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

Optimal. Leaf size=74 \[ -\frac{c \sqrt{1-a^2 x^2}}{a}+\frac{c \sqrt{1-a^2 x^2}}{a^2 x}-\frac{3 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{a}-\frac{3 c \sin ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.207967, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {6157, 6149, 1807, 1809, 844, 216, 266, 63, 208} \[ -\frac{c \sqrt{1-a^2 x^2}}{a}+\frac{c \sqrt{1-a^2 x^2}}{a^2 x}-\frac{3 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{a}-\frac{3 c \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 844

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

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]

Rule 266

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right ) \, dx &=-\frac{c \int \frac{e^{-3 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )}{x^2} \, dx}{a^2}\\ &=-\frac{c \int \frac{(1-a x)^3}{x^2 \sqrt{1-a^2 x^2}} \, dx}{a^2}\\ &=\frac{c \sqrt{1-a^2 x^2}}{a^2 x}+\frac{c \int \frac{3 a-3 a^2 x+a^3 x^2}{x \sqrt{1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac{c \sqrt{1-a^2 x^2}}{a}+\frac{c \sqrt{1-a^2 x^2}}{a^2 x}-\frac{c \int \frac{-3 a^3+3 a^4 x}{x \sqrt{1-a^2 x^2}} \, dx}{a^4}\\ &=-\frac{c \sqrt{1-a^2 x^2}}{a}+\frac{c \sqrt{1-a^2 x^2}}{a^2 x}-(3 c) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx+\frac{(3 c) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{a}\\ &=-\frac{c \sqrt{1-a^2 x^2}}{a}+\frac{c \sqrt{1-a^2 x^2}}{a^2 x}-\frac{3 c \sin ^{-1}(a x)}{a}+\frac{(3 c) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{2 a}\\ &=-\frac{c \sqrt{1-a^2 x^2}}{a}+\frac{c \sqrt{1-a^2 x^2}}{a^2 x}-\frac{3 c \sin ^{-1}(a x)}{a}-\frac{(3 c) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{a^3}\\ &=-\frac{c \sqrt{1-a^2 x^2}}{a}+\frac{c \sqrt{1-a^2 x^2}}{a^2 x}-\frac{3 c \sin ^{-1}(a x)}{a}-\frac{3 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0628591, size = 57, normalized size = 0.77 \[ -\frac{c \left (\sqrt{1-a^2 x^2} (a x-1)+3 a x \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+3 a x \sin ^{-1}(a x)\right )}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.057, size = 266, normalized size = 3.6 \begin{align*}{\frac{c}{{a}^{2}x} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+cx \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}+{\frac{3\,cx}{2}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{3\,c}{2}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-2\,{\frac{c \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{{a}^{3} \left ( x+{a}^{-1} \right ) ^{2}}}-3\,{\frac{c \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{3/2}}{a}}-{\frac{9\,cx}{2}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{9\,c}{2}\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{c}{a} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-3\,{\frac{c{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }{a}}+3\,{\frac{c\sqrt{-{a}^{2}{x}^{2}+1}}{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c/a^2/x*(-a^2*x^2+1)^(5/2)+c*x*(-a^2*x^2+1)^(3/2)+3/2*c*x*(-a^2*x^2+1)^(1/2)+3/2*c/(a^2)^(1/2)*arctan((a^2)^(1
/2)*x/(-a^2*x^2+1)^(1/2))-2*c/a^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-3*c/a*(-a^2*(x+1/a)^2+2*a*(x+1/
a))^(3/2)-9/2*c*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-9/2*c/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*
a*(x+1/a))^(1/2))+c*(-a^2*x^2+1)^(3/2)/a-3*c/a*arctanh(1/(-a^2*x^2+1)^(1/2))+3*c*(-a^2*x^2+1)^(1/2)/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 (c - \frac{c}{a^{2} x^{2}}\right )}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.1854, size = 190, normalized size = 2.57 \begin{align*} \frac{6 \, a c x \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + 3 \, a c x \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - a c x - \sqrt{-a^{2} x^{2} + 1}{\left (a c x - c\right )}}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*a*c*x*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + 3*a*c*x*log((sqrt(-a^2*x^2 + 1) - 1)/x) - a*c*x - sqrt(-a^2*
x^2 + 1)*(a*c*x - c))/(a^2*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25703, size = 176, normalized size = 2.38 \begin{align*} -\frac{a^{2} c x}{2 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}{\left | a \right |}} - \frac{3 \, c \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} - \frac{3 \, c \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1} c}{a} + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c}{2 \, a^{2} x{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2*c*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*abs(a)) - 3*c*arcsin(a*x)*sgn(a)/abs(a) - 3*c*log(1/2*abs(-2*sqr
t(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - sqrt(-a^2*x^2 + 1)*c/a + 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) +
a)*c/(a^2*x*abs(a))