3.863 \(\int e^{-3 \tanh ^{-1}(a+b x)} \, dx\)

Optimal. Leaf size=68 \[ -\frac {2 (-a-b x+1)^{3/2}}{b \sqrt {a+b x+1}}-\frac {3 \sqrt {a+b x+1} \sqrt {-a-b x+1}}{b}-\frac {3 \sin ^{-1}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6161, 47, 50, 53, 619, 216} \[ -\frac {2 (-a-b x+1)^{3/2}}{b \sqrt {a+b x+1}}-\frac {3 \sqrt {a+b x+1} \sqrt {-a-b x+1}}{b}-\frac {3 \sin ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(-3*ArcTanh[a + b*x]),x]

[Out]

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

Rule 47

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

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 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 6161

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 43, normalized size = 0.63 \[ \frac {\sqrt {1-(a+b x)^2} \left (-\frac {4}{a+b x+1}-1\right )}{b}-\frac {3 \sin ^{-1}(a+b x)}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(-3*ArcTanh[a + b*x]),x]

[Out]

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

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fricas [A]  time = 0.63, size = 101, normalized size = 1.49 \[ \frac {3 \, {\left (b x + a + 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a + 5\right )}}{b^{2} x + {\left (a + 1\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*(b*x + 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 + 5))/(b^2*x + (a + 1)*b)

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giac [A]  time = 0.21, size = 94, normalized size = 1.38 \[ \frac {3 \, \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{{\left | b \right |}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b} + \frac {8}{{\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} + 1\right )} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*arcsin(-b*x - a)*sgn(b)/abs(b) - sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/b + 8/(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 +
1)*abs(b) + b)/(b^2*x + a*b) + 1)*abs(b))

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maple [B]  time = 0.04, size = 264, normalized size = 3.88 \[ -\frac {\left (-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {5}{2}}}{b^{4} \left (x +\frac {1}{b}+\frac {a}{b}\right )^{3}}-\frac {2 \left (-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {5}{2}}}{b^{3} \left (x +\frac {1}{b}+\frac {a}{b}\right )^{2}}-\frac {2 \left (-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {3}{2}}}{b}-3 \sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}\, x -\frac {3 \sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}\, a}{b}-\frac {3 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right )}{\sqrt {b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^4/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)-2/b^3/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x
+(1+a)/b))^(5/2)-2/b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)-3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x
-3/b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a-3/(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 = 0.41, size = 104, normalized size = 1.53 \[ \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + 2 \, b^{2} x + 2 \, a b + b} - \frac {3 \, \arcsin \left (b x + a\right )}{b} - \frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{2} x + a b + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-{\left (a+b\,x\right )}^2\right )}^{3/2}}{{\left (a+b\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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