### 3.352 $$\int e^{-3 \coth ^{-1}(a x)} x \sqrt{c-a c x} \, dx$$

Optimal. Leaf size=182 $\frac{316 \sqrt{\frac{1}{a x}+1} \sqrt{c-a c x}}{15 a^2 \sqrt{1-\frac{1}{a x}}}-\frac{158 \sqrt{c-a c x}}{15 a^2 \sqrt{1-\frac{1}{a x}} \sqrt{\frac{1}{a x}+1}}+\frac{2 x^2 \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}} \sqrt{\frac{1}{a x}+1}}-\frac{32 x \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}} \sqrt{\frac{1}{a x}+1}}$

[Out]

(-158*Sqrt[c - a*c*x])/(15*a^2*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)]) + (316*Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x])/
(15*a^2*Sqrt[1 - 1/(a*x)]) - (32*x*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)]) + (2*x^2*Sqrt[c
- a*c*x])/(5*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)])

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Rubi [A]  time = 0.216173, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {6176, 6181, 89, 78, 45, 37} $\frac{316 \sqrt{\frac{1}{a x}+1} \sqrt{c-a c x}}{15 a^2 \sqrt{1-\frac{1}{a x}}}-\frac{158 \sqrt{c-a c x}}{15 a^2 \sqrt{1-\frac{1}{a x}} \sqrt{\frac{1}{a x}+1}}+\frac{2 x^2 \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}} \sqrt{\frac{1}{a x}+1}}-\frac{32 x \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}} \sqrt{\frac{1}{a x}+1}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-158*Sqrt[c - a*c*x])/(15*a^2*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)]) + (316*Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x])/
(15*a^2*Sqrt[1 - 1/(a*x)]) - (32*x*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)]) + (2*x^2*Sqrt[c
- a*c*x])/(5*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)])

Rule 6176

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^p/(x^p*(1 + c/(d
*x))^p), Int[u*x^p*(1 + c/(d*x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^
2, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6181

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Sub
st[Int[((1 + (d*x)/c)^p*(1 + x/a)^(n/2))/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x], x] /; FreeQ[{a, c, d, m, n,
p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[m]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

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

Rule 45

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int e^{-3 \coth ^{-1}(a x)} x \sqrt{c-a c x} \, dx &=\frac{\sqrt{c-a c x} \int e^{-3 \coth ^{-1}(a x)} \sqrt{1-\frac{1}{a x}} x^{3/2} \, dx}{\sqrt{1-\frac{1}{a x}} \sqrt{x}}\\ &=-\frac{\left (\sqrt{\frac{1}{x}} \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^2}{x^{7/2} \left (1+\frac{x}{a}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{\sqrt{1-\frac{1}{a x}}}\\ &=\frac{2 x^2 \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}-\frac{\left (2 \sqrt{\frac{1}{x}} \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{-\frac{8}{a}+\frac{5 x}{2 a^2}}{x^{5/2} \left (1+\frac{x}{a}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{5 \sqrt{1-\frac{1}{a x}}}\\ &=-\frac{32 x \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}+\frac{2 x^2 \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}-\frac{\left (79 \sqrt{\frac{1}{x}} \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{1}{x^{3/2} \left (1+\frac{x}{a}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{15 a^2 \sqrt{1-\frac{1}{a x}}}\\ &=-\frac{158 \sqrt{c-a c x}}{15 a^2 \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}-\frac{32 x \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}+\frac{2 x^2 \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}-\frac{\left (158 \sqrt{\frac{1}{x}} \sqrt{c-a c x}\right ) \operatorname{Subst}\left (\int \frac{1}{x^{3/2} \sqrt{1+\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )}{15 a^2 \sqrt{1-\frac{1}{a x}}}\\ &=-\frac{158 \sqrt{c-a c x}}{15 a^2 \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}+\frac{316 \sqrt{1+\frac{1}{a x}} \sqrt{c-a c x}}{15 a^2 \sqrt{1-\frac{1}{a x}}}-\frac{32 x \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}+\frac{2 x^2 \sqrt{c-a c x}}{5 \sqrt{1-\frac{1}{a x}} \sqrt{1+\frac{1}{a x}}}\\ \end{align*}

