3.245 \(\int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx\)

Optimal. Leaf size=89 \[ \frac{2 x \left (\frac{1}{a x}+1\right )^{5/2} (c-a c x)^{5/2}}{7 \left (1-\frac{1}{a x}\right )^{5/2}}-\frac{18 \left (\frac{1}{a x}+1\right )^{5/2} (c-a c x)^{5/2}}{35 a \left (1-\frac{1}{a x}\right )^{5/2}} \]

[Out]

(-18*(1 + 1/(a*x))^(5/2)*(c - a*c*x)^(5/2))/(35*a*(1 - 1/(a*x))^(5/2)) + (2*(1 + 1/(a*x))^(5/2)*x*(c - a*c*x)^
(5/2))/(7*(1 - 1/(a*x))^(5/2))

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Rubi [A]  time = 0.158009, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6176, 6181, 78, 37} \[ \frac{2 x \left (\frac{1}{a x}+1\right )^{5/2} (c-a c x)^{5/2}}{7 \left (1-\frac{1}{a x}\right )^{5/2}}-\frac{18 \left (\frac{1}{a x}+1\right )^{5/2} (c-a c x)^{5/2}}{35 a \left (1-\frac{1}{a x}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcCoth[a*x])*(c - a*c*x)^(5/2),x]

[Out]

(-18*(1 + 1/(a*x))^(5/2)*(c - a*c*x)^(5/2))/(35*a*(1 - 1/(a*x))^(5/2)) + (2*(1 + 1/(a*x))^(5/2)*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 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^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx &=\frac{(c-a c x)^{5/2} \int e^{3 \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 ) \left (1+\frac{x}{a}\right )^{3/2}}{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 )^{5/2} x (c-a c x)^{5/2}}{7 \left (1-\frac{1}{a x}\right )^{5/2}}+\frac{\left (9 \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 )^{3/2}}{x^{7/2}} \, dx,x,\frac{1}{x}\right )}{7 a \left (1-\frac{1}{a x}\right )^{5/2}}\\ &=-\frac{18 \left (1+\frac{1}{a x}\right )^{5/2} (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 )^{5/2} x (c-a c x)^{5/2}}{7 \left (1-\frac{1}{a x}\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0388098, size = 59, normalized size = 0.66 \[ \frac{2 \sqrt{\frac{1}{a x}+1} (5 a x-9) \sqrt{c-a c x} (a c x+c)^2}{35 a \sqrt{1-\frac{1}{a x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - a*c*x)^(5/2),x]

[Out]

(2*Sqrt[1 + 1/(a*x)]*(-9 + 5*a*x)*Sqrt[c - a*c*x]*(c + a*c*x)^2)/(35*a*Sqrt[1 - 1/(a*x)])

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Maple [A]  time = 0.117, size = 48, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,ax+2 \right ) \left ( 5\,ax-9 \right ) }{35\, \left ( ax-1 \right ) a} \left ( -acx+c \right ) ^{{\frac{5}{2}}} \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(a*x+1)*(5*a*x-9)*(-a*c*x+c)^(5/2)/a/(a*x-1)/((a*x-1)/(a*x+1))^(3/2)

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Maxima [A]  time = 1.08365, size = 100, normalized size = 1.12 \begin{align*} \frac{2 \,{\left (5 \, a^{3} \sqrt{-c} c^{2} x^{3} - 9 \, a^{2} \sqrt{-c} c^{2} x^{2} - 5 \, a \sqrt{-c} c^{2} x + 9 \, \sqrt{-c} c^{2}\right )}{\left (a x + 1\right )}^{\frac{3}{2}}}{35 \,{\left (a x - 1\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="maxima")

[Out]

2/35*(5*a^3*sqrt(-c)*c^2*x^3 - 9*a^2*sqrt(-c)*c^2*x^2 - 5*a*sqrt(-c)*c^2*x + 9*sqrt(-c)*c^2)*(a*x + 1)^(3/2)/(
(a*x - 1)*a)

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Fricas [A]  time = 1.50827, size = 177, normalized size = 1.99 \begin{align*} \frac{2 \,{\left (5 \, a^{4} c^{2} x^{4} + 6 \, a^{3} c^{2} x^{3} - 12 \, a^{2} c^{2} x^{2} - 22 \, a c^{2} x - 9 \, c^{2}\right )} \sqrt{-a c x + c} \sqrt{\frac{a x - 1}{a x + 1}}}{35 \,{\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))^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/35*(5*a^4*c^2*x^4 + 6*a^3*c^2*x^3 - 12*a^2*c^2*x^2 - 22*a*c^2*x - 9*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))**(3/2)*(-a*c*x+c)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.18758, size = 188, normalized size = 2.11 \begin{align*} -\frac{2 \,{\left (\frac{48 \, \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} + 14 \,{\left (3 \,{\left (a c x + c\right )}^{2} \sqrt{-a c x - c} + 10 \,{\left (-a c x - c\right )}^{\frac{3}{2}} c\right )} c}{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))^(3/2)*(-a*c*x+c)^(5/2),x, algorithm="giac")

[Out]

-2/105*(48*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 + 14*(3*(a*c*x + c)^2*sqrt(-a*c*x - c) + 10*(-a*c*x - c)^(3/2)*c)*c)/(c*sgn(-
a*c*x - c)))/a