3.394 \(\int e^{2 \tanh ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx\)

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]

10/3*(-a*c*x+c)^(3/2)/a^3/c-8/5*(-a*c*x+c)^(5/2)/a^3/c^2+2/7*(-a*c*x+c)^(7/2)/a^3/c^3-4*(-a*c*x+c)^(1/2)/a^3

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Rubi [A]  time = 0.16, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {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*ArcTanh[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 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]))))

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]
)

Rubi steps

\begin {align*} \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\\ &=c \int \frac {x^2 (1+a x)}{\sqrt {c-a c x}} \, dx\\ &=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\\ &=-\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*}

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Mathematica [A]  time = 0.05, size = 40, normalized size = 0.50 \[ -\frac {2 \left (15 a^3 x^3+39 a^2 x^2+52 a x+104\right ) \sqrt {c-a c x}}{105 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[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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fricas [A]  time = 0.42, size = 36, normalized size = 0.45 \[ -\frac {2 \, {\left (15 \, a^{3} x^{3} + 39 \, a^{2} x^{2} + 52 \, a x + 104\right )} \sqrt {-a c x + c}}{105 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+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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giac [B]  time = 0.17, size = 142, normalized size = 1.78 \[ -\frac {2 \, {\left (\frac {7 \, {\left (3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} - 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-a c x + c} c^{2}\right )}}{a^{2} c^{2}} + \frac {3 \, {\left (5 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} + 21 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} c - 35 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-a c x + c} c^{3}\right )}}{a^{2} c^{3}}\right )}}{105 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(7*(3*(a*c*x - c)^2*sqrt(-a*c*x + c) - 10*(-a*c*x + c)^(3/2)*c + 15*sqrt(-a*c*x + c)*c^2)/(a^2*c^2) + 3
*(5*(a*c*x - c)^3*sqrt(-a*c*x + c) + 21*(a*c*x - c)^2*sqrt(-a*c*x + c)*c - 35*(-a*c*x + c)^(3/2)*c^2 + 35*sqrt
(-a*c*x + c)*c^3)/(a^2*c^3))/a

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maple [A]  time = 0.03, size = 37, normalized size = 0.46 \[ -\frac {2 \sqrt {-a c x +c}\, \left (15 x^{3} a^{3}+39 a^{2} x^{2}+52 a x +104\right )}{105 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+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 = 0.31, size = 60, normalized size = 0.75 \[ \frac {2 \, {\left (15 \, {\left (-a c x + c\right )}^{\frac {7}{2}} - 84 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c + 175 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} - 210 \, \sqrt {-a c x + c} c^{3}\right )}}{105 \, a^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+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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mupad [B]  time = 0.06, size = 66, normalized size = 0.82 \[ \frac {10\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a^3\,c}-\frac {4\,\sqrt {c-a\,c\,x}}{a^3}-\frac {8\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a^3\,c^2}+\frac {2\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a^3\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(10*(c - a*c*x)^(3/2))/(3*a^3*c) - (4*(c - a*c*x)^(1/2))/a^3 - (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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sympy [A]  time = 8.68, size = 68, normalized size = 0.85 \[ - \frac {2 \left (2 c^{3} \sqrt {- a c x + c} - \frac {5 c^{2} \left (- a c x + c\right )^{\frac {3}{2}}}{3} + \frac {4 c \left (- a c x + c\right )^{\frac {5}{2}}}{5} - \frac {\left (- a c x + c\right )^{\frac {7}{2}}}{7}\right )}{a^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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