3.199 \(\int e^{-\coth ^{-1}(a x)} (c-a c x)^2 \, dx\)

Optimal. Leaf size=100 \[ \frac{1}{3} a^2 c^2 x^3 \sqrt{1-\frac{1}{a^2 x^2}}-\frac{3}{2} a c^2 x^2 \sqrt{1-\frac{1}{a^2 x^2}}+\frac{11}{3} c^2 x \sqrt{1-\frac{1}{a^2 x^2}}-\frac{5 c^2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a} \]

[Out]

(11*c^2*Sqrt[1 - 1/(a^2*x^2)]*x)/3 - (3*a*c^2*Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 + (a^2*c^2*Sqrt[1 - 1/(a^2*x^2)]*x^
3)/3 - (5*c^2*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(2*a)

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Rubi [A]  time = 0.286164, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {6175, 6178, 1807, 807, 266, 63, 208} \[ \frac{1}{3} a^2 c^2 x^3 \sqrt{1-\frac{1}{a^2 x^2}}-\frac{3}{2} a c^2 x^2 \sqrt{1-\frac{1}{a^2 x^2}}+\frac{11}{3} c^2 x \sqrt{1-\frac{1}{a^2 x^2}}-\frac{5 c^2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^2/E^ArcCoth[a*x],x]

[Out]

(11*c^2*Sqrt[1 - 1/(a^2*x^2)]*x)/3 - (3*a*c^2*Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 + (a^2*c^2*Sqrt[1 - 1/(a^2*x^2)]*x^
3)/3 - (5*c^2*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(2*a)

Rule 6175

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

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

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^{-\coth ^{-1}(a x)} (c-a c x)^2 \, dx &=\left (a^2 c^2\right ) \int e^{-\coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right )^2 x^2 \, dx\\ &=-\left (\left (a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^3}{x^4 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\right )\\ &=\frac{1}{3} a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^3+\frac{1}{3} \left (a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\frac{9}{a}-\frac{11 x}{a^2}+\frac{3 x^2}{a^3}}{x^3 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{3}{2} a c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{1}{3} a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^3-\frac{1}{6} \left (a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\frac{22}{a^2}-\frac{15 x}{a^3}}{x^2 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{11}{3} c^2 \sqrt{1-\frac{1}{a^2 x^2}} x-\frac{3}{2} a c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{1}{3} a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^3+\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=\frac{11}{3} c^2 \sqrt{1-\frac{1}{a^2 x^2}} x-\frac{3}{2} a c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{1}{3} a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^3+\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{4 a}\\ &=\frac{11}{3} c^2 \sqrt{1-\frac{1}{a^2 x^2}} x-\frac{3}{2} a c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{1}{3} a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^3-\frac{1}{2} \left (5 a c^2\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-a^2 x^2} \, dx,x,\sqrt{1-\frac{1}{a^2 x^2}}\right )\\ &=\frac{11}{3} c^2 \sqrt{1-\frac{1}{a^2 x^2}} x-\frac{3}{2} a c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{1}{3} a^2 c^2 \sqrt{1-\frac{1}{a^2 x^2}} x^3-\frac{5 c^2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.113812, size = 64, normalized size = 0.64 \[ \frac{c^2 \left (a x \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a^2 x^2-9 a x+22\right )-15 \log \left (a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )\right )}{6 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*(a*Sqrt[1 - 1/(a^2*x^2)]*x*(22 - 9*a*x + 2*a^2*x^2) - 15*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x]))/(6*a)

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Maple [B]  time = 0.132, size = 176, normalized size = 1.8 \begin{align*}{\frac{ \left ( ax+1 \right ){c}^{2}}{6\,a}\sqrt{{\frac{ax-1}{ax+1}}} \left ( 2\, \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}\sqrt{{a}^{2}}-9\,\sqrt{{a}^{2}}\sqrt{{a}^{2}{x}^{2}-1}xa+24\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }+9\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}}{\sqrt{{a}^{2}}}} \right ) a-24\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)*c^2*(2*((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)-9*(a^2)^(1/2)*(a^2*x^2-1)^(1/2)
*x*a+24*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)+9*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*a-24*a*ln(
(a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2)))/((a*x-1)*(a*x+1))^(1/2)/a/(a^2)^(1/2)

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Maxima [B]  time = 1.02074, size = 244, normalized size = 2.44 \begin{align*} -\frac{1}{6} \, a{\left (\frac{15 \, c^{2} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{15 \, c^{2} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac{2 \,{\left (33 \, c^{2} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{2}} - 40 \, c^{2} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} + 15 \, c^{2} \sqrt{\frac{a x - 1}{a x + 1}}\right )}}{\frac{3 \,{\left (a x - 1\right )} a^{2}}{a x + 1} - \frac{3 \,{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} + \frac{{\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*(15*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 15*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 + 2*(33*
c^2*((a*x - 1)/(a*x + 1))^(5/2) - 40*c^2*((a*x - 1)/(a*x + 1))^(3/2) + 15*c^2*sqrt((a*x - 1)/(a*x + 1)))/(3*(a
*x - 1)*a^2/(a*x + 1) - 3*(a*x - 1)^2*a^2/(a*x + 1)^2 + (a*x - 1)^3*a^2/(a*x + 1)^3 - a^2))

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Fricas [A]  time = 1.6813, size = 240, normalized size = 2.4 \begin{align*} -\frac{15 \, c^{2} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 15 \, c^{2} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) -{\left (2 \, a^{3} c^{2} x^{3} - 7 \, a^{2} c^{2} x^{2} + 13 \, a c^{2} x + 22 \, c^{2}\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{6 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(15*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 15*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1) - (2*a^3*c^2*x^3 -
7*a^2*c^2*x^2 + 13*a*c^2*x + 22*c^2)*sqrt((a*x - 1)/(a*x + 1)))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16438, size = 122, normalized size = 1.22 \begin{align*} \frac{5 \, c^{2} \log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} - 1} \right |}\right ) \mathrm{sgn}\left (a x + 1\right )}{2 \,{\left | a \right |}} + \frac{1}{6} \, \sqrt{a^{2} x^{2} - 1}{\left ({\left (2 \, a c^{2} x \mathrm{sgn}\left (a x + 1\right ) - 9 \, c^{2} \mathrm{sgn}\left (a x + 1\right )\right )} x + \frac{22 \, c^{2} \mathrm{sgn}\left (a x + 1\right )}{a}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*c^2*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sgn(a*x + 1)/abs(a) + 1/6*sqrt(a^2*x^2 - 1)*((2*a*c^2*x*sgn(a*
x + 1) - 9*c^2*sgn(a*x + 1))*x + 22*c^2*sgn(a*x + 1)/a)