\(\int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx\) [299]

Optimal result

Integrand size = 21, antiderivative size = 94 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx=\frac {2 \sqrt {1+\frac {1}{a x}} \sqrt {c-a c x}}{\sqrt {1-\frac {1}{a x}}}-\frac {2 \sqrt {\frac {1}{x}} \sqrt {c-a c x} \text {arcsinh}\left (\frac {\sqrt {\frac {1}{x}}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {1-\frac {1}{a x}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6311, 6316, 49, 56, 221} \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx=\frac {2 \sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}}{\sqrt {1-\frac {1}{a x}}}-\frac {2 \sqrt {\frac {1}{x}} \text {arcsinh}\left (\frac {\sqrt {\frac {1}{x}}}{\sqrt {a}}\right ) \sqrt {c-a c x}}{\sqrt {a} \sqrt {1-\frac {1}{a x}}} \]

[In]

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

[Out]

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

Rule 49

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 56

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 221

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 6311

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

Rule 6316

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx=\frac {2 \sqrt {c-a c x} \left (\sqrt {a} \sqrt {1+\frac {1}{a x}}-\sqrt {\frac {1}{x}} \text {arcsinh}\left (\frac {\sqrt {\frac {1}{x}}}{\sqrt {a}}\right )\right )}{\sqrt {a} \sqrt {1-\frac {1}{a x}}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74

method	result	size
default	\(-\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (\sqrt {c}\, \arctan \left (\frac {\sqrt {-c \left (a x +1\right )}}{\sqrt {c}}\right )-\sqrt {-c \left (a x +1\right )}\right )}{\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x +1\right )}}\)	\(70\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.20 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx=\left [\frac {{\left (a x - 1\right )} \sqrt {-c} \log \left (-\frac {a^{2} c x^{2} + a c x + 2 \, \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} - 2 \, c}{a x^{2} - x}\right ) + 2 \, \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a x - 1}, -\frac {2 \, {\left ({\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}}}{a c x - c}\right ) - \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{a x - 1}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx=\int \frac {\sqrt {- c \left (a x - 1\right )}}{x \sqrt {\frac {a x - 1}{a x + 1}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx=\int { \frac {\sqrt {-a c x + c}}{x \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx=\frac {2 \, c^{3} {\left (\frac {\arctan \left (\frac {\sqrt {-a c x - c}}{\sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {c \arctan \left (\frac {\sqrt {2} \sqrt {-c}}{\sqrt {c}}\right ) - \sqrt {2} \sqrt {-c} \sqrt {c}}{c^{\frac {5}{2}}} - \frac {\sqrt {-a c x - c}}{c^{2}}\right )}}{{\left | c \right |} \mathrm {sgn}\left (a x + 1\right )} \]

[In]

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

[Out]

2*c^3*(arctan(sqrt(-a*c*x - c)/sqrt(c))/c^(3/2) - (c*arctan(sqrt(2)*sqrt(-c)/sqrt(c)) - sqrt(2)*sqrt(-c)*sqrt(
c))/c^(5/2) - sqrt(-a*c*x - c)/c^2)/(abs(c)*sgn(a*x + 1))

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx=\int \frac {\sqrt {c-a\,c\,x}}{x\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

[In]

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

[Out]

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