### 3.229 $$\int e^{\coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx$$

Optimal. Leaf size=77 $\frac{2 a^2 c^2 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{5 \sqrt{c-a c x}}+\frac{8 a^2 c^3 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{15 (c-a c x)^{3/2}}$

[Out]

(8*a^2*c^3*(1 - 1/(a^2*x^2))^(3/2)*x^3)/(15*(c - a*c*x)^(3/2)) + (2*a^2*c^2*(1 - 1/(a^2*x^2))^(3/2)*x^3)/(5*Sq
rt[c - a*c*x])

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Rubi [A]  time = 0.15801, antiderivative size = 89, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {6176, 6181, 78, 37} $\frac{2 x \left (\frac{1}{a x}+1\right )^{3/2} (c-a c x)^{3/2}}{5 \left (1-\frac{1}{a x}\right )^{3/2}}-\frac{14 \left (\frac{1}{a x}+1\right )^{3/2} (c-a c x)^{3/2}}{15 a \left (1-\frac{1}{a x}\right )^{3/2}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-14*(1 + 1/(a*x))^(3/2)*(c - a*c*x)^(3/2))/(15*a*(1 - 1/(a*x))^(3/2)) + (2*(1 + 1/(a*x))^(3/2)*x*(c - a*c*x)^
(3/2))/(5*(1 - 1/(a*x))^(3/2))

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 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 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^{\coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx &=\frac{(c-a c x)^{3/2} \int e^{\coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right )^{3/2} x^{3/2} \, dx}{\left (1-\frac{1}{a x}\right )^{3/2} x^{3/2}}\\ &=-\frac{\left (\left (\frac{1}{x}\right )^{3/2} (c-a c x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right ) \sqrt{1+\frac{x}{a}}}{x^{7/2}} \, dx,x,\frac{1}{x}\right )}{\left (1-\frac{1}{a x}\right )^{3/2}}\\ &=\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x (c-a c x)^{3/2}}{5 \left (1-\frac{1}{a x}\right )^{3/2}}+\frac{\left (7 \left (\frac{1}{x}\right )^{3/2} (c-a c x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x}{a}}}{x^{5/2}} \, dx,x,\frac{1}{x}\right )}{5 a \left (1-\frac{1}{a x}\right )^{3/2}}\\ &=-\frac{14 \left (1+\frac{1}{a x}\right )^{3/2} (c-a c x)^{3/2}}{15 a \left (1-\frac{1}{a x}\right )^{3/2}}+\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x (c-a c x)^{3/2}}{5 \left (1-\frac{1}{a x}\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0297813, size = 57, normalized size = 0.74 $-\frac{2 c \sqrt{\frac{1}{a x}+1} (a x+1) (3 a x-7) \sqrt{c-a c x}}{15 a \sqrt{1-\frac{1}{a x}}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*c*Sqrt[1 + 1/(a*x)]*(1 + a*x)*(-7 + 3*a*x)*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - 1/(a*x)])

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Maple [A]  time = 0.039, size = 48, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,ax+2 \right ) \left ( 3\,ax-7 \right ) }{15\, \left ( ax-1 \right ) a} \left ( -acx+c \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(a*x+1)*(3*a*x-7)*(-a*c*x+c)^(3/2)/a/(a*x-1)/((a*x-1)/(a*x+1))^(1/2)

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Maxima [A]  time = 1.11792, size = 61, normalized size = 0.79 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} \sqrt{-c} c x^{2} - 4 \, a \sqrt{-c} c x - 7 \, \sqrt{-c} c\right )} \sqrt{a x + 1}}{15 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^2*sqrt(-c)*c*x^2 - 4*a*sqrt(-c)*c*x - 7*sqrt(-c)*c)*sqrt(a*x + 1)/a

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Fricas [A]  time = 1.6449, size = 142, normalized size = 1.84 \begin{align*} -\frac{2 \,{\left (3 \, a^{3} c x^{3} - a^{2} c x^{2} - 11 \, a c x - 7 \, c\right )} \sqrt{-a c x + c} \sqrt{\frac{a x - 1}{a x + 1}}}{15 \,{\left (a^{2} x - a\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^3*c*x^3 - a^2*c*x^2 - 11*a*c*x - 7*c)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^2*x - a)

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

[Out]

Timed out

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Giac [A]  time = 1.13195, size = 100, normalized size = 1.3 \begin{align*} \frac{2 \,{\left (\frac{8 \, \sqrt{2} \sqrt{-c} c}{\mathrm{sgn}\left (c\right )} - \frac{3 \,{\left (a c x + c\right )}^{2} \sqrt{-a c x - c} + 10 \,{\left (-a c x - c\right )}^{\frac{3}{2}} c}{c \mathrm{sgn}\left (-a c x - c\right )}\right )}}{15 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(8*sqrt(2)*sqrt(-c)*c/sgn(c) - (3*(a*c*x + c)^2*sqrt(-a*c*x - c) + 10*(-a*c*x - c)^(3/2)*c)/(c*sgn(-a*c*x
- c)))/a