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

Optimal. Leaf size=177 \[ \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {3 a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6192, 6193, 72} \[ \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {3 a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3)/(2*(1 - a*x)*(c - a^2*c*x^2)^(3/2)) + (a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[x
])/(c - a^2*c*x^2)^(3/2) - (3*a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 - a*x])/(4*(c - a^2*c*x^2)^(3/2)) - (a^3*(
1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 + a*x])/(4*(c - a^2*c*x^2)^(3/2))

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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^4} \, 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 {1}{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}{x}+\frac {a}{2 (-1+a x)^2}-\frac {3 a}{4 (-1+a x)}-\frac {a}{4 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{3/2}}\\ &=\frac {a^3 \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^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {3 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (1+a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 68, normalized size = 0.38 \[ \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (\frac {1}{2-2 a x}-\frac {3}{4} \log (1-a x)-\frac {1}{4} \log (a x+1)+\log (x)\right )}{\left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3*((2 - 2*a*x)^(-1) + Log[x] - (3*Log[1 - a*x])/4 - Log[1 + a*x]/4))/(c - a^2*c
*x^2)^(3/2)

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fricas [A]  time = 0.50, size = 63, normalized size = 0.36 \[ -\frac {\sqrt {-a^{2} c} {\left ({\left (a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 4 \, {\left (a x - 1\right )} \log \relax (x) + 2\right )}}{4 \, {\left (a^{2} c^{2} x - a c^{2}\right )}} \]

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*sqrt(-a^2*c)*((a*x - 1)*log(a*x + 1) + 3*(a*x - 1)*log(a*x - 1) - 4*(a*x - 1)*log(x) + 2)/(a^2*c^2*x - a*
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 = 92, normalized size = 0.52 \[ \frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (4 a \ln \relax (x ) x -3 \ln \left (a x -1\right ) x a -a x \ln \left (a x +1\right )-4 \ln \relax (x )+3 \ln \left (a x -1\right )+\ln \left (a x +1\right )-2\right )}{4 \sqrt {\frac {a x -1}{a x +1}}\, \left (a^{2} x^{2}-1\right ) c^{2}} \]

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

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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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