3.432 \(\int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{c-a c x}}{x} \, dx\)

Optimal. Leaf size=107 \[ \frac{2 c^2 (1-a x)^{3/2} \sqrt{a x+1}}{(c-a c x)^{3/2}}+\frac{8 c^2 (1-a x)^{3/2}}{\sqrt{a x+1} (c-a c x)^{3/2}}-\frac{2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt{a x+1}\right )}{(c-a c x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.128615, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {6130, 23, 87, 63, 208} \[ \frac{2 c^2 (1-a x)^{3/2} \sqrt{a x+1}}{(c-a c x)^{3/2}}+\frac{8 c^2 (1-a x)^{3/2}}{\sqrt{a x+1} (c-a c x)^{3/2}}-\frac{2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt{a x+1}\right )}{(c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 23

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

Rule 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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

Mathematica [C]  time = 0.0347293, size = 51, normalized size = 0.48 \[ \frac{2 c \sqrt{1-a x} \left (\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},a x+1\right )+a x+4\right )}{\sqrt{a x+1} \sqrt{c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*c*Sqrt[1 - a*x]*(4 + a*x + Hypergeometric2F1[-1/2, 1, 1/2, 1 + a*x]))/(Sqrt[1 + a*x]*Sqrt[c - a*c*x])

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Maple [A]  time = 0.108, size = 78, normalized size = 0.7 \begin{align*} 2\,{\frac{\sqrt{-c \left ( ax-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}}{ \left ( ax-1 \right ) \left ( ax+1 \right ) c} \left ( \sqrt{c}{\it Artanh} \left ({\frac{\sqrt{c \left ( ax+1 \right ) }}{\sqrt{c}}} \right ) \sqrt{c \left ( ax+1 \right ) }-acx-5\,c \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-c*(a*x-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(c^(1/2)*arctanh((c*(a*x+1))^(1/2)/c^(1/2))*(c*(a*x+1))^(1/2)-a*c*x-5*
c)/(a*x-1)/c/(a*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \sqrt{-a c x + c}}{{\left (a x + 1\right )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*sqrt(-a*c*x + c)/((a*x + 1)^3*x), x)

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Fricas [A]  time = 1.60018, size = 466, normalized size = 4.36 \begin{align*} \left [\frac{{\left (a^{2} x^{2} - 1\right )} \sqrt{c} \log \left (-\frac{a^{2} c x^{2} + a c x + 2 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c} \sqrt{c} - 2 \, c}{a x^{2} - x}\right ) - 2 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}{\left (a x + 5\right )}}{a^{2} x^{2} - 1}, -\frac{2 \,{\left ({\left (a^{2} x^{2} - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) + \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}{\left (a x + 5\right )}\right )}}{a^{2} x^{2} - 1}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x,x, algorithm="fricas")

[Out]

[((a^2*x^2 - 1)*sqrt(c)*log(-(a^2*c*x^2 + a*c*x + 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/(a*x^2
- x)) - 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 5))/(a^2*x^2 - 1), -2*((a^2*x^2 - 1)*sqrt(-c)*arctan(sqrt
(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 5))/(a^
2*x^2 - 1)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- c \left (a x - 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}{x \left (a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/x,x)

[Out]

Integral(sqrt(-c*(a*x - 1))*(-(a*x - 1)*(a*x + 1))**(3/2)/(x*(a*x + 1)**3), x)

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Giac [A]  time = 1.22053, size = 127, normalized size = 1.19 \begin{align*} 2 \,{\left (\frac{\arctan \left (\frac{\sqrt{a c x + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} + \frac{4}{\sqrt{a c x + c}} + \frac{\sqrt{a c x + c}}{c}\right )}{\left | c \right |} - \frac{\sqrt{2}{\left (\sqrt{2} \sqrt{c}{\left | c \right |} \arctan \left (\frac{\sqrt{2} \sqrt{c}}{\sqrt{-c}}\right ) + 6 \, \sqrt{-c}{\left | c \right |}\right )}}{\sqrt{-c} \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x,x, algorithm="giac")

[Out]

2*(arctan(sqrt(a*c*x + c)/sqrt(-c))/sqrt(-c) + 4/sqrt(a*c*x + c) + sqrt(a*c*x + c)/c)*abs(c) - sqrt(2)*(sqrt(2
)*sqrt(c)*abs(c)*arctan(sqrt(2)*sqrt(c)/sqrt(-c)) + 6*sqrt(-c)*abs(c))/(sqrt(-c)*sqrt(c))