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

Optimal. Leaf size=89 \[ -\frac {2 \sqrt {1-a^2 x^2} \sqrt {c-\frac {c}{a x}}}{1-a x}-\frac {2 \sqrt {a} \sqrt {x} \sqrt {c-\frac {c}{a x}} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {1-a x}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6134, 6128, 879, 848, 54, 215} \[ -\frac {2 \sqrt {1-a^2 x^2} \sqrt {c-\frac {c}{a x}}}{1-a x}-\frac {2 \sqrt {a} \sqrt {x} \sqrt {c-\frac {c}{a x}} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 879

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(e*f
- d*g)*(d + e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + 1)*(e*f + d*g)), x] - Dist[(e*(e*f*
(p + 1) - d*g*(2*n + p + 3)))/(g*(n + 1)*(e*f + d*g)), Int[(d + e*x)^(m - 1)*(f + g*x)^(n + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&
EqQ[m + p - 1, 0] && LtQ[n, -1] && IntegerQ[2*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 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]

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 {(1-a x)^{3/2}}{x^{3/2} \sqrt {1-a^2 x^2}} \, dx}{\sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1-a^2 x^2}}{1-a x}-\frac {\left (a \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1-a x}}{\sqrt {x} \sqrt {1-a^2 x^2}} \, dx}{\sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1-a^2 x^2}}{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^2 x^2}}{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^2 x^2}}{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*}

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Mathematica [A]  time = 0.05, size = 60, normalized size = 0.67 \[ -\frac {2 \sqrt {c-\frac {c}{a x}} \left (\sqrt {a x+1}+\sqrt {a} \sqrt {x} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )\right )}{\sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[c - c/(a*x)]/(E^ArcTanh[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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fricas [A]  time = 0.63, size = 235, normalized size = 2.64 \[ \left [\frac {{\left (a x - 1\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a x - 1\right )}}, \frac {{\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{a x - 1}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)/(a*x+1)*(-a^2*x^2+1)^(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {c - \frac {c}{a x}}}{{\left (a x + 1\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 91, normalized size = 1.02 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (-\arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right ) x a +2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}\right ) \sqrt {-a^{2} x^{2}+1}}{\left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}\, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {c - \frac {c}{a x}}}{{\left (a x + 1\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c-\frac {c}{a\,x}}\,\sqrt {1-a^2\,x^2}}{x\,\left (a\,x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{x \left (a x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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