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

Optimal. Leaf size=173 \[ -\frac{23 a^2 (c-a c x)^{3/2} \tanh ^{-1}\left (\sqrt{a x+1}\right )}{4 c (1-a x)^{3/2}}+\frac{4 \sqrt{2} a^2 (c-a c x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a x+1}}{\sqrt{2}}\right )}{c (1-a x)^{3/2}}-\frac{\sqrt{a x+1} (c-a c x)^{3/2}}{2 c x^2 (1-a x)^{3/2}}-\frac{9 a \sqrt{a x+1} (c-a c x)^{3/2}}{4 c x (1-a x)^{3/2}} \]

[Out]

-(Sqrt[1 + a*x]*(c - a*c*x)^(3/2))/(2*c*x^2*(1 - a*x)^(3/2)) - (9*a*Sqrt[1 + a*x]*(c - a*c*x)^(3/2))/(4*c*x*(1
- a*x)^(3/2)) - (23*a^2*(c - a*c*x)^(3/2)*ArcTanh[Sqrt[1 + a*x]])/(4*c*(1 - a*x)^(3/2)) + (4*Sqrt[2]*a^2*(c -
a*c*x)^(3/2)*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]])/(c*(1 - a*x)^(3/2))

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Rubi [A]  time = 0.149217, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {6130, 23, 98, 151, 156, 63, 208} \[ -\frac{23 a^2 (c-a c x)^{3/2} \tanh ^{-1}\left (\sqrt{a x+1}\right )}{4 c (1-a x)^{3/2}}+\frac{4 \sqrt{2} a^2 (c-a c x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a x+1}}{\sqrt{2}}\right )}{c (1-a x)^{3/2}}-\frac{\sqrt{a x+1} (c-a c x)^{3/2}}{2 c x^2 (1-a x)^{3/2}}-\frac{9 a \sqrt{a x+1} (c-a c x)^{3/2}}{4 c x (1-a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 + a*x]*(c - a*c*x)^(3/2))/(2*c*x^2*(1 - a*x)^(3/2)) - (9*a*Sqrt[1 + a*x]*(c - a*c*x)^(3/2))/(4*c*x*(1
- a*x)^(3/2)) - (23*a^2*(c - a*c*x)^(3/2)*ArcTanh[Sqrt[1 + a*x]])/(4*c*(1 - a*x)^(3/2)) + (4*Sqrt[2]*a^2*(c -
a*c*x)^(3/2)*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]])/(c*(1 - a*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 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 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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]

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{(c-a c x)^{3/2} \int \frac{(1+a x)^{3/2}}{x^3 (c-a c x)} \, dx}{(1-a x)^{3/2}}\\ &=-\frac{\sqrt{1+a x} (c-a c x)^{3/2}}{2 c x^2 (1-a x)^{3/2}}-\frac{(c-a c x)^{3/2} \int \frac{-\frac{9 a c}{2}-\frac{7}{2} a^2 c x}{x^2 \sqrt{1+a x} (c-a c x)} \, dx}{2 c (1-a x)^{3/2}}\\ &=-\frac{\sqrt{1+a x} (c-a c x)^{3/2}}{2 c x^2 (1-a x)^{3/2}}-\frac{9 a \sqrt{1+a x} (c-a c x)^{3/2}}{4 c x (1-a x)^{3/2}}+\frac{(c-a c x)^{3/2} \int \frac{\frac{23 a^2 c^2}{4}+\frac{9}{4} a^3 c^2 x}{x \sqrt{1+a x} (c-a c x)} \, dx}{2 c^2 (1-a x)^{3/2}}\\ &=-\frac{\sqrt{1+a x} (c-a c x)^{3/2}}{2 c x^2 (1-a x)^{3/2}}-\frac{9 a \sqrt{1+a x} (c-a c x)^{3/2}}{4 c x (1-a x)^{3/2}}+\frac{\left (4 a^3 (c-a c x)^{3/2}\right ) \int \frac{1}{\sqrt{1+a x} (c-a c x)} \, dx}{(1-a x)^{3/2}}+\frac{\left (23 a^2 (c-a c x)^{3/2}\right ) \int \frac{1}{x \sqrt{1+a x}} \, dx}{8 c (1-a x)^{3/2}}\\ &=-\frac{\sqrt{1+a x} (c-a c x)^{3/2}}{2 c x^2 (1-a x)^{3/2}}-\frac{9 a \sqrt{1+a x} (c-a c x)^{3/2}}{4 c x (1-a x)^{3/2}}+\frac{\left (8 a^2 (c-a c x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{2 c-c x^2} \, dx,x,\sqrt{1+a x}\right )}{(1-a x)^{3/2}}+\frac{\left (23 a (c-a c 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 (1-a x)^{3/2}}\\ &=-\frac{\sqrt{1+a x} (c-a c x)^{3/2}}{2 c x^2 (1-a x)^{3/2}}-\frac{9 a \sqrt{1+a x} (c-a c x)^{3/2}}{4 c x (1-a x)^{3/2}}-\frac{23 a^2 (c-a c x)^{3/2} \tanh ^{-1}\left (\sqrt{1+a x}\right )}{4 c (1-a x)^{3/2}}+\frac{4 \sqrt{2} a^2 (c-a c x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{1+a x}}{\sqrt{2}}\right )}{c (1-a x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0536777, size = 92, normalized size = 0.53 \[ -\frac{\sqrt{c-a c x} \left (23 a^2 x^2 \tanh ^{-1}\left (\sqrt{a x+1}\right )-16 \sqrt{2} a^2 x^2 \tanh ^{-1}\left (\frac{\sqrt{a x+1}}{\sqrt{2}}\right )+\sqrt{a x+1} (9 a x+2)\right )}{4 x^2 \sqrt{1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c - a*c*x]*(Sqrt[1 + a*x]*(2 + 9*a*x) + 23*a^2*x^2*ArcTanh[Sqrt[1 + a*x]] - 16*Sqrt[2]*a^2*x^2*ArcTanh[
Sqrt[1 + a*x]/Sqrt[2]]))/(4*x^2*Sqrt[1 - a*x])

