### 3.346 $$\int \frac{e^{-2 \coth ^{-1}(a x)} \sqrt{c-a c x}}{x^2} \, dx$$

Optimal. Leaf size=78 $\frac{\sqrt{c-a c x}}{x}-5 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right )+4 \sqrt{2} a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )$

[Out]

Sqrt[c - a*c*x]/x - 5*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]] + 4*Sqrt[2]*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]
/(Sqrt[2]*Sqrt[c])]

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Rubi [A]  time = 0.226114, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.348, Rules used = {6167, 6130, 21, 98, 156, 63, 208, 206} $\frac{\sqrt{c-a c x}}{x}-5 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right )+4 \sqrt{2} a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Sqrt[c - a*c*x]/x - 5*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]] + 4*Sqrt[2]*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]
/(Sqrt[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 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 98

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

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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\\ &=-\frac{\int \frac{(c-a c x)^{3/2}}{x^2 (1+a x)} \, dx}{c}\\ &=\frac{\sqrt{c-a c x}}{x}+\frac{\int \frac{\frac{5 a c^2}{2}-\frac{3}{2} a^2 c^2 x}{x (1+a x) \sqrt{c-a c x}} \, dx}{c}\\ &=\frac{\sqrt{c-a c x}}{x}+\frac{1}{2} (5 a c) \int \frac{1}{x \sqrt{c-a c x}} \, dx-\left (4 a^2 c\right ) \int \frac{1}{(1+a x) \sqrt{c-a c x}} \, dx\\ &=\frac{\sqrt{c-a c x}}{x}-5 \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a}-\frac{x^2}{a c}} \, dx,x,\sqrt{c-a c x}\right )+(8 a) \operatorname{Subst}\left (\int \frac{1}{2-\frac{x^2}{c}} \, dx,x,\sqrt{c-a c x}\right )\\ &=\frac{\sqrt{c-a c x}}{x}-5 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right )+4 \sqrt{2} a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0411239, size = 78, normalized size = 1. $\frac{\sqrt{c-a c x}}{x}-5 a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right )+4 \sqrt{2} a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Sqrt[c - a*c*x]/x - 5*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]] + 4*Sqrt[2]*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]
/(Sqrt[2]*Sqrt[c])]

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Maple [A]  time = 0.051, size = 71, normalized size = 0.9 \begin{align*} -2\,ac \left ( -1/2\,{\frac{\sqrt{-acx+c}}{acx}}+5/2\,{\frac{1}{\sqrt{c}}{\it Artanh} \left ({\frac{\sqrt{-acx+c}}{\sqrt{c}}} \right ) }-2\,{\frac{\sqrt{2}}{\sqrt{c}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-acx+c}\sqrt{2}}{\sqrt{c}}} \right ) } \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*c*(-1/2*(-a*c*x+c)^(1/2)/x/a/c+5/2/c^(1/2)*arctanh((-a*c*x+c)^(1/2)/c^(1/2))-2*2^(1/2)/c^(1/2)*arctanh(1/
2*(-a*c*x+c)^(1/2)*2^(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((-a*c*x+c)^(1/2)*(a*x-1)/(a*x+1)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32769, size = 455, normalized size = 5.83 \begin{align*} \left [\frac{4 \, \sqrt{2} a \sqrt{c} x \log \left (\frac{a c x - 2 \, \sqrt{2} \sqrt{-a c x + c} \sqrt{c} - 3 \, c}{a x + 1}\right ) + 5 \, 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{4 \, \sqrt{2} a \sqrt{-c} x \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c} \sqrt{-c}}{2 \, c}\right ) - 5 \, 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((-a*c*x+c)^(1/2)*(a*x-1)/(a*x+1)/x^2,x, algorithm="fricas")

[Out]

[1/2*(4*sqrt(2)*a*sqrt(c)*x*log((a*c*x - 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + 5*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, -(4*sqrt(2)*a*sqrt(-c)*x*arctan(1/2
*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) - 5*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 = 6.87575, size = 162, normalized size = 2.08 \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{6 a c \operatorname{atan}{\left (\frac{\sqrt{- a c x + c}}{\sqrt{- c}} \right )}}{\sqrt{- c}} - \frac{4 \sqrt{2} a c \operatorname{atan}{\left (\frac{\sqrt{2} \sqrt{- a c x + c}}{2 \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((-a*c*x+c)**(1/2)*(a*x-1)/(a*x+1)/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 + 6*a*c*atan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - 4*sqrt(2)*a*c*atan(sqrt(2)*sqrt(-
a*c*x + c)/(2*sqrt(-c)))/sqrt(-c) + sqrt(-a*c*x + c)/x

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Giac [A]  time = 1.15494, size = 96, normalized size = 1.23 \begin{align*} -\frac{4 \, \sqrt{2} a c \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c}}{2 \, \sqrt{-c}}\right )}{\sqrt{-c}} + \frac{5 \, 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((-a*c*x+c)^(1/2)*(a*x-1)/(a*x+1)/x^2,x, algorithm="giac")

[Out]

-4*sqrt(2)*a*c*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) + 5*a*c*arctan(sqrt(-a*c*x + c)/sqrt(-c)
)/sqrt(-c) + sqrt(-a*c*x + c)/x