3.9.45 \(\int e^{-\tanh ^{-1}(a+b x)} x \, dx\) [845]

Optimal. Leaf size=84 \[ -\frac {(1+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}-\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}-\frac {(1+2 a) \text {ArcSin}(a+b x)}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6298, 81, 52, 55, 633, 222} \begin {gather*} -\frac {(2 a+1) \text {ArcSin}(a+b x)}{2 b^2}-\frac {\sqrt {a+b x+1} (-a-b x+1)^{3/2}}{2 b^2}-\frac {(2 a+1) \sqrt {a+b x+1} \sqrt {-a-b x+1}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/E^ArcTanh[a + b*x],x]

[Out]

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

Rule 52

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 55

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 81

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 222

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 633

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 6298

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 \, dx &=\int \frac {x \sqrt {1-a-b x}}{\sqrt {1+a+b x}} \, dx\\ &=-\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}-\frac {(1+2 a) \int \frac {\sqrt {1-a-b x}}{\sqrt {1+a+b x}} \, dx}{2 b}\\ &=-\frac {(1+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}-\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}-\frac {(1+2 a) \int \frac {1}{\sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{2 b}\\ &=-\frac {(1+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}-\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}-\frac {(1+2 a) \int \frac {1}{\sqrt {(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{2 b}\\ &=-\frac {(1+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}-\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}+\frac {(1+2 a) \text {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^3}\\ &=-\frac {(1+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}-\frac {(1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}-\frac {(1+2 a) \sin ^{-1}(a+b x)}{2 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 99, normalized size = 1.18 \begin {gather*} \frac {\sqrt {1+a+b x} \left (-2+a+a^2+3 b x-b^2 x^2\right )}{2 b^2 \sqrt {1-a-b x}}+\frac {(1+2 a) \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )}{(-b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/E^ArcTanh[a + b*x],x]

[Out]

(Sqrt[1 + a + b*x]*(-2 + a + a^2 + 3*b*x - b^2*x^2))/(2*b^2*Sqrt[1 - a - b*x]) + ((1 + 2*a)*Sqrt[b]*ArcSinh[(S
qrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])])/(-b)^(5/2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(211\) vs. \(2(70)=140\).
time = 0.08, size = 212, normalized size = 2.52

method	result	size
risch	\(\frac {\left (-b x +a +2\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b \sqrt {b^{2}}}-\frac {a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}\)	\(146\)
default	\(\frac {-\frac {\left (-2 b^{2} x -2 b a \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{4 b^{2}}-\frac {\left (-4 b^{2} \left (-a^{2}+1\right )-4 b^{2} a^{2}\right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{8 b^{2} \sqrt {b^{2}}}}{b}+\frac {\left (-1-a \right ) \left (\sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}+\frac {b \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right )}{\sqrt {b^{2}}}\right )}{b^{2}}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/4*(-2*b^2*x-2*a*b)/b^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/8*(-4*b^2*(-a^2+1)-4*b^2*a^2)/b^2/(b^2)^(1/2)*
arctan((b^2)^(1/2)*(x+1/b*a)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)))+(-1-a)/b^2*((-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/b))
^(1/2)+b/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/b))^(1/2)))

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Maxima [A]
time = 0.47, size = 107, normalized size = 1.27 \begin {gather*} \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{2 \, b} - \frac {a \arcsin \left (b x + a\right )}{b^{2}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{2 \, b^{2}} - \frac {\arcsin \left (b x + a\right )}{2 \, b^{2}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 91, normalized size = 1.08 \begin {gather*} \frac {{\left (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 ) + \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x - a - 2\right )}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}{a + b x + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 68, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (\frac {x}{b} - \frac {a b + 2 \, b}{b^{3}}\right )} + \frac {{\left (2 \, a + 1\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{2 \, b {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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