### 3.296 $$\int e^{\coth ^{-1}(a x)} x^2 \sqrt{c-a c x} \, dx$$

Optimal. Leaf size=140 $\frac{16 x \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{105 a^2 \sqrt{1-\frac{1}{a x}}}+\frac{2 x^3 \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{7 \sqrt{1-\frac{1}{a x}}}-\frac{8 x^2 \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{35 a \sqrt{1-\frac{1}{a x}}}$

[Out]

(16*(1 + 1/(a*x))^(3/2)*x*Sqrt[c - a*c*x])/(105*a^2*Sqrt[1 - 1/(a*x)]) - (8*(1 + 1/(a*x))^(3/2)*x^2*Sqrt[c - a
*c*x])/(35*a*Sqrt[1 - 1/(a*x)]) + (2*(1 + 1/(a*x))^(3/2)*x^3*Sqrt[c - a*c*x])/(7*Sqrt[1 - 1/(a*x)])

________________________________________________________________________________________

Rubi [A]  time = 0.215309, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.19, Rules used = {6176, 6181, 45, 37} $\frac{16 x \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{105 a^2 \sqrt{1-\frac{1}{a x}}}+\frac{2 x^3 \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{7 \sqrt{1-\frac{1}{a x}}}-\frac{8 x^2 \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{35 a \sqrt{1-\frac{1}{a x}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCoth[a*x]*x^2*Sqrt[c - a*c*x],x]

[Out]

(16*(1 + 1/(a*x))^(3/2)*x*Sqrt[c - a*c*x])/(105*a^2*Sqrt[1 - 1/(a*x)]) - (8*(1 + 1/(a*x))^(3/2)*x^2*Sqrt[c - a
*c*x])/(35*a*Sqrt[1 - 1/(a*x)]) + (2*(1 + 1/(a*x))^(3/2)*x^3*Sqrt[c - a*c*x])/(7*Sqrt[1 - 1/(a*x)])

Rule 6176

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

Rule 6181

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int e^{\coth ^{-1}(a x)} x^2 \sqrt{c-a c x} \, dx &=\frac{\sqrt{c-a c x} \int e^{\coth ^{-1}(a x)} \sqrt{1-\frac{1}{a x}} x^{5/2} \, dx}{\sqrt{1-\frac{1}{a x}} \sqrt{x}}\\ &=-\frac{\left (\sqrt{\frac{1}{x}} \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x}{a}}}{x^{9/2}} \, dx,x,\frac{1}{x}\right )}{\sqrt{1-\frac{1}{a x}}}\\ &=\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x^3 \sqrt{c-a c x}}{7 \sqrt{1-\frac{1}{a x}}}+\frac{\left (4 \sqrt{\frac{1}{x}} \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x}{a}}}{x^{7/2}} \, dx,x,\frac{1}{x}\right )}{7 a \sqrt{1-\frac{1}{a x}}}\\ &=-\frac{8 \left (1+\frac{1}{a x}\right )^{3/2} x^2 \sqrt{c-a c x}}{35 a \sqrt{1-\frac{1}{a x}}}+\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x^3 \sqrt{c-a c x}}{7 \sqrt{1-\frac{1}{a x}}}-\frac{\left (8 \sqrt{\frac{1}{x}} \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x}{a}}}{x^{5/2}} \, dx,x,\frac{1}{x}\right )}{35 a^2 \sqrt{1-\frac{1}{a x}}}\\ &=\frac{16 \left (1+\frac{1}{a x}\right )^{3/2} x \sqrt{c-a c x}}{105 a^2 \sqrt{1-\frac{1}{a x}}}-\frac{8 \left (1+\frac{1}{a x}\right )^{3/2} x^2 \sqrt{c-a c x}}{35 a \sqrt{1-\frac{1}{a x}}}+\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x^3 \sqrt{c-a c x}}{7 \sqrt{1-\frac{1}{a x}}}\\ \end{align*}

Mathematica [A]  time = 0.0331096, size = 64, normalized size = 0.46 $\frac{2 \sqrt{\frac{1}{a x}+1} (a x+1) \left (15 a^2 x^2-12 a x+8\right ) \sqrt{c-a c x}}{105 a^3 \sqrt{1-\frac{1}{a x}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^ArcCoth[a*x]*x^2*Sqrt[c - a*c*x],x]

[Out]

(2*Sqrt[1 + 1/(a*x)]*(1 + a*x)*Sqrt[c - a*c*x]*(8 - 12*a*x + 15*a^2*x^2))/(105*a^3*Sqrt[1 - 1/(a*x)])

________________________________________________________________________________________

Maple [A]  time = 0.043, size = 49, normalized size = 0.4 \begin{align*}{\frac{ \left ( 2\,ax+2 \right ) \left ( 15\,{a}^{2}{x}^{2}-12\,ax+8 \right ) }{105\,{a}^{3}}\sqrt{-acx+c}{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(a*x+1)*(15*a^2*x^2-12*a*x+8)*(-a*c*x+c)^(1/2)/a^3/((a*x-1)/(a*x+1))^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 1.10896, size = 74, normalized size = 0.53 \begin{align*} \frac{2 \,{\left (15 \, a^{3} \sqrt{-c} x^{3} + 3 \, a^{2} \sqrt{-c} x^{2} - 4 \, a \sqrt{-c} x + 8 \, \sqrt{-c}\right )} \sqrt{a x + 1}}{105 \, a^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^3*sqrt(-c)*x^3 + 3*a^2*sqrt(-c)*x^2 - 4*a*sqrt(-c)*x + 8*sqrt(-c))*sqrt(a*x + 1)/a^3

________________________________________________________________________________________

Fricas [A]  time = 1.58551, size = 151, normalized size = 1.08 \begin{align*} \frac{2 \,{\left (15 \, a^{4} x^{4} + 18 \, a^{3} x^{3} - a^{2} x^{2} + 4 \, a x + 8\right )} \sqrt{-a c x + c} \sqrt{\frac{a x - 1}{a x + 1}}}{105 \,{\left (a^{4} x - a^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^4*x^4 + 18*a^3*x^3 - a^2*x^2 + 4*a*x + 8)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^4*x - a^3)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.24733, size = 196, normalized size = 1.4 \begin{align*} \frac{2 \,{\left (\frac{22 \, \sqrt{2} \sqrt{-c} c}{a^{2} \mathrm{sgn}\left (c\right )} + \frac{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} + 14 \,{\left (3 \,{\left (a c x + c\right )}^{2} \sqrt{-a c x - c} + 10 \,{\left (-a c x - c\right )}^{\frac{3}{2}} c\right )} c}{a^{2} c^{2} \mathrm{sgn}\left (-a c x - c\right )}\right )}}{105 \, a c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(22*sqrt(2)*sqrt(-c)*c/(a^2*sgn(c)) + (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 + 14*(3*(a*c*x + c)^2*sqrt(-a*c*x - c) + 10*(-a*c*x - c)^(3/2)*c)*c)/(a^2*
c^2*sgn(-a*c*x - c)))/(a*c)