### 3.228 $$\int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx$$

Optimal. Leaf size=115 $\frac{64 a^2 c^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{105 (c-a c x)^{3/2}}+\frac{16 a^2 c^3 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{35 \sqrt{c-a c x}}+\frac{2}{7} a^2 c^2 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \sqrt{c-a c x}$

[Out]

(64*a^2*c^4*(1 - 1/(a^2*x^2))^(3/2)*x^3)/(105*(c - a*c*x)^(3/2)) + (16*a^2*c^3*(1 - 1/(a^2*x^2))^(3/2)*x^3)/(3
5*Sqrt[c - a*c*x]) + (2*a^2*c^2*(1 - 1/(a^2*x^2))^(3/2)*x^3*Sqrt[c - a*c*x])/7

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Rubi [A]  time = 0.171192, antiderivative size = 137, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 5, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.278, Rules used = {6176, 6181, 89, 78, 37} $\frac{142 \left (\frac{1}{a x}+1\right )^{3/2} (c-a c x)^{5/2}}{105 a^2 x \left (1-\frac{1}{a x}\right )^{5/2}}-\frac{36 \left (\frac{1}{a x}+1\right )^{3/2} (c-a c x)^{5/2}}{35 a \left (1-\frac{1}{a x}\right )^{5/2}}+\frac{2 x \left (\frac{1}{a x}+1\right )^{3/2} (c-a c x)^{5/2}}{7 \left (1-\frac{1}{a x}\right )^{5/2}}$

Warning: Unable to verify antiderivative.

[In]

Int[E^ArcCoth[a*x]*(c - a*c*x)^(5/2),x]

[Out]

(-36*(1 + 1/(a*x))^(3/2)*(c - a*c*x)^(5/2))/(35*a*(1 - 1/(a*x))^(5/2)) + (142*(1 + 1/(a*x))^(3/2)*(c - a*c*x)^
(5/2))/(105*a^2*(1 - 1/(a*x))^(5/2)*x) + (2*(1 + 1/(a*x))^(3/2)*x*(c - a*c*x)^(5/2))/(7*(1 - 1/(a*x))^(5/2))

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 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 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)^{5/2} \, dx &=\frac{(c-a c x)^{5/2} \int e^{\coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right )^{5/2} x^{5/2} \, dx}{\left (1-\frac{1}{a x}\right )^{5/2} x^{5/2}}\\ &=-\frac{\left (\left (\frac{1}{x}\right )^{5/2} (c-a c x)^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^2 \sqrt{1+\frac{x}{a}}}{x^{9/2}} \, dx,x,\frac{1}{x}\right )}{\left (1-\frac{1}{a x}\right )^{5/2}}\\ &=\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x (c-a c x)^{5/2}}{7 \left (1-\frac{1}{a x}\right )^{5/2}}-\frac{\left (2 \left (\frac{1}{x}\right )^{5/2} (c-a c x)^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{9}{a}+\frac{7 x}{2 a^2}\right ) \sqrt{1+\frac{x}{a}}}{x^{7/2}} \, dx,x,\frac{1}{x}\right )}{7 \left (1-\frac{1}{a x}\right )^{5/2}}\\ &=-\frac{36 \left (1+\frac{1}{a x}\right )^{3/2} (c-a c x)^{5/2}}{35 a \left (1-\frac{1}{a x}\right )^{5/2}}+\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x (c-a c x)^{5/2}}{7 \left (1-\frac{1}{a x}\right )^{5/2}}-\frac{\left (71 \left (\frac{1}{x}\right )^{5/2} (c-a c x)^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x}{a}}}{x^{5/2}} \, dx,x,\frac{1}{x}\right )}{35 a^2 \left (1-\frac{1}{a x}\right )^{5/2}}\\ &=-\frac{36 \left (1+\frac{1}{a x}\right )^{3/2} (c-a c x)^{5/2}}{35 a \left (1-\frac{1}{a x}\right )^{5/2}}+\frac{142 \left (1+\frac{1}{a x}\right )^{3/2} (c-a c x)^{5/2}}{105 a^2 \left (1-\frac{1}{a x}\right )^{5/2} x}+\frac{2 \left (1+\frac{1}{a x}\right )^{3/2} x (c-a c x)^{5/2}}{7 \left (1-\frac{1}{a x}\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0369356, size = 67, normalized size = 0.58 $\frac{2 c^2 \sqrt{\frac{1}{a x}+1} (a x+1) \left (15 a^2 x^2-54 a x+71\right ) \sqrt{c-a c x}}{105 a \sqrt{1-\frac{1}{a x}}}$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[a*x]*(c - a*c*x)^(5/2),x]

