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

Optimal. Leaf size=103 \[ \frac{2 (c-a c x)^{5/2}}{3 a c^2 \sqrt{1-a^2 x^2}}+\frac{16 (c-a c x)^{3/2}}{3 a c \sqrt{1-a^2 x^2}}-\frac{64 \sqrt{c-a c x}}{3 a \sqrt{1-a^2 x^2}} \]

[Out]

(-64*Sqrt[c - a*c*x])/(3*a*Sqrt[1 - a^2*x^2]) + (16*(c - a*c*x)^(3/2))/(3*a*c*Sqrt[1 - a^2*x^2]) + (2*(c - a*c
*x)^(5/2))/(3*a*c^2*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.083532, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6127, 657, 649} \[ \frac{2 (c-a c x)^{5/2}}{3 a c^2 \sqrt{1-a^2 x^2}}+\frac{16 (c-a c x)^{3/2}}{3 a c \sqrt{1-a^2 x^2}}-\frac{64 \sqrt{c-a c x}}{3 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*Sqrt[c - a*c*x])/(3*a*Sqrt[1 - a^2*x^2]) + (16*(c - a*c*x)^(3/2))/(3*a*c*Sqrt[1 - a^2*x^2]) + (2*(c - a*c
*x)^(5/2))/(3*a*c^2*Sqrt[1 - a^2*x^2])

Rule 6127

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

Rule 657

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*Simplify[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, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rule 649

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*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} \sqrt{c-a c x} \, dx &=\frac{\int \frac{(c-a c x)^{7/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac{2 (c-a c x)^{5/2}}{3 a c^2 \sqrt{1-a^2 x^2}}+\frac{8 \int \frac{(c-a c x)^{5/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=\frac{16 (c-a c x)^{3/2}}{3 a c \sqrt{1-a^2 x^2}}+\frac{2 (c-a c x)^{5/2}}{3 a c^2 \sqrt{1-a^2 x^2}}+\frac{32 \int \frac{(c-a c x)^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c}\\ &=-\frac{64 \sqrt{c-a c x}}{3 a \sqrt{1-a^2 x^2}}+\frac{16 (c-a c x)^{3/2}}{3 a c \sqrt{1-a^2 x^2}}+\frac{2 (c-a c x)^{5/2}}{3 a c^2 \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0281885, size = 51, normalized size = 0.5 \[ \frac{2 c \sqrt{1-a x} \left (a^2 x^2-10 a x-23\right )}{3 a \sqrt{a x+1} \sqrt{c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*c*Sqrt[1 - a*x]*(-23 - 10*a*x + a^2*x^2))/(3*a*Sqrt[1 + a*x]*Sqrt[c - a*c*x])

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Maple [A]  time = 0.031, size = 54, normalized size = 0.5 \begin{align*}{\frac{2\,{a}^{2}{x}^{2}-20\,ax-46}{3\, \left ( ax+1 \right ) ^{2} \left ( ax-1 \right ) ^{2}a} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}\sqrt{-acx+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(1/2)*(a^2*x^2-10*a*x-23)/(a*x+1)^2/(a*x-1)^2/a

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Maxima [A]  time = 1.04341, size = 68, normalized size = 0.66 \begin{align*} \frac{2 \,{\left (a^{2} \sqrt{c} x^{2} - 10 \, a \sqrt{c} x - 23 \, \sqrt{c}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{3 \,{\left (a^{3} x^{2} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a^2*sqrt(c)*x^2 - 10*a*sqrt(c)*x - 23*sqrt(c))*sqrt(a*x + 1)*(a*x - 1)/(a^3*x^2 - a)

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Fricas [A]  time = 1.59835, size = 108, normalized size = 1.05 \begin{align*} -\frac{2 \,{\left (a^{2} x^{2} - 10 \, a x - 23\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{3 \,{\left (a^{3} x^{2} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(a^2*x^2 - 10*a*x - 23)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(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 (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}{\left (a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.25188, size = 77, normalized size = 0.75 \begin{align*} \frac{32 \, \sqrt{2}{\left | c \right |}}{3 \, a \sqrt{c}} + \frac{2 \,{\left ({\left (a c x + c\right )}^{\frac{3}{2}} - 12 \, \sqrt{a c x + c} c - \frac{12 \, c^{2}}{\sqrt{a c x + c}}\right )}{\left | c \right |}}{3 \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/3*sqrt(2)*abs(c)/(a*sqrt(c)) + 2/3*((a*c*x + c)^(3/2) - 12*sqrt(a*c*x + c)*c - 12*c^2/sqrt(a*c*x + c))*abs(
c)/(a*c^2)