3.1237 \(\int e^{-2 \tanh ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=112 \[ \frac {2 x^2 \sqrt {c-a^2 c x^2}}{3 a}-\frac {1}{4} x^3 \sqrt {c-a^2 c x^2}+\frac {(32-21 a x) \sqrt {c-a^2 c x^2}}{24 a^3}+\frac {7 \sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{8 a^3} \]

[Out]

7/8*arctan(a*x*c^(1/2)/(-a^2*c*x^2+c)^(1/2))*c^(1/2)/a^3+2/3*x^2*(-a^2*c*x^2+c)^(1/2)/a-1/4*x^3*(-a^2*c*x^2+c)
^(1/2)+1/24*(-21*a*x+32)*(-a^2*c*x^2+c)^(1/2)/a^3

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Rubi [A]  time = 0.28, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6152, 1809, 833, 780, 217, 203} \[ -\frac {1}{4} x^3 \sqrt {c-a^2 c x^2}+\frac {2 x^2 \sqrt {c-a^2 c x^2}}{3 a}+\frac {(32-21 a x) \sqrt {c-a^2 c x^2}}{24 a^3}+\frac {7 \sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^2*Sqrt[c - a^2*c*x^2])/(3*a) - (x^3*Sqrt[c - a^2*c*x^2])/4 + ((32 - 21*a*x)*Sqrt[c - a^2*c*x^2])/(24*a^3)
 + (7*Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(8*a^3)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
 + 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
 Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
 - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
 && PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 6152

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/c^(n/2), Int[(x^m
*(c + d*x^2)^(p + n/2))/(1 - a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p
] || GtQ[c, 0]) && ILtQ[n/2, 0]

Rubi steps

\begin {align*} \int e^{-2 \tanh ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx &=c \int \frac {x^2 (1-a x)^2}{\sqrt {c-a^2 c x^2}} \, dx\\ &=-\frac {1}{4} x^3 \sqrt {c-a^2 c x^2}-\frac {\int \frac {x^2 \left (-7 a^2 c+8 a^3 c x\right )}{\sqrt {c-a^2 c x^2}} \, dx}{4 a^2}\\ &=\frac {2 x^2 \sqrt {c-a^2 c x^2}}{3 a}-\frac {1}{4} x^3 \sqrt {c-a^2 c x^2}+\frac {\int \frac {x \left (-16 a^3 c^2+21 a^4 c^2 x\right )}{\sqrt {c-a^2 c x^2}} \, dx}{12 a^4 c}\\ &=\frac {2 x^2 \sqrt {c-a^2 c x^2}}{3 a}-\frac {1}{4} x^3 \sqrt {c-a^2 c x^2}+\frac {(32-21 a x) \sqrt {c-a^2 c x^2}}{24 a^3}+\frac {(7 c) \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx}{8 a^2}\\ &=\frac {2 x^2 \sqrt {c-a^2 c x^2}}{3 a}-\frac {1}{4} x^3 \sqrt {c-a^2 c x^2}+\frac {(32-21 a x) \sqrt {c-a^2 c x^2}}{24 a^3}+\frac {(7 c) \operatorname {Subst}\left (\int \frac {1}{1+a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c-a^2 c x^2}}\right )}{8 a^2}\\ &=\frac {2 x^2 \sqrt {c-a^2 c x^2}}{3 a}-\frac {1}{4} x^3 \sqrt {c-a^2 c x^2}+\frac {(32-21 a x) \sqrt {c-a^2 c x^2}}{24 a^3}+\frac {7 \sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{8 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 88, normalized size = 0.79 \[ \frac {\left (-6 a^3 x^3+16 a^2 x^2-21 a x+32\right ) \sqrt {c-a^2 c x^2}-21 \sqrt {c} \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )}{24 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - a^2*c*x^2]*(32 - 21*a*x + 16*a^2*x^2 - 6*a^3*x^3) - 21*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqr
t[c]*(-1 + a^2*x^2))])/(24*a^3)

