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

Optimal. Leaf size=189 \[ -\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}+\frac{22 (1-a x)^2}{21 a c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{478 (1-a x)}{105 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{a c^4}-\frac{4 (630-431 a x)}{315 a c^4 \sqrt{1-a^2 x^2}}+\frac{2 (1155-829 a x)}{315 a c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{3 \sin ^{-1}(a x)}{a c^4} \]

[Out]

-(1 - a*x)^3/(9*a*c^4*(1 - a^2*x^2)^(9/2)) + (22*(1 - a*x)^2)/(21*a*c^4*(1 - a^2*x^2)^(7/2)) - (478*(1 - a*x))
/(105*a*c^4*(1 - a^2*x^2)^(5/2)) + (2*(1155 - 829*a*x))/(315*a*c^4*(1 - a^2*x^2)^(3/2)) - (4*(630 - 431*a*x))/
(315*a*c^4*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(a*c^4) - (3*ArcSin[a*x])/(a*c^4)

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Rubi [A]  time = 0.636255, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}+\frac{22 (1-a x)^2}{21 a c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{478 (1-a x)}{105 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{a c^4}-\frac{4 (630-431 a x)}{315 a c^4 \sqrt{1-a^2 x^2}}+\frac{2 (1155-829 a x)}{315 a c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{3 \sin ^{-1}(a x)}{a c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - a*x)^3/(9*a*c^4*(1 - a^2*x^2)^(9/2)) + (22*(1 - a*x)^2)/(21*a*c^4*(1 - a^2*x^2)^(7/2)) - (478*(1 - a*x))
/(105*a*c^4*(1 - a^2*x^2)^(5/2)) + (2*(1155 - 829*a*x))/(315*a*c^4*(1 - a^2*x^2)^(3/2)) - (4*(630 - 431*a*x))/
(315*a*c^4*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(a*c^4) - (3*ArcSin[a*x])/(a*c^4)

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 )^4} \, dx &=\frac{a^8 \int \frac{e^{-3 \tanh ^{-1}(a x)} x^8}{\left (1-a^2 x^2\right )^4} \, dx}{c^4}\\ &=\frac{a^8 \int \frac{x^8 (1-a x)^3}{\left (1-a^2 x^2\right )^{11/2}} \, dx}{c^4}\\ &=-\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}-\frac{a^8 \int \frac{(1-a x)^2 \left (\frac{3}{a^8}-\frac{9 x}{a^7}+\frac{9 x^2}{a^6}-\frac{9 x^3}{a^5}+\frac{9 x^4}{a^4}-\frac{9 x^5}{a^3}+\frac{9 x^6}{a^2}-\frac{9 x^7}{a}\right )}{\left (1-a^2 x^2\right )^{9/2}} \, dx}{9 c^4}\\ &=-\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}+\frac{22 (1-a x)^2}{21 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac{a^8 \int \frac{(1-a x) \left (\frac{111}{a^8}-\frac{378 x}{a^7}+\frac{315 x^2}{a^6}-\frac{252 x^3}{a^5}+\frac{189 x^4}{a^4}-\frac{126 x^5}{a^3}+\frac{63 x^6}{a^2}\right )}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{63 c^4}\\ &=-\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}+\frac{22 (1-a x)^2}{21 a c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{478 (1-a x)}{105 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{a^8 \int \frac{\frac{879}{a^8}-\frac{4725 x}{a^7}+\frac{3150 x^2}{a^6}-\frac{1890 x^3}{a^5}+\frac{945 x^4}{a^4}-\frac{315 x^5}{a^3}}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{315 c^4}\\ &=-\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}+\frac{22 (1-a x)^2}{21 a c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{478 (1-a x)}{105 a c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{2 (1155-829 a x)}{315 a c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^8 \int \frac{\frac{2337}{a^8}-\frac{6615 x}{a^7}+\frac{2835 x^2}{a^6}-\frac{945 x^3}{a^5}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{945 c^4}\\ &=-\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}+\frac{22 (1-a x)^2}{21 a c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{478 (1-a x)}{105 a c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{2 (1155-829 a x)}{315 a c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{4 (630-431 a x)}{315 a c^4 \sqrt{1-a^2 x^2}}-\frac{a^8 \int \frac{\frac{2835}{a^8}-\frac{945 x}{a^7}}{\sqrt{1-a^2 x^2}} \, dx}{945 c^4}\\ &=-\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}+\frac{22 (1-a x)^2}{21 a c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{478 (1-a x)}{105 a c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{2 (1155-829 a x)}{315 a c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{4 (630-431 a x)}{315 a c^4 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c^4}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^4}\\ &=-\frac{(1-a x)^3}{9 a c^4 \left (1-a^2 x^2\right )^{9/2}}+\frac{22 (1-a x)^2}{21 a c^4 \left (1-a^2 x^2\right )^{7/2}}-\frac{478 (1-a x)}{105 a c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac{2 (1155-829 a x)}{315 a c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac{4 (630-431 a x)}{315 a c^4 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c^4}-\frac{3 \sin ^{-1}(a x)}{a c^4}\\ \end{align*}

