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

Optimal. Leaf size=80 \[ \frac{2 (c-a c x)^{7/2}}{7 a^3 c^3}-\frac{8 (c-a c x)^{5/2}}{5 a^3 c^2}+\frac{10 (c-a c x)^{3/2}}{3 a^3 c}-\frac{4 \sqrt{c-a c x}}{a^3} \]

[Out]

(-4*Sqrt[c - a*c*x])/a^3 + (10*(c - a*c*x)^(3/2))/(3*a^3*c) - (8*(c - a*c*x)^(5/2))/(5*a^3*c^2) + (2*(c - a*c*
x)^(7/2))/(7*a^3*c^3)

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Rubi [A]  time = 0.162893, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6130, 21, 77} \[ \frac{2 (c-a c x)^{7/2}}{7 a^3 c^3}-\frac{8 (c-a c x)^{5/2}}{5 a^3 c^2}+\frac{10 (c-a c x)^{3/2}}{3 a^3 c}-\frac{4 \sqrt{c-a c x}}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[c - a*c*x])/a^3 + (10*(c - a*c*x)^(3/2))/(3*a^3*c) - (8*(c - a*c*x)^(5/2))/(5*a^3*c^2) + (2*(c - a*c*
x)^(7/2))/(7*a^3*c^3)

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^2 \sqrt{c-a c x} \, dx &=\int \frac{x^2 (1+a x) \sqrt{c-a c x}}{1-a x} \, dx\\ &=c \int \frac{x^2 (1+a x)}{\sqrt{c-a c x}} \, dx\\ &=c \int \left (\frac{2}{a^2 \sqrt{c-a c x}}-\frac{5 \sqrt{c-a c x}}{a^2 c}+\frac{4 (c-a c x)^{3/2}}{a^2 c^2}-\frac{(c-a c x)^{5/2}}{a^2 c^3}\right ) \, dx\\ &=-\frac{4 \sqrt{c-a c x}}{a^3}+\frac{10 (c-a c x)^{3/2}}{3 a^3 c}-\frac{8 (c-a c x)^{5/2}}{5 a^3 c^2}+\frac{2 (c-a c x)^{7/2}}{7 a^3 c^3}\\ \end{align*}

Mathematica [A]  time = 0.052234, size = 40, normalized size = 0.5 \[ -\frac{2 \left (15 a^3 x^3+39 a^2 x^2+52 a x+104\right ) \sqrt{c-a c x}}{105 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - a*c*x]*(104 + 52*a*x + 39*a^2*x^2 + 15*a^3*x^3))/(105*a^3)

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Maple [A]  time = 0.031, size = 37, normalized size = 0.5 \begin{align*} -{\frac{30\,{x}^{3}{a}^{3}+78\,{a}^{2}{x}^{2}+104\,ax+208}{105\,{a}^{3}}\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^2*(-a*c*x+c)^(1/2),x)

[Out]

-2/105*(-a*c*x+c)^(1/2)*(15*a^3*x^3+39*a^2*x^2+52*a*x+104)/a^3

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Maxima [A]  time = 0.963379, size = 81, normalized size = 1.01 \begin{align*} \frac{2 \,{\left (15 \,{\left (-a c x + c\right )}^{\frac{7}{2}} - 84 \,{\left (-a c x + c\right )}^{\frac{5}{2}} c + 175 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c^{2} - 210 \, \sqrt{-a c x + c} c^{3}\right )}}{105 \, a^{3} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*(-a*c*x + c)^(7/2) - 84*(-a*c*x + c)^(5/2)*c + 175*(-a*c*x + c)^(3/2)*c^2 - 210*sqrt(-a*c*x + c)*c^3
)/(a^3*c^3)

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Fricas [A]  time = 1.80303, size = 95, normalized size = 1.19 \begin{align*} -\frac{2 \,{\left (15 \, a^{3} x^{3} + 39 \, a^{2} x^{2} + 52 \, a x + 104\right )} \sqrt{-a c x + c}}{105 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*a^3*x^3 + 39*a^2*x^2 + 52*a*x + 104)*sqrt(-a*c*x + c)/a^3

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Sympy [A]  time = 8.01316, size = 68, normalized size = 0.85 \begin{align*} - \frac{2 \left (2 c^{3} \sqrt{- a c x + c} - \frac{5 c^{2} \left (- a c x + c\right )^{\frac{3}{2}}}{3} + \frac{4 c \left (- a c x + c\right )^{\frac{5}{2}}}{5} - \frac{\left (- a c x + c\right )^{\frac{7}{2}}}{7}\right )}{a^{3} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*c**3*sqrt(-a*c*x + c) - 5*c**2*(-a*c*x + c)**(3/2)/3 + 4*c*(-a*c*x + c)**(5/2)/5 - (-a*c*x + c)**(7/2)/7
)/(a**3*c**3)

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Giac [A]  time = 1.2239, size = 108, normalized size = 1.35 \begin{align*} -\frac{2 \,{\left (15 \,{\left (a c x - c\right )}^{3} \sqrt{-a c x + c} + 84 \,{\left (a c x - c\right )}^{2} \sqrt{-a c x + c} c - 175 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c^{2} + 210 \, \sqrt{-a c x + c} c^{3}\right )}}{105 \, a^{3} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*(a*c*x - c)^3*sqrt(-a*c*x + c) + 84*(a*c*x - c)^2*sqrt(-a*c*x + c)*c - 175*(-a*c*x + c)^(3/2)*c^2 +
210*sqrt(-a*c*x + c)*c^3)/(a^3*c^3)