### 3.236 $$\int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx$$

Optimal. Leaf size=40 $\frac{4 (c-a c x)^{5/2}}{5 a}-\frac{2 (c-a c x)^{7/2}}{7 a c}$

[Out]

(4*(c - a*c*x)^(5/2))/(5*a) - (2*(c - a*c*x)^(7/2))/(7*a*c)

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Rubi [A]  time = 0.0876669, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {6167, 6130, 21, 43} $\frac{4 (c-a c x)^{5/2}}{5 a}-\frac{2 (c-a c x)^{7/2}}{7 a c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*(c - a*c*x)^(5/2))/(5*a) - (2*(c - a*c*x)^(7/2))/(7*a*c)

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx &=-\int e^{2 \tanh ^{-1}(a x)} (c-a c x)^{5/2} \, dx\\ &=-\int \frac{(1+a x) (c-a c x)^{5/2}}{1-a x} \, dx\\ &=-\left (c \int (1+a x) (c-a c x)^{3/2} \, dx\right )\\ &=-\left (c \int \left (2 (c-a c x)^{3/2}-\frac{(c-a c x)^{5/2}}{c}\right ) \, dx\right )\\ &=\frac{4 (c-a c x)^{5/2}}{5 a}-\frac{2 (c-a c x)^{7/2}}{7 a c}\\ \end{align*}

Mathematica [A]  time = 0.039523, size = 34, normalized size = 0.85 $\frac{2 c^2 (a x-1)^2 (5 a x+9) \sqrt{c-a c x}}{35 a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 21, normalized size = 0.5 \begin{align*}{\frac{10\,ax+18}{35\,a} \left ( -acx+c \right ) ^{{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00252, size = 43, normalized size = 1.08 \begin{align*} -\frac{2 \,{\left (5 \,{\left (-a c x + c\right )}^{\frac{7}{2}} - 14 \,{\left (-a c x + c\right )}^{\frac{5}{2}} c\right )}}{35 \, a c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(5*(-a*c*x + c)^(7/2) - 14*(-a*c*x + c)^(5/2)*c)/(a*c)

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Fricas [A]  time = 1.60886, size = 103, normalized size = 2.58 \begin{align*} \frac{2 \,{\left (5 \, a^{3} c^{2} x^{3} - a^{2} c^{2} x^{2} - 13 \, a c^{2} x + 9 \, c^{2}\right )} \sqrt{-a c x + c}}{35 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*a^3*c^2*x^3 - a^2*c^2*x^2 - 13*a*c^2*x + 9*c^2)*sqrt(-a*c*x + c)/a

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Sympy [A]  time = 12.5798, size = 80, normalized size = 2. \begin{align*} \begin{cases} \frac{- c^{2} \left (\begin{cases} 0 & \text{for}\: c = 0 \\- \frac{2 \left (- a c x + c\right )^{\frac{3}{2}}}{3 c} & \text{otherwise} \end{cases}\right ) - \frac{2 \left (\frac{c^{2} \left (- a c x + c\right )^{\frac{3}{2}}}{3} - \frac{2 c \left (- a c x + c\right )^{\frac{5}{2}}}{5} + \frac{\left (- a c x + c\right )^{\frac{7}{2}}}{7}\right )}{c}}{a} & \text{for}\: a \neq 0 \\- c^{\frac{5}{2}} x & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)*(-a*c*x+c)**(5/2),x)

[Out]

Piecewise(((-c**2*Piecewise((0, Eq(c, 0)), (-2*(-a*c*x + c)**(3/2)/(3*c), True)) - 2*(c**2*(-a*c*x + c)**(3/2)
/3 - 2*c*(-a*c*x + c)**(5/2)/5 + (-a*c*x + c)**(7/2)/7)/c)/a, Ne(a, 0)), (-c**(5/2)*x, True))

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Giac [B]  time = 1.12242, size = 108, normalized size = 2.7 \begin{align*} \frac{2 \,{\left (35 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c + \frac{15 \,{\left (a c x - c\right )}^{3} \sqrt{-a c x + c} + 42 \,{\left (a c x - c\right )}^{2} \sqrt{-a c x + c} c - 35 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c^{2}}{c}\right )}}{105 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(35*(-a*c*x + c)^(3/2)*c + (15*(a*c*x - c)^3*sqrt(-a*c*x + c) + 42*(a*c*x - c)^2*sqrt(-a*c*x + c)*c - 35
*(-a*c*x + c)^(3/2)*c^2)/c)/a