\(\int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx\) [301]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {4 \sqrt {c-a c x}}{a^4}-\frac {14 (c-a c x)^{3/2}}{3 a^4 c}+\frac {18 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {10 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4} \]

[Out]

-14/3*(-a*c*x+c)^(3/2)/a^4/c+18/5*(-a*c*x+c)^(5/2)/a^4/c^2-10/7*(-a*c*x+c)^(7/2)/a^4/c^3+2/9*(-a*c*x+c)^(9/2)/
a^4/c^4+4*(-a*c*x+c)^(1/2)/a^4

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 101, 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} \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4}-\frac {10 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {18 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {14 (c-a c x)^{3/2}}{3 a^4 c}+\frac {4 \sqrt {c-a c x}}{a^4} \]

[In]

Int[E^(2*ArcCoth[a*x])*x^3*Sqrt[c - a*c*x],x]

[Out]

(4*Sqrt[c - a*c*x])/a^4 - (14*(c - a*c*x)^(3/2))/(3*a^4*c) + (18*(c - a*c*x)^(5/2))/(5*a^4*c^2) - (10*(c - a*c
*x)^(7/2))/(7*a^4*c^3) + (2*(c - a*c*x)^(9/2))/(9*a^4*c^4)

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^3 \sqrt {c-a c x} \, dx \\ & = -\int \frac {x^3 (1+a x) \sqrt {c-a c x}}{1-a x} \, dx \\ & = -\left (c \int \frac {x^3 (1+a x)}{\sqrt {c-a c x}} \, dx\right ) \\ & = -\left (c \int \left (\frac {2}{a^3 \sqrt {c-a c x}}-\frac {7 \sqrt {c-a c x}}{a^3 c}+\frac {9 (c-a c x)^{3/2}}{a^3 c^2}-\frac {5 (c-a c x)^{5/2}}{a^3 c^3}+\frac {(c-a c x)^{7/2}}{a^3 c^4}\right ) \, dx\right ) \\ & = \frac {4 \sqrt {c-a c x}}{a^4}-\frac {14 (c-a c x)^{3/2}}{3 a^4 c}+\frac {18 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {10 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.48 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {2 \sqrt {c-a c x} \left (272+136 a x+102 a^2 x^2+85 a^3 x^3+35 a^4 x^4\right )}{315 a^4} \]

[In]

Integrate[E^(2*ArcCoth[a*x])*x^3*Sqrt[c - a*c*x],x]

[Out]

(2*Sqrt[c - a*c*x]*(272 + 136*a*x + 102*a^2*x^2 + 85*a^3*x^3 + 35*a^4*x^4))/(315*a^4)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.45

method	result	size
gosper	\(\frac {2 \sqrt {-a c x +c}\, \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right )}{315 a^{4}}\)	\(45\)
trager	\(\frac {2 \sqrt {-a c x +c}\, \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right )}{315 a^{4}}\)	\(45\)
pseudoelliptic	\(\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right )}{315 a^{4}}\)	\(46\)
risch	\(-\frac {2 c \left (35 a^{4} x^{4}+85 a^{3} x^{3}+102 a^{2} x^{2}+136 a x +272\right ) \left (a x -1\right )}{315 a^{4} \sqrt {-c \left (a x -1\right )}}\)	\(52\)
derivativedivides	\(\frac {\frac {2 \left (-a c x +c \right )^{\frac {9}{2}}}{9}-\frac {10 c \left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {18 c^{2} \left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {14 c^{3} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+4 c^{4} \sqrt {-a c x +c}}{a^{4} c^{4}}\)	\(75\)
default	\(\frac {\frac {2 \left (-a c x +c \right )^{\frac {9}{2}}}{9}-\frac {10 c \left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {18 c^{2} \left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {14 c^{3} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+4 c^{4} \sqrt {-a c x +c}}{a^{4} c^{4}}\)	\(75\)

[In]

int(1/(a*x-1)*(a*x+1)*x^3*(-a*c*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/315*(-a*c*x+c)^(1/2)*(35*a^4*x^4+85*a^3*x^3+102*a^2*x^2+136*a*x+272)/a^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.44 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {2 \, {\left (35 \, a^{4} x^{4} + 85 \, a^{3} x^{3} + 102 \, a^{2} x^{2} + 136 \, a x + 272\right )} \sqrt {-a c x + c}}{315 \, a^{4}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*x^3*(-a*c*x+c)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*a^4*x^4 + 85*a^3*x^3 + 102*a^2*x^2 + 136*a*x + 272)*sqrt(-a*c*x + c)/a^4

