\(\int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx\) [302]

Optimal result

Integrand size = 23, antiderivative size = 80 \[ \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {4 \sqrt {c-a c x}}{a^3}-\frac {10 (c-a c x)^{3/2}}{3 a^3 c}+\frac {8 (c-a c x)^{5/2}}{5 a^3 c^2}-\frac {2 (c-a c x)^{7/2}}{7 a^3 c^3} \]

[Out]

-10/3*(-a*c*x+c)^(3/2)/a^3/c+8/5*(-a*c*x+c)^(5/2)/a^3/c^2-2/7*(-a*c*x+c)^(7/2)/a^3/c^3+4*(-a*c*x+c)^(1/2)/a^3

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6302, 6265, 21, 78} \[ \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx=-\frac {2 (c-a c x)^{7/2}}{7 a^3 c^3}+\frac {8 (c-a c x)^{5/2}}{5 a^3 c^2}-\frac {10 (c-a c x)^{3/2}}{3 a^3 c}+\frac {4 \sqrt {c-a c x}}{a^3} \]

[In]

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

[Out]

(4*Sqrt[c - a*c*x])/a^3 - (10*(c - a*c*x)^(3/2))/(3*a^3*c) + (8*(c - a*c*x)^(5/2))/(5*a^3*c^2) - (2*(c - a*c*x
)^(7/2))/(7*a^3*c^3)

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^2 \sqrt {c-a c x} \, dx \\ & = -\int \frac {x^2 (1+a x) \sqrt {c-a c x}}{1-a x} \, dx \\ & = -\left (c \int \frac {x^2 (1+a x)}{\sqrt {c-a c x}} \, dx\right ) \\ & = -\left (c \int \left (\frac {2}{a^2 \sqrt {c-a c x}}-\frac {5 \sqrt {c-a c x}}{a^2 c}+\frac {4 (c-a c x)^{3/2}}{a^2 c^2}-\frac {(c-a c x)^{5/2}}{a^2 c^3}\right ) \, dx\right ) \\ & = \frac {4 \sqrt {c-a c x}}{a^3}-\frac {10 (c-a c x)^{3/2}}{3 a^3 c}+\frac {8 (c-a c x)^{5/2}}{5 a^3 c^2}-\frac {2 (c-a c x)^{7/2}}{7 a^3 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.50 \[ \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c-a c x} \left (104+52 a x+39 a^2 x^2+15 a^3 x^3\right )}{105 a^3} \]

[In]

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

[Out]

(2*Sqrt[c - a*c*x]*(104 + 52*a*x + 39*a^2*x^2 + 15*a^3*x^3))/(105*a^3)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.46

method	result	size
gosper	\(\frac {2 \sqrt {-a c x +c}\, \left (15 a^{3} x^{3}+39 a^{2} x^{2}+52 a x +104\right )}{105 a^{3}}\)	\(37\)
trager	\(\frac {2 \sqrt {-a c x +c}\, \left (15 a^{3} x^{3}+39 a^{2} x^{2}+52 a x +104\right )}{105 a^{3}}\)	\(37\)
pseudoelliptic	\(\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (15 a^{3} x^{3}+39 a^{2} x^{2}+52 a x +104\right )}{105 a^{3}}\)	\(38\)
risch	\(-\frac {2 c \left (15 a^{3} x^{3}+39 a^{2} x^{2}+52 a x +104\right ) \left (a x -1\right )}{105 a^{3} \sqrt {-c \left (a x -1\right )}}\)	\(44\)
derivativedivides	\(-\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {7}{2}}}{7}-\frac {4 c \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {5 c^{2} \left (-a c x +c \right )^{\frac {3}{2}}}{3}-2 c^{3} \sqrt {-a c x +c}\right )}{c^{3} a^{3}}\)	\(61\)
default	\(\frac {-\frac {2 \left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {8 c \left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {10 c^{2} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+4 c^{3} \sqrt {-a c x +c}}{a^{3} c^{3}}\)	\(61\)