Mathematica [A]  time = 0.148053, size = 122, normalized size = 0.65 \[ \frac{315 a^7 x^7+2669 a^6 x^6+2967 a^5 x^5-4029 a^4 x^4-7399 a^3 x^3-339 a^2 x^2-945 (a x-1) (a x+1)^4 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)+4047 a x+1664}{315 a (a x-1) \sqrt{1-a^2 x^2} (a c x+c)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1664 + 4047*a*x - 339*a^2*x^2 - 7399*a^3*x^3 - 4029*a^4*x^4 + 2967*a^5*x^5 + 2669*a^6*x^6 + 315*a^7*x^7 - 945
*(-1 + a*x)*(1 + a*x)^4*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(315*a*(-1 + a*x)*(c + a*c*x)^4*Sqrt[1 - a^2*x^2])

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Maple [B]  time = 0.095, size = 576, 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)^4,x)

[Out]

-1723/10080/a^6/c^4/(x+1/a)^5*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-31/256/a/c^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/
2)+1/384/a^5/c^4/(x-1/a)^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)-1/144/a^8/c^4/(x+1/a)^7*(-a^2*(x+1/a)^2+2*a*(x+1
/a))^(5/2)+13/252/a^7/c^4/(x+1/a)^6*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+35/96/a^5/c^4/(x+1/a)^4*(-a^2*(x+1/a)^2
+2*a*(x+1/a))^(5/2)-769/768/a^4/c^4/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-811/384/a^3/c^4/(x+1/a)^2*(-a
^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-1629/512/c^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-1629/512/c^4/(a^2)^(1/2)*arc
tan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))+29/768/a^4/c^4/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5
/2)-25/192/a^3/c^4/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)+93/512/c^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*
x+93/512/c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))-543/256/a/c^4*(-a^2*(x+1/a)^
2+2*a*(x+1/a))^(3/2)

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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 x + 1\right )}^{3}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{4}}\,{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)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.63303, size = 647, normalized size = 3.42 \begin{align*} -\frac{1664 \, a^{7} x^{7} + 4992 \, a^{6} x^{6} + 1664 \, a^{5} x^{5} - 8320 \, a^{4} x^{4} - 8320 \, a^{3} x^{3} + 1664 \, a^{2} x^{2} + 4992 \, a x - 1890 \,{\left (a^{7} x^{7} + 3 \, a^{6} x^{6} + a^{5} x^{5} - 5 \, a^{4} x^{4} - 5 \, a^{3} x^{3} + a^{2} x^{2} + 3 \, a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (315 \, a^{7} x^{7} + 2669 \, a^{6} x^{6} + 2967 \, a^{5} x^{5} - 4029 \, a^{4} x^{4} - 7399 \, a^{3} x^{3} - 339 \, a^{2} x^{2} + 4047 \, a x + 1664\right )} \sqrt{-a^{2} x^{2} + 1} + 1664}{315 \,{\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)/(c-c/a^2/x^2)^4,x, algorithm="fricas")

[Out]

-1/315*(1664*a^7*x^7 + 4992*a^6*x^6 + 1664*a^5*x^5 - 8320*a^4*x^4 - 8320*a^3*x^3 + 1664*a^2*x^2 + 4992*a*x - 1
890*(a^7*x^7 + 3*a^6*x^6 + a^5*x^5 - 5*a^4*x^4 - 5*a^3*x^3 + a^2*x^2 + 3*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) -
1)/(a*x)) + (315*a^7*x^7 + 2669*a^6*x^6 + 2967*a^5*x^5 - 4029*a^4*x^4 - 7399*a^3*x^3 - 339*a^2*x^2 + 4047*a*x
+ 1664)*sqrt(-a^2*x^2 + 1) + 1664)/(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)/(c-c/a**2/x**2)**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 x + 1\right )}^{3}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{4}}\,{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)^4,x, algorithm="giac")

[Out]

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