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

Optimal. Leaf size=163 \[ \frac{47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt{a x+1} (c-a c x)^{3/2}}-\frac{47 a^2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt{a x+1}\right )}{4 (c-a c x)^{3/2}}-\frac{c^2 (1-a x)^{3/2}}{2 x^2 \sqrt{a x+1} (c-a c x)^{3/2}}+\frac{13 a c^2 (1-a x)^{3/2}}{4 x \sqrt{a x+1} (c-a c x)^{3/2}} \]

[Out]

(47*a^2*c^2*(1 - a*x)^(3/2))/(4*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (c^2*(1 - a*x)^(3/2))/(2*x^2*Sqrt[1 + a*x]*
(c - a*c*x)^(3/2)) + (13*a*c^2*(1 - a*x)^(3/2))/(4*x*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (47*a^2*c^2*(1 - a*x)^
(3/2)*ArcTanh[Sqrt[1 + a*x]])/(4*(c - a*c*x)^(3/2))

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Rubi [A]  time = 0.129531, antiderivative size = 163, 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 = {6130, 23, 89, 78, 51, 63, 208} \[ \frac{47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt{a x+1} (c-a c x)^{3/2}}-\frac{47 a^2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt{a x+1}\right )}{4 (c-a c x)^{3/2}}-\frac{c^2 (1-a x)^{3/2}}{2 x^2 \sqrt{a x+1} (c-a c x)^{3/2}}+\frac{13 a c^2 (1-a x)^{3/2}}{4 x \sqrt{a x+1} (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(47*a^2*c^2*(1 - a*x)^(3/2))/(4*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (c^2*(1 - a*x)^(3/2))/(2*x^2*Sqrt[1 + a*x]*
(c - a*c*x)^(3/2)) + (13*a*c^2*(1 - a*x)^(3/2))/(4*x*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (47*a^2*c^2*(1 - a*x)^
(3/2)*ArcTanh[Sqrt[1 + a*x]])/(4*(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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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^{-3 \tanh ^{-1}(a x)} \sqrt{c-a c x}}{x^3} \, dx &=\int \frac{(1-a x)^{3/2} \sqrt{c-a c x}}{x^3 (1+a x)^{3/2}} \, dx\\ &=\frac{(1-a x)^{3/2} \int \frac{(c-a c x)^2}{x^3 (1+a x)^{3/2}} \, dx}{(c-a c x)^{3/2}}\\ &=-\frac{c^2 (1-a x)^{3/2}}{2 x^2 \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{(1-a x)^{3/2} \int \frac{-\frac{13 a c^2}{2}+2 a^2 c^2 x}{x^2 (1+a x)^{3/2}} \, dx}{2 (c-a c x)^{3/2}}\\ &=-\frac{c^2 (1-a x)^{3/2}}{2 x^2 \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{13 a c^2 (1-a x)^{3/2}}{4 x \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{\left (47 a^2 c^2 (1-a x)^{3/2}\right ) \int \frac{1}{x (1+a x)^{3/2}} \, dx}{8 (c-a c x)^{3/2}}\\ &=\frac{47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt{1+a x} (c-a c x)^{3/2}}-\frac{c^2 (1-a x)^{3/2}}{2 x^2 \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{13 a c^2 (1-a x)^{3/2}}{4 x \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{\left (47 a^2 c^2 (1-a x)^{3/2}\right ) \int \frac{1}{x \sqrt{1+a x}} \, dx}{8 (c-a c x)^{3/2}}\\ &=\frac{47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt{1+a x} (c-a c x)^{3/2}}-\frac{c^2 (1-a x)^{3/2}}{2 x^2 \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{13 a c^2 (1-a x)^{3/2}}{4 x \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{\left (47 a 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 )}{4 (c-a c x)^{3/2}}\\ &=\frac{47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt{1+a x} (c-a c x)^{3/2}}-\frac{c^2 (1-a x)^{3/2}}{2 x^2 \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{13 a c^2 (1-a x)^{3/2}}{4 x \sqrt{1+a x} (c-a c x)^{3/2}}-\frac{47 a^2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt{1+a x}\right )}{4 (c-a c x)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0270787, size = 65, normalized size = 0.4 \[ \frac{c \sqrt{1-a x} \left (47 a^2 x^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},a x+1\right )+13 a x-2\right )}{4 x^2 \sqrt{a x+1} \sqrt{c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*Sqrt[1 - a*x]*(-2 + 13*a*x + 47*a^2*x^2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + a*x]))/(4*x^2*Sqrt[1 + a*x]*Sqr
t[c - a*c*x])

