3.844 \(\int e^{-\tanh ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=130 \[ \frac {\left (2 a^2+2 a+1\right ) \sqrt {a+b x+1} \sqrt {-a-b x+1}}{2 b^3}+\frac {\left (2 a^2+2 a+1\right ) \sin ^{-1}(a+b x)}{2 b^3}+\frac {(4 a+1) \sqrt {a+b x+1} (-a-b x+1)^{3/2}}{6 b^3}-\frac {x \sqrt {a+b x+1} (-a-b x+1)^{3/2}}{3 b^2} \]

[Out]

1/2*(2*a^2+2*a+1)*arcsin(b*x+a)/b^3+1/6*(1+4*a)*(-b*x-a+1)^(3/2)*(b*x+a+1)^(1/2)/b^3-1/3*x*(-b*x-a+1)^(3/2)*(b
*x+a+1)^(1/2)/b^2+1/2*(2*a^2+2*a+1)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/b^3

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Rubi [A]  time = 0.15, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6163, 90, 80, 50, 53, 619, 216} \[ \frac {\left (2 a^2+2 a+1\right ) \sqrt {a+b x+1} \sqrt {-a-b x+1}}{2 b^3}+\frac {\left (2 a^2+2 a+1\right ) \sin ^{-1}(a+b x)}{2 b^3}-\frac {x \sqrt {a+b x+1} (-a-b x+1)^{3/2}}{3 b^2}+\frac {(4 a+1) \sqrt {a+b x+1} (-a-b x+1)^{3/2}}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/E^ArcTanh[a + b*x],x]

[Out]

((1 + 2*a + 2*a^2)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(2*b^3) + ((1 + 4*a)*(1 - a - b*x)^(3/2)*Sqrt[1 + a +
b*x])/(6*b^3) - (x*(1 - a - b*x)^(3/2)*Sqrt[1 + a + b*x])/(3*b^2) + ((1 + 2*a + 2*a^2)*ArcSin[a + b*x])/(2*b^3
)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 53

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

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 216

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

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 6163

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1
+ a*c + b*c*x)^(n/2))/(1 - a*c - b*c*x)^(n/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rubi steps

\begin {align*} \int e^{-\tanh ^{-1}(a+b x)} x^2 \, dx &=\int \frac {x^2 \sqrt {1-a-b x}}{\sqrt {1+a+b x}} \, dx\\ &=-\frac {x (1-a-b x)^{3/2} \sqrt {1+a+b x}}{3 b^2}-\frac {\int \frac {\sqrt {1-a-b x} \left (-1+a^2+(1+4 a) b x\right )}{\sqrt {1+a+b x}} \, dx}{3 b^2}\\ &=\frac {(1+4 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{6 b^3}-\frac {x (1-a-b x)^{3/2} \sqrt {1+a+b x}}{3 b^2}+\frac {\left (1+2 a+2 a^2\right ) \int \frac {\sqrt {1-a-b x}}{\sqrt {1+a+b x}} \, dx}{2 b^2}\\ &=\frac {\left (1+2 a+2 a^2\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^3}+\frac {(1+4 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{6 b^3}-\frac {x (1-a-b x)^{3/2} \sqrt {1+a+b x}}{3 b^2}+\frac {\left (1+2 a+2 a^2\right ) \int \frac {1}{\sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{2 b^2}\\ &=\frac {\left (1+2 a+2 a^2\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^3}+\frac {(1+4 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{6 b^3}-\frac {x (1-a-b x)^{3/2} \sqrt {1+a+b x}}{3 b^2}+\frac {\left (1+2 a+2 a^2\right ) \int \frac {1}{\sqrt {(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{2 b^2}\\ &=\frac {\left (1+2 a+2 a^2\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^3}+\frac {(1+4 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{6 b^3}-\frac {x (1-a-b x)^{3/2} \sqrt {1+a+b x}}{3 b^2}-\frac {\left (1+2 a+2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{4 b^2}}} \, dx,x,-2 a b-2 b^2 x\right )}{4 b^4}\\ &=\frac {\left (1+2 a+2 a^2\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^3}+\frac {(1+4 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{6 b^3}-\frac {x (1-a-b x)^{3/2} \sqrt {1+a+b x}}{3 b^2}+\frac {\left (1+2 a+2 a^2\right ) \sin ^{-1}(a+b x)}{2 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 126, normalized size = 0.97 \[ \frac {\left (2 a^2+2 a+1\right ) \sqrt {-b} \sinh ^{-1}\left (\frac {\sqrt {-b} \sqrt {-a-b x+1}}{\sqrt {2} \sqrt {b}}\right )}{b^{7/2}}-\frac {\sqrt {a+b x+1} \left (2 a^3+7 a^2+a (8 b x-5)+2 b^3 x^3-5 b^2 x^2+7 b x-4\right )}{6 b^3 \sqrt {-a-b x+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/E^ArcTanh[a + b*x],x]

