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

Optimal. Leaf size=111 \[ -\frac{c^2 \sqrt{1-a^2 x^2}}{a}-\frac{c^2 \sqrt{1-a^2 x^2}}{a^2 x}-\frac{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}+\frac{5 c^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{a}-\frac{5 c^2 \sin ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.295462, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {6131, 6128, 1805, 1807, 1809, 844, 216, 266, 63, 208} \[ -\frac{c^2 \sqrt{1-a^2 x^2}}{a}-\frac{c^2 \sqrt{1-a^2 x^2}}{a^2 x}-\frac{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}+\frac{5 c^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{a}-\frac{5 c^2 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6131

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

Rule 6128

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*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[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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 x}\right )^2 \, dx &=\frac{c^2 \int \frac{e^{-3 \tanh ^{-1}(a x)} (1-a x)^2}{x^2} \, dx}{a^2}\\ &=\frac{c^2 \int \frac{(1-a x)^5}{x^2 \left (1-a^2 x^2\right )^{3/2}} \, dx}{a^2}\\ &=-\frac{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}-\frac{c^2 \int \frac{-1+5 a x+5 a^2 x^2-a^3 x^3}{x^2 \sqrt{1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}-\frac{c^2 \sqrt{1-a^2 x^2}}{a^2 x}+\frac{c^2 \int \frac{-5 a-5 a^2 x+a^3 x^2}{x \sqrt{1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}-\frac{c^2 \sqrt{1-a^2 x^2}}{a}-\frac{c^2 \sqrt{1-a^2 x^2}}{a^2 x}-\frac{c^2 \int \frac{5 a^3+5 a^4 x}{x \sqrt{1-a^2 x^2}} \, dx}{a^4}\\ &=-\frac{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}-\frac{c^2 \sqrt{1-a^2 x^2}}{a}-\frac{c^2 \sqrt{1-a^2 x^2}}{a^2 x}-\left (5 c^2\right ) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx-\frac{\left (5 c^2\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{a}\\ &=-\frac{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}-\frac{c^2 \sqrt{1-a^2 x^2}}{a}-\frac{c^2 \sqrt{1-a^2 x^2}}{a^2 x}-\frac{5 c^2 \sin ^{-1}(a x)}{a}-\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{2 a}\\ &=-\frac{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}-\frac{c^2 \sqrt{1-a^2 x^2}}{a}-\frac{c^2 \sqrt{1-a^2 x^2}}{a^2 x}-\frac{5 c^2 \sin ^{-1}(a x)}{a}+\frac{\left (5 c^2\right ) \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{16 c^2 (1-a x)}{a \sqrt{1-a^2 x^2}}-\frac{c^2 \sqrt{1-a^2 x^2}}{a}-\frac{c^2 \sqrt{1-a^2 x^2}}{a^2 x}-\frac{5 c^2 \sin ^{-1}(a x)}{a}+\frac{5 c^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.216785, size = 81, normalized size = 0.73 \[ \frac{c^2 \left (-\frac{\sqrt{1-a^2 x^2} \left (a^2 x^2+18 a x+1\right )}{a x (a x+1)}+5 \log \left (\sqrt{1-a^2 x^2}+1\right )-5 \log (a x)-5 \sin ^{-1}(a x)\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*(-((Sqrt[1 - a^2*x^2]*(1 + 18*a*x + a^2*x^2))/(a*x*(1 + a*x))) - 5*ArcSin[a*x] - 5*Log[a*x] + 5*Log[1 + S
qrt[1 - a^2*x^2]]))/a

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Maple [B]  time = 0.052, size = 329, normalized size = 3. \begin{align*} -4\,{\frac{{c}^{2} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{{a}^{4} \left ( x+{a}^{-1} \right ) ^{3}}}-4\,{\frac{{c}^{2} \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}}}-{\frac{7\,{c}^{2}}{3\,a} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{7\,x{c}^{2}}{2}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{7\,{c}^{2}}{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}^{2}}{{a}^{2}x} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{c}^{2}x \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}-{\frac{3\,x{c}^{2}}{2}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{3\,{c}^{2}}{2}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{5\,{c}^{2}}{3\,a} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-5\,{\frac{{c}^{2}\sqrt{-{a}^{2}{x}^{2}+1}}{a}}+5\,{\frac{{c}^{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*c^2/a^4/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-4*c^2/a^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)
-7/3*c^2/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-7/2*c^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-7/2*c^2/(a^2)^(1/2)
*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))-c^2/a^2/x*(-a^2*x^2+1)^(5/2)-c^2*x*(-a^2*x^2+1)^(3/2
)-3/2*c^2*x*(-a^2*x^2+1)^(1/2)-3/2*c^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-5/3*c^2*(-a^2*x^2+
1)^(3/2)/a-5*c^2*(-a^2*x^2+1)^(1/2)/a+5*c^2/a*arctanh(1/(-a^2*x^2+1)^(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 (c - \frac{c}{a x}\right )}^{2}}{{\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/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*x))^2/(a*x + 1)^3, x)

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Fricas [A]  time = 2.47556, size = 313, normalized size = 2.82 \begin{align*} -\frac{17 \, a^{2} c^{2} x^{2} + 17 \, a c^{2} x - 10 \,{\left (a^{2} c^{2} x^{2} + a c^{2} x\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + 5 \,{\left (a^{2} c^{2} x^{2} + a c^{2} x\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) +{\left (a^{2} c^{2} x^{2} + 18 \, a c^{2} x + c^{2}\right )} \sqrt{-a^{2} x^{2} + 1}}{a^{3} x^{2} + a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(17*a^2*c^2*x^2 + 17*a*c^2*x - 10*(a^2*c^2*x^2 + a*c^2*x)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + 5*(a^2*c^2
*x^2 + a*c^2*x)*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (a^2*c^2*x^2 + 18*a*c^2*x + c^2)*sqrt(-a^2*x^2 + 1))/(a^3*x^
2 + a^2*x)

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

[Out]

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

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Giac [A]  time = 1.20234, size = 266, normalized size = 2.4 \begin{align*} -\frac{5 \, c^{2} \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} + \frac{5 \, c^{2} \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^{2}}{a} + \frac{{\left (c^{2} + \frac{65 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c^{2}}{a^{2} x}\right )} a^{2} x}{2 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} + 1\right )}{\left | a \right |}} - \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c^{2}}{2 \, a^{2} x{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*c^2*arcsin(a*x)*sgn(a)/abs(a) + 5*c^2*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a)
- sqrt(-a^2*x^2 + 1)*c^2/a + 1/2*(c^2 + 65*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^2/(a^2*x))*a^2*x/((sqrt(-a^2*x^2
+ 1)*abs(a) + a)*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a)) - 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c
^2/(a^2*x*abs(a))