### 3.343 $$\int e^{-2 \coth ^{-1}(a x)} x \sqrt{c-a c x} \, dx$$

Optimal. Leaf size=97 $\frac{2 (c-a c x)^{5/2}}{5 a^2 c^2}+\frac{2 (c-a c x)^{3/2}}{3 a^2 c}+\frac{4 \sqrt{c-a c x}}{a^2}-\frac{4 \sqrt{2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{a^2}$

[Out]

(4*Sqrt[c - a*c*x])/a^2 + (2*(c - a*c*x)^(3/2))/(3*a^2*c) + (2*(c - a*c*x)^(5/2))/(5*a^2*c^2) - (4*Sqrt[2]*Sqr
t[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/a^2

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Rubi [A]  time = 0.170344, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {6167, 6130, 21, 80, 50, 63, 206} $\frac{2 (c-a c x)^{5/2}}{5 a^2 c^2}+\frac{2 (c-a c x)^{3/2}}{3 a^2 c}+\frac{4 \sqrt{c-a c x}}{a^2}-\frac{4 \sqrt{2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{a^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*Sqrt[c - a*c*x])/a^2 + (2*(c - a*c*x)^(3/2))/(3*a^2*c) + (2*(c - a*c*x)^(5/2))/(5*a^2*c^2) - (4*Sqrt[2]*Sqr
t[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/a^2

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]

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 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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

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 63

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 206

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

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

Mathematica [A]  time = 0.0818633, size = 70, normalized size = 0.72 $\frac{2 \left (3 a^2 x^2-11 a x+38\right ) \sqrt{c-a c x}-60 \sqrt{2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c-a c x}}{\sqrt{2} \sqrt{c}}\right )}{15 a^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[c - a*c*x]*(38 - 11*a*x + 3*a^2*x^2) - 60*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/(Sqrt[2]*Sqrt[c])])/
(15*a^2)

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Maple [A]  time = 0.043, size = 73, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{{a}^{2}{c}^{2}} \left ( 1/5\, \left ( -acx+c \right ) ^{5/2}+1/3\,c \left ( -acx+c \right ) ^{3/2}+2\,\sqrt{-acx+c}{c}^{2}-2\,{c}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-acx+c}\sqrt{2}}{\sqrt{c}}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64144, size = 366, normalized size = 3.77 \begin{align*} \left [\frac{2 \,{\left (15 \, \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 (3 \, a^{2} x^{2} - 11 \, a x + 38\right )} \sqrt{-a c x + c}\right )}}{15 \, a^{2}}, \frac{2 \,{\left (30 \, \sqrt{2} \sqrt{-c} \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c} \sqrt{-c}}{2 \, c}\right ) +{\left (3 \, a^{2} x^{2} - 11 \, a x + 38\right )} \sqrt{-a c x + c}\right )}}{15 \, a^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/15*(15*sqrt(2)*sqrt(c)*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + (3*a^2*x^2 - 11*
a*x + 38)*sqrt(-a*c*x + c))/a^2, 2/15*(30*sqrt(2)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) + (
3*a^2*x^2 - 11*a*x + 38)*sqrt(-a*c*x + c))/a^2]

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Sympy [A]  time = 4.81473, size = 92, normalized size = 0.95 \begin{align*} \frac{2 \left (\frac{2 \sqrt{2} c^{3} \operatorname{atan}{\left (\frac{\sqrt{2} \sqrt{- a c x + c}}{2 \sqrt{- c}} \right )}}{\sqrt{- c}} + 2 c^{2} \sqrt{- a c x + c} + \frac{c \left (- a c x + c\right )^{\frac{3}{2}}}{3} + \frac{\left (- a c x + c\right )^{\frac{5}{2}}}{5}\right )}{a^{2} c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*sqrt(2)*c**3*atan(sqrt(2)*sqrt(-a*c*x + c)/(2*sqrt(-c)))/sqrt(-c) + 2*c**2*sqrt(-a*c*x + c) + c*(-a*c*x +
c)**(3/2)/3 + (-a*c*x + c)**(5/2)/5)/(a**2*c**2)

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Giac [A]  time = 1.12735, size = 142, normalized size = 1.46 \begin{align*} \frac{4 \, \sqrt{2} c \arctan \left (\frac{\sqrt{2} \sqrt{-a c x + c}}{2 \, \sqrt{-c}}\right )}{a^{2} \sqrt{-c}} + \frac{2 \,{\left (3 \,{\left (a c x - c\right )}^{2} \sqrt{-a c x + c} a^{8} c^{8} + 5 \,{\left (-a c x + c\right )}^{\frac{3}{2}} a^{8} c^{9} + 30 \, \sqrt{-a c x + c} a^{8} c^{10}\right )}}{15 \, a^{10} c^{10}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-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^2*sqrt(-c)) + 2/15*(3*(a*c*x - c)^2*sqrt(-a*c*x +
c)*a^8*c^8 + 5*(-a*c*x + c)^(3/2)*a^8*c^9 + 30*sqrt(-a*c*x + c)*a^8*c^10)/(a^10*c^10)