3.476 \(\int \frac{e^{-2 \coth ^{-1}(a x)}}{(c-\frac{c}{a x})^{3/2}} \, dx\)

Optimal. Leaf size=94 \[ \frac{x \sqrt{c-\frac{c}{a x}}}{c^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{3/2}}+\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{a c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.21546, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6167, 6133, 25, 514, 375, 103, 21, 83, 63, 208} \[ \frac{x \sqrt{c-\frac{c}{a x}}}{c^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{3/2}}+\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{a c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^(3/2)),x]

[Out]

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

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 6133

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, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &
&  !GtQ[c, 0]

Rule 25

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

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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 83

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

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

Mathematica [A]  time = 0.0517848, size = 94, normalized size = 1. \[ \frac{x \sqrt{c-\frac{c}{a x}}}{c^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a c^{3/2}}+\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{2} \sqrt{c}}\right )}{a c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.17, size = 134, normalized size = 1.4 \begin{align*} -{\frac{x}{2\,{c}^{2}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -2\,\sqrt{ \left ( ax-1 \right ) x}{a}^{3/2}\sqrt{{a}^{-1}}+\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1 \right ){\frac{1}{\sqrt{a}}}} \right ) a\sqrt{{a}^{-1}}+\sqrt{2}\ln \left ({\frac{1}{ax+1} \left ( 2\,\sqrt{2}\sqrt{{a}^{-1}}\sqrt{ \left ( ax-1 \right ) x}a-3\,ax+1 \right ) } \right ) \sqrt{a} \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{a}^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.98137, size = 531, normalized size = 5.65 \begin{align*} \left [\frac{2 \, a x \sqrt{\frac{a c x - c}{a x}} + \sqrt{2} \sqrt{c} \log \left (-\frac{\frac{2 \, \sqrt{2} a x \sqrt{\frac{a c x - c}{a x}}}{\sqrt{c}} + 3 \, a x - 1}{a x + 1}\right ) + \sqrt{c} \log \left (-2 \, a c x + 2 \, a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + c\right )}{2 \, a c^{2}}, -\frac{\sqrt{2} c \sqrt{-\frac{1}{c}} \arctan \left (\frac{\sqrt{2} a x \sqrt{-\frac{1}{c}} \sqrt{\frac{a c x - c}{a x}}}{a x - 1}\right ) - a x \sqrt{\frac{a c x - c}{a x}} - \sqrt{-c} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right )}{a c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*a*x*sqrt((a*c*x - c)/(a*x)) + sqrt(2)*sqrt(c)*log(-(2*sqrt(2)*a*x*sqrt((a*c*x - c)/(a*x))/sqrt(c) + 3*
a*x - 1)/(a*x + 1)) + sqrt(c)*log(-2*a*c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c))/(a*c^2), -(sqrt(2)*c*
sqrt(-1/c)*arctan(sqrt(2)*a*x*sqrt(-1/c)*sqrt((a*c*x - c)/(a*x))/(a*x - 1)) - a*x*sqrt((a*c*x - c)/(a*x)) - sq
rt(-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c))/(a*c^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25617, size = 176, normalized size = 1.87 \begin{align*} -a c{\left (\frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a c x - c}{a x}}}{2 \, \sqrt{-c}}\right )}{a^{2} \sqrt{-c} c^{2}} - \frac{\arctan \left (\frac{\sqrt{\frac{a c x - c}{a x}}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c^{2}} - \frac{\sqrt{\frac{a c x - c}{a x}}}{a^{2}{\left (c - \frac{a c x - c}{a x}\right )} c^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*c*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt((a*c*x - c)/(a*x))/sqrt(-c))/(a^2*sqrt(-c)*c^2) - arctan(sqrt((a*c*x - c
)/(a*x))/sqrt(-c))/(a^2*sqrt(-c)*c^2) - sqrt((a*c*x - c)/(a*x))/(a^2*(c - (a*c*x - c)/(a*x))*c^2))