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

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

[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 = 0.0472998, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6130, 21, 43} \[ \frac{2 (c-a c x)^{7/2}}{7 a c}-\frac{4 (c-a c x)^{5/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[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 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 \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\\ &=c \int (1+a x) (c-a c x)^{3/2} \, dx\\ &=c \int \left (2 (c-a c x)^{3/2}-\frac{(c-a c x)^{5/2}}{c}\right ) \, dx\\ &=-\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 = 0.0366126, 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*ArcTanh[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 = 0.028, 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)^2/(-a^2*x^2+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 = 0.947635, 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((a*x+1)^2/(-a^2*x^2+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 = 2.22103, size = 104, normalized size = 2.6 \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((a*x+1)^2/(-a^2*x^2+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 = 12.5871, size = 76, normalized size = 1.9 \begin{align*} c^{2} \left (\begin{cases} \sqrt{c} x & \text{for}\: a = 0 \\0 & \text{for}\: c = 0 \\- \frac{2 \left (- a c x + c\right )^{\frac{3}{2}}}{3 a 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 )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*Piecewise((sqrt(c)*x, Eq(a, 0)), (0, Eq(c, 0)), (-2*(-a*c*x + c)**(3/2)/(3*a*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)/(a*c)

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Giac [B]  time = 1.19518, 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((a*x+1)^2/(-a^2*x^2+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 - 3
5*(-a*c*x + c)^(3/2)*c^2)/c)/a