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

Optimal. Leaf size=40 \[ \frac{4 (c-a c x)^{3/2}}{3 a}-\frac{2 (c-a c x)^{5/2}}{5 a c} \]

[Out]

(4*(c - a*c*x)^(3/2))/(3*a) - (2*(c - a*c*x)^(5/2))/(5*a*c)

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Rubi [A]  time = 0.0891647, 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} \[ \frac{4 (c-a c x)^{3/2}}{3 a}-\frac{2 (c-a c x)^{5/2}}{5 a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0301593, size = 30, normalized size = 0.75 \[ -\frac{2 c (a x-1) (3 a x+7) \sqrt{c-a c x}}{15 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 21, normalized size = 0.5 \begin{align*}{\frac{6\,ax+14}{15\,a} \left ( -acx+c \right ) ^{{\frac{3}{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)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.01969, size = 43, normalized size = 1.08 \begin{align*} -\frac{2 \,{\left (3 \,{\left (-a c x + c\right )}^{\frac{5}{2}} - 10 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c\right )}}{15 \, 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)^(3/2),x, algorithm="maxima")

[Out]

-2/15*(3*(-a*c*x + c)^(5/2) - 10*(-a*c*x + c)^(3/2)*c)/(a*c)

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Fricas [A]  time = 1.50783, size = 76, normalized size = 1.9 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} c x^{2} + 4 \, a c x - 7 \, c\right )} \sqrt{-a c x + c}}{15 \, 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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 10.8728, size = 61, normalized size = 1.52 \begin{align*} \begin{cases} - \frac{c \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 \left (- a c x + c\right )^{\frac{3}{2}}}{3} + \frac{\left (- a c x + c\right )^{\frac{5}{2}}}{5}\right )}{c}}{a} & \text{for}\: a \neq 0 \\- c^{\frac{3}{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)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.10256, size = 76, normalized size = 1.9 \begin{align*} \frac{2 \,{\left (5 \,{\left (-a c x + c\right )}^{\frac{3}{2}} - \frac{3 \,{\left (a c x - c\right )}^{2} \sqrt{-a c x + c} - 5 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c}{c}\right )}}{15 \, 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)^(3/2),x, algorithm="giac")

[Out]

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