3.503 $$\int \frac{e^{2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^2} \, dx$$

Optimal. Leaf size=42 $4 a \sqrt{c-\frac{c}{a x}}-\frac{2 a \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}$

[Out]

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

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Rubi [A]  time = 0.32396, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {6167, 6133, 25, 514, 444, 43} $4 a \sqrt{c-\frac{c}{a x}}-\frac{2 a \left (c-\frac{c}{a x}\right )^{3/2}}{3 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0385772, size = 28, normalized size = 0.67 $\frac{2 (5 a x+1) \sqrt{c-\frac{c}{a x}}}{3 x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.122, size = 27, normalized size = 0.6 \begin{align*}{\frac{10\,ax+2}{3\,x}\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)/(a*x-1)*(c-c/a/x)^(1/2)/x^2,x)

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.54706, size = 58, normalized size = 1.38 \begin{align*} \frac{2 \,{\left (5 \, a x + 1\right )} \sqrt{\frac{a c x - c}{a x}}}{3 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError