### 3.678 $$\int \frac{e^{2 \coth ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^2} \, dx$$

Optimal. Leaf size=82 $\frac{\sqrt{c-a^2 c x^2}}{x}-a \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+2 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )$

[Out]

Sqrt[c - a^2*c*x^2]/x - a*Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]] + 2*a*Sqrt[c]*ArcTanh[Sqrt[c - a^2
*c*x^2]/Sqrt[c]]

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Rubi [A]  time = 0.347903, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {6167, 6151, 1807, 844, 217, 203, 266, 63, 208} $\frac{\sqrt{c-a^2 c x^2}}{x}-a \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+2 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Sqrt[c - a^2*c*x^2]/x - a*Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]] + 2*a*Sqrt[c]*ArcTanh[Sqrt[c - a^2
*c*x^2]/Sqrt[c]]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6151

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^(n/2), Int[x^m*(c
+ d*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] ||
GtQ[c, 0]) && IGtQ[n/2, 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 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 \frac{e^{2 \coth ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^2} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^2} \, dx\\ &=-\left (c \int \frac{(1+a x)^2}{x^2 \sqrt{c-a^2 c x^2}} \, dx\right )\\ &=\frac{\sqrt{c-a^2 c x^2}}{x}+\int \frac{-2 a c-a^2 c x}{x \sqrt{c-a^2 c x^2}} \, dx\\ &=\frac{\sqrt{c-a^2 c x^2}}{x}-(2 a c) \int \frac{1}{x \sqrt{c-a^2 c x^2}} \, dx-\left (a^2 c\right ) \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx\\ &=\frac{\sqrt{c-a^2 c x^2}}{x}-(a c) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-a^2 c x}} \, dx,x,x^2\right )-\left (a^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{1+a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c-a^2 c x^2}}\right )\\ &=\frac{\sqrt{c-a^2 c x^2}}{x}-a \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+\frac{2 \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2 c}} \, dx,x,\sqrt{c-a^2 c x^2}\right )}{a}\\ &=\frac{\sqrt{c-a^2 c x^2}}{x}-a \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+2 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0913365, size = 104, normalized size = 1.27 $\frac{\sqrt{c-a^2 c x^2}}{x}+2 a \sqrt{c} \log \left (\sqrt{c} \sqrt{c-a^2 c x^2}+c\right )+a \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )-2 a \sqrt{c} \log (x)$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[c - a^2*c*x^2]/x + a*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))] - 2*a*Sqrt[c]*Log
[x] + 2*a*Sqrt[c]*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2]]

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Maple [B]  time = 0.058, size = 208, normalized size = 2.5 \begin{align*}{\frac{1}{cx} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{a}^{2}x\sqrt{-{a}^{2}c{x}^{2}+c}+{{a}^{2}c\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}+2\,\sqrt{c}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{-{a}^{2}c{x}^{2}+c}}{x}} \right ) a-2\,\sqrt{-{a}^{2}c{x}^{2}+c}a+2\,a\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }-2\,{\frac{{a}^{2}c}{\sqrt{{a}^{2}c}}\arctan \left ({\sqrt{{a}^{2}c}x{\frac{1}{\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c/x*(-a^2*c*x^2+c)^(3/2)+a^2*x*(-a^2*c*x^2+c)^(1/2)+a^2*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c
)^(1/2))+2*c^(1/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)*a-2*(-a^2*c*x^2+c)^(1/2)*a+2*a*(-c*a^2*(x-1/a)^2
-2*a*c*(x-1/a))^(1/2)-2*a^2*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.60041, size = 477, normalized size = 5.82 \begin{align*} \left [\frac{a \sqrt{c} x \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right ) + a \sqrt{c} x \log \left (-\frac{a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{c} - 2 \, c}{x^{2}}\right ) + \sqrt{-a^{2} c x^{2} + c}}{x}, \frac{4 \, a \sqrt{-c} x \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) + a \sqrt{-c} x \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right ) + 2 \, \sqrt{-a^{2} c x^{2} + c}}{2 \, x}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(a*sqrt(c)*x*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + a*sqrt(c)*x*log(-(a^2*c*x^2 - 2*sqrt(
-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) + sqrt(-a^2*c*x^2 + c))/x, 1/2*(4*a*sqrt(-c)*x*arctan(sqrt(-a^2*c*x^2 + c)
*sqrt(-c)/(a^2*c*x^2 - c)) + a*sqrt(-c)*x*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) + 2*sqrt(
-a^2*c*x^2 + c))/x]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19055, size = 181, normalized size = 2.21 \begin{align*} -\frac{4 \, a c \arctan \left (-\frac{\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} - \frac{a^{2} \sqrt{-c} \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}} - \frac{2 \, a^{2} \sqrt{-c} c}{{\left ({\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{2} - c\right )}{\left | a \right |}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a*c*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) - a^2*sqrt(-c)*log(abs(-sqrt(-a^2*c)
*x + sqrt(-a^2*c*x^2 + c)))/abs(a) - 2*a^2*sqrt(-c)*c/(((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c)*abs(a))