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

Optimal. Leaf size=97 \[ \frac{(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{a c}-\frac{3 \sin ^{-1}(a x)}{a c} \]

[Out]

(1 - a*x)^3/(3*a*c*(1 - a^2*x^2)^(3/2)) - (2*(1 - a*x)^2)/(a*c*Sqrt[1 - a^2*x^2]) - (3*Sqrt[1 - a^2*x^2])/(a*c
) - (3*ArcSin[a*x])/(a*c)

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Rubi [A]  time = 0.233686, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6157, 6149, 1635, 21, 669, 641, 216} \[ \frac{(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{a c}-\frac{3 \sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - a*x)^3/(3*a*c*(1 - a^2*x^2)^(3/2)) - (2*(1 - a*x)^2)/(a*c*Sqrt[1 - a^2*x^2]) - (3*Sqrt[1 - a^2*x^2])/(a*c
) - (3*ArcSin[a*x])/(a*c)

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 669

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

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)}}{c-\frac{c}{a^2 x^2}} \, dx &=-\frac{a^2 \int \frac{e^{-3 \tanh ^{-1}(a x)} x^2}{1-a^2 x^2} \, dx}{c}\\ &=-\frac{a^2 \int \frac{x^2 (1-a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c}\\ &=\frac{(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 \int \frac{\left (\frac{3}{a^2}-\frac{3 x}{a}\right ) (1-a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c}\\ &=\frac{(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}+\frac{\int \frac{(1-a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=\frac{(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{3 \int \frac{1-a x}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=\frac{(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{a c}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=\frac{(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{3 \sqrt{1-a^2 x^2}}{a c}-\frac{3 \sin ^{-1}(a x)}{a c}\\ \end{align*}

Mathematica [A]  time = 0.0794545, size = 78, normalized size = 0.8 \[ \frac{3 a^3 x^3+16 a^2 x^2-9 (a x+1) \sqrt{1-a^2 x^2} \sin ^{-1}(a x)-5 a x-14}{3 a c (a x+1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14 - 5*a*x + 16*a^2*x^2 + 3*a^3*x^3 - 9*(1 + a*x)*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(3*a*c*(1 + a*x)*Sqrt[1 - a
^2*x^2])

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Maple [B]  time = 0.062, size = 330, normalized size = 3.4 \begin{align*}{\frac{1}{6\,{a}^{5}c \left ( x+{a}^{-1} \right ) ^{4}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{11}{12\,{a}^{4}c \left ( x+{a}^{-1} \right ) ^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{47}{24\,{a}^{3}c \left ( x+{a}^{-1} \right ) ^{2}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{95}{48\,ac} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{95\,x}{32\,c}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{95}{32\,c}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{48\,ac} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{x}{32\,c}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}-{\frac{1}{32\,c}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{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)/(c-c/a^2/x^2),x)

[Out]

1/6/a^5/c/(x+1/a)^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-11/12/a^4/c/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2
)-47/24/a^3/c/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-95/48/c/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-95/32/
c*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-95/32/c/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(
1/2))+1/48/a/c*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)-1/32/c*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x-1/32/c/(a^2)^(1/
2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))

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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 )}}\,{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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.15619, size = 238, normalized size = 2.45 \begin{align*} -\frac{14 \, a^{2} x^{2} + 28 \, a x - 18 \,{\left (a^{2} x^{2} + 2 \, a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (3 \, a^{2} x^{2} + 19 \, a x + 14\right )} \sqrt{-a^{2} x^{2} + 1} + 14}{3 \,{\left (a^{3} c x^{2} + 2 \, a^{2} c x + a c\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),x, algorithm="fricas")

[Out]

-1/3*(14*a^2*x^2 + 28*a*x - 18*(a^2*x^2 + 2*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^2*x^2 + 19*
a*x + 14)*sqrt(-a^2*x^2 + 1) + 14)/(a^3*c*x^2 + 2*a^2*c*x + a*c)

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

[Out]

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

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Giac [A]  time = 1.24862, size = 171, normalized size = 1.76 \begin{align*} -\frac{3 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a c} + \frac{2 \,{\left (\frac{24 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} + \frac{9 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + 11\right )}}{3 \, c{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} + 1\right )}^{3}{\left | a \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),x, algorithm="giac")

[Out]

-3*arcsin(a*x)*sgn(a)/(c*abs(a)) - sqrt(-a^2*x^2 + 1)/(a*c) + 2/3*(24*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x)
+ 9*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 11)/(c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)^3*abs(a
))