3.9.62 \(\int e^{-3 \tanh ^{-1}(a+b x)} x \, dx\) [862]

Optimal. Leaf size=119 \[ \frac {(1+a) (1-a-b x)^{5/2}}{b^2 \sqrt {1+a+b x}}+\frac {3 (3+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}+\frac {(3+2 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}+\frac {3 (3+2 a) \text {ArcSin}(a+b x)}{2 b^2} \]

[Out]

3/2*(3+2*a)*arcsin(b*x+a)/b^2+(1+a)*(-b*x-a+1)^(5/2)/b^2/(b*x+a+1)^(1/2)+1/2*(3+2*a)*(-b*x-a+1)^(3/2)*(b*x+a+1
)^(1/2)/b^2+3/2*(3+2*a)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/b^2

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Rubi [A]
time = 0.06, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6298, 79, 52, 55, 633, 222} \begin {gather*} \frac {3 (2 a+3) \text {ArcSin}(a+b x)}{2 b^2}+\frac {(a+1) (-a-b x+1)^{5/2}}{b^2 \sqrt {a+b x+1}}+\frac {(2 a+3) \sqrt {a+b x+1} (-a-b x+1)^{3/2}}{2 b^2}+\frac {3 (2 a+3) \sqrt {a+b x+1} \sqrt {-a-b x+1}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/E^(3*ArcTanh[a + b*x]),x]

[Out]

((1 + a)*(1 - a - b*x)^(5/2))/(b^2*Sqrt[1 + a + b*x]) + (3*(3 + 2*a)*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(2*b
^2) + ((3 + 2*a)*(1 - a - b*x)^(3/2)*Sqrt[1 + a + b*x])/(2*b^2) + (3*(3 + 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 79

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

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^{-3 \tanh ^{-1}(a+b x)} x \, dx &=\int \frac {x (1-a-b x)^{3/2}}{(1+a+b x)^{3/2}} \, dx\\ &=\frac {(1+a) (1-a-b x)^{5/2}}{b^2 \sqrt {1+a+b x}}+\frac {(3+2 a) \int \frac {(1-a-b x)^{3/2}}{\sqrt {1+a+b x}} \, dx}{b}\\ &=\frac {(1+a) (1-a-b x)^{5/2}}{b^2 \sqrt {1+a+b x}}+\frac {(3+2 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}+\frac {(3 (3+2 a)) \int \frac {\sqrt {1-a-b x}}{\sqrt {1+a+b x}} \, dx}{2 b}\\ &=\frac {(1+a) (1-a-b x)^{5/2}}{b^2 \sqrt {1+a+b x}}+\frac {3 (3+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}+\frac {(3+2 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}+\frac {(3 (3+2 a)) \int \frac {1}{\sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{2 b}\\ &=\frac {(1+a) (1-a-b x)^{5/2}}{b^2 \sqrt {1+a+b x}}+\frac {3 (3+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}+\frac {(3+2 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}+\frac {(3 (3+2 a)) \int \frac {1}{\sqrt {(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{2 b}\\ &=\frac {(1+a) (1-a-b x)^{5/2}}{b^2 \sqrt {1+a+b x}}+\frac {3 (3+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}+\frac {(3+2 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}-\frac {(3 (3+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+a) (1-a-b x)^{5/2}}{b^2 \sqrt {1+a+b x}}+\frac {3 (3+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}+\frac {(3+2 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}+\frac {3 (3+2 a) \sin ^{-1}(a+b x)}{2 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 157, normalized size = 1.32 \begin {gather*} \frac {\sqrt {-b} \left (14-a^3-9 b x-6 b^2 x^2+b^3 x^3-a^2 (14+b x)+a \left (1-20 b x+b^2 x^2\right )\right )+6 (3+2 a) \sqrt {b} \sqrt {1-a^2-2 a b x-b^2 x^2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {-b}}\right )}{2 (-b)^{5/2} \sqrt {-((-1+a+b x) (1+a+b x))}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x/E^(3*ArcTanh[a + b*x]),x]

[Out]

