3.1159 \(\int e^{3 \tanh ^{-1}(a x)} \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=111 \[ -\frac{a x^2 \sqrt{c-a^2 c x^2}}{2 \sqrt{1-a^2 x^2}}-\frac{3 x \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}-\frac{4 \sqrt{c-a^2 c x^2} \log (1-a x)}{a \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0845421, antiderivative size = 111, 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 = {6143, 6140, 43} \[ -\frac{a x^2 \sqrt{c-a^2 c x^2}}{2 \sqrt{1-a^2 x^2}}-\frac{3 x \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}}-\frac{4 \sqrt{c-a^2 c x^2} \log (1-a x)}{a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6143

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d*x^2)^Frac
Part[p])/(1 - a^2*x^2)^FracPart[p], Int[(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x
] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0])

Rule 6140

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])

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

Mathematica [A]  time = 0.0259945, size = 54, normalized size = 0.49 \[ \frac{\sqrt{c-a^2 c x^2} \left (-\frac{a x^2}{2}-\frac{4 \log (1-a x)}{a}-3 x\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.085, size = 63, normalized size = 0.6 \begin{align*}{\frac{{a}^{2}{x}^{2}+6\,ax+8\,\ln \left ( ax-1 \right ) }{ \left ( 2\,{a}^{2}{x}^{2}-2 \right ) a}\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.91532, size = 734, normalized size = 6.61 \begin{align*} \left [\frac{4 \,{\left (a^{2} x^{2} - 1\right )} \sqrt{c} \log \left (\frac{a^{6} c x^{6} - 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} + 4 \, a c x +{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} - 2 \, c}{a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1}\right ) + \sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} + 6 \, a x\right )} \sqrt{-a^{2} x^{2} + 1}}{2 \,{\left (a^{3} x^{2} - a\right )}}, -\frac{8 \,{\left (a^{2} x^{2} - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} - 2 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c}}{a^{4} c x^{4} - 2 \, a^{3} c x^{3} - a^{2} c x^{2} + 2 \, a c x}\right ) - \sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} + 6 \, a x\right )} \sqrt{-a^{2} x^{2} + 1}}{2 \,{\left (a^{3} x^{2} - a\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(4*(a^2*x^2 - 1)*sqrt(c)*log((a^6*c*x^6 - 4*a^5*c*x^5 + 5*a^4*c*x^4 - 4*a^2*c*x^2 + 4*a*c*x + (a^4*x^4 -
4*a^3*x^3 + 6*a^2*x^2 - 4*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - 2*c)/(a^4*x^4 - 2*a^3*x^3 + 2
*a*x - 1)) + sqrt(-a^2*c*x^2 + c)*(a^2*x^2 + 6*a*x)*sqrt(-a^2*x^2 + 1))/(a^3*x^2 - a), -1/2*(8*(a^2*x^2 - 1)*s
qrt(-c)*arctan(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 - 2*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt(-c)/(a^4*c*x^4 - 2*a^3*c*x^3
- a^2*c*x^2 + 2*a*c*x)) - sqrt(-a^2*c*x^2 + c)*(a^2*x^2 + 6*a*x)*sqrt(-a^2*x^2 + 1))/(a^3*x^2 - a)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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