Sympy [A] (verification not implemented)

Time = 2.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.33 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\begin {cases} \frac {2 \cdot \left (2 c^{4} \sqrt {- a c x + c} - \frac {7 c^{3} \left (- a c x + c\right )^{\frac {3}{2}}}{3} + \frac {9 c^{2} \left (- a c x + c\right )^{\frac {5}{2}}}{5} - \frac {5 c \left (- a c x + c\right )^{\frac {7}{2}}}{7} + \frac {\left (- a c x + c\right )^{\frac {9}{2}}}{9}\right )}{a^{4} c^{4}} & \text {for}\: a c \neq 0 \\\sqrt {c} \left (\frac {x^{4}}{4} + \frac {2 x^{3}}{3 a} + \frac {x^{2}}{a^{2}} + \frac {2 x}{a^{3}} + \frac {2 \left (\begin {cases} - x & \text {for}\: a = 0 \\\frac {\log {\left (a x - 1 \right )}}{a} & \text {otherwise} \end {cases}\right )}{a^{3}}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*x**3*(-a*c*x+c)**(1/2),x)

[Out]

Piecewise((2*(2*c**4*sqrt(-a*c*x + c) - 7*c**3*(-a*c*x + c)**(3/2)/3 + 9*c**2*(-a*c*x + c)**(5/2)/5 - 5*c*(-a*
c*x + c)**(7/2)/7 + (-a*c*x + c)**(9/2)/9)/(a**4*c**4), Ne(a*c, 0)), (sqrt(c)*(x**4/4 + 2*x**3/(3*a) + x**2/a*
*2 + 2*x/a**3 + 2*Piecewise((-x, Eq(a, 0)), (log(a*x - 1)/a, True))/a**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.73 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {2 \, {\left (35 \, {\left (-a c x + c\right )}^{\frac {9}{2}} - 225 \, {\left (-a c x + c\right )}^{\frac {7}{2}} c + 567 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c^{2} - 735 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{3} + 630 \, \sqrt {-a c x + c} c^{4}\right )}}{315 \, a^{4} c^{4}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*x^3*(-a*c*x+c)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*(-a*c*x + c)^(9/2) - 225*(-a*c*x + c)^(7/2)*c + 567*(-a*c*x + c)^(5/2)*c^2 - 735*(-a*c*x + c)^(3/2)*
c^3 + 630*sqrt(-a*c*x + c)*c^4)/(a^4*c^4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (83) = 166\).

Time = 0.27 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.87 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {2 \, {\left (\frac {9 \, {\left (5 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} + 21 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} c - 35 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-a c x + c} c^{3}\right )}}{a^{3} c^{3}} + \frac {35 \, {\left (a c x - c\right )}^{4} \sqrt {-a c x + c} + 180 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} c + 378 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} c^{2} - 420 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {-a c x + c} c^{4}}{a^{3} c^{4}}\right )}}{315 \, a} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*x^3*(-a*c*x+c)^(1/2),x, algorithm="giac")

[Out]

2/315*(9*(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^3*c^3) + (35*(a*c*x - c)^4*sqrt(-a*c*x + c) + 180*(a*c*x - c)^3*sqrt(-a*c*x + c)*
c + 378*(a*c*x - c)^2*sqrt(-a*c*x + c)*c^2 - 420*(-a*c*x + c)^(3/2)*c^3 + 315*sqrt(-a*c*x + c)*c^4)/(a^3*c^4))
/a

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {4\,\sqrt {c-a\,c\,x}}{a^4}-\frac {14\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a^4\,c}+\frac {18\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a^4\,c^2}-\frac {10\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a^4\,c^3}+\frac {2\,{\left (c-a\,c\,x\right )}^{9/2}}{9\,a^4\,c^4} \]

[In]

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

[Out]

(4*(c - a*c*x)^(1/2))/a^4 - (14*(c - a*c*x)^(3/2))/(3*a^4*c) + (18*(c - a*c*x)^(5/2))/(5*a^4*c^2) - (10*(c - a
*c*x)^(7/2))/(7*a^4*c^3) + (2*(c - a*c*x)^(9/2))/(9*a^4*c^4)