### 3.302 $$\int e^{2 \coth ^{-1}(a x)} x^2 \sqrt{c-a c x} \, dx$$

Optimal. Leaf size=80 $-\frac{2 (c-a c x)^{7/2}}{7 a^3 c^3}+\frac{8 (c-a c x)^{5/2}}{5 a^3 c^2}-\frac{10 (c-a c x)^{3/2}}{3 a^3 c}+\frac{4 \sqrt{c-a c x}}{a^3}$

[Out]

(4*Sqrt[c - a*c*x])/a^3 - (10*(c - a*c*x)^(3/2))/(3*a^3*c) + (8*(c - a*c*x)^(5/2))/(5*a^3*c^2) - (2*(c - a*c*x
)^(7/2))/(7*a^3*c^3)

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Rubi [A]  time = 0.224154, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.174, Rules used = {6167, 6130, 21, 77} $-\frac{2 (c-a c x)^{7/2}}{7 a^3 c^3}+\frac{8 (c-a c x)^{5/2}}{5 a^3 c^2}-\frac{10 (c-a c x)^{3/2}}{3 a^3 c}+\frac{4 \sqrt{c-a c x}}{a^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*Sqrt[c - a*c*x])/a^3 - (10*(c - a*c*x)^(3/2))/(3*a^3*c) + (8*(c - a*c*x)^(5/2))/(5*a^3*c^2) - (2*(c - a*c*x
)^(7/2))/(7*a^3*c^3)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int e^{2 \coth ^{-1}(a x)} x^2 \sqrt{c-a c x} \, dx &=-\int e^{2 \tanh ^{-1}(a x)} x^2 \sqrt{c-a c x} \, dx\\ &=-\int \frac{x^2 (1+a x) \sqrt{c-a c x}}{1-a x} \, dx\\ &=-\left (c \int \frac{x^2 (1+a x)}{\sqrt{c-a c x}} \, dx\right )\\ &=-\left (c \int \left (\frac{2}{a^2 \sqrt{c-a c x}}-\frac{5 \sqrt{c-a c x}}{a^2 c}+\frac{4 (c-a c x)^{3/2}}{a^2 c^2}-\frac{(c-a c x)^{5/2}}{a^2 c^3}\right ) \, dx\right )\\ &=\frac{4 \sqrt{c-a c x}}{a^3}-\frac{10 (c-a c x)^{3/2}}{3 a^3 c}+\frac{8 (c-a c x)^{5/2}}{5 a^3 c^2}-\frac{2 (c-a c x)^{7/2}}{7 a^3 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0533907, size = 40, normalized size = 0.5 $\frac{2 \left (15 a^3 x^3+39 a^2 x^2+52 a x+104\right ) \sqrt{c-a c x}}{105 a^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[c - a*c*x]*(104 + 52*a*x + 39*a^2*x^2 + 15*a^3*x^3))/(105*a^3)

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Maple [A]  time = 0.044, size = 37, normalized size = 0.5 \begin{align*}{\frac{30\,{x}^{3}{a}^{3}+78\,{a}^{2}{x}^{2}+104\,ax+208}{105\,{a}^{3}}\sqrt{-acx+c}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(-a*c*x+c)^(1/2)*(15*a^3*x^3+39*a^2*x^2+52*a*x+104)/a^3

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Maxima [A]  time = 1.02344, size = 81, normalized size = 1.01 \begin{align*} -\frac{2 \,{\left (15 \,{\left (-a c x + c\right )}^{\frac{7}{2}} - 84 \,{\left (-a c x + c\right )}^{\frac{5}{2}} c + 175 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c^{2} - 210 \, \sqrt{-a c x + c} c^{3}\right )}}{105 \, a^{3} c^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*(-a*c*x + c)^(7/2) - 84*(-a*c*x + c)^(5/2)*c + 175*(-a*c*x + c)^(3/2)*c^2 - 210*sqrt(-a*c*x + c)*c^
3)/(a^3*c^3)

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Fricas [A]  time = 1.53695, size = 93, normalized size = 1.16 \begin{align*} \frac{2 \,{\left (15 \, a^{3} x^{3} + 39 \, a^{2} x^{2} + 52 \, a x + 104\right )} \sqrt{-a c x + c}}{105 \, a^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^3*x^3 + 39*a^2*x^2 + 52*a*x + 104)*sqrt(-a*c*x + c)/a^3

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Sympy [A]  time = 11.4617, size = 68, normalized size = 0.85 \begin{align*} - \frac{2 \left (- 2 c^{3} \sqrt{- a c x + c} + \frac{5 c^{2} \left (- a c x + c\right )^{\frac{3}{2}}}{3} - \frac{4 c \left (- a c x + c\right )^{\frac{5}{2}}}{5} + \frac{\left (- a c x + c\right )^{\frac{7}{2}}}{7}\right )}{a^{3} c^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-2*c**3*sqrt(-a*c*x + c) + 5*c**2*(-a*c*x + c)**(3/2)/3 - 4*c*(-a*c*x + c)**(5/2)/5 + (-a*c*x + c)**(7/2)/
7)/(a**3*c**3)

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Giac [A]  time = 1.155, size = 108, normalized size = 1.35 \begin{align*} \frac{2 \,{\left (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 - 175 \,{\left (-a c x + c\right )}^{\frac{3}{2}} c^{2} + 210 \, \sqrt{-a c x + c} c^{3}\right )}}{105 \, a^{3} c^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*(a*c*x - c)^3*sqrt(-a*c*x + c) + 84*(a*c*x - c)^2*sqrt(-a*c*x + c)*c - 175*(-a*c*x + c)^(3/2)*c^2 +
210*sqrt(-a*c*x + c)*c^3)/(a^3*c^3)