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

Optimal. Leaf size=152 \[ \frac{x^3 \sqrt{c-\frac{c}{a^2 x^2}}}{3 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{3 x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{2 a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}}}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{a^3 \sqrt{1-\frac{1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.286066, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6197, 6193, 77} \[ \frac{x^3 \sqrt{c-\frac{c}{a^2 x^2}}}{3 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{3 x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{2 a \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}}}{a^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 \sqrt{c-\frac{c}{a^2 x^2}} \log (1-a x)}{a^3 \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0425586, size = 63, normalized size = 0.41 \[ \frac{\sqrt{c-\frac{c}{a^2 x^2}} \left (a x \left (2 a^2 x^2+9 a x+24\right )+24 \log (1-a x)\right )}{6 a^3 \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*(a*x*(24 + 9*a*x + 2*a^2*x^2) + 24*Log[1 - a*x]))/(6*a^3*Sqrt[1 - 1/(a^2*x^2)])

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Maple [A]  time = 0.21, size = 82, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,{x}^{3}{a}^{3}+9\,{a}^{2}{x}^{2}+24\,ax+24\,\ln \left ( ax-1 \right ) \right ) x \left ( ax-1 \right ) }{6\,{a}^{2} \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)*x^2*(c-c/a^2/x^2)^(1/2),x)

[Out]

1/6*(2*x^3*a^3+9*a^2*x^2+24*a*x+24*ln(a*x-1))*x*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*(a*x-1)/a^2/(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}}} 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)*x^2*(c-c/a^2/x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.61875, size = 97, normalized size = 0.64 \begin{align*} \frac{{\left (2 \, a^{3} x^{3} + 9 \, a^{2} x^{2} + 24 \, a x + 24 \, \log \left (a x - 1\right )\right )} \sqrt{a^{2} c}}{6 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a^3*x^3 + 9*a^2*x^2 + 24*a*x + 24*log(a*x - 1))*sqrt(a^2*c)/a^4

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

[Out]

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