### 3.842 $$\int \frac{e^{2 \coth ^{-1}(a x)}}{\sqrt{c-\frac{c}{a^2 x^2}}} \, dx$$

Optimal. Leaf size=111 $-\frac{(a x+1)^2}{a^2 x \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{2 (1-a x) (a x+1)}{a^2 x \sqrt{c-\frac{c}{a^2 x^2}}}+\frac{2 \sqrt{1-a x} \sqrt{a x+1} \sin ^{-1}(a x)}{a^2 x \sqrt{c-\frac{c}{a^2 x^2}}}$

[Out]

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

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Rubi [A]  time = 0.307312, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.292, Rules used = {6167, 6159, 6129, 78, 50, 41, 216} $-\frac{(a x+1)^2}{a^2 x \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{2 (1-a x) (a x+1)}{a^2 x \sqrt{c-\frac{c}{a^2 x^2}}}+\frac{2 \sqrt{1-a x} \sqrt{a x+1} \sin ^{-1}(a x)}{a^2 x \sqrt{c-\frac{c}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 6159

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

Rule 6129

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

Rule 78

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 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]

Rubi steps

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

Mathematica [A]  time = 0.0708311, size = 68, normalized size = 0.61 $\frac{a^2 x^2+2 \sqrt{a^2 x^2-1} \log \left (\sqrt{a^2 x^2-1}+a x\right )-2 a x-3}{a^2 x \sqrt{c-\frac{c}{a^2 x^2}}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.184, size = 177, normalized size = 1.6 \begin{align*}{\frac{1}{x \left ( ax-1 \right ) a}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}} \left ( \sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}\sqrt{c}x{a}^{2}+2\,\ln \left ( x\sqrt{c}+\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}} \right ) xac-\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}a\sqrt{c}-2\,a\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}\sqrt{c}-2\,\ln \left ( x\sqrt{c}+\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}} \right ) c \right ){\frac{1}{\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}}}}{c}^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.76919, size = 448, normalized size = 4.04 \begin{align*} \left [\frac{{\left (a x - 1\right )} \sqrt{c} \log \left (2 \, a^{2} c x^{2} + 2 \, a^{2} \sqrt{c} x^{2} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right ) +{\left (a^{2} x^{2} - 3 \, a x\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x - a c}, -\frac{2 \,{\left (a x - 1\right )} \sqrt{-c} \arctan \left (\frac{a^{2} \sqrt{-c} x^{2} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right ) -{\left (a^{2} x^{2} - 3 \, a x\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{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)*(a*x+1)/(c-c/a^2/x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{{\left (a x - 1\right )} \sqrt{c - \frac{c}{a^{2} 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)/(c-c/a^2/x^2)^(1/2),x, algorithm="giac")

[Out]

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