Mathematica [A]  time = 0.0302175, size = 57, normalized size = 0.31 $\frac{2 \left (3 a^3 x^3-16 a^2 x^2+79 a x+158\right ) \sqrt{c-a c x}}{15 a^3 x \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[c - a*c*x]*(158 + 79*a*x - 16*a^2*x^2 + 3*a^3*x^3))/(15*a^3*Sqrt[1 - 1/(a^2*x^2)]*x)

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Maple [A]  time = 0.043, size = 64, normalized size = 0.4 \begin{align*}{\frac{ \left ( 2\,ax+2 \right ) \left ( 3\,{x}^{3}{a}^{3}-16\,{a}^{2}{x}^{2}+79\,ax+158 \right ) }{15\,{a}^{2} \left ( ax-1 \right ) ^{2}}\sqrt{-acx+c} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}} \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))^(3/2),x)

[Out]

2/15*(a*x+1)*(3*a^3*x^3-16*a^2*x^2+79*a*x+158)*(-a*c*x+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/a^2/(a*x-1)^2

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Maxima [A]  time = 1.11408, size = 123, normalized size = 0.68 \begin{align*} \frac{2 \,{\left (3 \, a^{4} \sqrt{-c} x^{4} - 13 \, a^{3} \sqrt{-c} x^{3} + 63 \, a^{2} \sqrt{-c} x^{2} + 237 \, a \sqrt{-c} x + 158 \, \sqrt{-c}\right )}{\left (a x - 1\right )}^{2}}{15 \,{\left (a^{4} x^{2} - 2 \, a^{3} x + a^{2}\right )}{\left (a x + 1\right )}^{\frac{3}{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))^(3/2),x, algorithm="maxima")

[Out]

2/15*(3*a^4*sqrt(-c)*x^4 - 13*a^3*sqrt(-c)*x^3 + 63*a^2*sqrt(-c)*x^2 + 237*a*sqrt(-c)*x + 158*sqrt(-c))*(a*x -
1)^2/((a^4*x^2 - 2*a^3*x + a^2)*(a*x + 1)^(3/2))

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Fricas [A]  time = 1.39988, size = 139, normalized size = 0.76 \begin{align*} \frac{2 \,{\left (3 \, a^{3} x^{3} - 16 \, a^{2} x^{2} + 79 \, a x + 158\right )} \sqrt{-a c x + c} \sqrt{\frac{a x - 1}{a x + 1}}}{15 \,{\left (a^{3} x - 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))^(3/2),x, algorithm="fricas")

[Out]

2/15*(3*a^3*x^3 - 16*a^2*x^2 + 79*a*x + 158)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^3*x - a^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.19304, size = 113, normalized size = 0.62 \begin{align*} -\frac{2 \,{\left (3 \,{\left (a c x + c\right )}^{2} \sqrt{-a c x - c}{\left | c \right |} + 25 \,{\left (-a c x - c\right )}^{\frac{3}{2}} c{\left | c \right |} + 120 \, \sqrt{-a c x - c} c^{2}{\left | c \right |} - \frac{60 \, c^{3}{\left | c \right |}}{\sqrt{-a c x - c}}\right )}}{15 \, a^{2} c^{3}} \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))^(3/2),x, algorithm="giac")

[Out]

-2/15*(3*(a*c*x + c)^2*sqrt(-a*c*x - c)*abs(c) + 25*(-a*c*x - c)^(3/2)*c*abs(c) + 120*sqrt(-a*c*x - c)*c^2*abs
(c) - 60*c^3*abs(c)/sqrt(-a*c*x - c))/(a^2*c^3)