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

Optimal. Leaf size=87 \[ \frac {a x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac {a x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \tanh ^{-1}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6192, 6193, 77, 207} \[ \frac {a x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac {a x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \tanh ^{-1}(a x)}{2 \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 207

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

Rule 6192

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

Rule 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), 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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 59, normalized size = 0.68 \[ \frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left ((a x-1) \tanh ^{-1}(a x)+1\right )}{2 a c (a x-1) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 86, normalized size = 0.99 \[ -\frac {{\left (a^{2} x - a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) + 2 \, \sqrt {-a^{2} c}}{4 \, {\left (a^{4} c^{2} x - a^{3} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.06, size = 84, normalized size = 0.97 \[ \frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (\ln \left (a x -1\right ) x a -a x \ln \left (a x +1\right )-\ln \left (a x -1\right )+\ln \left (a x +1\right )-2\right )}{4 \sqrt {\frac {a x -1}{a x +1}}\, \left (a^{2} x^{2}-1\right ) c^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/((a*x-1)/(a*x+1))^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(ln(a*x-1)*x*a-a*x*ln(a*x+1)-ln(a*x-1)+ln(a*x+1)-2)/(a^2*x^
2-1)/c^2/a^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {\frac {a x - 1}{a x + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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