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

Optimal. Leaf size=50 \[ x \sqrt{c-\frac{c}{a x}}+\frac{3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.158676, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6167, 6133, 25, 514, 375, 78, 63, 208} \[ x \sqrt{c-\frac{c}{a x}}+\frac{3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0298758, size = 50, normalized size = 1. \[ x \sqrt{c-\frac{c}{a x}}+\frac{3 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-\frac{c}{a x}}}{\sqrt{c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.166, size = 118, normalized size = 2.4 \begin{align*}{\frac{x}{2}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( 4\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}-2\,\sqrt{a{x}^{2}-x}\sqrt{a}+\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{a{x}^{2}-x}\sqrt{a}+2\,ax-1 \right ){\frac{1}{\sqrt{a}}}} \right ) +2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax-1 \right ) x}\sqrt{a}+2\,ax-1}{\sqrt{a}}} \right ) \right ){\frac{1}{\sqrt{ \left ( ax-1 \right ) x}}}{\frac{1}{\sqrt{a}}}} \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)

[Out]

1/2*(c*(a*x-1)/a/x)^(1/2)*x*(4*((a*x-1)*x)^(1/2)*a^(1/2)-2*(a*x^2-x)^(1/2)*a^(1/2)+ln(1/2*(2*(a*x^2-x)^(1/2)*a
^(1/2)+2*a*x-1)/a^(1/2))+2*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2)))/((a*x-1)*x)^(1/2)/a^(1/2)

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

[Out]

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

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Fricas [A]  time = 1.94038, size = 273, normalized size = 5.46 \begin{align*} \left [\frac{2 \, a x \sqrt{\frac{a c x - c}{a x}} + 3 \, \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}, \frac{a x \sqrt{\frac{a c x - c}{a x}} - 3 \, \sqrt{-c} \arctan \left (\frac{\sqrt{-c} \sqrt{\frac{a c x - c}{a x}}}{c}\right )}{a}\right ] \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, algorithm="fricas")

[Out]

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

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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 )}{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)*(c-c/a/x)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.29285, size = 131, normalized size = 2.62 \begin{align*} \frac{3 \, \sqrt{c} \log \left ({\left | a \right |} \sqrt{{\left | c \right |}}\right ) \mathrm{sgn}\left (x\right )}{2 \, a} - \frac{3 \, \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 )}{2 \, a \mathrm{sgn}\left (x\right )} + \frac{\sqrt{a^{2} c x^{2} - a c x}{\left | a \right |}}{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)*(c-c/a/x)^(1/2),x, algorithm="giac")

[Out]

3/2*sqrt(c)*log(abs(a)*sqrt(abs(c)))*sgn(x)/a - 3/2*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*sgn(x)) + sqrt(a^2*c*x^2 - a*c*x)*abs(a)/(a^2*sgn(x))