### 3.500 $$\int e^{2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}} x \, dx$$

Optimal. Leaf size=80 $\frac{7 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{4 a^2}+\frac{1}{2} x^2 \sqrt{c-\frac{c}{a x}}+\frac{7 x \sqrt{c-\frac{c}{a x}}}{4 a}$

[Out]

(7*Sqrt[c - c/(a*x)]*x)/(4*a) + (Sqrt[c - c/(a*x)]*x^2)/2 + (7*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(4*
a^2)

________________________________________________________________________________________

Rubi [A]  time = 0.240854, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.36, Rules used = {6167, 6133, 25, 434, 446, 78, 51, 63, 208} $\frac{7 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{4 a^2}+\frac{1}{2} x^2 \sqrt{c-\frac{c}{a x}}+\frac{7 x \sqrt{c-\frac{c}{a x}}}{4 a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(7*Sqrt[c - c/(a*x)]*x)/(4*a) + (Sqrt[c - c/(a*x)]*x^2)/2 + (7*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(4*
a^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 434

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[((a + b*x^n)^p*(d + c*x
^n)^q)/x^(n*q), x] /; FreeQ[{a, b, c, d, 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 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 e^{2 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}} x \, dx &=-\int e^{2 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a x}} x \, dx\\ &=-\int \frac{\sqrt{c-\frac{c}{a x}} x (1+a x)}{1-a x} \, dx\\ &=\frac{c \int \frac{1+a x}{\sqrt{c-\frac{c}{a x}}} \, dx}{a}\\ &=\frac{c \int \frac{\left (a+\frac{1}{x}\right ) x}{\sqrt{c-\frac{c}{a x}}} \, dx}{a}\\ &=-\frac{c \operatorname{Subst}\left (\int \frac{a+x}{x^3 \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\frac{1}{2} \sqrt{c-\frac{c}{a x}} x^2-\frac{(7 c) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{4 a}\\ &=\frac{7 \sqrt{c-\frac{c}{a x}} x}{4 a}+\frac{1}{2} \sqrt{c-\frac{c}{a x}} x^2-\frac{(7 c) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )}{8 a^2}\\ &=\frac{7 \sqrt{c-\frac{c}{a x}} x}{4 a}+\frac{1}{2} \sqrt{c-\frac{c}{a x}} x^2+\frac{7 \operatorname{Subst}\left (\int \frac{1}{a-\frac{a x^2}{c}} \, dx,x,\sqrt{c-\frac{c}{a x}}\right )}{4 a}\\ &=\frac{7 \sqrt{c-\frac{c}{a x}} x}{4 a}+\frac{1}{2} \sqrt{c-\frac{c}{a x}} x^2+\frac{7 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{4 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0713571, size = 77, normalized size = 0.96 $\frac{\sqrt{c-\frac{c}{a x}} \left (a x \sqrt{1-\frac{1}{a x}} (2 a x+7)+7 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a x}}\right )\right )}{4 a^2 \sqrt{1-\frac{1}{a x}}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c - c/(a*x)]*(a*Sqrt[1 - 1/(a*x)]*x*(7 + 2*a*x) + 7*ArcTanh[Sqrt[1 - 1/(a*x)]]))/(4*a^2*Sqrt[1 - 1/(a*x)
])

________________________________________________________________________________________

Maple [B]  time = 0.161, size = 139, normalized size = 1.7 \begin{align*}{\frac{x}{8}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( 4\,\sqrt{a{x}^{2}-x}{a}^{5/2}x-2\,\sqrt{a{x}^{2}-x}{a}^{3/2}+16\,{a}^{3/2}\sqrt{ \left ( ax-1 \right ) x}+8\,a\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) -\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{a{x}^{2}-x}\sqrt{a}+2\,ax-1 \right ){\frac{1}{\sqrt{a}}}} \right ) a \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{a}^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(c*(a*x-1)/a/x)^(1/2)*x*(4*(a*x^2-x)^(1/2)*a^(5/2)*x-2*(a*x^2-x)^(1/2)*a^(3/2)+16*a^(3/2)*((a*x-1)*x)^(1/2
)+8*a*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))-ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2)
)*a)/((a*x-1)*x)^(1/2)/a^(5/2)

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.54527, size = 327, normalized size = 4.09 \begin{align*} \left [\frac{2 \,{\left (2 \, a^{2} x^{2} + 7 \, a x\right )} \sqrt{\frac{a c x - c}{a x}} + 7 \, \sqrt{c} \log \left (-2 \, a c x - 2 \, a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}} + c\right )}{8 \, a^{2}}, \frac{{\left (2 \, a^{2} x^{2} + 7 \, a x\right )} \sqrt{\frac{a c x - c}{a x}} - 7 \, \sqrt{-c} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right )}{4 \, a^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.24579, size = 153, normalized size = 1.91 \begin{align*} \frac{1}{4} \, \sqrt{a^{2} c x^{2} - a c x}{\left (\frac{2 \, x{\left | a \right |}}{a^{2} \mathrm{sgn}\left (x\right )} + \frac{7 \,{\left | a \right |}}{a^{3} \mathrm{sgn}\left (x\right )}\right )} + \frac{7 \, \sqrt{c} \log \left ({\left | a \right |} \sqrt{{\left | c \right |}}\right ) \mathrm{sgn}\left (x\right )}{8 \, a^{2}} - \frac{7 \, \sqrt{c} \log \left ({\left | -2 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} - a c x}\right )}{\left | a \right |} + a \sqrt{c} \right |}\right )}{8 \, a^{2} \mathrm{sgn}\left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(a^2*c*x^2 - a*c*x)*(2*x*abs(a)/(a^2*sgn(x)) + 7*abs(a)/(a^3*sgn(x))) + 7/8*sqrt(c)*log(abs(a)*sqrt(ab
s(c)))*sgn(x)/a^2 - 7/8*sqrt(c)*log(abs(-2*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*abs(a) + a*sqrt(c)))/(a^2
*sgn(x))