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

Optimal. Leaf size=97 \[ -\frac{\sqrt{\frac{1}{a x}+1} \sqrt{c-a c x}}{x \sqrt{1-\frac{1}{a x}}}-\frac{\sqrt{a} \sqrt{\frac{1}{x}} \sqrt{c-a c x} \sinh ^{-1}\left (\frac{\sqrt{\frac{1}{x}}}{\sqrt{a}}\right )}{\sqrt{1-\frac{1}{a x}}} \]

[Out]

-((Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x])/(Sqrt[1 - 1/(a*x)]*x)) - (Sqrt[a]*Sqrt[x^(-1)]*Sqrt[c - a*c*x]*ArcSinh[S
qrt[x^(-1)]/Sqrt[a]])/Sqrt[1 - 1/(a*x)]

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Rubi [A]  time = 0.194801, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {6176, 6181, 50, 54, 215} \[ -\frac{\sqrt{\frac{1}{a x}+1} \sqrt{c-a c x}}{x \sqrt{1-\frac{1}{a x}}}-\frac{\sqrt{a} \sqrt{\frac{1}{x}} \sqrt{c-a c x} \sinh ^{-1}\left (\frac{\sqrt{\frac{1}{x}}}{\sqrt{a}}\right )}{\sqrt{1-\frac{1}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x])/(Sqrt[1 - 1/(a*x)]*x)) - (Sqrt[a]*Sqrt[x^(-1)]*Sqrt[c - a*c*x]*ArcSinh[S
qrt[x^(-1)]/Sqrt[a]])/Sqrt[1 - 1/(a*x)]

Rule 6176

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

Rule 6181

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

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 54

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

Rule 215

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

Rubi steps

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

Mathematica [A]  time = 0.0387791, size = 76, normalized size = 0.78 \[ -\frac{\sqrt{\frac{1}{x}} \sqrt{c-a c x} \left (\sqrt{\frac{1}{x}} \sqrt{\frac{1}{a x}+1}+\sqrt{a} \sinh ^{-1}\left (\frac{\sqrt{\frac{1}{x}}}{\sqrt{a}}\right )\right )}{\sqrt{1-\frac{1}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.144, size = 78, normalized size = 0.8 \begin{align*} -{\frac{1}{x} \left ( \arctan \left ({\sqrt{-c \left ( ax+1 \right ) }{\frac{1}{\sqrt{c}}}} \right ) xac+\sqrt{-c \left ( ax+1 \right ) }\sqrt{c} \right ) \sqrt{-c \left ( ax-1 \right ) }{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}{\frac{1}{\sqrt{-c \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.57579, size = 514, normalized size = 5.3 \begin{align*} \left [\frac{{\left (a^{2} x^{2} - a x\right )} \sqrt{-c} \log \left (-\frac{a^{2} c x^{2} + a c x + 2 \, \sqrt{-a c x + c}{\left (a x + 1\right )} \sqrt{-c} \sqrt{\frac{a x - 1}{a x + 1}} - 2 \, c}{a x^{2} - x}\right ) - 2 \, \sqrt{-a c x + c}{\left (a x + 1\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{2 \,{\left (a x^{2} - x\right )}}, -\frac{{\left (a^{2} x^{2} - a x\right )} \sqrt{c} \arctan \left (\frac{\sqrt{-a c x + c} \sqrt{c} \sqrt{\frac{a x - 1}{a x + 1}}}{a c x - c}\right ) + \sqrt{-a c x + c}{\left (a x + 1\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{a x^{2} - x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [C]  time = 1.19978, size = 162, normalized size = 1.67 \begin{align*} -\frac{a^{2} c^{2}{\left (\frac{\arctan \left (\frac{\sqrt{-a c x - c}}{\sqrt{c}}\right )}{\sqrt{c} \mathrm{sgn}\left (-a c x - c\right )} + \frac{\sqrt{-a c x - c}}{a c x \mathrm{sgn}\left (-a c x - c\right )}\right )} - \frac{-i \, a^{2} \sqrt{-c} c \arctan \left (-i \, \sqrt{2}\right ) - \sqrt{2} a^{2} \sqrt{-c} c}{\mathrm{sgn}\left (c\right )}}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*c^2*(arctan(sqrt(-a*c*x - c)/sqrt(c))/(sqrt(c)*sgn(-a*c*x - c)) + sqrt(-a*c*x - c)/(a*c*x*sgn(-a*c*x - c
))) - (-I*a^2*sqrt(-c)*c*arctan(-I*sqrt(2)) - sqrt(2)*a^2*sqrt(-c)*c)/sgn(c))/(a*c)