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

Optimal. Leaf size=157 \[ \frac{2 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^2 (c-a c x)^{3/2}}-\frac{10 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^2 (c-a c x)^{3/2}}+\frac{16 c^2 (1-a x)^{3/2} \sqrt{a x+1}}{a^2 (c-a c x)^{3/2}}+\frac{8 c^2 (1-a x)^{3/2}}{a^2 \sqrt{a x+1} (c-a c x)^{3/2}} \]

[Out]

(8*c^2*(1 - a*x)^(3/2))/(a^2*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) + (16*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(a^2*(c
- a*c*x)^(3/2)) - (10*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2))/(3*a^2*(c - a*c*x)^(3/2)) + (2*c^2*(1 - a*x)^(3/2)
*(1 + a*x)^(5/2))/(5*a^2*(c - a*c*x)^(3/2))

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Rubi [A]  time = 0.105492, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {6130, 23, 77} \[ \frac{2 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^2 (c-a c x)^{3/2}}-\frac{10 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^2 (c-a c x)^{3/2}}+\frac{16 c^2 (1-a x)^{3/2} \sqrt{a x+1}}{a^2 (c-a c x)^{3/2}}+\frac{8 c^2 (1-a x)^{3/2}}{a^2 \sqrt{a x+1} (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*c^2*(1 - a*x)^(3/2))/(a^2*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) + (16*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(a^2*(c
- a*c*x)^(3/2)) - (10*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2))/(3*a^2*(c - a*c*x)^(3/2)) + (2*c^2*(1 - a*x)^(3/2)
*(1 + a*x)^(5/2))/(5*a^2*(c - a*c*x)^(3/2))

Rule 6130

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

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

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 \tanh ^{-1}(a x)} x \sqrt{c-a c x} \, dx &=\int \frac{x (1-a x)^{3/2} \sqrt{c-a c x}}{(1+a x)^{3/2}} \, dx\\ &=\frac{(1-a x)^{3/2} \int \frac{x (c-a c x)^2}{(1+a x)^{3/2}} \, dx}{(c-a c x)^{3/2}}\\ &=\frac{(1-a x)^{3/2} \int \left (-\frac{4 c^2}{a (1+a x)^{3/2}}+\frac{8 c^2}{a \sqrt{1+a x}}-\frac{5 c^2 \sqrt{1+a x}}{a}+\frac{c^2 (1+a x)^{3/2}}{a}\right ) \, dx}{(c-a c x)^{3/2}}\\ &=\frac{8 c^2 (1-a x)^{3/2}}{a^2 \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{16 c^2 (1-a x)^{3/2} \sqrt{1+a x}}{a^2 (c-a c x)^{3/2}}-\frac{10 c^2 (1-a x)^{3/2} (1+a x)^{3/2}}{3 a^2 (c-a c x)^{3/2}}+\frac{2 c^2 (1-a x)^{3/2} (1+a x)^{5/2}}{5 a^2 (c-a c x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0331603, size = 60, normalized size = 0.38 \[ \frac{2 c \sqrt{1-a x} \left (3 a^3 x^3-16 a^2 x^2+79 a x+158\right )}{15 a^2 \sqrt{a x+1} \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[1 - a*x]*(158 + 79*a*x - 16*a^2*x^2 + 3*a^3*x^3))/(15*a^2*Sqrt[1 + a*x]*Sqrt[c - a*c*x])

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Maple [A]  time = 0.033, size = 63, normalized size = 0.4 \begin{align*}{\frac{6\,{x}^{3}{a}^{3}-32\,{a}^{2}{x}^{2}+158\,ax+316}{15\, \left ( ax+1 \right ) ^{2} \left ( ax-1 \right ) ^{2}{a}^{2}} \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(x*(-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

2/15*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(1/2)*(3*a^3*x^3-16*a^2*x^2+79*a*x+158)/(a*x+1)^2/(a*x-1)^2/a^2

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Maxima [A]  time = 1.02841, size = 86, normalized size = 0.55 \begin{align*} \frac{2 \,{\left (3 \, a^{3} \sqrt{c} x^{3} - 16 \, a^{2} \sqrt{c} x^{2} + 79 \, a \sqrt{c} x + 158 \, \sqrt{c}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{15 \,{\left (a^{4} x^{2} - a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^3*sqrt(c)*x^3 - 16*a^2*sqrt(c)*x^2 + 79*a*sqrt(c)*x + 158*sqrt(c))*sqrt(a*x + 1)*(a*x - 1)/(a^4*x^2
- a^2)

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Fricas [A]  time = 1.60425, size = 134, normalized size = 0.85 \begin{align*} -\frac{2 \,{\left (3 \, a^{3} x^{3} - 16 \, a^{2} x^{2} + 79 \, a x + 158\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{15 \,{\left (a^{4} x^{2} - a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^3*x^3 - 16*a^2*x^2 + 79*a*x + 158)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^4*x^2 - a^2)

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

[Out]

Integral(x*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 = 1.26133, size = 105, normalized size = 0.67 \begin{align*} -\frac{224 \, \sqrt{2}{\left | c \right |}}{15 \, a^{2} \sqrt{c}} + \frac{2 \,{\left (3 \,{\left (a c x + c\right )}^{\frac{5}{2}}{\left | c \right |} - 25 \,{\left (a c x + c\right )}^{\frac{3}{2}} c{\left | c \right |} + 120 \, \sqrt{a c x + c} c^{2}{\left | c \right |} + \frac{60 \, c^{3}{\left | c \right |}}{\sqrt{a c x + c}}\right )}}{15 \, a^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-224/15*sqrt(2)*abs(c)/(a^2*sqrt(c)) + 2/15*(3*(a*c*x + c)^(5/2)*abs(c) - 25*(a*c*x + c)^(3/2)*c*abs(c) + 120*
sqrt(a*c*x + c)*c^2*abs(c) + 60*c^3*abs(c)/sqrt(a*c*x + c))/(a^2*c^3)