(Sqrt[-b]*(14 - a^3 - 9*b*x - 6*b^2*x^2 + b^3*x^3 - a^2*(14 + b*x) + a*(1 - 20*b*x + b^2*x^2)) + 6*(3 + 2*a)*S
qrt[b]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-b])])/(2*(-b)^(5/2
)*Sqrt[-((-1 + a + b*x)*(1 + a + b*x))])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(465\) vs. \(2(101)=202\).
time = 0.08, size = 466, normalized size = 3.92

method	result	size
risch	\(-\frac {\left (-b x +a +6\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3 a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b \sqrt {b^{2}}}+\frac {9 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b \sqrt {b^{2}}}+\frac {4 \sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}\, a}{b^{3} \left (x +\frac {1}{b}+\frac {a}{b}\right )}+\frac {4 \sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}{b^{3} \left (x +\frac {1}{b}+\frac {a}{b}\right )}\)	\(243\)
default	\(\frac {\frac {\left (-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x +\frac {1+a}{b}\right )^{2}}+3 b \left (\frac {\left (-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {3}{2}}}{3}+b \left (-\frac {\left (-2 b^{2} \left (x +\frac {1+a}{b}\right )+2 b \right ) \sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}{4 b^{2}}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right )}{2 \sqrt {b^{2}}}\right )\right )}{b^{3}}+\frac {\left (-1-a \right ) \left (-\frac {\left (-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x +\frac {1+a}{b}\right )^{3}}-2 b \left (\frac {\left (-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x +\frac {1+a}{b}\right )^{2}}+3 b \left (\frac {\left (-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )\right )^{\frac {3}{2}}}{3}+b \left (-\frac {\left (-2 b^{2} \left (x +\frac {1+a}{b}\right )+2 b \right ) \sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}{4 b^{2}}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-b^{2} \left (x +\frac {1+a}{b}\right )^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right )}{2 \sqrt {b^{2}}}\right )\right )\right )\right )}{b^{4}}\)	\(466\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/b/(x+(1+a)/b)^2*(-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/b))^(5/2)+3*b*(1/3*(-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/
b))^(3/2)+b*(-1/4*(-2*b^2*(x+(1+a)/b)+2*b)/b^2*(-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/b))^(1/2)+1/2/(b^2)^(1/2)*arct
an((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)))))+(-1-a)/b^4*(-1/b/(x+(1+a)/b)^3*(
-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/b))^(5/2)-2*b*(1/b/(x+(1+a)/b)^2*(-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/b))^(5/2)+3*
b*(1/3*(-b^2*(x+(1+a)/b)^2+2*b*(x+(1+a)/b))^(3/2)+b*(-1/4*(-2*b^2*(x+(1+a)/b)+2*b)/b^2*(-b^2*(x+(1+a)/b)^2+2*b
*(x+(1+a)/b))^(1/2)+1/2/(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (101) = 202\).
time = 0.48, size = 299, normalized size = 2.51 \begin {gather*} -\frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2} + 2 \, b^{3} x + 2 \, a b^{2} + b^{2}} - \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2} + 2 \, b^{3} x + 2 \, a b^{2} + b^{2}} + \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{2 \, {\left (b^{3} x + a b^{2} + b^{2}\right )}} + \frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{3} x + a b^{2} + b^{2}} + \frac {3 \, a \arcsin \left (b x + a\right )}{b^{2}} + \frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{3} x + a b^{2} + b^{2}} + \frac {9 \, \arcsin \left (b x + a\right )}{2 \, b^{2}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 0.37, size = 130, normalized size = 1.09 \begin {gather*} -\frac {3 \, {\left ({\left (2 \, a + 3\right )} b x + 2 \, a^{2} + 5 \, a + 3\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 (b^{2} x^{2} - a^{2} - 5 \, b x - 15 \, a - 14\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, {\left (b^{3} x + {\left (a + 1\right )} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*((2*a + 3)*b*x + 2*a^2 + 5*a + 3)*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)) + (b^2*x^2 - a^2 - 5*b*x - 15*a - 14)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/(b^3*x + (a + 1)*b^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 127, normalized size = 1.07 \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} + 6 \, b^{2}}{b^{4}}\right )} - \frac {3 \, {\left (2 \, a + 3\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{2 \, b {\left | b \right |}} - \frac {8 \, {\left (a + 1\right )}}{b {\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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(x/b - (a*b^2 + 6*b^2)/b^4) - 3/2*(2*a + 3)*arcsin(-b*x - a)*sgn(b)/(b
*abs(b)) - 8*(a + 1)/(b*((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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (1-{\left (a+b\,x\right )}^2\right )}^{3/2}}{{\left (a+b\,x+1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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