3.297 \(\int e^{\coth ^{-1}(a x)} x \sqrt{c-a c x} \, dx\)

Optimal. Leaf size=92 \[ \frac{2 x^2 \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}}}-\frac{4 x \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}}} \]

[Out]

(-4*(1 + 1/(a*x))^(3/2)*x*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - 1/(a*x)]) + (2*(1 + 1/(a*x))^(3/2)*x^2*Sqrt[c - a*c*
x])/(5*Sqrt[1 - 1/(a*x)])

________________________________________________________________________________________

Rubi [A]  time = 0.172632, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {6176, 6181, 45, 37} \[ \frac{2 x^2 \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}}}-\frac{4 x \left (\frac{1}{a x}+1\right )^{3/2} \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(1 + 1/(a*x))^(3/2)*x*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - 1/(a*x)]) + (2*(1 + 1/(a*x))^(3/2)*x^2*Sqrt[c - a*c*
x])/(5*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 \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^{3/2} \, 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^{7/2}} \, dx,x,\frac{1}{x}\right )}{\sqrt{1-\frac{1}{a x}}}\\ &=\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x^2 \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}}}+\frac{\left (2 \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 )}{5 a \sqrt{1-\frac{1}{a x}}}\\ &=-\frac{4 \left (1+\frac{1}{a x}\right )^{3/2} x \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}}}+\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x^2 \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}}}\\ \end{align*}

Mathematica [A]  time = 0.025395, size = 56, normalized size = 0.61 \[ \frac{2 \sqrt{\frac{1}{a x}+1} (a x+1) (3 a x-2) \sqrt{c-a c x}}{15 a^2 \sqrt{1-\frac{1}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.043, size = 41, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,ax+2 \right ) \left ( 3\,ax-2 \right ) }{15\,{a}^{2}}\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*(-a*c*x+c)^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.0592, size = 55, normalized size = 0.6 \begin{align*} \frac{2 \,{\left (3 \, a^{2} \sqrt{-c} x^{2} + a \sqrt{-c} x - 2 \, \sqrt{-c}\right )} \sqrt{a x + 1}}{15 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.57287, size = 131, normalized size = 1.42 \begin{align*} \frac{2 \,{\left (3 \, a^{3} x^{3} + 4 \, a^{2} x^{2} - a x - 2\right )} \sqrt{-a c x + c} \sqrt{\frac{a x - 1}{a x + 1}}}{15 \,{\left (a^{3} x - a^{2}\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*(-a*c*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.18634, size = 101, normalized size = 1.1 \begin{align*} \frac{2 \,{\left (\frac{2 \, \sqrt{2} \sqrt{-c} c^{2}}{\mathrm{sgn}\left (c\right )} + \frac{3 \,{\left (a c x + c\right )}^{2} \sqrt{-a c x - c} + 5 \,{\left (-a c x - c\right )}^{\frac{3}{2}} c}{\mathrm{sgn}\left (-a c x - c\right )}\right )}}{15 \, a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(2*sqrt(2)*sqrt(-c)*c^2/sgn(c) + (3*(a*c*x + c)^2*sqrt(-a*c*x - c) + 5*(-a*c*x - c)^(3/2)*c)/sgn(-a*c*x -
c))/(a^2*c^2)