### 3.307 $$\int \frac{e^{2 \coth ^{-1}(a x)} \sqrt{c-a c x}}{x^3} \, dx$$

Optimal. Leaf size=68 $\frac{7}{4} a^2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right )+\frac{\sqrt{c-a c x}}{2 x^2}+\frac{7 a \sqrt{c-a c x}}{4 x}$

[Out]

Sqrt[c - a*c*x]/(2*x^2) + (7*a*Sqrt[c - a*c*x])/(4*x) + (7*a^2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/4

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Rubi [A]  time = 0.210414, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.304, Rules used = {6167, 6130, 21, 78, 51, 63, 208} $\frac{7}{4} a^2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right )+\frac{\sqrt{c-a c x}}{2 x^2}+\frac{7 a \sqrt{c-a c x}}{4 x}$

Antiderivative was successfully veriﬁed.

[In]

Int[(E^(2*ArcCoth[a*x])*Sqrt[c - a*c*x])/x^3,x]

[Out]

Sqrt[c - a*c*x]/(2*x^2) + (7*a*Sqrt[c - a*c*x])/(4*x) + (7*a^2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/4

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 51

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

Rubi steps

\begin{align*} \int \frac{e^{2 \coth ^{-1}(a x)} \sqrt{c-a c x}}{x^3} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)} \sqrt{c-a c x}}{x^3} \, dx\\ &=-\int \frac{(1+a x) \sqrt{c-a c x}}{x^3 (1-a x)} \, dx\\ &=-\left (c \int \frac{1+a x}{x^3 \sqrt{c-a c x}} \, dx\right )\\ &=\frac{\sqrt{c-a c x}}{2 x^2}-\frac{1}{4} (7 a c) \int \frac{1}{x^2 \sqrt{c-a c x}} \, dx\\ &=\frac{\sqrt{c-a c x}}{2 x^2}+\frac{7 a \sqrt{c-a c x}}{4 x}-\frac{1}{8} \left (7 a^2 c\right ) \int \frac{1}{x \sqrt{c-a c x}} \, dx\\ &=\frac{\sqrt{c-a c x}}{2 x^2}+\frac{7 a \sqrt{c-a c x}}{4 x}+\frac{1}{4} (7 a) \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}}{2 x^2}+\frac{7 a \sqrt{c-a c x}}{4 x}+\frac{7}{4} a^2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0472039, size = 55, normalized size = 0.81 $\frac{7}{4} a^2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{c}}\right )+\frac{(7 a x+2) \sqrt{c-a c x}}{4 x^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^(2*ArcCoth[a*x])*Sqrt[c - a*c*x])/x^3,x]

[Out]

((2 + 7*a*x)*Sqrt[c - a*c*x])/(4*x^2) + (7*a^2*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]])/4

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Maple [A]  time = 0.052, size = 65, normalized size = 1. \begin{align*} 2\,{a}^{2}{c}^{2} \left ({\frac{1}{{a}^{2}{x}^{2}{c}^{2}} \left ( -{\frac{7\, \left ( -acx+c \right ) ^{3/2}}{8\,c}}+{\frac{9\,\sqrt{-acx+c}}{8}} \right ) }+{\frac{7}{8\,{c}^{3/2}}{\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^3,x)

[Out]

2*a^2*c^2*((-7/8/c*(-a*c*x+c)^(3/2)+9/8*(-a*c*x+c)^(1/2))/x^2/a^2/c^2+7/8/c^(3/2)*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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59502, size = 289, normalized size = 4.25 \begin{align*} \left [\frac{7 \, a^{2} \sqrt{c} x^{2} \log \left (\frac{a c x - 2 \, \sqrt{-a c x + c} \sqrt{c} - 2 \, c}{x}\right ) + 2 \, \sqrt{-a c x + c}{\left (7 \, a x + 2\right )}}{8 \, x^{2}}, -\frac{7 \, a^{2} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{-a c x + c} \sqrt{-c}}{c}\right ) - \sqrt{-a c x + c}{\left (7 \, a x + 2\right )}}{4 \, x^{2}}\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^3,x, algorithm="fricas")

[Out]

[1/8*(7*a^2*sqrt(c)*x^2*log((a*c*x - 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x) + 2*sqrt(-a*c*x + c)*(7*a*x + 2))/x^
2, -1/4*(7*a^2*sqrt(-c)*x^2*arctan(sqrt(-a*c*x + c)*sqrt(-c)/c) - sqrt(-a*c*x + c)*(7*a*x + 2))/x^2]

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Sympy [B]  time = 23.2499, size = 270, normalized size = 3.97 \begin{align*} \frac{10 a^{2} c^{4} \sqrt{- a c x + c}}{16 a c^{4} x - 8 c^{4} + 8 c^{2} \left (- a c x + c\right )^{2}} - \frac{6 a^{2} c^{3} \left (- a c x + c\right )^{\frac{3}{2}}}{16 a c^{4} x - 8 c^{4} + 8 c^{2} \left (- a c x + c\right )^{2}} - \frac{3 a^{2} c^{3} \sqrt{\frac{1}{c^{5}}} \log{\left (- c^{3} \sqrt{\frac{1}{c^{5}}} + \sqrt{- a c x + c} \right )}}{8} + \frac{3 a^{2} c^{3} \sqrt{\frac{1}{c^{5}}} \log{\left (c^{3} \sqrt{\frac{1}{c^{5}}} + \sqrt{- a c x + c} \right )}}{8} - \frac{a^{2} 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^{2} 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 \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**3,x)

[Out]

10*a**2*c**4*sqrt(-a*c*x + c)/(16*a*c**4*x - 8*c**4 + 8*c**2*(-a*c*x + c)**2) - 6*a**2*c**3*(-a*c*x + c)**(3/2
)/(16*a*c**4*x - 8*c**4 + 8*c**2*(-a*c*x + c)**2) - 3*a**2*c**3*sqrt(c**(-5))*log(-c**3*sqrt(c**(-5)) + sqrt(-
a*c*x + c))/8 + 3*a**2*c**3*sqrt(c**(-5))*log(c**3*sqrt(c**(-5)) + sqrt(-a*c*x + c))/8 - a**2*c**2*sqrt(c**(-3
))*log(-c**2*sqrt(c**(-3)) + sqrt(-a*c*x + c))/2 + a**2*c**2*sqrt(c**(-3))*log(c**2*sqrt(c**(-3)) + sqrt(-a*c*
x + c))/2 + a*sqrt(-a*c*x + c)/x

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Giac [A]  time = 1.1418, size = 97, normalized size = 1.43 \begin{align*} -\frac{7 \, a^{2} c \arctan \left (\frac{\sqrt{-a c x + c}}{\sqrt{-c}}\right )}{4 \, \sqrt{-c}} - \frac{7 \,{\left (-a c x + c\right )}^{\frac{3}{2}} a^{2} c - 9 \, \sqrt{-a c x + c} a^{2} c^{2}}{4 \, a^{2} c^{2} x^{2}} \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^3,x, algorithm="giac")

[Out]

-7/4*a^2*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - 1/4*(7*(-a*c*x + c)^(3/2)*a^2*c - 9*sqrt(-a*c*x + c)*a
^2*c^2)/(a^2*c^2*x^2)