3.1239 \(\int e^{-2 \tanh ^{-1}(a x)} \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=87 \[ \frac{(1-a x) \sqrt{c-a^2 c x^2}}{2 a}+\frac{3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0753709, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {6142, 671, 641, 217, 203} \[ \frac{(1-a x) \sqrt{c-a^2 c x^2}}{2 a}+\frac{3 \sqrt{c-a^2 c x^2}}{2 a}+\frac{3 \sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6142

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

Rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0506014, size = 99, normalized size = 1.14 \[ \frac{\sqrt{c-a^2 c x^2} \left (\sqrt{a x+1} \left (a^2 x^2-5 a x+4\right )-6 \sqrt{1-a x} \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )\right )}{2 a \sqrt{1-a x} \sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c - a^2*c*x^2]*(Sqrt[1 + a*x]*(4 - 5*a*x + a^2*x^2) - 6*Sqrt[1 - a*x]*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(2
*a*Sqrt[1 - a*x]*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.034, size = 126, normalized size = 1.5 \begin{align*} -{\frac{x}{2}\sqrt{-{a}^{2}c{x}^{2}+c}}-{\frac{c}{2}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}+2\,{\frac{\sqrt{-c{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,ac \left ( x+{a}^{-1} \right ) }}{a}}+2\,{\frac{c}{\sqrt{{a}^{2}c}}\arctan \left ({\frac{\sqrt{{a}^{2}c}x}{\sqrt{-c{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,ac \left ( x+{a}^{-1} \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x*(-a^2*c*x^2+c)^(1/2)-1/2*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))+2/a*(-c*a^2*(x+1/
a)^2+2*a*c*(x+1/a))^(1/2)+2*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-c*a^2*(x+1/a)^2+2*a*c*(x+1/a))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.43303, size = 309, normalized size = 3.55 \begin{align*} \left [-\frac{2 \, \sqrt{-a^{2} c x^{2} + c}{\left (a x - 4\right )} - 3 \, \sqrt{-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right )}{4 \, a}, -\frac{\sqrt{-a^{2} c x^{2} + c}{\left (a x - 4\right )} + 3 \, \sqrt{c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right )}{2 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20922, size = 177, normalized size = 2.03 \begin{align*} -\frac{{\left (12 \, a^{3} c^{\frac{3}{2}} \arctan \left (\frac{\sqrt{-c + \frac{2 \, c}{a x + 1}}}{\sqrt{c}}\right ) \mathrm{sgn}\left (\frac{1}{a x + 1}\right ) \mathrm{sgn}\left (a\right ) - \frac{{\left (3 \, a^{3} c^{3} \sqrt{-c + \frac{2 \, c}{a x + 1}} \mathrm{sgn}\left (\frac{1}{a x + 1}\right ) \mathrm{sgn}\left (a\right ) + 5 \, a^{3} c^{2}{\left (-c + \frac{2 \, c}{a x + 1}\right )}^{\frac{3}{2}} \mathrm{sgn}\left (\frac{1}{a x + 1}\right ) \mathrm{sgn}\left (a\right )\right )}{\left (a x + 1\right )}^{2}}{c^{2}}\right )}{\left | a \right |}}{4 \, a^{5} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(12*a^3*c^(3/2)*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(c))*sgn(1/(a*x + 1))*sgn(a) - (3*a^3*c^3*sqrt(-c + 2
*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) + 5*a^3*c^2*(-c + 2*c/(a*x + 1))^(3/2)*sgn(1/(a*x + 1))*sgn(a))*(a*x + 1
)^2/c^2)*abs(a)/(a^5*c)