\(\int e^{-\coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{3/2} \, dx\) [857]

Optimal result

Integrand size = 24, antiderivative size = 147 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=-\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6332, 6328, 76} \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {c x \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 x \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

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

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 6328

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

Rule 6332

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.43 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{3/2} \left (-\frac {3}{2 a}-\frac {1}{2 a^3 x^2}+\frac {1}{a^2 x}+x-\frac {\log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.54

method	result	size
default	\(-\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {3}{2}} \sqrt {\frac {a x -1}{a x +1}}\, x \left (-2 a^{3} x^{3}+2 a^{2} \ln \left (x \right ) x^{2}-2 a x +1\right )}{2 \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.30 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx=\frac {{\left (2 \, a^{3} c x^{3} - 2 \, a^{2} c x^{2} \log \left (x\right ) + 2 \, a c x - c\right )} \sqrt {a^{2} c}}{2 \, a^{4} x^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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