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]

-2*(-a*c*x+c)^(3/2)/a^2/c+2/5*(-a*c*x+c)^(5/2)/a^2/c^2+4*(-a*c*x+c)^(1/2)/a^2

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Rubi [A]  time = 0.15, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]))))

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 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]

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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.50, size = 27, normalized size = 0.47 \[ \frac {2 \, {\left (a^{2} x^{2} + 3 \, a x + 6\right )} \sqrt {-a c x + c}}{5 \, a^{2}} \]

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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giac [A]  time = 0.13, size = 92, normalized size = 1.61 \[ -\frac {2 \, {\left (\frac {5 \, {\left ({\left (-a c x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-a c x + c} c\right )}}{a c} - \frac {3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} - 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-a c x + c} c^{2}}{a c^{2}}\right )}}{15 \, a} \]

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]

-2/15*(5*((-a*c*x + c)^(3/2) - 3*sqrt(-a*c*x + c)*c)/(a*c) - (3*(a*c*x - c)^2*sqrt(-a*c*x + c) - 10*(-a*c*x +
c)^(3/2)*c + 15*sqrt(-a*c*x + c)*c^2)/(a*c^2))/a

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maple [A]  time = 0.03, size = 28, normalized size = 0.49 \[ \frac {2 \sqrt {-a c x +c}\, \left (a^{2} x^{2}+3 a x +6\right )}{5 a^{2}} \]

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 = 0.31, size = 44, normalized size = 0.77 \[ \frac {2 \, {\left ({\left (-a c x + c\right )}^{\frac {5}{2}} - 5 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 10 \, \sqrt {-a c x + c} c^{2}\right )}}{5 \, a^{2} c^{2}} \]

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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mupad [B]  time = 0.05, size = 46, normalized size = 0.81 \[ \frac {2\,{\left (c-a\,c\,x\right )}^{5/2}-10\,c\,{\left (c-a\,c\,x\right )}^{3/2}+20\,c^2\,\sqrt {c-a\,c\,x}}{5\,a^2\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(c - a*c*x)^(5/2) - 10*c*(c - a*c*x)^(3/2) + 20*c^2*(c - a*c*x)^(1/2))/(5*a^2*c^2)

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sympy [A]  time = 3.89, size = 48, normalized size = 0.84 \[ \frac {2 \left (2 c^{2} \sqrt {- a c x + c} - c \left (- a c x + c\right )^{\frac {3}{2}} + \frac {\left (- a c x + c\right )^{\frac {5}{2}}}{5}\right )}{a^{2} c^{2}} \]

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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