### 3.888 $$\int e^{2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}} x \, dx$$

Optimal. Leaf size=98 $\frac{x (a x+1) \sqrt{c-\frac{c}{a^2 x^2}}}{2 a}+\frac{3 x \sqrt{c-\frac{c}{a^2 x^2}}}{2 a}-\frac{3 x \sqrt{c-\frac{c}{a^2 x^2}} \sin ^{-1}(a x)}{2 a \sqrt{1-a x} \sqrt{a x+1}}$

[Out]

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

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Rubi [A]  time = 0.310332, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.24, Rules used = {6167, 6159, 6129, 50, 41, 216} $\frac{x (a x+1) \sqrt{c-\frac{c}{a^2 x^2}}}{2 a}+\frac{3 x \sqrt{c-\frac{c}{a^2 x^2}}}{2 a}-\frac{3 x \sqrt{c-\frac{c}{a^2 x^2}} \sin ^{-1}(a x)}{2 a \sqrt{1-a x} \sqrt{a x+1}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0635101, size = 77, normalized size = 0.79 $\frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (\sqrt{1-a^2 x^2} (a x+4)+6 \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )\right )}{2 a \sqrt{1-a^2 x^2}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x*((4 + a*x)*Sqrt[1 - a^2*x^2] + 6*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(2*a*Sqrt[1 - a^2*x^
2])

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Maple [A]  time = 0.206, size = 147, normalized size = 1.5 \begin{align*} -{\frac{x}{2\,{a}^{2}}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}} \left ( -x\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}{a}^{2}+\sqrt{c}\ln \left ( x\sqrt{c}+\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}} \right ) -4\,\sqrt{c}\ln \left ({\frac{1}{\sqrt{c}} \left ( \sqrt{c}\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}+cx \right ) } \right ) -4\,\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}a \right ){\frac{1}{\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x*(-x*(c*(a^2*x^2-1)/a^2)^(1/2)*a^2+c^(1/2)*ln(x*c^(1/2)+(c*(a^2*x^2-1)/a^2
)^(1/2))-4*c^(1/2)*ln((c^(1/2)*((a*x-1)*(a*x+1)*c/a^2)^(1/2)+c*x)/c^(1/2))-4*((a*x-1)*(a*x+1)*c/a^2)^(1/2)*a)/
(c*(a^2*x^2-1)/a^2)^(1/2)/a^2

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.75855, size = 404, normalized size = 4.12 \begin{align*} \left [\frac{2 \,{\left (a^{2} x^{2} + 4 \, a x\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} + 3 \, \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 )}{4 \, a^{2}}, \frac{{\left (a^{2} x^{2} + 4 \, a x\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - 3 \, \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 )}{2 \, a^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(a^2*x^2 + 4*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) + 3*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, 1/2*((a^2*x^2 + 4*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - 3*sqrt(-c)*arc
tan(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]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14294, size = 143, normalized size = 1.46 \begin{align*} \frac{1}{4} \,{\left (2 \, \sqrt{a^{2} c x^{2} - c}{\left (\frac{x \mathrm{sgn}\left (x\right )}{a^{2}} + \frac{4 \, \mathrm{sgn}\left (x\right )}{a^{3}}\right )} - \frac{6 \, \sqrt{c} \log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} - c} \right |}\right ) \mathrm{sgn}\left (x\right )}{a^{2}{\left | a \right |}} + \frac{{\left (3 \, a \sqrt{c} \log \left ({\left | c \right |}\right ) - 8 \, \sqrt{-c}{\left | a \right |}\right )} \mathrm{sgn}\left (x\right )}{a^{3}{\left | a \right |}}\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)*x*(c-c/a^2/x^2)^(1/2),x, algorithm="giac")

[Out]

1/4*(2*sqrt(a^2*c*x^2 - c)*(x*sgn(x)/a^2 + 4*sgn(x)/a^3) - 6*sqrt(c)*log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 -
c)))*sgn(x)/(a^2*abs(a)) + (3*a*sqrt(c)*log(abs(c)) - 8*sqrt(-c)*abs(a))*sgn(x)/(a^3*abs(a)))*abs(a)