\(\int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx\) [258]

Optimal result

Integrand size = 20, antiderivative size = 76 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} \left (1-\frac {1}{a x}\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {1+\frac {1}{a x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6311, 6316, 95, 212} \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} \left (1-\frac {1}{a x}\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {\frac {1}{a x}+1}}\right )}{\left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}} \]

[In]

Int[1/(E^ArcCoth[a*x]*(c - a*c*x)^(3/2)),x]

[Out]

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

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} \left (1-\frac {1}{a x}\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {1+\frac {1}{a x}}}\right )}{\left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}} \]

[In]

Integrate[1/(E^ArcCoth[a*x]*(c - a*c*x)^(3/2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03

method	result	size
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \sqrt {-c \left (a x -1\right )}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {-c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\left (a x -1\right ) \sqrt {-c \left (a x +1\right )}\, c^{\frac {3}{2}} a}\)	\(78\)

[In]

int(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.86 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=\left [\frac {\sqrt {2} \sqrt {-\frac {1}{c}} \log \left (-\frac {a^{2} x^{2} + 2 \, \sqrt {2} \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {-\frac {1}{c}} + 2 \, a x - 3}{a^{2} x^{2} - 2 \, a x + 1}\right )}{2 \, a c}, -\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{{\left (a x - 1\right )} \sqrt {c}}\right )}{a c^{\frac {3}{2}}}\right ] \]

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=\int \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{\left (- c \left (a x - 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(((a*x-1)/(a*x+1))**(1/2)/(-a*c*x+c)**(3/2),x)

[Out]

Integral(sqrt((a*x - 1)/(a*x + 1))/(-c*(a*x - 1))**(3/2), x)

Maxima [F]

\[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=\int { \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (-a c x + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt((a*x - 1)/(a*x + 1))/(-a*c*x + c)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=\frac {{\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x - c}}{2 \, \sqrt {c}}\right )}{a \sqrt {c}} - \frac {\sqrt {2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {c}}\right )}{a \sqrt {c}}\right )} {\left | c \right |} \mathrm {sgn}\left (a x + 1\right )}{c^{2}} \]

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=\int \frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{{\left (c-a\,c\,x\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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