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Maple [A]  time = 0.11, size = 131, normalized size = 0.8 \begin{align*} -{\frac{1}{ \left ( 4\,ax-4 \right ){x}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ( ax-1 \right ) } \left ( 16\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( ax+1 \right ) }\sqrt{2}}{\sqrt{c}}} \right ){x}^{2}{a}^{2}c-23\,c{\it Artanh} \left ({\frac{\sqrt{c \left ( ax+1 \right ) }}{\sqrt{c}}} \right ){x}^{2}{a}^{2}-9\,xa\sqrt{c \left ( ax+1 \right ) }\sqrt{c}-2\,\sqrt{c \left ( ax+1 \right ) }\sqrt{c} \right ){\frac{1}{\sqrt{c \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(16*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x^2*a^2*
c-23*c*arctanh((c*(a*x+1))^(1/2)/c^(1/2))*x^2*a^2-9*x*a*(c*(a*x+1))^(1/2)*c^(1/2)-2*(c*(a*x+1))^(1/2)*c^(1/2))
/c^(1/2)/(a*x-1)/(c*(a*x+1))^(1/2)/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.99908, size = 869, normalized size = 5.02 \begin{align*} \left [\frac{16 \, \sqrt{2}{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \sqrt{c} \log \left (-\frac{a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt{2} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c} \sqrt{c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 23 \,{\left (a^{3} x^{3} - 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 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}{\left (9 \, a x + 2\right )}}{8 \,{\left (a x^{3} - x^{2}\right )}}, \frac{16 \, \sqrt{2}{\left (a^{3} x^{3} - a^{2} x^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{2} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) - 23 \,{\left (a^{3} x^{3} - 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 ) + \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}{\left (9 \, a x + 2\right )}}{4 \,{\left (a x^{3} - x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(16*sqrt(2)*(a^3*x^3 - a^2*x^2)*sqrt(c)*log(-(a^2*c*x^2 + 2*a*c*x - 2*sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*
c*x + c)*sqrt(c) - 3*c)/(a^2*x^2 - 2*a*x + 1)) + 23*(a^3*x^3 - a^2*x^2)*sqrt(c)*log(-(a^2*c*x^2 + a*c*x + 2*sq
rt(-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)*(9*a*x +
2))/(a*x^3 - x^2), 1/4*(16*sqrt(2)*(a^3*x^3 - a^2*x^2)*sqrt(-c)*arctan(sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x
+ c)*sqrt(-c)/(a^2*c*x^2 - c)) - 23*(a^3*x^3 - a^2*x^2)*sqrt(-c)*arctan(sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*s
qrt(-c)/(a^2*c*x^2 - c)) + sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(9*a*x + 2))/(a*x^3 - 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 (a x + 1\right )^{3}}{x^{3} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.3226, size = 244, normalized size = 1.41 \begin{align*} -\frac{1}{4} \, a^{2} c^{3}{\left (\frac{16 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a c x + c}}{2 \, \sqrt{-c}}\right )}{\sqrt{-c} c{\left | c \right |}} - \frac{23 \, \arctan \left (\frac{\sqrt{a c x + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c{\left | c \right |}} + \frac{9 \,{\left (a c x + c\right )}^{\frac{3}{2}} - 7 \, \sqrt{a c x + c} c}{a^{2} c^{3} x^{2}{\left | c \right |}}\right )} - \frac{\sqrt{2}{\left (23 \, \sqrt{2} a^{2} c^{2} \arctan \left (\frac{\sqrt{2} \sqrt{c}}{\sqrt{-c}}\right ) - 32 \, a^{2} c^{2} \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) - 22 \, a^{2} \sqrt{-c} c^{\frac{3}{2}}\right )}}{8 \, \sqrt{-c}{\left | c \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*c^3*(16*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c*abs(c)) - 23*arctan(sqrt(a*c
*x + c)/sqrt(-c))/(sqrt(-c)*c*abs(c)) + (9*(a*c*x + c)^(3/2) - 7*sqrt(a*c*x + c)*c)/(a^2*c^3*x^2*abs(c))) - 1/
8*sqrt(2)*(23*sqrt(2)*a^2*c^2*arctan(sqrt(2)*sqrt(c)/sqrt(-c)) - 32*a^2*c^2*arctan(sqrt(c)/sqrt(-c)) - 22*a^2*
sqrt(-c)*c^(3/2))/(sqrt(-c)*abs(c))