\(\int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx\) [890]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 27, antiderivative size = 117 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\sqrt {c-\frac {c}{a^2 x^2}}-\frac {a \sqrt {c-\frac {c}{a^2 x^2}} x \arcsin (a x)}{\sqrt {1-a x} \sqrt {1+a x}}+\frac {2 a \sqrt {c-\frac {c}{a^2 x^2}} x \text {arctanh}\left (\sqrt {1-a x} \sqrt {1+a x}\right )}{\sqrt {1-a x} \sqrt {1+a x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6302, 6294, 6264, 100, 163, 41, 222, 94, 214} \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=-\frac {a x \arcsin (a x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a x} \sqrt {a x+1}}+\frac {2 a x \text {arctanh}\left (\sqrt {1-a x} \sqrt {a x+1}\right ) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a x} \sqrt {a x+1}}+\sqrt {c-\frac {c}{a^2 x^2}} \]

[In]

Int[(E^(2*ArcCoth[a*x])*Sqrt[c - c/(a^2*x^2)])/x,x]

[Out]

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 94

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

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
 FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
 :> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 222

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

Rule 6264

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

Rule 6294

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

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (\sqrt {-1+a^2 x^2}-2 a x \arctan \left (\frac {1}{\sqrt {-1+a^2 x^2}}\right )+a x \log \left (a x+\sqrt {-1+a^2 x^2}\right )\right )}{\sqrt {-1+a^2 x^2}} \]

[In]

Integrate[(E^(2*ArcCoth[a*x])*Sqrt[c - c/(a^2*x^2)])/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.24

method result size
risch \(\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}+\frac {\left (\frac {a^{2} \ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c}\right )}{\sqrt {a^{2} c}}-\frac {2 a \ln \left (\frac {-2 c +2 \sqrt {-c}\, \sqrt {a^{2} c \,x^{2}-c}}{x}\right )}{\sqrt {-c}}\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {c \left (a^{2} x^{2}-1\right )}\, x}{a^{2} x^{2}-1}\) \(145\)
default \(-\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (-\sqrt {-\frac {c}{a^{2}}}\, \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{3} c \,x^{2}+a^{3} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} \sqrt {-\frac {c}{a^{2}}}+c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x +\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\right ) \sqrt {-\frac {c}{a^{2}}}\, a x -2 c^{\frac {3}{2}} \sqrt {-\frac {c}{a^{2}}}\, \ln \left (\frac {\sqrt {c}\, \sqrt {\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}}+c x}{\sqrt {c}}\right ) a x -2 \sqrt {-\frac {c}{a^{2}}}\, \sqrt {\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}}\, a^{2} c x +2 \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, c x \,a^{2} \sqrt {-\frac {c}{a^{2}}}+2 \ln \left (\frac {2 \sqrt {-\frac {c}{a^{2}}}\, \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{2}-2 c}{a^{2} x}\right ) c^{2} x \right )}{a \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, c \sqrt {-\frac {c}{a^{2}}}}\) \(306\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.15 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\left [-\sqrt {-c} \arctan \left (\frac {a^{2} \sqrt {-c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right ) + \sqrt {-c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a \sqrt {-c} x \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 2 \, c}{x^{2}}\right ) + \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}, -2 \, \sqrt {c} \arctan \left (\frac {a \sqrt {c} x \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right ) + \frac {1}{2} \, \sqrt {c} \log \left (2 \, a^{2} c x^{2} + 2 \, a^{2} \sqrt {c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right ) + \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{x \left (a x - 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{{\left (a x - 1\right )} x} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx={\left (\frac {4 \, \sqrt {c} \arctan \left (-\frac {\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - c}}{\sqrt {c}}\right ) \mathrm {sgn}\left (x\right )}{a} - \frac {\sqrt {c} \log \left ({\left | -\sqrt {a^{2} c} x + \sqrt {a^{2} c x^{2} - c} \right |}\right ) \mathrm {sgn}\left (x\right )}{{\left | a \right |}} + \frac {2 \, c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )}{{\left ({\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - c}\right )}^{2} + c\right )} {\left | a \right |}}\right )} {\left | a \right |} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,\left (a\,x+1\right )}{x\,\left (a\,x-1\right )} \,d x \]

[In]

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

[Out]

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