3.429 \(\int e^{-3 \tanh ^{-1}(a x)} x^2 \sqrt{c-a c x} \, dx\)

Optimal. Leaf size=197 \[ \frac{2 c^2 (1-a x)^{3/2} (a x+1)^{7/2}}{7 a^3 (c-a c x)^{3/2}}-\frac{12 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^3 (c-a c x)^{3/2}}+\frac{26 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^3 (c-a c x)^{3/2}}-\frac{24 c^2 (1-a x)^{3/2} \sqrt{a x+1}}{a^3 (c-a c x)^{3/2}}-\frac{8 c^2 (1-a x)^{3/2}}{a^3 \sqrt{a x+1} (c-a c x)^{3/2}} \]

[Out]

(-8*c^2*(1 - a*x)^(3/2))/(a^3*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (24*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(a^3*(
c - a*c*x)^(3/2)) + (26*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2))/(3*a^3*(c - a*c*x)^(3/2)) - (12*c^2*(1 - a*x)^(3/
2)*(1 + a*x)^(5/2))/(5*a^3*(c - a*c*x)^(3/2)) + (2*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(7/2))/(7*a^3*(c - a*c*x)^(3/
2))

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Rubi [A]  time = 0.149603, antiderivative size = 197, 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, 23, 88} \[ \frac{2 c^2 (1-a x)^{3/2} (a x+1)^{7/2}}{7 a^3 (c-a c x)^{3/2}}-\frac{12 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^3 (c-a c x)^{3/2}}+\frac{26 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^3 (c-a c x)^{3/2}}-\frac{24 c^2 (1-a x)^{3/2} \sqrt{a x+1}}{a^3 (c-a c x)^{3/2}}-\frac{8 c^2 (1-a x)^{3/2}}{a^3 \sqrt{a x+1} (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sqrt[c - a*c*x])/E^(3*ArcTanh[a*x]),x]

[Out]

(-8*c^2*(1 - a*x)^(3/2))/(a^3*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (24*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(a^3*(
c - a*c*x)^(3/2)) + (26*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2))/(3*a^3*(c - a*c*x)^(3/2)) - (12*c^2*(1 - a*x)^(3/
2)*(1 + a*x)^(5/2))/(5*a^3*(c - a*c*x)^(3/2)) + (2*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(7/2))/(7*a^3*(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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} x^2 \sqrt{c-a c x} \, dx &=\int \frac{x^2 (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^2 (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^2 (1+a x)^{3/2}}-\frac{12 c^2}{a^2 \sqrt{1+a x}}+\frac{13 c^2 \sqrt{1+a x}}{a^2}-\frac{6 c^2 (1+a x)^{3/2}}{a^2}+\frac{c^2 (1+a x)^{5/2}}{a^2}\right ) \, dx}{(c-a c x)^{3/2}}\\ &=-\frac{8 c^2 (1-a x)^{3/2}}{a^3 \sqrt{1+a x} (c-a c x)^{3/2}}-\frac{24 c^2 (1-a x)^{3/2} \sqrt{1+a x}}{a^3 (c-a c x)^{3/2}}+\frac{26 c^2 (1-a x)^{3/2} (1+a x)^{3/2}}{3 a^3 (c-a c x)^{3/2}}-\frac{12 c^2 (1-a x)^{3/2} (1+a x)^{5/2}}{5 a^3 (c-a c x)^{3/2}}+\frac{2 c^2 (1-a x)^{3/2} (1+a x)^{7/2}}{7 a^3 (c-a c x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0425967, size = 68, normalized size = 0.35 \[ \frac{2 c \sqrt{1-a x} \left (15 a^4 x^4-66 a^3 x^3+167 a^2 x^2-668 a x-1336\right )}{105 a^3 \sqrt{a x+1} \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[c - a*c*x])/E^(3*ArcTanh[a*x]),x]

[Out]

(2*c*Sqrt[1 - a*x]*(-1336 - 668*a*x + 167*a^2*x^2 - 66*a^3*x^3 + 15*a^4*x^4))/(105*a^3*Sqrt[1 + a*x]*Sqrt[c -
a*c*x])

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Maple [A]  time = 0.032, size = 71, normalized size = 0.4 \begin{align*}{\frac{30\,{x}^{4}{a}^{4}-132\,{x}^{3}{a}^{3}+334\,{a}^{2}{x}^{2}-1336\,ax-2672}{105\, \left ( ax+1 \right ) ^{2} \left ( ax-1 \right ) ^{2}{a}^{3}} \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^2*(-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

2/105*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(1/2)*(15*a^4*x^4-66*a^3*x^3+167*a^2*x^2-668*a*x-1336)/(a*x+1)^2/(a*x-1)^2
/a^3

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Maxima [A]  time = 1.03741, size = 101, normalized size = 0.51 \begin{align*} \frac{2 \,{\left (15 \, a^{4} \sqrt{c} x^{4} - 66 \, a^{3} \sqrt{c} x^{3} + 167 \, a^{2} \sqrt{c} x^{2} - 668 \, a \sqrt{c} x - 1336 \, \sqrt{c}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{105 \,{\left (a^{5} x^{2} - a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^4*sqrt(c)*x^4 - 66*a^3*sqrt(c)*x^3 + 167*a^2*sqrt(c)*x^2 - 668*a*sqrt(c)*x - 1336*sqrt(c))*sqrt(a*
x + 1)*(a*x - 1)/(a^5*x^2 - a^3)

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Fricas [A]  time = 1.53482, size = 158, normalized size = 0.8 \begin{align*} -\frac{2 \,{\left (15 \, a^{4} x^{4} - 66 \, a^{3} x^{3} + 167 \, a^{2} x^{2} - 668 \, a x - 1336\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{105 \,{\left (a^{5} x^{2} - a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*a^4*x^4 - 66*a^3*x^3 + 167*a^2*x^2 - 668*a*x - 1336)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^5*x^2 -
a^3)

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

[Out]

Integral(x**2*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 = 1.32868, size = 126, normalized size = 0.64 \begin{align*} \frac{1888 \, \sqrt{2}{\left | c \right |}}{105 \, a^{3} \sqrt{c}} + \frac{2 \,{\left (15 \,{\left (a c x + c\right )}^{\frac{7}{2}}{\left | c \right |} - 126 \,{\left (a c x + c\right )}^{\frac{5}{2}} c{\left | c \right |} + 455 \,{\left (a c x + c\right )}^{\frac{3}{2}} c^{2}{\left | c \right |} - 1260 \, \sqrt{a c x + c} c^{3}{\left | c \right |} - \frac{420 \, c^{4}{\left | c \right |}}{\sqrt{a c x + c}}\right )}}{105 \, a^{3} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1888/105*sqrt(2)*abs(c)/(a^3*sqrt(c)) + 2/105*(15*(a*c*x + c)^(7/2)*abs(c) - 126*(a*c*x + c)^(5/2)*c*abs(c) +
455*(a*c*x + c)^(3/2)*c^2*abs(c) - 1260*sqrt(a*c*x + c)*c^3*abs(c) - 420*c^4*abs(c)/sqrt(a*c*x + c))/(a^3*c^4)