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

Optimal. Leaf size=57 \[ -\frac{2 (c-a c x)^{5/2}}{5 a^2 c^2}+\frac{2 (c-a c x)^{3/2}}{a^2 c}-\frac{4 \sqrt{c-a c x}}{a^2} \]

[Out]

(-4*Sqrt[c - a*c*x])/a^2 + (2*(c - a*c*x)^(3/2))/(a^2*c) - (2*(c - a*c*x)^(5/2))/(5*a^2*c^2)

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Rubi [A]  time = 0.0857842, antiderivative size = 57, 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, 21, 77} \[ -\frac{2 (c-a c x)^{5/2}}{5 a^2 c^2}+\frac{2 (c-a c x)^{3/2}}{a^2 c}-\frac{4 \sqrt{c-a c x}}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[c - a*c*x])/a^2 + (2*(c - a*c*x)^(3/2))/(a^2*c) - (2*(c - a*c*x)^(5/2))/(5*a^2*c^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 21

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

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

Mathematica [A]  time = 0.0404044, size = 31, normalized size = 0.54 \[ -\frac{2 \left (a^2 x^2+3 a x+6\right ) \sqrt{c-a c x}}{5 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - a*c*x]*(6 + 3*a*x + a^2*x^2))/(5*a^2)

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Maple [A]  time = 0.031, size = 28, normalized size = 0.5 \begin{align*} -{\frac{2\,{a}^{2}{x}^{2}+6\,ax+12}{5\,{a}^{2}}\sqrt{-acx+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(-a*c*x+c)^(1/2)*(a^2*x^2+3*a*x+6)/a^2

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Maxima [A]  time = 0.94846, size = 59, normalized size = 1.04 \begin{align*} -\frac{2 \,{\left ({\left (-a c x + c\right )}^{\frac{5}{2}} - 5 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c + 10 \, \sqrt{-a c x + c} c^{2}\right )}}{5 \, a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*((-a*c*x + c)^(5/2) - 5*(-a*c*x + c)^(3/2)*c + 10*sqrt(-a*c*x + c)*c^2)/(a^2*c^2)

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Fricas [A]  time = 1.90873, size = 66, normalized size = 1.16 \begin{align*} -\frac{2 \,{\left (a^{2} x^{2} + 3 \, a x + 6\right )} \sqrt{-a c x + c}}{5 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(a^2*x^2 + 3*a*x + 6)*sqrt(-a*c*x + c)/a^2

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Sympy [A]  time = 6.45034, size = 48, normalized size = 0.84 \begin{align*} \frac{2 \left (- 2 c^{2} \sqrt{- a c x + c} + c \left (- a c x + c\right )^{\frac{3}{2}} - \frac{\left (- a c x + c\right )^{\frac{5}{2}}}{5}\right )}{a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-2*c**2*sqrt(-a*c*x + c) + c*(-a*c*x + c)**(3/2) - (-a*c*x + c)**(5/2)/5)/(a**2*c**2)

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Giac [A]  time = 1.22635, size = 74, normalized size = 1.3 \begin{align*} -\frac{2 \,{\left ({\left (a c x - c\right )}^{2} \sqrt{-a c x + c} - 5 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c + 10 \, \sqrt{-a c x + c} c^{2}\right )}}{5 \, a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*((a*c*x - c)^2*sqrt(-a*c*x + c) - 5*(-a*c*x + c)^(3/2)*c + 10*sqrt(-a*c*x + c)*c^2)/(a^2*c^2)