### 3.877 $$\int \frac{e^{-3 \coth ^{-1}(a x)}}{(c-\frac{c}{a^2 x^2})^{3/2}} \, dx$$

Optimal. Leaf size=168 $\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{3 \sqrt{1-\frac{1}{a^2 x^2}}}{a c (a x+1) \sqrt{c-\frac{c}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}}}{2 a c (a x+1)^2 \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{3 \sqrt{1-\frac{1}{a^2 x^2}} \log (a x+1)}{a c \sqrt{c-\frac{c}{a^2 x^2}}}$

[Out]

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

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Rubi [A]  time = 0.140652, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {6197, 6193, 43} $\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{3 \sqrt{1-\frac{1}{a^2 x^2}}}{a c (a x+1) \sqrt{c-\frac{c}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}}}{2 a c (a x+1)^2 \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{3 \sqrt{1-\frac{1}{a^2 x^2}} \log (a x+1)}{a c \sqrt{c-\frac{c}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6197

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])

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]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0731578, size = 63, normalized size = 0.38 $\frac{\left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (2 a x+\frac{-6 a x-5}{(a x+1)^2}-6 \log (a x+1)\right )}{2 a \left (c-\frac{c}{a^2 x^2}\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1 - 1/(a^2*x^2))^(3/2)*(2*a*x + (-5 - 6*a*x)/(1 + a*x)^2 - 6*Log[1 + a*x]))/(2*a*(c - c/(a^2*x^2))^(3/2))

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Maple [A]  time = 0.232, size = 102, normalized size = 0.6 \begin{align*} -{\frac{ \left ( ax+1 \right ) \left ( -2\,{x}^{3}{a}^{3}+6\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}-4\,{a}^{2}{x}^{2}+12\,ax\ln \left ( ax+1 \right ) +4\,ax+6\,\ln \left ( ax+1 \right ) +5 \right ) }{2\,{a}^{4}{x}^{3}} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}} \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a*x-1)/(a*x+1))^(3/2)*(a*x+1)*(-2*x^3*a^3+6*ln(a*x+1)*a^2*x^2-4*a^2*x^2+12*a*x*ln(a*x+1)+4*a*x+6*ln(a*x
+1)+5)/a^4/x^3/(c*(a^2*x^2-1)/a^2/x^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}{{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.75579, size = 176, normalized size = 1.05 \begin{align*} \frac{{\left (2 \, a^{3} x^{3} + 4 \, a^{2} x^{2} - 4 \, a x - 6 \,{\left (a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x + 1\right ) - 5\right )} \sqrt{a^{2} c}}{2 \,{\left (a^{4} c^{2} x^{2} + 2 \, a^{3} c^{2} x + a^{2} c^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^3*x^3 + 4*a^2*x^2 - 4*a*x - 6*(a^2*x^2 + 2*a*x + 1)*log(a*x + 1) - 5)*sqrt(a^2*c)/(a^4*c^2*x^2 + 2*a^
3*c^2*x + a^2*c^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}{{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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