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Maple [A]  time = 0.113, size = 100, normalized size = 0.6 \begin{align*}{\frac{1}{ \left ( 4\,ax-4 \right ) \left ( ax+1 \right ){x}^{2}}\sqrt{-c \left ( ax-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1} \left ( 47\,{\it Artanh} \left ({\frac{\sqrt{c \left ( ax+1 \right ) }}{\sqrt{c}}} \right ){x}^{2}{a}^{2}\sqrt{c \left ( ax+1 \right ) }-47\,{x}^{2}{a}^{2}\sqrt{c}-13\,xa\sqrt{c}+2\,\sqrt{c} \right ){\frac{1}{\sqrt{c}}}} \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^3,x)

[Out]

1/4*(-c*(a*x-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(47*arctanh((c*(a*x+1))^(1/2)/c^(1/2))*x^2*a^2*(c*(a*x+1))^(1/2)-47*
x^2*a^2*c^(1/2)-13*x*a*c^(1/2)+2*c^(1/2))/c^(1/2)/(a*x-1)/(a*x+1)/x^2

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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^{3}}\,{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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.84219, size = 547, normalized size = 3.36 \begin{align*} \left [\frac{47 \,{\left (a^{4} x^{4} - a^{2} x^{2}\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 \,{\left (47 \, a^{2} x^{2} + 13 \, a x - 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{8 \,{\left (a^{2} x^{4} - x^{2}\right )}}, -\frac{47 \,{\left (a^{4} x^{4} - a^{2} x^{2}\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 ) +{\left (47 \, a^{2} x^{2} + 13 \, a x - 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{4 \,{\left (a^{2} x^{4} - x^{2}\right )}}\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^3,x, algorithm="fricas")

[Out]

[1/8*(47*(a^4*x^4 - a^2*x^2)*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*(47*a^2*x^2 + 13*a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^2*x^4 - x^2), -1/4*(4
7*(a^4*x^4 - a^2*x^2)*sqrt(-c)*arctan(sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + (47*a^2*
x^2 + 13*a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^2*x^4 - x^2)]

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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^{3} \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**3,x)

[Out]

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

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Giac [A]  time = 1.23052, size = 178, normalized size = 1.09 \begin{align*} \frac{1}{4} \, a^{2} c{\left (\frac{47 \, \arctan \left (\frac{\sqrt{a c x + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c} + \frac{32}{\sqrt{a c x + c} c} + \frac{15 \,{\left (a c x + c\right )}^{\frac{3}{2}} - 17 \, \sqrt{a c x + c} c}{a^{2} c^{3} x^{2}}\right )}{\left | c \right |} - \frac{\sqrt{2}{\left (47 \, \sqrt{2} a^{2} \sqrt{c}{\left | c \right |} \arctan \left (\frac{\sqrt{2} \sqrt{c}}{\sqrt{-c}}\right ) + 58 \, a^{2} \sqrt{-c}{\left | c \right |}\right )}}{8 \, \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^3,x, algorithm="giac")

[Out]

1/4*a^2*c*(47*arctan(sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c) + 32/(sqrt(a*c*x + c)*c) + (15*(a*c*x + c)^(3/2) -
17*sqrt(a*c*x + c)*c)/(a^2*c^3*x^2))*abs(c) - 1/8*sqrt(2)*(47*sqrt(2)*a^2*sqrt(c)*abs(c)*arctan(sqrt(2)*sqrt(
c)/sqrt(-c)) + 58*a^2*sqrt(-c)*abs(c))/(sqrt(-c)*sqrt(c))