (a x+1)^3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)-423 a x-176}{35 a \sqrt{1-a^2 x^2} (a c x+c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-176 - 423*a*x - 125*a^2*x^2 + 368*a^3*x^3 + 286*a^4*x^4 + 35*a^5*x^5 - 105*(1 + a*x)^3*Sqrt[1 - a^2*x^2]*Arc
Sin[a*x])/(35*a*(c + a*c*x)^3*Sqrt[1 - a^2*x^2])

________________________________________________________________________________________

Maple [B]  time = 0.076, size = 494, normalized size = 3.1 \begin{align*} \text{result too large to display} \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)/(c-c/a^2/x^2)^3,x)

[Out]

1/64/a^4/c^3/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)-5/64/a^3/c^3/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^
(5/2)+51/512/c^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x+51/512/c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a
)^2-2*a*(x-1/a))^(1/2))-17/256/a/c^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+1/56/a^7/c^3/(x+1/a)^6*(-a^2*(x+1/a)^2
+2*a*(x+1/a))^(5/2)-61/560/a^6/c^3/(x+1/a)^5*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+5/16/a^5/c^3/(x+1/a)^4*(-a^2*(
x+1/a)^2+2*a*(x+1/a))^(5/2)-31/32/a^4/c^3/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-263/128/a^3/c^3/(x+1/a)
^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-1587/512/c^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-1587/512/c^3/(a^2)^(1/
2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))-529/256/a/c^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)

________________________________________________________________________________________

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 x + 1\right )}^{3}{\left (c - \frac{c}{a^{2} x^{2}}\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)/(c-c/a^2/x^2)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.63076, size = 483, normalized size = 3.04 \begin{align*} -\frac{176 \, a^{5} x^{5} + 528 \, a^{4} x^{4} + 352 \, a^{3} x^{3} - 352 \, a^{2} x^{2} - 528 \, a x - 210 \,{\left (a^{5} x^{5} + 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} - 3 \, a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (35 \, a^{5} x^{5} + 286 \, a^{4} x^{4} + 368 \, a^{3} x^{3} - 125 \, a^{2} x^{2} - 423 \, a x - 176\right )} \sqrt{-a^{2} x^{2} + 1} - 176}{35 \,{\left (a^{6} c^{3} x^{5} + 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x - a c^{3}\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)/(c-c/a^2/x^2)^3,x, algorithm="fricas")

[Out]

-1/35*(176*a^5*x^5 + 528*a^4*x^4 + 352*a^3*x^3 - 352*a^2*x^2 - 528*a*x - 210*(a^5*x^5 + 3*a^4*x^4 + 2*a^3*x^3
- 2*a^2*x^2 - 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (35*a^5*x^5 + 286*a^4*x^4 + 368*a^3*x^3 - 12
5*a^2*x^2 - 423*a*x - 176)*sqrt(-a^2*x^2 + 1) - 176)/(a^6*c^3*x^5 + 3*a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^3*c^3*
x^2 - 3*a^2*c^3*x - a*c^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{6} \left (\int \frac{x^{6} \sqrt{- a^{2} x^{2} + 1}}{a^{9} x^{9} + 3 a^{8} x^{8} - 8 a^{6} x^{6} - 6 a^{5} x^{5} + 6 a^{4} x^{4} + 8 a^{3} x^{3} - 3 a x - 1}\, dx + \int - \frac{a^{2} x^{8} \sqrt{- a^{2} x^{2} + 1}}{a^{9} x^{9} + 3 a^{8} x^{8} - 8 a^{6} x^{6} - 6 a^{5} x^{5} + 6 a^{4} x^{4} + 8 a^{3} x^{3} - 3 a x - 1}\, dx\right )}{c^{3}} \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)/(c-c/a**2/x**2)**3,x)

[Out]

a**6*(Integral(x**6*sqrt(-a**2*x**2 + 1)/(a**9*x**9 + 3*a**8*x**8 - 8*a**6*x**6 - 6*a**5*x**5 + 6*a**4*x**4 +
8*a**3*x**3 - 3*a*x - 1), x) + Integral(-a**2*x**8*sqrt(-a**2*x**2 + 1)/(a**9*x**9 + 3*a**8*x**8 - 8*a**6*x**6
- 6*a**5*x**5 + 6*a**4*x**4 + 8*a**3*x**3 - 3*a*x - 1), x))/c**3

________________________________________________________________________________________

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 x + 1\right )}^{3}{\left (c - \frac{c}{a^{2} x^{2}}\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)/(c-c/a^2/x^2)^3,x, algorithm="giac")

[Out]

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