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

Optimal. Leaf size=57 \[ \frac{2 (c-a c x)^{5/2}}{5 a^2 c^2}-\frac{2 (c-a c x)^{3/2}}{a^2 c}+\frac{4 \sqrt{c-a c x}}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.152234, antiderivative size = 57, 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 = {6167, 6130, 21, 77} \[ \frac{2 (c-a c x)^{5/2}}{5 a^2 c^2}-\frac{2 (c-a c x)^{3/2}}{a^2 c}+\frac{4 \sqrt{c-a c x}}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/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 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 \coth ^{-1}(a x)} x \sqrt{c-a c x} \, dx &=-\int e^{2 \tanh ^{-1}(a x)} x \sqrt{c-a c x} \, dx\\ &=-\int \frac{x (1+a x) \sqrt{c-a c x}}{1-a x} \, dx\\ &=-\left (c \int \frac{x (1+a x)}{\sqrt{c-a c x}} \, dx\right )\\ &=-\left (c \int \left (\frac{2}{a \sqrt{c-a c x}}-\frac{3 \sqrt{c-a c x}}{a c}+\frac{(c-a c x)^{3/2}}{a c^2}\right ) \, dx\right )\\ &=\frac{4 \sqrt{c-a c x}}{a^2}-\frac{2 (c-a c x)^{3/2}}{a^2 c}+\frac{2 (c-a c x)^{5/2}}{5 a^2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0429672, size = 31, normalized size = 0.54 \[ \frac{2 \left (a^2 x^2+3 a x+6\right ) \sqrt{c-a c x}}{5 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c - a*c*x]*(6 + 3*a*x + a^2*x^2))/(5*a^2)

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Maple [A]  time = 0.041, size = 28, normalized size = 0.5 \begin{align*}{\frac{2\,{a}^{2}{x}^{2}+6\,ax+12}{5\,{a}^{2}}\sqrt{-acx+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(-a*c*x+c)^(1/2)*(a^2*x^2+3*a*x+6)/a^2

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Maxima [A]  time = 1.01281, size = 59, normalized size = 1.04 \begin{align*} \frac{2 \,{\left ({\left (-a c x + c\right )}^{\frac{5}{2}} - 5 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c + 10 \, \sqrt{-a c x + c} c^{2}\right )}}{5 \, 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)*x*(-a*c*x+c)^(1/2),x, algorithm="maxima")

[Out]

2/5*((-a*c*x + c)^(5/2) - 5*(-a*c*x + c)^(3/2)*c + 10*sqrt(-a*c*x + c)*c^2)/(a^2*c^2)

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Fricas [A]  time = 1.53086, size = 65, normalized size = 1.14 \begin{align*} \frac{2 \,{\left (a^{2} x^{2} + 3 \, a x + 6\right )} \sqrt{-a c x + c}}{5 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(a^2*x^2 + 3*a*x + 6)*sqrt(-a*c*x + c)/a^2

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Sympy [A]  time = 6.18236, size = 48, normalized size = 0.84 \begin{align*} \frac{2 \left (2 c^{2} \sqrt{- a c x + c} - c \left (- a c x + c\right )^{\frac{3}{2}} + \frac{\left (- a c x + c\right )^{\frac{5}{2}}}{5}\right )}{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)*x*(-a*c*x+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.17294, size = 74, normalized size = 1.3 \begin{align*} \frac{2 \,{\left ({\left (a c x - c\right )}^{2} \sqrt{-a c x + c} - 5 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c + 10 \, \sqrt{-a c x + c} c^{2}\right )}}{5 \, 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)*x*(-a*c*x+c)^(1/2),x, algorithm="giac")

[Out]

2/5*((a*c*x - c)^2*sqrt(-a*c*x + c) - 5*(-a*c*x + c)^(3/2)*c + 10*sqrt(-a*c*x + c)*c^2)/(a^2*c^2)