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

Optimal. Leaf size=86 \[ \frac{2 \sqrt{a} \sqrt{x} \sqrt{c-\frac{c}{a x}} \sinh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{\sqrt{1-a x}}-\frac{2 \sqrt{a x+1} \sqrt{c-\frac{c}{a x}}}{\sqrt{1-a x}} \]

[Out]

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

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Rubi [A]  time = 0.248922, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {6134, 6128, 848, 47, 54, 215} \[ \frac{2 \sqrt{a} \sqrt{x} \sqrt{c-\frac{c}{a x}} \sinh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{\sqrt{1-a x}}-\frac{2 \sqrt{a x+1} \sqrt{c-\frac{c}{a x}}}{\sqrt{1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6134

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Dist[(x^p*(c + d/x)^p)/(1 + (c*
x)/d)^p, Int[(u*(1 + (c*x)/d)^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*
d^2, 0] &&  !IntegerQ[p]

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && IntegerQ[2*p]

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0366927, size = 61, normalized size = 0.71 \[ -\frac{2 \sqrt{c-\frac{c}{a x}} \left (\sqrt{a x+1}-\sqrt{a} \sqrt{x} \sinh ^{-1}\left (\sqrt{a} \sqrt{x}\right )\right )}{\sqrt{1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.134, size = 90, normalized size = 1.1 \begin{align*}{\frac{1}{ax-1}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}}\sqrt{-{a}^{2}{x}^{2}+1} \left ( \arctan \left ({\frac{2\,ax+1}{2}{\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{- \left ( ax+1 \right ) x}}}} \right ) xa+2\,\sqrt{a}\sqrt{- \left ( ax+1 \right ) x} \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{- \left ( ax+1 \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(a*x-1)/a/x)^(1/2)*(-a^2*x^2+1)^(1/2)*(arctan(1/2/a^(1/2)*(2*a*x+1)/(-(a*x+1)*x)^(1/2))*x*a+2*a^(1/2)*(-(a*
x+1)*x)^(1/2))/(a*x-1)/(-(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}}}{\sqrt{-a^{2} x^{2} + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.27025, size = 506, normalized size = 5.88 \begin{align*} \left [\frac{{\left (a x - 1\right )} \sqrt{-c} \log \left (-\frac{8 \, a^{3} c x^{3} - 7 \, a c x + 4 \,{\left (2 \, a^{2} x^{2} + a x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c} \sqrt{\frac{a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a c x - c}{a x}}}{2 \,{\left (a x - 1\right )}}, -\frac{{\left (a x - 1\right )} \sqrt{c} \arctan \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1} a \sqrt{c} x \sqrt{\frac{a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a c x - c}{a x}}}{a x - 1}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a*x - 1)*sqrt(-c)*log(-(8*a^3*c*x^3 - 7*a*c*x + 4*(2*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a
*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a*x - 1), -((a*x - 1)*sqrt(c
)*arctan(2*sqrt(-a^2*x^2 + 1)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) - 2*sqrt(-a^2*x^2
+ 1)*sqrt((a*c*x - c)/(a*x)))/(a*x - 1)]

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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 \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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