[Out]

(2*c^2*Sqrt[1 + 1/(a*x)]*(1 + a*x)*Sqrt[c - a*c*x]*(71 - 54*a*x + 15*a^2*x^2))/(105*a*Sqrt[1 - 1/(a*x)])

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Maple [A]  time = 0.043, size = 56, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,ax+2 \right ) \left ( 15\,{a}^{2}{x}^{2}-54\,ax+71 \right ) }{105\, \left ( ax-1 \right ) ^{2}a} \left ( -acx+c \right ) ^{{\frac{5}{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)^(5/2),x)

[Out]

2/105*(a*x+1)*(15*a^2*x^2-54*a*x+71)*(-a*c*x+c)^(5/2)/a/(a*x-1)^2/((a*x-1)/(a*x+1))^(1/2)

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Maxima [A]  time = 1.10341, size = 90, normalized size = 0.78 \begin{align*} \frac{2 \,{\left (15 \, a^{3} \sqrt{-c} c^{2} x^{3} - 39 \, a^{2} \sqrt{-c} c^{2} x^{2} + 17 \, a \sqrt{-c} c^{2} x + 71 \, \sqrt{-c} c^{2}\right )} \sqrt{a x + 1}}{105 \, 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)^(5/2),x, algorithm="maxima")

[Out]

2/105*(15*a^3*sqrt(-c)*c^2*x^3 - 39*a^2*sqrt(-c)*c^2*x^2 + 17*a*sqrt(-c)*c^2*x + 71*sqrt(-c)*c^2)*sqrt(a*x + 1
)/a

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Fricas [A]  time = 1.60649, size = 182, normalized size = 1.58 \begin{align*} \frac{2 \,{\left (15 \, a^{4} c^{2} x^{4} - 24 \, a^{3} c^{2} x^{3} - 22 \, a^{2} c^{2} x^{2} + 88 \, a c^{2} x + 71 \, c^{2}\right )} \sqrt{-a c x + c} \sqrt{\frac{a x - 1}{a x + 1}}}{105 \,{\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)^(5/2),x, algorithm="fricas")

[Out]

2/105*(15*a^4*c^2*x^4 - 24*a^3*c^2*x^3 - 22*a^2*c^2*x^2 + 88*a*c^2*x + 71*c^2)*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)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.18411, size = 134, normalized size = 1.17 \begin{align*} \frac{2 \,{\left (\frac{64 \, \sqrt{2} \sqrt{-c} c^{2}}{\mathrm{sgn}\left (c\right )} + \frac{15 \,{\left (a c x + c\right )}^{3} \sqrt{-a c x - c} - 84 \,{\left (a c x + c\right )}^{2} \sqrt{-a c x - c} c - 140 \,{\left (-a c x - c\right )}^{\frac{3}{2}} c^{2}}{c \mathrm{sgn}\left (-a c x - c\right )}\right )}}{105 \, 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)^(5/2),x, algorithm="giac")

[Out]

2/105*(64*sqrt(2)*sqrt(-c)*c^2/sgn(c) + (15*(a*c*x + c)^3*sqrt(-a*c*x - c) - 84*(a*c*x + c)^2*sqrt(-a*c*x - c)
*c - 140*(-a*c*x - c)^(3/2)*c^2)/(c*sgn(-a*c*x - c)))/a