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fricas [A]  time = 0.73, size = 168, normalized size = 1.50 \[ \left [-\frac {2 \, {\left (6 \, a^{3} x^{3} - 16 \, a^{2} x^{2} + 21 \, a x - 32\right )} \sqrt {-a^{2} c x^{2} + c} - 21 \, \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right )}{48 \, a^{3}}, -\frac {{\left (6 \, a^{3} x^{3} - 16 \, a^{2} x^{2} + 21 \, a x - 32\right )} \sqrt {-a^{2} c x^{2} + c} + 21 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right )}{24 \, a^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(2*(6*a^3*x^3 - 16*a^2*x^2 + 21*a*x - 32)*sqrt(-a^2*c*x^2 + c) - 21*sqrt(-c)*log(2*a^2*c*x^2 + 2*sqrt(-
a^2*c*x^2 + c)*a*sqrt(-c)*x - c))/a^3, -1/24*((6*a^3*x^3 - 16*a^2*x^2 + 21*a*x - 32)*sqrt(-a^2*c*x^2 + c) + 21
*sqrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)))/a^3]

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giac [B]  time = 0.41, size = 221, normalized size = 1.97 \[ -\frac {{\left (336 \, a^{5} \sqrt {c} \arctan \left (\frac {\sqrt {-c + \frac {2 \, c}{a x + 1}}}{\sqrt {c}}\right ) \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) + \frac {{\left (75 \, a^{5} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{3} c \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) - 83 \, a^{5} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{2} c^{2} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) - 21 \, a^{5} c^{4} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) - 77 \, a^{5} c^{3} {\left (-c + \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a)\right )} {\left (a x + 1\right )}^{4}}{c^{4}}\right )} {\left | a \right |}}{192 \, a^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(336*a^5*sqrt(c)*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(c))*sgn(1/(a*x + 1))*sgn(a) + (75*a^5*(c - 2*c/(a
*x + 1))^3*c*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) - 83*a^5*(c - 2*c/(a*x + 1))^2*c^2*sqrt(-c + 2*c
/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) - 21*a^5*c^4*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) - 77*a^5*c^3
*(-c + 2*c/(a*x + 1))^(3/2)*sgn(1/(a*x + 1))*sgn(a))*(a*x + 1)^4/c^4)*abs(a)/a^9

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maple [A]  time = 0.04, size = 178, normalized size = 1.59 \[ \frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4 a^{2} c}-\frac {9 x \sqrt {-a^{2} c \,x^{2}+c}}{8 a^{2}}-\frac {9 c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{8 a^{2} \sqrt {a^{2} c}}-\frac {2 \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3 a^{3} c}+\frac {2 \sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 a c \left (x +\frac {1}{a}\right )}}{a^{3}}+\frac {2 c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 a c \left (x +\frac {1}{a}\right )}}\right )}{a^{2} \sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(-a^2*c*x^2+c)^(3/2)/a^2/c-9/8/a^2*x*(-a^2*c*x^2+c)^(1/2)-9/8/a^2*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x
/(-a^2*c*x^2+c)^(1/2))-2/3/a^3*(-a^2*c*x^2+c)^(3/2)/c+2/a^3*(-(x+1/a)^2*a^2*c+2*a*c*(x+1/a))^(1/2)+2/a^2*c/(a^
2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-(x+1/a)^2*a^2*c+2*a*c*(x+1/a))^(1/2))

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maxima [A]  time = 0.43, size = 93, normalized size = 0.83 \[ -\frac {9 \, \sqrt {-a^{2} c x^{2} + c} x}{8 \, a^{2}} + \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x}{4 \, a^{2} c} + \frac {7 \, \sqrt {c} \arcsin \left (a x\right )}{8 \, a^{3}} + \frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{3}} - \frac {2 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{3 \, a^{3} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/8*sqrt(-a^2*c*x^2 + c)*x/a^2 + 1/4*(-a^2*c*x^2 + c)^(3/2)*x/(a^2*c) + 7/8*sqrt(c)*arcsin(a*x)/a^3 + 2*sqrt(
-a^2*c*x^2 + c)/a^3 - 2/3*(-a^2*c*x^2 + c)^(3/2)/(a^3*c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {x^2\,\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )}{{\left (a\,x+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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