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

Optimal. Leaf size=79 \[ \frac {c x \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6177, 879, 875, 208} \[ \frac {c x \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[1 - 1/(a^2*x^2)]*x)/Sqrt[c - c/(a*x)] - (3*Sqrt[c]*ArcTanh[(Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)])/Sqrt[c - c/
(a*x)]])/a

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]

Rule 875

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e^2, Subst[I
nt[1/(c*(e*f + d*g) + e^2*g*x^2), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x] &&
 NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 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 6177

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 65, normalized size = 0.82 \[ \frac {\sqrt {c-\frac {c}{a x}} \left (x \sqrt {\frac {1}{a x}+1}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {1}{a x}+1}\right )}{a}\right )}{\sqrt {1-\frac {1}{a x}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.64, size = 297, normalized size = 3.76 \[ \left [\frac {3 \, {\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{4 \, {\left (a^{2} x - a\right )}}, \frac {3 \, {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} x - a\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 101, normalized size = 1.28 \[ \frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}-3 \ln \left (\frac {2 \sqrt {\left (a x +1\right ) x}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right )\right )}{2 \left (a x -1\right ) \sqrt {\left (a x +1\right ) x}\, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________