3.429 \(\int e^{-3 \tanh ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx\)

Optimal. Leaf size=197 \[ \frac {2 c^2 (1-a x)^{3/2} (a x+1)^{7/2}}{7 a^3 (c-a c x)^{3/2}}-\frac {12 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^3 (c-a c x)^{3/2}}+\frac {26 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^3 (c-a c x)^{3/2}}-\frac {24 c^2 (1-a x)^{3/2} \sqrt {a x+1}}{a^3 (c-a c x)^{3/2}}-\frac {8 c^2 (1-a x)^{3/2}}{a^3 \sqrt {a x+1} (c-a c x)^{3/2}} \]

[Out]

26/3*c^2*(-a*x+1)^(3/2)*(a*x+1)^(3/2)/a^3/(-a*c*x+c)^(3/2)-12/5*c^2*(-a*x+1)^(3/2)*(a*x+1)^(5/2)/a^3/(-a*c*x+c
)^(3/2)+2/7*c^2*(-a*x+1)^(3/2)*(a*x+1)^(7/2)/a^3/(-a*c*x+c)^(3/2)-8*c^2*(-a*x+1)^(3/2)/a^3/(-a*c*x+c)^(3/2)/(a
*x+1)^(1/2)-24*c^2*(-a*x+1)^(3/2)*(a*x+1)^(1/2)/a^3/(-a*c*x+c)^(3/2)

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Rubi [A]  time = 0.15, antiderivative size = 197, 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, 23, 88} \[ \frac {2 c^2 (1-a x)^{3/2} (a x+1)^{7/2}}{7 a^3 (c-a c x)^{3/2}}-\frac {12 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^3 (c-a c x)^{3/2}}+\frac {26 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^3 (c-a c x)^{3/2}}-\frac {24 c^2 (1-a x)^{3/2} \sqrt {a x+1}}{a^3 (c-a c x)^{3/2}}-\frac {8 c^2 (1-a x)^{3/2}}{a^3 \sqrt {a x+1} (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*c^2*(1 - a*x)^(3/2))/(a^3*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (24*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(a^3*(
c - a*c*x)^(3/2)) + (26*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2))/(3*a^3*(c - a*c*x)^(3/2)) - (12*c^2*(1 - a*x)^(3/
2)*(1 + a*x)^(5/2))/(5*a^3*(c - a*c*x)^(3/2)) + (2*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(7/2))/(7*a^3*(c - a*c*x)^(3/
2))

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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^{-3 \tanh ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx &=\int \frac {x^2 (1-a x)^{3/2} \sqrt {c-a c x}}{(1+a x)^{3/2}} \, dx\\ &=\frac {(1-a x)^{3/2} \int \frac {x^2 (c-a c x)^2}{(1+a x)^{3/2}} \, dx}{(c-a c x)^{3/2}}\\ &=\frac {(1-a x)^{3/2} \int \left (\frac {4 c^2}{a^2 (1+a x)^{3/2}}-\frac {12 c^2}{a^2 \sqrt {1+a x}}+\frac {13 c^2 \sqrt {1+a x}}{a^2}-\frac {6 c^2 (1+a x)^{3/2}}{a^2}+\frac {c^2 (1+a x)^{5/2}}{a^2}\right ) \, dx}{(c-a c x)^{3/2}}\\ &=-\frac {8 c^2 (1-a x)^{3/2}}{a^3 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {24 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{a^3 (c-a c x)^{3/2}}+\frac {26 c^2 (1-a x)^{3/2} (1+a x)^{3/2}}{3 a^3 (c-a c x)^{3/2}}-\frac {12 c^2 (1-a x)^{3/2} (1+a x)^{5/2}}{5 a^3 (c-a c x)^{3/2}}+\frac {2 c^2 (1-a x)^{3/2} (1+a x)^{7/2}}{7 a^3 (c-a c x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 68, normalized size = 0.35 \[ \frac {2 c \sqrt {1-a x} \left (15 a^4 x^4-66 a^3 x^3+167 a^2 x^2-668 a x-1336\right )}{105 a^3 \sqrt {a x+1} \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[1 - a*x]*(-1336 - 668*a*x + 167*a^2*x^2 - 66*a^3*x^3 + 15*a^4*x^4))/(105*a^3*Sqrt[1 + a*x]*Sqrt[c -
a*c*x])

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fricas [A]  time = 0.50, size = 68, normalized size = 0.35 \[ -\frac {2 \, {\left (15 \, a^{4} x^{4} - 66 \, a^{3} x^{3} + 167 \, a^{2} x^{2} - 668 \, a x - 1336\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{105 \, {\left (a^{5} x^{2} - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

-2/105*(15*a^4*x^4 - 66*a^3*x^3 + 167*a^2*x^2 - 668*a*x - 1336)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^5*x^2 -
a^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.03, size = 71, normalized size = 0.36 \[ \frac {2 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \sqrt {-a c x +c}\, \left (15 x^{4} a^{4}-66 x^{3} a^{3}+167 a^{2} x^{2}-668 a x -1336\right )}{105 \left (a x +1\right )^{2} \left (a x -1\right )^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

2/105*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(1/2)*(15*a^4*x^4-66*a^3*x^3+167*a^2*x^2-668*a*x-1336)/(a*x+1)^2/(a*x-1)^2
/a^3

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maxima [A]  time = 0.37, size = 75, normalized size = 0.38 \[ \frac {2 \, {\left (15 \, a^{4} \sqrt {c} x^{4} - 66 \, a^{3} \sqrt {c} x^{3} + 167 \, a^{2} \sqrt {c} x^{2} - 668 \, a \sqrt {c} x - 1336 \, \sqrt {c}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{105 \, {\left (a^{5} x^{2} - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

2/105*(15*a^4*sqrt(c)*x^4 - 66*a^3*sqrt(c)*x^3 + 167*a^2*sqrt(c)*x^2 - 668*a*sqrt(c)*x - 1336*sqrt(c))*sqrt(a*
x + 1)*(a*x - 1)/(a^5*x^2 - a^3)

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mupad [B]  time = 1.04, size = 107, normalized size = 0.54 \[ \frac {1888\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{105\,a^3\,\left (a\,x-1\right )}-\frac {2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}\,\left (15\,a^2\,x^2-66\,a\,x+182\right )}{105\,a^3}-\frac {4\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{a^3\,\left (a\,x+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(1 - a^2*x^2)^(3/2)*(c - a*c*x)^(1/2))/(a*x + 1)^3,x)

[Out]

(1888*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2))/(105*a^3*(a*x - 1)) - (2*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2)*(1
5*a^2*x^2 - 66*a*x + 182))/(105*a^3) - (4*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2))/(a^3*(a*x + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {- c \left (a x - 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-a*c*x+c)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)

[Out]

Integral(x**2*sqrt(-c*(a*x - 1))*(-(a*x - 1)*(a*x + 1))**(3/2)/(a*x + 1)**3, x)

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