### 3.685 $$\int e^{3 \coth ^{-1}(a x)} \sqrt{c-a^2 c x^2} \, dx$$

Optimal. Leaf size=113 $\frac{x \sqrt{c-a^2 c x^2}}{2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{3 \sqrt{c-a^2 c x^2}}{a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-a^2 c x^2} \log (1-a x)}{a^2 x \sqrt{1-\frac{1}{a^2 x^2}}}$

[Out]

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

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Rubi [A]  time = 0.133175, antiderivative size = 113, 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 = {6192, 6193, 43} $\frac{x \sqrt{c-a^2 c x^2}}{2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{3 \sqrt{c-a^2 c x^2}}{a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-a^2 c x^2} \log (1-a x)}{a^2 x \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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]

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

Mathematica [A]  time = 0.0244481, size = 57, normalized size = 0.5 $\frac{\sqrt{c-a^2 c x^2} (a x (a x+6)+8 \log (1-a x))}{2 a^2 x \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[c - a^2*c*x^2]*(a*x*(6 + a*x) + 8*Log[1 - a*x]))/(2*a^2*Sqrt[1 - 1/(a^2*x^2)]*x)

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Maple [A]  time = 0.174, size = 67, normalized size = 0.6 \begin{align*}{\frac{ \left ({a}^{2}{x}^{2}+6\,ax+8\,\ln \left ( ax-1 \right ) \right ) \left ( ax-1 \right ) }{2\,a \left ( ax+1 \right ) ^{2}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) } \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*x^2+6*a*x+8*ln(a*x-1))*(-c*(a^2*x^2-1))^(1/2)*(a*x-1)/a/(a*x+1)^2/((a*x-1)/(a*x+1))^(3/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.53231, size = 77, normalized size = 0.68 \begin{align*} \frac{{\left (a^{2} x^{2} + 6 \, a x + 8 \, \log \left (a x - 1\right )\right )} \sqrt{-a^{2} c}}{2 \, a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*x^2 + 6*a*x + 8*log(a*x - 1))*sqrt(-a^2*c)/a^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(1/((a*x-1)/(a*x+1))**(3/2)*(-a**2*c*x**2+c)**(1/2),x)

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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