\(\int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx\) [341]

Optimal result

Integrand size = 23, antiderivative size = 139 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {4 \sqrt {c-a c x}}{a^4}+\frac {2 (c-a c x)^{3/2}}{3 a^4 c}+\frac {2 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {2 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4}-\frac {4 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6302, 6265, 21, 90, 52, 65, 212} \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=-\frac {4 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a^4}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4}-\frac {2 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{5/2}}{5 a^4 c^2}+\frac {2 (c-a c x)^{3/2}}{3 a^4 c}+\frac {4 \sqrt {c-a c x}}{a^4} \]

[In]

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

[Out]

(4*Sqrt[c - a*c*x])/a^4 + (2*(c - a*c*x)^(3/2))/(3*a^4*c) + (2*(c - a*c*x)^(5/2))/(5*a^4*c^2) - (2*(c - a*c*x)
^(7/2))/(7*a^4*c^3) + (2*(c - a*c*x)^(9/2))/(9*a^4*c^4) - (4*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*
Sqrt[c])])/a^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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 \\ & = -\frac {\int \frac {x^3 (c-a c x)^{3/2}}{1+a x} \, dx}{c} \\ & = -\frac {\int \left (\frac {(c-a c x)^{3/2}}{a^3}-\frac {(c-a c x)^{3/2}}{a^3 (1+a x)}-\frac {(c-a c x)^{5/2}}{a^3 c}+\frac {(c-a c x)^{7/2}}{a^3 c^2}\right ) \, dx}{c} \\ & = \frac {2 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {2 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4}+\frac {\int \frac {(c-a c x)^{3/2}}{1+a x} \, dx}{a^3 c} \\ & = \frac {2 (c-a c x)^{3/2}}{3 a^4 c}+\frac {2 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {2 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4}+\frac {2 \int \frac {\sqrt {c-a c x}}{1+a x} \, dx}{a^3} \\ & = \frac {4 \sqrt {c-a c x}}{a^4}+\frac {2 (c-a c x)^{3/2}}{3 a^4 c}+\frac {2 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {2 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4}+\frac {(4 c) \int \frac {1}{(1+a x) \sqrt {c-a c x}} \, dx}{a^3} \\ & = \frac {4 \sqrt {c-a c x}}{a^4}+\frac {2 (c-a c x)^{3/2}}{3 a^4 c}+\frac {2 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {2 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4}-\frac {8 \text {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )}{a^4} \\ & = \frac {4 \sqrt {c-a c x}}{a^4}+\frac {2 (c-a c x)^{3/2}}{3 a^4 c}+\frac {2 (c-a c x)^{5/2}}{5 a^4 c^2}-\frac {2 (c-a c x)^{7/2}}{7 a^4 c^3}+\frac {2 (c-a c x)^{9/2}}{9 a^4 c^4}-\frac {4 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )}{a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.61 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {2 \left (\sqrt {c-a c x} \left (788-236 a x+138 a^2 x^2-95 a^3 x^3+35 a^4 x^4\right )-630 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )\right )}{315 a^4} \]

[In]

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

[Out]

(2*(Sqrt[c - a*c*x]*(788 - 236*a*x + 138*a^2*x^2 - 95*a^3*x^3 + 35*a^4*x^4) - 630*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt
[c - a*c*x]/(Sqrt[2]*Sqrt[c])]))/(315*a^4)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54

method	result	size
pseudoelliptic	\(\frac {\frac {2 \left (35 a^{4} x^{4}-95 a^{3} x^{3}+138 a^{2} x^{2}-236 a x +788\right ) \sqrt {-c \left (a x -1\right )}}{315}-4 \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a^{4}}\)	\(75\)
risch	\(-\frac {2 \left (35 a^{4} x^{4}-95 a^{3} x^{3}+138 a^{2} x^{2}-236 a x +788\right ) \left (a x -1\right ) c}{315 a^{4} \sqrt {-c \left (a x -1\right )}}-\frac {4 \,\operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sqrt {c}}{a^{4}}\)	\(82\)
derivativedivides	\(\frac {\frac {2 \left (-a c x +c \right )^{\frac {9}{2}}}{9}-\frac {2 c \left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {2 c^{2} \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {2 c^{3} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+4 c^{4} \sqrt {-a c x +c}-4 c^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a^{4} c^{4}}\)	\(101\)
default	\(\frac {\frac {2 \left (-a c x +c \right )^{\frac {9}{2}}}{9}-\frac {2 c \left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {2 c^{2} \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {2 c^{3} \left (-a c x +c \right )^{\frac {3}{2}}}{3}+4 c^{4} \sqrt {-a c x +c}-4 c^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{a^{4} c^{4}}\)	\(101\)

[In]

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

[Out]

