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

Optimal. Leaf size=135 \[ -\frac{2 c x^3 \sqrt{1-a^2 x^2}}{7 \sqrt{c-a c x}}+\frac{26 c x^2 \sqrt{1-a^2 x^2}}{35 a \sqrt{c-a c x}}+\frac{104 \sqrt{1-a^2 x^2} \sqrt{c-a c x}}{105 a^3}+\frac{104 c \sqrt{1-a^2 x^2}}{105 a^3 \sqrt{c-a c x}} \]

[Out]

(104*c*Sqrt[1 - a^2*x^2])/(105*a^3*Sqrt[c - a*c*x]) + (26*c*x^2*Sqrt[1 - a^2*x^2])/(35*a*Sqrt[c - a*c*x]) - (2
*c*x^3*Sqrt[1 - a^2*x^2])/(7*Sqrt[c - a*c*x]) + (104*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(105*a^3)

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Rubi [A]  time = 0.21296, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {6128, 881, 871, 795, 649} \[ -\frac{2 c x^3 \sqrt{1-a^2 x^2}}{7 \sqrt{c-a c x}}+\frac{26 c x^2 \sqrt{1-a^2 x^2}}{35 a \sqrt{c-a c x}}+\frac{104 \sqrt{1-a^2 x^2} \sqrt{c-a c x}}{105 a^3}+\frac{104 c \sqrt{1-a^2 x^2}}{105 a^3 \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(104*c*Sqrt[1 - a^2*x^2])/(105*a^3*Sqrt[c - a*c*x]) + (26*c*x^2*Sqrt[1 - a^2*x^2])/(35*a*Sqrt[c - a*c*x]) - (2
*c*x^3*Sqrt[1 - a^2*x^2])/(7*Sqrt[c - a*c*x]) + (104*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(105*a^3)

Rule 6128

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

Rule 881

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(d +
e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + p + 2)), x] - Dist[(e*f*(p + 1) - d*g*(2*n + p
+ 3))/(g*(n + p + 2)), Int[(d + e*x)^(m - 1)*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m,
n, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1]
&& IntegerQ[2*p]

Rule 871

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d +
e*x)^(m - 1)*(f + g*x)^n*(a + c*x^2)^(p + 1))/(c*(m - n - 1)), x] - Dist[(n*(e*f + d*g))/(e*(m - n - 1)), Int[
(d + e*x)^m*(f + g*x)^(n - 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0]
&& EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] && GtQ[n, 0] && NeQ[m - n - 1, 0] && (IntegerQ[2*p
] || IntegerQ[n])

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

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^{-\tanh ^{-1}(a x)} x^2 \sqrt{c-a c x} \, dx &=\frac{\int \frac{x^2 (c-a c x)^{3/2}}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=-\frac{2 c x^3 \sqrt{1-a^2 x^2}}{7 \sqrt{c-a c x}}+\frac{13}{7} \int \frac{x^2 \sqrt{c-a c x}}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{26 c x^2 \sqrt{1-a^2 x^2}}{35 a \sqrt{c-a c x}}-\frac{2 c x^3 \sqrt{1-a^2 x^2}}{7 \sqrt{c-a c x}}-\frac{52 \int \frac{x \sqrt{c-a c x}}{\sqrt{1-a^2 x^2}} \, dx}{35 a}\\ &=\frac{26 c x^2 \sqrt{1-a^2 x^2}}{35 a \sqrt{c-a c x}}-\frac{2 c x^3 \sqrt{1-a^2 x^2}}{7 \sqrt{c-a c x}}+\frac{104 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}{105 a^3}+\frac{52 \int \frac{\sqrt{c-a c x}}{\sqrt{1-a^2 x^2}} \, dx}{105 a^2}\\ &=\frac{104 c \sqrt{1-a^2 x^2}}{105 a^3 \sqrt{c-a c x}}+\frac{26 c x^2 \sqrt{1-a^2 x^2}}{35 a \sqrt{c-a c x}}-\frac{2 c x^3 \sqrt{1-a^2 x^2}}{7 \sqrt{c-a c x}}+\frac{104 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}{105 a^3}\\ \end{align*}

Mathematica [A]  time = 0.036173, size = 55, normalized size = 0.41 \[ -\frac{2 c \sqrt{1-a^2 x^2} \left (15 a^3 x^3-39 a^2 x^2+52 a x-104\right )}{105 a^3 \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 56, normalized size = 0.4 \begin{align*}{\frac{30\,{x}^{3}{a}^{3}-78\,{a}^{2}{x}^{2}+104\,ax-208}{ \left ( 105\,ax-105 \right ){a}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}\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)*(-a^2*x^2+1)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.02985, size = 84, normalized size = 0.62 \begin{align*} -\frac{2 \,{\left (15 \, a^{3} \sqrt{c} x^{3} - 39 \, a^{2} \sqrt{c} x^{2} + 52 \, a \sqrt{c} x - 104 \, \sqrt{c}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{105 \,{\left (a^{4} x - 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)*(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-2/105*(15*a^3*sqrt(c)*x^3 - 39*a^2*sqrt(c)*x^2 + 52*a*sqrt(c)*x - 104*sqrt(c))*sqrt(a*x + 1)*(a*x - 1)/(a^4*x
- a^3)

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Fricas [A]  time = 1.74107, size = 132, normalized size = 0.98 \begin{align*} \frac{2 \,{\left (15 \, a^{3} x^{3} - 39 \, a^{2} x^{2} + 52 \, a x - 104\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{105 \,{\left (a^{4} x - 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)*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

2/105*(15*a^3*x^3 - 39*a^2*x^2 + 52*a*x - 104)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^4*x - 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 )} \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}{a x + 1}\, 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)*(-a**2*x**2+1)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.18971, size = 100, normalized size = 0.74 \begin{align*} -\frac{2 \,{\left (\frac{76 \, \sqrt{2} c^{\frac{3}{2}}}{a^{3}} + \frac{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}}{a^{3} c^{2}}\right )}{\left | c \right |}}{105 \, c^{2}} \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)*(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

-2/105*(76*sqrt(2)*c^(3/2)/a^3 + (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^2))*abs(c)/c^2