[Out]

-1/6*(Sqrt[1 + a + b*x]*(-4 + 7*a^2 + 2*a^3 + 7*b*x - 5*b^2*x^2 + 2*b^3*x^3 + a*(-5 + 8*b*x)))/(b^3*Sqrt[1 - a
- b*x]) + ((1 + 2*a + 2*a^2)*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])])/b^(7/2)

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fricas [A]  time = 0.48, size = 117, normalized size = 0.90 \[ -\frac {3 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - {\left (2 \, b^{2} x^{2} - {\left (2 \, a + 3\right )} b x + 2 \, a^{2} + 9 \, a + 4\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(2*a^2 + 2*a + 1)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 - 1)) -
(2*b^2*x^2 - (2*a + 3)*b*x + 2*a^2 + 9*a + 4)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/b^3

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giac [A]  time = 0.84, size = 106, normalized size = 0.82 \[ \frac {1}{6} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (x {\left (\frac {2 \, x}{b} - \frac {2 \, a b^{3} + 3 \, b^{3}}{b^{5}}\right )} + \frac {2 \, a^{2} b^{2} + 9 \, a b^{2} + 4 \, b^{2}}{b^{5}}\right )} - \frac {{\left (2 \, a^{2} + 2 \, a + 1\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{2 \, b^{2} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(x*(2*x/b - (2*a*b^3 + 3*b^3)/b^5) + (2*a^2*b^2 + 9*a*b^2 + 4*b^2)/b^5)
- 1/2*(2*a^2 + 2*a + 1)*arcsin(-b*x - a)*sgn(b)/(b^2*abs(b))

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maple [B]  time = 0.04, size = 535, normalized size = 4.12 \[ -\frac {\left (-b^{2} x^{2}-2 a b x -a^{2}+1\right )^{\frac {3}{2}}}{3 b^{3}}-\frac {a x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b^{2}}-\frac {a^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b^{3}}-\frac {a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b^{2} \sqrt {b^{2}}}-\frac {x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 b^{2}}-\frac {a \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 b^{3}}-\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b^{2} \sqrt {b^{2}}}+\frac {\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}\, a^{2}}{b^{3}}+\frac {2 \sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}\, a}{b^{3}}+\frac {\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}{b^{3}}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right ) a^{2}}{b^{2} \sqrt {b^{2}}}+\frac {2 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right ) a}{b^{2} \sqrt {b^{2}}}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right )}{b^{2} \sqrt {b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/b^3*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)-a/b^2*x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-a^2/b^3*(-b^2*x^2-2*a*b*x-a^2+1
)^(1/2)-a/b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))-1/2*x/b^2*(-b^2*x^2-2*a*b
*x-a^2+1)^(1/2)-1/2*a/b^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2/b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*
x^2-2*a*b*x-a^2+1)^(1/2))+1/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a^2+2/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(
x+(1+a)/b))^(1/2)*a+1/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)+1/b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(
1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))*a^2+2/b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/
b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))*a+1/b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+
a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))

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maxima [A]  time = 0.43, size = 174, normalized size = 1.34 \[ -\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a x}{b^{2}} + \frac {a^{2} \arcsin \left (b x + a\right )}{b^{3}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{2 \, b^{2}} + \frac {a \arcsin \left (b x + a\right )}{b^{3}} - \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, b^{3}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{2 \, b^{3}} + \frac {\arcsin \left (b x + a\right )}{2 \, b^{3}} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a*x/b^2 + a^2*arcsin(b*x + a)/b^3 - 1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)
*x/b^2 + a*arcsin(b*x + a)/b^3 - 1/3*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/b^3 + 3/2*sqrt(-b^2*x^2 - 2*a*b*x -
a^2 + 1)*a/b^3 + 1/2*arcsin(b*x + a)/b^3 + sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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