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

Optimal. Leaf size=90 \[ \frac {3 \sqrt {x} \sqrt {c-\frac {c}{a x}} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {a} \sqrt {1-a x}}-\frac {x \sqrt {1-a^2 x^2} \sqrt {c-\frac {c}{a x}}}{1-a x} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6134, 6128, 881, 848, 54, 215} \[ \frac {3 \sqrt {x} \sqrt {c-\frac {c}{a x}} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {a} \sqrt {1-a x}}-\frac {x \sqrt {1-a^2 x^2} \sqrt {c-\frac {c}{a x}}}{1-a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[c - c/(a*x)]*x*Sqrt[1 - a^2*x^2])/(1 - a*x)) + (3*Sqrt[c - c/(a*x)]*Sqrt[x]*ArcSinh[Sqrt[a]*Sqrt[x]])/
(Sqrt[a]*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 881

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(d +
e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + p + 2)), x] - Dist[(e*f*(p + 1) - d*g*(2*n + p
+ 3))/(g*(n + p + 2)), Int[(d + e*x)^(m - 1)*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m,
n, 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 e^{-\tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} \, dx &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {e^{-\tanh ^{-1}(a x)} \sqrt {1-a x}}{\sqrt {x}} \, dx}{\sqrt {1-a x}}\\ &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {(1-a x)^{3/2}}{\sqrt {x} \sqrt {1-a^2 x^2}} \, dx}{\sqrt {1-a x}}\\ &=-\frac {\sqrt {c-\frac {c}{a x}} x \sqrt {1-a^2 x^2}}{1-a x}+\frac {\left (3 \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1-a x}}{\sqrt {x} \sqrt {1-a^2 x^2}} \, dx}{2 \sqrt {1-a x}}\\ &=-\frac {\sqrt {c-\frac {c}{a x}} x \sqrt {1-a^2 x^2}}{1-a x}+\frac {\left (3 \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+a x}} \, dx}{2 \sqrt {1-a x}}\\ &=-\frac {\sqrt {c-\frac {c}{a x}} x \sqrt {1-a^2 x^2}}{1-a x}+\frac {\left (3 \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 {\sqrt {c-\frac {c}{a x}} x \sqrt {1-a^2 x^2}}{1-a x}+\frac {3 \sqrt {c-\frac {c}{a x}} \sqrt {x} \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {a} \sqrt {1-a x}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 67, normalized size = 0.74 \[ \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (\frac {3 \sinh ^{-1}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {a}}-\sqrt {x} \sqrt {a x+1}\right )}{\sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 250, normalized size = 2.78 \[ \left [\frac {4 \, \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}} + 3 \, {\left (a x - 1\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right )}{4 \, {\left (a^{2} x - a\right )}}, \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a x \sqrt {\frac {a c x - c}{a x}} - 3 \, {\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right )}{2 \, {\left (a^{2} x - a\right )}}\right ] \]

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, algorithm="fricas")

[Out]

[1/4*(4*sqrt(-a^2*x^2 + 1)*a*x*sqrt((a*c*x - c)/(a*x)) + 3*(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)))/(a^2*x - a), 1/2*(2*sqr
t(-a^2*x^2 + 1)*a*x*sqrt((a*c*x - c)/(a*x)) - 3*(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)))/(a^2*x - a)]

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

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, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 91, normalized size = 1.01 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \sqrt {-a^{2} x^{2}+1}\, \left (2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}+3 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-\left (a x +1\right ) x}}\right )\right )}{2 \left (a x -1\right ) \sqrt {-\left (a x +1\right ) x}\, \sqrt {a}} \]

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)

[Out]

1/2*(c*(a*x-1)/a/x)^(1/2)*x*(-a^2*x^2+1)^(1/2)*(2*a^(1/2)*(-(a*x+1)*x)^(1/2)+3*arctan(1/2/a^(1/2)*(2*a*x+1)/(-
(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {c - \frac {c}{a x}}}{a x + 1}\,{d x} \]

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, algorithm="maxima")

[Out]

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

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

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))/(a*x + 1),x)

[Out]

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

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

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)

[Out]

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

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