3.579 \(\int \frac{e^{2 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^3} \, dx\)

Optimal. Leaf size=69 \[ -\frac{2 a^2 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}+\frac{2 a^2 \left (c-\frac{c}{a x}\right )^{3/2}}{c}-4 a^2 \sqrt{c-\frac{c}{a x}} \]

[Out]

-4*a^2*Sqrt[c - c/(a*x)] + (2*a^2*(c - c/(a*x))^(3/2))/c - (2*a^2*(c - c/(a*x))^(5/2))/(5*c^2)

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Rubi [A]  time = 0.221703, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6133, 25, 514, 446, 77} \[ -\frac{2 a^2 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}+\frac{2 a^2 \left (c-\frac{c}{a x}\right )^{3/2}}{c}-4 a^2 \sqrt{c-\frac{c}{a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-4*a^2*Sqrt[c - c/(a*x)] + (2*a^2*(c - c/(a*x))^(3/2))/c - (2*a^2*(c - c/(a*x))^(5/2))/(5*c^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 446

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{e^{2 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^3} \, dx &=\int \frac{\sqrt{c-\frac{c}{a x}} (1+a x)}{x^3 (1-a x)} \, dx\\ &=-\frac{c \int \frac{1+a x}{\sqrt{c-\frac{c}{a x}} x^4} \, dx}{a}\\ &=-\frac{c \int \frac{a+\frac{1}{x}}{\sqrt{c-\frac{c}{a x}} x^3} \, dx}{a}\\ &=\frac{c \operatorname{Subst}\left (\int \frac{x (a+x)}{\sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\frac{c \operatorname{Subst}\left (\int \left (\frac{2 a^2}{\sqrt{c-\frac{c x}{a}}}-\frac{3 a^2 \sqrt{c-\frac{c x}{a}}}{c}+\frac{a^2 \left (c-\frac{c x}{a}\right )^{3/2}}{c^2}\right ) \, dx,x,\frac{1}{x}\right )}{a}\\ &=-4 a^2 \sqrt{c-\frac{c}{a x}}+\frac{2 a^2 \left (c-\frac{c}{a x}\right )^{3/2}}{c}-\frac{2 a^2 \left (c-\frac{c}{a x}\right )^{5/2}}{5 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0413682, size = 36, normalized size = 0.52 \[ -\frac{2 \left (6 a^2 x^2+3 a x+1\right ) \sqrt{c-\frac{c}{a x}}}{5 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - c/(a*x)]*(1 + 3*a*x + 6*a^2*x^2))/(5*x^2)

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Maple [A]  time = 0.086, size = 35, normalized size = 0.5 \begin{align*} -{\frac{12\,{a}^{2}{x}^{2}+6\,ax+2}{5\,{x}^{2}}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(c*(a*x-1)/a/x)^(1/2)*(6*a^2*x^2+3*a*x+1)/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.82501, size = 78, normalized size = 1.13 \begin{align*} -\frac{2 \,{\left (6 \, a^{2} x^{2} + 3 \, a x + 1\right )} \sqrt{\frac{a c x - c}{a x}}}{5 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(6*a^2*x^2 + 3*a*x + 1)*sqrt((a*c*x - c)/(a*x))/x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\sqrt{c - \frac{c}{a x}}}{a x^{4} - x^{3}}\, dx - \int \frac{a x \sqrt{c - \frac{c}{a x}}}{a x^{4} - x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(sqrt(c - c/(a*x))/(a*x**4 - x**3), x) - Integral(a*x*sqrt(c - c/(a*x))/(a*x**4 - x**3), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError