\(\int e^{\coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\) [775]

Optimal result

Integrand size = 18, antiderivative size = 107 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {2 c \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}{a}+c \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/2} x+\frac {c \csc ^{-1}(a x)}{a}+\frac {c \text {arctanh}\left (\sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}\right )}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6329, 99, 159, 21, 132, 41, 222, 94, 214} \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c \text {arctanh}\left (\sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}\right )}{a}+c x \sqrt {1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/2}-\frac {2 c \sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}}{a}+\frac {c \csc ^{-1}(a x)}{a} \]

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[b*d^(m
+ n)*f^p, Int[(a + b*x)^(m - 1)/(c + d*x)^m, x], x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandTo
Sum[(a + b*x)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n,
-1]))

Rule 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

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 6329

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.50 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c \left (\sqrt {1-\frac {1}{a^2 x^2}} (-1+a x)+\arcsin \left (\frac {1}{a x}\right )+\log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )\right )}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.19

method	result	size
risch	\(-\frac {\left (a x -1\right ) c}{x \,a^{2} \sqrt {\frac {a x -1}{a x +1}}}+\frac {\left (\frac {a \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{\sqrt {a^{2}}}+\sqrt {\left (a x -1\right ) \left (a x +1\right )}+\arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )\right ) c \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{a \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\)	\(127\)
default	\(\frac {\left (a x -1\right ) c \left (-\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{2} x^{2}+\left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}+\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x +\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{2} x +a x \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )\right )}{\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} x \sqrt {a^{2}}}\)	\(163\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {2 \, a c x \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - a c x \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + a c x \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (a^{2} c x^{2} - c\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.09 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-{\left (\frac {4 \, c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{\frac {{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} + \frac {2 \, c \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} - \frac {c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac {c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} a \]

[In]

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

[Out]

-(4*c*((a*x - 1)/(a*x + 1))^(3/2)/((a*x - 1)^2*a^2/(a*x + 1)^2 - a^2) + 2*c*arctan(sqrt((a*x - 1)/(a*x + 1)))/
a^2 - c*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 + c*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2)*a

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.21 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {2 \, c \arctan \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1}\right )}{a \mathrm {sgn}\left (a x + 1\right )} - \frac {c \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right )}{{\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} + \frac {\sqrt {a^{2} x^{2} - 1} c}{a \mathrm {sgn}\left (a x + 1\right )} - \frac {2 \, c}{{\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} + 1\right )} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} \]

[In]

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

[Out]

-2*c*arctan(-x*abs(a) + sqrt(a^2*x^2 - 1))/(a*sgn(a*x + 1)) - c*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))/(abs(a
)*sgn(a*x + 1)) + sqrt(a^2*x^2 - 1)*c/(a*sgn(a*x + 1)) - 2*c/(((x*abs(a) - sqrt(a^2*x^2 - 1))^2 + 1)*abs(a)*sg
n(a*x + 1))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79 \[ \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {2\,c\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}-\frac {2\,c\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}+\frac {4\,c\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{a-\frac {a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}} \]

[In]

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

[Out]

(2*c*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/a - (2*c*atan(((a*x - 1)/(a*x + 1))^(1/2)))/a + (4*c*((a*x - 1)/(a*x
+ 1))^(3/2))/(a - (a*(a*x - 1)^2)/(a*x + 1)^2)