### 3.898 $$\int e^{3 \coth ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}} \, dx$$

Optimal. Leaf size=109 $\frac{x \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{a \sqrt{1-\frac{1}{a^2 x^2}}}$

[Out]

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

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Rubi [A]  time = 0.110757, antiderivative size = 109, 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, 72} $\frac{x \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{a \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0291231, size = 50, normalized size = 0.46 $\frac{\sqrt{c-\frac{c}{a^2 x^2}} (a x+4 \log (1-a x)-\log (x))}{a \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.227, size = 65, normalized size = 0.6 \begin{align*} -{\frac{ \left ( -ax+\ln \left ( x \right ) -4\,\ln \left ( ax-1 \right ) \right ) x \left ( ax-1 \right ) }{ \left ( ax+1 \right ) ^{2}}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}} \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)*(c-c/a^2/x^2)^(1/2),x)

[Out]

-(-a*x+ln(x)-4*ln(a*x-1))*x*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*(a*x-1)/(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{c - \frac{c}{a^{2} x^{2}}}}{\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)*(c-c/a^2/x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.70914, size = 66, normalized size = 0.61 \begin{align*} \frac{\sqrt{a^{2} c}{\left (a x + 4 \, \log \left (a x - 1\right ) - \log \left (x\right )\right )}}{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)*(c-c/a^2/x^2)^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c - \frac{c}{a^{2} x^{2}}}}{\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)*(c-c/a^2/x^2)^(1/2),x, algorithm="giac")

[Out]

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