[In]

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

[Out]

2/105*(-a*c*x+c)^(1/2)*(15*a^3*x^3+39*a^2*x^2+52*a*x+104)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.45 \[ \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {2 \, {\left (15 \, a^{3} x^{3} + 39 \, a^{2} x^{2} + 52 \, a x + 104\right )} \sqrt {-a c x + c}}{105 \, a^{3}} \]

[In]

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

[Out]

2/105*(15*a^3*x^3 + 39*a^2*x^2 + 52*a*x + 104)*sqrt(-a*c*x + c)/a^3

Sympy [A] (verification not implemented)

Time = 2.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.36 \[ \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx=\begin {cases} - \frac {2 \left (- 2 c^{3} \sqrt {- a c x + c} + \frac {5 c^{2} \left (- a c x + c\right )^{\frac {3}{2}}}{3} - \frac {4 c \left (- a c x + c\right )^{\frac {5}{2}}}{5} + \frac {\left (- a c x + c\right )^{\frac {7}{2}}}{7}\right )}{a^{3} c^{3}} & \text {for}\: a c \neq 0 \\\sqrt {c} \left (\frac {x^{3}}{3} + \frac {x^{2}}{a} + \frac {2 x}{a^{2}} + \frac {2 \left (\begin {cases} - x & \text {for}\: a = 0 \\\frac {\log {\left (a x - 1 \right )}}{a} & \text {otherwise} \end {cases}\right )}{a^{2}}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*(-2*c**3*sqrt(-a*c*x + c) + 5*c**2*(-a*c*x + c)**(3/2)/3 - 4*c*(-a*c*x + c)**(5/2)/5 + (-a*c*x +
c)**(7/2)/7)/(a**3*c**3), Ne(a*c, 0)), (sqrt(c)*(x**3/3 + x**2/a + 2*x/a**2 + 2*Piecewise((-x, Eq(a, 0)), (lo
g(a*x - 1)/a, True))/a**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.75 \[ \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx=-\frac {2 \, {\left (15 \, {\left (-a c x + c\right )}^{\frac {7}{2}} - 84 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c + 175 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} - 210 \, \sqrt {-a c x + c} c^{3}\right )}}{105 \, a^{3} c^{3}} \]

[In]

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

[Out]

-2/105*(15*(-a*c*x + c)^(7/2) - 84*(-a*c*x + c)^(5/2)*c + 175*(-a*c*x + c)^(3/2)*c^2 - 210*sqrt(-a*c*x + c)*c^
3)/(a^3*c^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (66) = 132\).

Time = 0.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.78 \[ \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {2 \, {\left (\frac {7 \, {\left (3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} - 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-a c x + c} c^{2}\right )}}{a^{2} c^{2}} + \frac {3 \, {\left (5 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} + 21 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} c - 35 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-a c x + c} c^{3}\right )}}{a^{2} c^{3}}\right )}}{105 \, a} \]

[In]

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

[Out]

2/105*(7*(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^2*c^2) + 3*
(5*(a*c*x - c)^3*sqrt(-a*c*x + c) + 21*(a*c*x - c)^2*sqrt(-a*c*x + c)*c - 35*(-a*c*x + c)^(3/2)*c^2 + 35*sqrt(
-a*c*x + c)*c^3)/(a^2*c^3))/a

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.82 \[ \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx=\frac {4\,\sqrt {c-a\,c\,x}}{a^3}-\frac {10\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a^3\,c}+\frac {8\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a^3\,c^2}-\frac {2\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a^3\,c^3} \]

[In]

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

[Out]

(4*(c - a*c*x)^(1/2))/a^3 - (10*(c - a*c*x)^(3/2))/(3*a^3*c) + (8*(c - a*c*x)^(5/2))/(5*a^3*c^2) - (2*(c - a*c
*x)^(7/2))/(7*a^3*c^3)