\(\int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx\) [303]

Optimal result

Integrand size = 21, antiderivative size = 57 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {4 \sqrt {c-a c x}}{a^2}-\frac {2 (c-a c x)^{3/2}}{a^2 c}+\frac {2 (c-a c x)^{5/2}}{5 a^2 c^2} \]

[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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6302, 6265, 21, 78} \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {2 (c-a c x)^{5/2}}{5 a^2 c^2}-\frac {2 (c-a c x)^{3/2}}{a^2 c}+\frac {4 \sqrt {c-a c x}}{a^2} \]

[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 78

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 6265

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 6302

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*} \text {integral}& = -\int e^{2 \text {arctanh}(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] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.54 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c-a c x} \left (6+3 a x+a^2 x^2\right )}{5 a^2} \]

[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)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.49

method	result	size
gosper	\(\frac {2 \sqrt {-a c x +c}\, \left (a^{2} x^{2}+3 a x +6\right )}{5 a^{2}}\)	\(28\)
trager	\(\frac {2 \sqrt {-a c x +c}\, \left (a^{2} x^{2}+3 a x +6\right )}{5 a^{2}}\)	\(28\)
pseudoelliptic	\(\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (a^{2} x^{2}+3 a x +6\right )}{5 a^{2}}\)	\(29\)
risch	\(-\frac {2 c \left (a^{2} x^{2}+3 a x +6\right ) \left (a x -1\right )}{5 a^{2} \sqrt {-c \left (a x -1\right )}}\)	\(35\)
derivativedivides	\(\frac {\frac {2 \left (-a c x +c \right )^{\frac {5}{2}}}{5}-2 c \left (-a c x +c \right )^{\frac {3}{2}}+4 c^{2} \sqrt {-a c x +c}}{a^{2} c^{2}}\)	\(47\)
default	\(\frac {\frac {2 \left (-a c x +c \right )^{\frac {5}{2}}}{5}-2 c \left (-a c x +c \right )^{\frac {3}{2}}+4 c^{2} \sqrt {-a c x +c}}{a^{2} c^{2}}\)	\(47\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.47 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {2 \, {\left (a^{2} x^{2} + 3 \, a x + 6\right )} \sqrt {-a c x + c}}{5 \, a^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 2.53 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.40 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\begin {cases} \frac {2 \cdot \left (2 c^{2} \sqrt {- a c x + c} - c \left (- a c x + c\right )^{\frac {3}{2}} + \frac {\left (- a c x + c\right )^{\frac {5}{2}}}{5}\right )}{a^{2} c^{2}} & \text {for}\: a c \neq 0 \\\sqrt {c} \left (\frac {x^{2}}{2} + \frac {2 x}{a} + \frac {2 \left (\begin {cases} - x & \text {for}\: a = 0 \\\frac {\log {\left (a x - 1 \right )}}{a} & \text {otherwise} \end {cases}\right )}{a}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((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), Ne(a*c, 0)
), (sqrt(c)*(x**2/2 + 2*x/a + 2*Piecewise((-x, Eq(a, 0)), (log(a*x - 1)/a, True))/a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {2 \, {\left ({\left (-a c x + c\right )}^{\frac {5}{2}} - 5 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 10 \, \sqrt {-a c x + c} c^{2}\right )}}{5 \, a^{2} c^{2}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.61 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=-\frac {2 \, {\left (\frac {5 \, {\left ({\left (-a c x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-a c x + c} c\right )}}{a c} - \frac {3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} - 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-a c x + c} c^{2}}{a c^{2}}\right )}}{15 \, a} \]

[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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a c x} \, dx=\frac {2\,{\left (c-a\,c\,x\right )}^{5/2}-10\,c\,{\left (c-a\,c\,x\right )}^{3/2}+20\,c^2\,\sqrt {c-a\,c\,x}}{5\,a^2\,c^2} \]

[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)