2/315*((35*a^4*x^4-95*a^3*x^3+138*a^2*x^2-236*a*x+788)*(-c*(a*x-1))^(1/2)-630*c^(1/2)*2^(1/2)*arctanh(1/2*(-c*
(a*x-1))^(1/2)*2^(1/2)/c^(1/2)))/a^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.21 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\left [\frac {2 \, {\left (315 \, \sqrt {2} \sqrt {c} \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + {\left (35 \, a^{4} x^{4} - 95 \, a^{3} x^{3} + 138 \, a^{2} x^{2} - 236 \, a x + 788\right )} \sqrt {-a c x + c}\right )}}{315 \, a^{4}}, \frac {2 \, {\left (630 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) + {\left (35 \, a^{4} x^{4} - 95 \, a^{3} x^{3} + 138 \, a^{2} x^{2} - 236 \, a x + 788\right )} \sqrt {-a c x + c}\right )}}{315 \, a^{4}}\right ] \]

[In]

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

[Out]

[2/315*(315*sqrt(2)*sqrt(c)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (35*a^4*x^4 -
95*a^3*x^3 + 138*a^2*x^2 - 236*a*x + 788)*sqrt(-a*c*x + c))/a^4, 2/315*(630*sqrt(2)*sqrt(-c)*arctan(1/2*sqrt(2
)*sqrt(-a*c*x + c)*sqrt(-c)/c) + (35*a^4*x^4 - 95*a^3*x^3 + 138*a^2*x^2 - 236*a*x + 788)*sqrt(-a*c*x + c))/a^4
]

Sympy [A] (verification not implemented)

Time = 4.14 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.26 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\begin {cases} \frac {2 \cdot \left (\frac {2 \sqrt {2} c^{5} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} + 2 c^{4} \sqrt {- a c x + c} + \frac {c^{3} \left (- a c x + c\right )^{\frac {3}{2}}}{3} + \frac {c^{2} \left (- a c x + c\right )^{\frac {5}{2}}}{5} - \frac {c \left (- a c x + c\right )^{\frac {7}{2}}}{7} + \frac {\left (- a c x + c\right )^{\frac {9}{2}}}{9}\right )}{a^{4} c^{4}} & \text {for}\: a c \neq 0 \\\sqrt {c} \left (\frac {x^{4}}{4} - \frac {2 x^{3}}{3 a} + \frac {x^{2}}{a^{2}} - \frac {2 x}{a^{3}} + \frac {2 \left (\begin {cases} x & \text {for}\: a = 0 \\\frac {\log {\left (a x + 1 \right )}}{a} & \text {otherwise} \end {cases}\right )}{a^{3}}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.88 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {2 \, {\left (315 \, \sqrt {2} c^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right ) + 35 \, {\left (-a c x + c\right )}^{\frac {9}{2}} - 45 \, {\left (-a c x + c\right )}^{\frac {7}{2}} c + 63 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c^{2} + 105 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{3} + 630 \, \sqrt {-a c x + c} c^{4}\right )}}{315 \, a^{4} c^{4}} \]

[In]

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

[Out]

2/315*(315*sqrt(2)*c^(9/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c))) + 3
5*(-a*c*x + c)^(9/2) - 45*(-a*c*x + c)^(7/2)*c + 63*(-a*c*x + c)^(5/2)*c^2 + 105*(-a*c*x + c)^(3/2)*c^3 + 630*
sqrt(-a*c*x + c)*c^4)/(a^4*c^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {4 \, \sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{a^{4} \sqrt {-c}} + \frac {2 \, {\left (35 \, {\left (a c x - c\right )}^{4} \sqrt {-a c x + c} a^{32} c^{32} + 45 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} a^{32} c^{33} + 63 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a^{32} c^{34} + 105 \, {\left (-a c x + c\right )}^{\frac {3}{2}} a^{32} c^{35} + 630 \, \sqrt {-a c x + c} a^{32} c^{36}\right )}}{315 \, a^{36} c^{36}} \]

[In]

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

[Out]

4*sqrt(2)*c*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/(a^4*sqrt(-c)) + 2/315*(35*(a*c*x - c)^4*sqrt(-a*c*x
+ c)*a^32*c^32 + 45*(a*c*x - c)^3*sqrt(-a*c*x + c)*a^32*c^33 + 63*(a*c*x - c)^2*sqrt(-a*c*x + c)*a^32*c^34 +
105*(-a*c*x + c)^(3/2)*a^32*c^35 + 630*sqrt(-a*c*x + c)*a^32*c^36)/(a^36*c^36)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.82 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \sqrt {c-a c x} \, dx=\frac {4\,\sqrt {c-a\,c\,x}}{a^4}+\frac {2\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a^4\,c}+\frac {2\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a^4\,c^2}-\frac {2\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a^4\,c^3}+\frac {2\,{\left (c-a\,c\,x\right )}^{9/2}}{9\,a^4\,c^4}+\frac {\sqrt {2}\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,4{}\mathrm {i}}{a^4} \]

[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 + (2*(c - a*c*x)^(3/2))/(3*a^4*c) + (2*(c - a*c*x)^(5/2))/(5*a^4*c^2) - (2*(c - a*c*
x)^(7/2))/(7*a^4*c^3) + (2*(c - a*c*x)^(9/2))/(9*a^4*c^4) + (2^(1/2)*c^(1/2)*atan((2^(1/2)*(c - a*c*x)^(1/2)*1
i)/(2*c^(1/2)))*4i)/a^4