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

Optimal. Leaf size=42 \[ \frac{\sqrt{c-a c x}}{x}+3 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right ) \]

[Out]

Sqrt[c - a*c*x]/x + 3*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]]

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Rubi [A]  time = 0.203317, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {6167, 6130, 21, 78, 63, 208} \[ \frac{\sqrt{c-a c x}}{x}+3 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c - a*c*x]/x + 3*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/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 6130

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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

Mathematica [A]  time = 0.0310648, size = 42, normalized size = 1. \[ \frac{\sqrt{c-a c x}}{x}+3 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c - a*c*x]/x + 3*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]]

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Maple [A]  time = 0.052, size = 45, normalized size = 1.1 \begin{align*} -2\,ac \left ( -1/2\,{\frac{\sqrt{-acx+c}}{acx}}-3/2\,{\frac{1}{\sqrt{c}}{\it Artanh} \left ({\frac{\sqrt{-acx+c}}{\sqrt{c}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58873, size = 235, normalized size = 5.6 \begin{align*} \left [\frac{3 \, a \sqrt{c} x \log \left (\frac{a c x - 2 \, \sqrt{-a c x + c} \sqrt{c} - 2 \, c}{x}\right ) + 2 \, \sqrt{-a c x + c}}{2 \, x}, -\frac{3 \, a \sqrt{-c} x \arctan \left (\frac{\sqrt{-a c x + c} \sqrt{-c}}{c}\right ) - \sqrt{-a c x + c}}{x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 12.1982, size = 119, normalized size = 2.83 \begin{align*} - \frac{a c^{2} \sqrt{\frac{1}{c^{3}}} \log{\left (- c^{2} \sqrt{\frac{1}{c^{3}}} + \sqrt{- a c x + c} \right )}}{2} + \frac{a c^{2} \sqrt{\frac{1}{c^{3}}} \log{\left (c^{2} \sqrt{\frac{1}{c^{3}}} + \sqrt{- a c x + c} \right )}}{2} - \frac{2 a c \operatorname{atan}{\left (\frac{\sqrt{- a c x + c}}{\sqrt{- c}} \right )}}{\sqrt{- c}} + \frac{\sqrt{- a c x + c}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*c**2*sqrt(c**(-3))*log(-c**2*sqrt(c**(-3)) + sqrt(-a*c*x + c))/2 + a*c**2*sqrt(c**(-3))*log(c**2*sqrt(c**(-
3)) + sqrt(-a*c*x + c))/2 - 2*a*c*atan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) + sqrt(-a*c*x + c)/x

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Giac [A]  time = 1.12742, size = 53, normalized size = 1.26 \begin{align*} -\frac{3 \, a c \arctan \left (\frac{\sqrt{-a c x + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} + \frac{\sqrt{-a c x + c}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) + sqrt(-a*c*x + c)/x